Chapter 6 - Chemical Composition

01:09:45
https://www.youtube.com/watch?v=Qn4xTMvMpvk

Ringkasan

TLDRThis video provides a comprehensive overview of Chapter 6 from the book 'Introductory Chemistry,' which delves into the chemical composition. Key learning objectives include understanding the conversion between moles and atoms, grams and moles, and applying mass percent composition as a conversion factor. The content also highlights sodium's role in diet and the importance of monitoring its intake due to health concerns. It discusses the environmental impact of chlorofluorocarbons (CFCs) and their effect on ozone depletion, emphasizing the ban on CFCs to protect the atmosphere. Additionally, the video covers the methods to determine empirical formulas from experimental data and compute molecular formulas from empirical data based on molecular mass. The concept of the mole is thoroughly explained, including Avogadro's number and its application in quantifying atoms. By the end, learners should be able to apply these concepts to real-world questions related to chemical compositions and conversions, enhancing their grasp of chemistry fundamentals.

Takeaways

  • 🔢 Understand the concept of 'mole' and its application in chemistry calculations.
  • ⚖️ Master the conversion between moles, grams, and number of atoms.
  • 🧮 Utilize Avogadro's number to convert between moles and particles.
  • 🍽️ Recognize sodium's role in health and dietary guidelines.
  • 🌎 Acknowledge the impact of CFCs on the ozone layer and environment.
  • 📊 Calculate empirical and molecular formulas from experimental data.
  • ♻️ Use mass percent composition in quantitative analysis.
  • ⚛️ Grasp the relationship between chemical formulas and molar masses.
  • 🌐 Appreciate the conversion factors from chemical formulas to determine compositions.
  • 🔍 Evaluate empirical data to derive chemical compositions.

Garis waktu

  • 00:00:00 - 00:05:00

    Chapter six of the book "Introductory Chemistry" focuses on chemical composition, covering conversions between moles, atoms, grams, and molecules. Key skills include using mass percent composition and determining empirical and molecular formulas. Sodium's dietary importance is highlighted with FDA recommendations.

  • 00:05:00 - 00:10:00

    The calculation of constituent elements in compounds requires knowledge of atomic and formula masses; for example, the mass of sodium versus sodium chloride. Examples illustrate why it's practical to use grams instead of number of atoms in everyday situations like buying nails.

  • 00:10:00 - 00:15:00

    Using a solution map, conversion from pounds to dozens or nails is explained. Conversion factors are determined by relationships like "one dozen weighs 50 pounds." Emphasizing unit cancellation, steps are shown to calculate everything from pounds to number of nails.

  • 00:15:00 - 00:20:00

    The mole is introduced as a count of 6.022 * 10^23 particles, similar to a 'dozen', but much larger to encompass atomic scales. Its historical context ties to Avogadro's number, and its size is linked to reasonable quantities like a mole of helium in balloons or copper in pennies.

  • 00:20:00 - 00:25:00

    Converting between moles and atoms uses Avogadro's number, while converting moles to atoms requires reversing the conversion process. Examples include calculating atoms from moles or vice versa, demonstrating usage of Avogadro's number as a conversion factor.

  • 00:25:00 - 00:30:00

    Discussion on atomic and molar mass includes conversion from grams to moles using atomic mass and highlights differences in weights, drawing comparisons with dozen groups of different-sized animals. Examples illustrate this in a step-by-step conversion process.

  • 00:30:00 - 00:35:00

    More examples show converting grams to moles and atoms, emphasizing solution maps and conversion factors between mass and atoms. Aluminum and compound formulas are used to illustrate methods, including using the periodic table for molar mass calculations.

  • 00:35:00 - 00:40:00

    Transitioning to molecules in compounds, the text explains calculating the mass from moles using molar mass, tying into chemical formula-derived conversion factors. Practical examples such as determining mass percent in compounds like nitrogen oxide are demonstrated.

  • 00:40:00 - 00:45:00

    From chemical formulas, conversion factors are used to find moles in constituent elements. Practical examples involve calcium carbonate converting to moles of oxygen. Further illustration includes calculating mass from moles using sodium chloride as a case study.

  • 00:45:00 - 00:50:00

    Chemical composition and CFCs are discussed regarding environmental impact, especially ozone layer depletion. Mass percent composition of compounds serves as conversion factors in practical health-related examples like sodium intake from salt matching FDA guidelines.

  • 00:50:00 - 00:55:00

    Mass percent composition, like for sodium in sodium chloride, is explained with examples of calculating intake limits. Calculations link compound composition to real-world dietary restrictions, e.g., converting sodium intake recommendations to salt consumption.

  • 00:55:00 - 01:00:00

    Empirical formulas, derived from experimentation, provide element ratios in compounds, sometimes matching molecular formulas. Methodologies for deriving empirical formulas from experimental data, like water breakdown, involve converting mass to moles and creating pseudoformulas.

  • 01:00:00 - 01:09:45

    Chapter review emphasizes mole concepts, chemical formulas, and deriving empirical/molecular formulas from lab data. Understanding mole as a counting unit, conversion between grams and moles, and leveraging chemical formulas for compositions are key learnings.

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Peta Pikiran

Mind Map

Pertanyaan yang Sering Diajukan

  • What is a mole in chemistry?

    A mole is a unit that represents 6.022 x 10^23 particles of a substance, including atoms, molecules, or ions.

  • What is the relationship between moles and grams?

    The molar mass, which is the mass of one mole of an element or compound, relates moles to grams. One mole of an element's mass in grams is equal to its atomic mass.

  • How can you calculate the number of moles from grams?

    To calculate the number of moles from grams, divide the mass of the substance by its molar mass.

  • What is Avogadro's number?

    Avogadro's number, 6.022 x 10^23, is the number of atoms, molecules, or particles in one mole of a substance.

  • How do you find the empirical formula from mass percent composition?

    Convert the mass of each element to moles, write a pseudoformula using the mole ratios, and then divide by the smallest number of moles to obtain whole-number subscripts for the empirical formula.

  • What is the difference between empirical and molecular formulas?

    The empirical formula shows the simplest whole-number ratio of atoms in a compound, whereas the molecular formula shows the actual number of atoms in one molecule of the compound.

  • What is the impact of CFCs on the environment?

    CFCs release chlorine atoms that destroy the ozone layer, which protects Earth from harmful ultraviolet radiation.

  • How does sodium impact health?

    Sodium is essential for fluid regulation in the body but consuming too much can lead to high blood pressure.

  • What is the molar mass?

    Molar mass is the mass of one mole of a substance and it is expressed in grams per mole.

  • What does the chemical formula indicate?

    A chemical formula indicates the types and relative numbers of atoms that constitute a compound.

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Teks
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Gulir Otomatis:
  • 00:00:01
    [Music]
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    here we're going to discuss chapter six
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    of the book introductory chemistry this
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    chapter is named the chemical
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    composition by the end of this chapter
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    you will be able to convert between
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    moles and the number of
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    atoms also to convert between grams and
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    moles
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    conversion between grams and the number
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    of atoms or
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    molecules also you'll be able to convert
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    between moles of a compound and mole of
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    a constituent
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    element as a same way we're going to be
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    able to convert between grams of a
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    compound and grams of a constituent
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    element you will be able to use the mass
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    percent composition as a conversion
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    factor the DET the mass percent
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    composition from a chemical formula also
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    you will be able to determine the an
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    empirical formula from experimental data
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    and finally you will be able to uh to
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    calculate a molecular formula from an
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    empirical formula and the molar
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    mass so sodium is an important dietary
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    mineral that we eat in our food
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    primarily as sodium chloride also known
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    as a table
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    salt sodium is involved in the
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    regulation of body fluids and eating too
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    much of it can lead to high blood
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    pressure so that's why it's really
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    important to control the amount of how
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    much sodium we eat because it is
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    necessarily to to our work to to our
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    body to work properly but also could be
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    a little bit detrimental if we ate too
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    much
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    so the FDA has some recommendations
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    about how much sodium you can eat and
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    you basically can eat 2.4 grams or 24 um
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    100 milligrams of sodium per
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    day the mass of sodium that we eat is
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    not the same as the mass of sodium
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    chloride that we eat how many grams of
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    sodium chloride we can consume and still
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    stay below the FDA recommendation for
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    sodium
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    the chemical composition of sodium
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    chloride is given by this formula na na
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    for sodium and cl for
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    chloride so as we can see there there is
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    one sodium ion to every chloride ion so
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    there a one to one that's why you have
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    here uh you don't put it but you assume
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    that there's a one and also here is a
  • 00:02:54
    one since the masses of sodium and
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    chlorine are different the relationship
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    between the mass of sodium and the mass
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    of sodium chloride is not clear from the
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    chemical formula alone that's why we
  • 00:03:08
    need to calculate the amount of a
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    constituent element in a given amount of
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    a
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    compound the information in a chemical
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    formula along with an atomic and formula
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    masses can be used to calculate the
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    amount of a constituent element in a
  • 00:03:26
    compound so for example how much iron is
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    an is in a given amount of iron ore or
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    how much chlorine is in a given amount
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    of a Chlor
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    chlorocarbon so this questions could be
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    answered by F by by having all this
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    information the chemical formula the
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    atomic and formula masses with those
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    information we can calculate and we can
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    look for the answer for this uh
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    questions
  • 00:03:55
    so some hardware stores sales Nails by
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    pound
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    which is easier than selling them by
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    mail so when you go to the any hardware
  • 00:04:05
    store you look for three pounds of nails
  • 00:04:08
    you don't ask for 50
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    Nails this problem is similar to asking
  • 00:04:14
    how many atoms are in a given mass of
  • 00:04:16
    that element so it's more easy to ask
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    for 8.25 gram for a specific mass
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    instead of asking for thousands of atoms
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    or of or of a specific um element or a
  • 00:04:30
    thousand of molecules for a compound so
  • 00:04:33
    that's why we use uh grams instead of
  • 00:04:36
    specific uh amount of
  • 00:04:41
    items a customer buys 2.6 pounds of a
  • 00:04:44
    mediumsized nails and a dozen of these
  • 00:04:47
    nails weigh
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    50 pounds how many nails did the
  • 00:04:53
    customer buy so to try to answer this
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    kind of questions we need to create what
  • 00:04:59
    we called a solution map this is going
  • 00:05:03
    to give us like a big picture of how we
  • 00:05:07
    can go from one point to another from A
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    to B okay for example here this is the
  • 00:05:14
    map we start with pounds of nail because
  • 00:05:17
    that is the information that we have he
  • 00:05:20
    wants to buy 2.6 pounds of this of of of
  • 00:05:24
    of nails so we start with that then we
  • 00:05:26
    need to get to the number of nails but
  • 00:05:29
    in the middle we have we need to look
  • 00:05:32
    for something that help us to change to
  • 00:05:35
    transform the units of pounds to the
  • 00:05:38
    number of nails and this is a definition
  • 00:05:41
    that we have also here in the question
  • 00:05:44
    but it says that one dozen of these
  • 00:05:47
    nails weigh
  • 00:05:49
    50 pounds so this is a relationship that
  • 00:05:52
    we can create one dozen
  • 00:05:56
    over150 pounds or we can also we arrange
  • 00:06:00
    this equation and put the the
  • 00:06:02
    denominator in denominator and
  • 00:06:04
    denominator in the denominator to use it
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    also remember that for each definition
  • 00:06:08
    we always produce two conversion factors
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    and we need to use the one that cancel
  • 00:06:14
    the unit that we need to cancel for
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    example this one we need to cancel the
  • 00:06:18
    the pounds of nails so we're going to
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    use this conversion factor instead of
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    the one that had the Dozen in the
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    denominator and the pounds in the
  • 00:06:26
    denominator okay so that's why we are
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    using this one here so by using this
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    we're going to have we're going to
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    change pounds to dozens to number of
  • 00:06:36
    nails okay um and then the doen in one
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    doen we have 12 nails so now we can
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    determine the number of nails so these
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    are the two conversion factor that we
  • 00:06:47
    need to use so this is point a this is
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    point B and this is the intermediate
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    point so we need to transformation
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    between this two and then another factor
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    to change from Doc to number of nails
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    from here to here so these are the two
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    conversion factor so we now convert the
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    pounds to the number of nail by
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    following this we start with 2.6 pounds
  • 00:07:11
    of nails times 1 do of nails is equal
  • 00:07:15
    to5 pounds so we cancel the uh pounds
  • 00:07:20
    with pounds and we have here the units
  • 00:07:22
    of Doon and then we cancel this unit by
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    using the second pericial factor that we
  • 00:07:27
    need to use to goes from dozens to
  • 00:07:30
    number of nails and this will be 12
  • 00:07:32
    nails in one dozen we cancel this dozen
  • 00:07:36
    and dozen units and we have here the
  • 00:07:38
    nails unit the ones that we need how
  • 00:07:40
    many nails did the customer buy so we
  • 00:07:42
    have now the nails and we then multiply
  • 00:07:45
    2.16 60 * 1 * 12 divided by .150 pounds
  • 00:07:51
    and this will be equal to
  • 00:07:54
    28
  • 00:07:57
    nails so the conversion factor for the
  • 00:08:01
    first part is the weight per dozen Nails
  • 00:08:04
    okay so this is definition that was in
  • 00:08:07
    the problem and as a definition remember
  • 00:08:10
    that we can create two conversion
  • 00:08:12
    factors we can put this in the
  • 00:08:14
    denominator and this one in the
  • 00:08:15
    denominator or this one in the
  • 00:08:18
    denominator and this one in the
  • 00:08:20
    denominator creating two conversion
  • 00:08:21
    factor and remember that we need to use
  • 00:08:23
    the one that the unit that we need to
  • 00:08:26
    cancel is in the denominator that's the
  • 00:08:29
    conversion factor that we need to use
  • 00:08:31
    the second conversion factor for the
  • 00:08:33
    second part was the number of nails in
  • 00:08:35
    one doen so one do of nails is equal to
  • 00:08:38
    12 Nails that's another definition okay
  • 00:08:42
    and also from here we can obtain two
  • 00:08:44
    conversion factor but we're just going
  • 00:08:47
    to use the one that help us to cancel
  • 00:08:50
    the unit that we are interested to
  • 00:08:51
    cancel and produce the other unit
  • 00:08:54
    eventually that will be at the end to
  • 00:08:56
    help us to uh find uh our unit
  • 00:09:00
    for answer the
  • 00:09:03
    question now with atoms we must use
  • 00:09:06
    their mass as a way to count
  • 00:09:08
    them atoms are too small and too
  • 00:09:12
    numerous to count them
  • 00:09:14
    individually even if you could see atoms
  • 00:09:17
    and count counted them 24 hours a day as
  • 00:09:22
    long as you live you would barely begin
  • 00:09:25
    to count the number of atoms in
  • 00:09:27
    something as small as a grain of
  • 00:09:32
    sand with nails we use a doen as a con
  • 00:09:36
    as a convenient number in our conver
  • 00:09:38
    conversation a dozen is too small to use
  • 00:09:41
    with atoms we need a larger number
  • 00:09:44
    because atoms are so small we can we
  • 00:09:47
    can't use the you know the word one
  • 00:09:50
    dozen of atoms but honestly that's too
  • 00:09:52
    too small as compared with it with the
  • 00:09:55
    males so in chemist our doen is called
  • 00:09:58
    the mole
  • 00:10:00
    the definition of mole is one MO is
  • 00:10:03
    equal to 6.022 * of 10 power of 23rd
  • 00:10:09
    particles so one mole is equal to this
  • 00:10:12
    amount of particles and those particles
  • 00:10:15
    could be atoms could be compounds one
  • 00:10:18
    mole of nails will be 6.022 * 10 to the
  • 00:10:21
    23r power Nails one mole of cows is
  • 00:10:27
    6.022 * 10 power of 23rd Cals so this is
  • 00:10:32
    the definition as a sway of one of one
  • 00:10:35
    dozen is equal to 12 particles so one
  • 00:10:39
    mole is equal to 6.022 * 10 to the^ of
  • 00:10:44
    23rd this number is known as the
  • 00:10:47
    avocados number one mole of anything is
  • 00:10:51
    6.022 * 10 power 23rd units of that
  • 00:10:56
    thing this number is called avad number
  • 00:10:59
    name uh You2 am Mario avagadro that was
  • 00:11:02
    the ones that defin it and established
  • 00:11:05
    this number so one mole of marbl
  • 00:11:08
    correspond to 6.022 * 10 to the power of
  • 00:11:13
    23rd marbles in the same way one mole of
  • 00:11:17
    sand grains corresponds to
  • 00:11:21
    6.022 time 10 to the power of 23rd sand
  • 00:11:26
    grain so one mole of atoms ions
  • 00:11:29
    molecules generally makes up objects of
  • 00:11:32
    reasonable sizes for example 22 real
  • 00:11:35
    copper pannies contains about one mole
  • 00:11:38
    of copper atoms so one mole of copper
  • 00:11:41
    atoms can be represented by 22
  • 00:11:46
    panes also two large helium balloons
  • 00:11:49
    contain approximately one mole of helium
  • 00:11:53
    atoms so that's the amount of helium
  • 00:11:56
    that you need to have one mole ofum
  • 00:12:00
    atoms the size of the mole is a measure
  • 00:12:03
    quantity the numerical value of the mole
  • 00:12:06
    is defined as being equal to the number
  • 00:12:09
    of atoms in exactly 12 gram of pure
  • 00:12:12
    carbon
  • 00:12:14
    12 this definition of the mole
  • 00:12:17
    establishes a relationship between the
  • 00:12:19
    mass grams of carbons and the number of
  • 00:12:22
    atoms the avocados number okay so that's
  • 00:12:24
    where comes all the definition of the
  • 00:12:27
    avocados number this Rel relationship
  • 00:12:29
    allow us to count atoms by waning
  • 00:12:35
    them so let's convert moles to numbers
  • 00:12:38
    of atoms Convert 3.5 moles of helium to
  • 00:12:41
    the number of helium atoms so here we
  • 00:12:43
    have in this problem the data the the
  • 00:12:46
    first information is that we have
  • 00:12:48
    3.45 moles of helium okay and we need to
  • 00:12:52
    find the atoms of human so we need to
  • 00:12:56
    look for something that help us to go
  • 00:12:58
    from MS to atom so we create a solution
  • 00:13:02
    map we need to we need a pericial factor
  • 00:13:04
    to goes from the MS of helium to atoms
  • 00:13:08
    and we have one definition that is the
  • 00:13:10
    avocados number that goes from moles to
  • 00:13:13
    number of particles remember that
  • 00:13:16
    avocados number is a relationship
  • 00:13:18
    between one mole and number of particles
  • 00:13:20
    in this case those particles are the
  • 00:13:23
    atoms okay so it's not like it's one Mo
  • 00:13:27
    than 6.022 * 10 power3 atoms all the
  • 00:13:31
    time no one mole is equal to that number
  • 00:13:35
    of particles and in this case the atoms
  • 00:13:38
    are those particles so we have one mole
  • 00:13:41
    of helium is equal to
  • 00:13:43
    6.022 time 10^ 33rd atoms of helium so
  • 00:13:48
    this is the conversion factor that we
  • 00:13:50
    need and because we start with moles of
  • 00:13:52
    helium we put the one mole in the
  • 00:13:55
    denominator so we can then cancel that
  • 00:13:58
    and have the atoms of helium and
  • 00:14:00
    nominator so here we have the solution
  • 00:14:03
    3.5 moles that is in the problem the
  • 00:14:07
    definition that we I mean the conversion
  • 00:14:09
    factor that we can obtain from the
  • 00:14:11
    definition of mole we cancel the mole of
  • 00:14:13
    helium and then we have the atoms of
  • 00:14:17
    helium so 3.5 time the avad number
  • 00:14:20
    divided by one is equal to 2.1 time 10
  • 00:14:23
    to the power of 24 the number of atoms
  • 00:14:27
    of helium
  • 00:14:30
    the same thing we can convert from atoms
  • 00:14:33
    to moles this is basically the reverse
  • 00:14:36
    uh way to use the avocados number so
  • 00:14:40
    what we have first is the atoms of
  • 00:14:44
    silver so we need to go from atoms of
  • 00:14:46
    silver to moles of silver and basically
  • 00:14:50
    the same uh we need to use a solution
  • 00:14:53
    map but now it goes from atoms from
  • 00:14:57
    particles to moles and as we can see
  • 00:15:00
    here the conversion factor that we need
  • 00:15:02
    is the avocados number the definition of
  • 00:15:04
    avocados number here so one mole
  • 00:15:08
    basically of silver is equal to
  • 00:15:11
    6.022 * 10^ 23rd atoms of of silver so
  • 00:15:16
    in this way by using this we can goes
  • 00:15:19
    from the atoms to the
  • 00:15:21
    mole so
  • 00:15:23
    1.1 * 10 ^ of 22 uh atoms of silver
  • 00:15:29
    divided by the avocados number that
  • 00:15:31
    would give us the number of moles of
  • 00:15:37
    silver that is 1.8 * 10^
  • 00:15:42
    of-2 the moles of
  • 00:15:46
    cylinder now these pictures have the
  • 00:15:49
    same number of nails both of them we
  • 00:15:52
    have one doen of large Nails we have one
  • 00:15:55
    dozen of small Nails the weight of one
  • 00:15:59
    dozen Nails change for different Nails
  • 00:16:02
    okay because the size of each
  • 00:16:07
    nail this picture have the same number
  • 00:16:09
    of atoms the weight of one mole of atoms
  • 00:16:12
    changed for different elements so we
  • 00:16:14
    have here the same number of atoms as
  • 00:16:17
    here but here the mass of this is 32.7 G
  • 00:16:22
    while the mass of this is 12.01 G but
  • 00:16:25
    the number of atoms is the same in both
  • 00:16:32
    so the atomic mass unit is defined as
  • 00:16:34
    112 of the mass of car of a carbon to
  • 00:16:38
    atom the molar mass as of any element is
  • 00:16:41
    the mass of one mle of atom of the
  • 00:16:43
    element is equal to the atomic mass of
  • 00:16:46
    the ele of that element expressed in
  • 00:16:48
    atomic mass un one copper atom has an
  • 00:16:52
    atomic mass of
  • 00:16:56
    6355 atomic mass unit
  • 00:16:59
    and one mole of copper atoms has the
  • 00:17:02
    mass of
  • 00:17:03
    6335 G of
  • 00:17:08
    copper the molar mass of copper is 63.5
  • 00:17:12
    55 G of uh by mole okay so this
  • 00:17:17
    basically is the molar mass of copper
  • 00:17:19
    and this one is the atomic mass so when
  • 00:17:22
    we use the atomic mass we're going to
  • 00:17:23
    use the units of atomic mass unit and
  • 00:17:26
    when we're looking for the molar mass
  • 00:17:29
    we're going to use the 6355 grams per
  • 00:17:32
    mole those units because this one is
  • 00:17:34
    going to give us the relationship
  • 00:17:36
    between grams or mass and
  • 00:17:42
    moles so the mass of one mole of atom of
  • 00:17:45
    an element is its molar mass the mass of
  • 00:17:48
    one mole of atoms change for different
  • 00:17:50
    elements for example 32.7 G of f sulfur
  • 00:17:55
    is equal to one more mole of sulfur and
  • 00:17:58
    the the number of aalas atoms of
  • 00:18:01
    sulfur 12.01 gram of carbon is equal to
  • 00:18:06
    one Mo of carbon and also the same name
  • 00:18:09
    the same number of atoms and 6.94 gram
  • 00:18:13
    of lithium is equal to 1 mole of lithium
  • 00:18:16
    and also the same number of atoms this
  • 00:18:18
    is equal if we can think about the doen
  • 00:18:21
    a dozen of cows a dozen of mice and a
  • 00:18:25
    dozen of whales you have a dozen of each
  • 00:18:28
    one but each group is going to have a
  • 00:18:31
    different weight because of the weight
  • 00:18:33
    of the individual animal so this is the
  • 00:18:36
    same thing we have the same number of
  • 00:18:38
    atoms for each of the these three three
  • 00:18:41
    different elements okay and that amount
  • 00:18:44
    of atoms is going to gave a different
  • 00:18:47
    weight okay for each of the element the
  • 00:18:51
    lighter the atom the less mass in one
  • 00:18:54
    mole of that atom okay so you have a
  • 00:18:58
    lighter at is going to have the less
  • 00:19:00
    less Mass because it's the lightest so
  • 00:19:03
    that's why you can see that here if
  • 00:19:06
    sulfur is a bigger one so it's going to
  • 00:19:08
    have a bigger mass for the same number
  • 00:19:11
    of
  • 00:19:15
    atoms so let's convert between grams and
  • 00:19:20
    moles and grams and
  • 00:19:23
    moles you no uh conversion factor or a
  • 00:19:27
    definition that we told before that is
  • 00:19:29
    the m Mass okay so that's the conversion
  • 00:19:31
    factor basic that we need to use when
  • 00:19:33
    who wants to go from Mass to moles or
  • 00:19:37
    grams okay the units of of mass the most
  • 00:19:40
    GRS so calculate the number of moles of
  • 00:19:44
    carbon in 58 grams of diamond so what we
  • 00:19:48
    have is that first information 58 gram
  • 00:19:51
    of
  • 00:19:52
    diamond we need look to look for the
  • 00:19:54
    moles of carbon so we can create this
  • 00:19:57
    solution map that goes goes from grams
  • 00:19:59
    of carbon to moles of carbon and here we
  • 00:20:02
    can see that the our conversion factor
  • 00:20:04
    now is the molar mass of carbon the ones
  • 00:20:07
    that can relate mass and mole because we
  • 00:20:10
    know that one mole of carbon is equal to
  • 00:20:14
    12.01 grams of carbon this comes from
  • 00:20:16
    the periodic table okay remember that
  • 00:20:19
    the atomic mass can be transformed to
  • 00:20:22
    the molar mass by using the aous
  • 00:20:26
    number so that's the relationship that
  • 00:20:28
    we're going to use use in this case and
  • 00:20:30
    that will be our conversion factor so we
  • 00:20:32
    start with 58 G of carbon and we need to
  • 00:20:35
    put the grams in the denominator as we
  • 00:20:37
    can see here so there going to be
  • 00:20:38
    basically these two factors that we're
  • 00:20:40
    going to use 58 or both data divided by
  • 00:20:45
    12.01 grams of carbon and we have the
  • 00:20:48
    moles of carbon in the denominator
  • 00:20:50
    that's why is the unit for our result
  • 00:20:53
    and here we have how many moles we have
  • 00:20:55
    in 58 G of carbon that is 4. * 10^
  • 00:21:01
    of-2 m of
  • 00:21:04
    carbon so what about if we wants to know
  • 00:21:07
    the number of atoms that we have in0
  • 00:21:10
    58 so we have this and this one we have
  • 00:21:13
    a mole so what we need to use to Cher
  • 00:21:16
    moles to number of atoms of carbon to
  • 00:21:20
    number of particles we need to use then
  • 00:21:24
    the avos number perfect so and now we
  • 00:21:27
    can create another
  • 00:21:29
    um another map here okay solution map we
  • 00:21:33
    need to start from the mass of carbon
  • 00:21:36
    goes through the moles of carbon and
  • 00:21:39
    then finally to the number of carbon
  • 00:21:41
    because we don't we don't have right now
  • 00:21:44
    a uh conversion factor that can goes
  • 00:21:46
    directly from mass of carbon to number
  • 00:21:48
    of atoms we need to go first through the
  • 00:21:51
    moles of carbon so to do this from ma
  • 00:21:54
    from the mass of carbon to the moles we
  • 00:21:56
    use the molar mass and to transform from
  • 00:21:59
    the moles to the number of atoms of or
  • 00:22:01
    number of particles we need to use the
  • 00:22:04
    avad number so now we add this part so
  • 00:22:07
    the one uh solution before that we have
  • 00:22:10
    in the previous slide okay so that's we
  • 00:22:12
    have now we have the AV number here we
  • 00:22:15
    now can cancel the moles of carbon and
  • 00:22:17
    now we have the atoms of carbon here so
  • 00:22:21
    58 gram of diamond is equal to 2.9
  • 00:22:26
    time 10^ 22nd
  • 00:22:29
    um atoms of
  • 00:22:35
    carbons so converting now between grams
  • 00:22:38
    and the number of atoms how many Al
  • 00:22:41
    aluminum aluminum atoms are in an
  • 00:22:44
    aluminum can with a mass of 16.2 gram so
  • 00:22:48
    we have mass here and we these two go to
  • 00:22:51
    atoms but we don't have a a a specific
  • 00:22:55
    conversion factor that goes from Mass to
  • 00:22:57
    atoms but we can go from Mass to moles
  • 00:23:01
    and then from moles to atoms okay that
  • 00:23:04
    way we can create our solution map here
  • 00:23:08
    that goes from the mass of aluminum to
  • 00:23:11
    the moles of aluminum and from moles of
  • 00:23:13
    aluminum to the number of particles or
  • 00:23:16
    number of atoms in this case of aluminum
  • 00:23:19
    so to go from Mass to moles we can use
  • 00:23:21
    the molar mass to go from moles to
  • 00:23:24
    number of atoms we need to use our
  • 00:23:26
    definition of the's number so we need to
  • 00:23:29
    look for the molar mass of aluminum
  • 00:23:31
    because we are looking for the atoms of
  • 00:23:33
    aluminum and the initial data is 16.2
  • 00:23:36
    gram of
  • 00:23:37
    aluminum so we used our molar mass of
  • 00:23:40
    aluminum and we can then uh solve a
  • 00:23:45
    problem by using 16.2 gram of aluminum
  • 00:23:49
    divided by the marolar mass because
  • 00:23:51
    remember that we need to put the U Mass
  • 00:23:54
    in the denominator so that way we can
  • 00:23:56
    cancel the mass of aluminum of of the
  • 00:23:59
    sample okay and now we have here moles
  • 00:24:02
    of of aluminum so we need to cancel this
  • 00:24:05
    so that's why we need to use in this way
  • 00:24:07
    the have got's number we need to put the
  • 00:24:09
    m in the bottom because we can then
  • 00:24:11
    cancel this to this if we use the other
  • 00:24:13
    way we put the M on on the top and then
  • 00:24:15
    the uh the atoms of aluminum our result
  • 00:24:19
    will be mole Square divided by atoms so
  • 00:24:23
    that's not what we're looking for so
  • 00:24:24
    that's why we need to invert and have
  • 00:24:27
    have it like this okay that so that way
  • 00:24:29
    we can cancel the unit of moles and we
  • 00:24:32
    will have here the unit of aluminum
  • 00:24:34
    atoms so that's way 6 16.2 G of aluminum
  • 00:24:40
    at 3.62 * 10 ^ of 23 atoms of
  • 00:24:50
    aluminum so counting now the molecules
  • 00:24:53
    by the gram for elements the M mass is
  • 00:24:56
    the mass of one mole of a atoms of that
  • 00:25:00
    element for compounds the M mass is the
  • 00:25:03
    mass of one mole of molecules or formula
  • 00:25:07
    units if we talking about ionic
  • 00:25:09
    compounds of that
  • 00:25:12
    compound ionic compounds do not contain
  • 00:25:16
    individual
  • 00:25:18
    molecules we convert between the mass of
  • 00:25:21
    a compound and moles of the compound and
  • 00:25:24
    then we calculate the number of
  • 00:25:26
    molecules or formula units for from the
  • 00:25:32
    moles so the M mass of a compound in
  • 00:25:35
    grams per mole is numerically equal to
  • 00:25:38
    the formula mass of the compound in
  • 00:25:40
    atomic mass units or ionic compounds the
  • 00:25:44
    formula mass a for compound is the sum
  • 00:25:48
    of the atomic masses of all of the atoms
  • 00:25:52
    in a chemical
  • 00:25:56
    formula so let's now convert between
  • 00:25:59
    grams and moles of acam so let's
  • 00:26:03
    calculate the mass in grams of 1.75
  • 00:26:07
    moles of water so we need to go from
  • 00:26:10
    moles to mass the only conversion factor
  • 00:26:14
    that we have is the molar mass so we
  • 00:26:19
    need to look for the molar mass of
  • 00:26:22
    water okay because we need to go from
  • 00:26:26
    moles to grams and the only
  • 00:26:28
    factor that has both unit is the m Mass
  • 00:26:32
    so we have 1.75 moles of water who wants
  • 00:26:35
    to determine the mass of water so we
  • 00:26:38
    need to goes from mole to mass and what
  • 00:26:44
    fact that's going to help us to
  • 00:26:46
    determine this one will be the molar
  • 00:26:50
    mass okay that is 18.02 gram of water in
  • 00:26:54
    one mle of water how can we found this
  • 00:26:57
    18
  • 00:26:59
    02 grams of water well we need to look
  • 00:27:02
    for each of the atomic mass okay of the
  • 00:27:05
    atoms that compose the uh the compound
  • 00:27:08
    and for example the mar mass for
  • 00:27:10
    hydrogen is one we have in the molecular
  • 00:27:13
    formula two hydrogen so we multiply two
  • 00:27:15
    by the atomic mass of hydrogen and for
  • 00:27:19
    oxygen we just have one oxygen here so
  • 00:27:21
    that's why we multiply one times the
  • 00:27:23
    atomic mass of oxygen this is 1 * 16 so
  • 00:27:27
    this going to be equal to
  • 00:27:29
    18.02 G per mole this is the molar mass
  • 00:27:34
    okay for water now we can use this to
  • 00:27:37
    convert from 1.75 moles of water to the
  • 00:27:41
    mass of water we have the converion
  • 00:27:46
    factor here we need to use this one
  • 00:27:48
    instead of the one that has the mole in
  • 00:27:49
    the denominator and the mass in the
  • 00:27:51
    denominator so here we can cancel the
  • 00:27:54
    mole of water and we have in the
  • 00:27:56
    denominator the units of mass the unit
  • 00:27:58
    of gr of water that is what we're
  • 00:28:00
    looking for so when we multiply this
  • 00:28:02
    we're going to have a total of
  • 00:28:05
    31.5 grams of
  • 00:28:11
    water so converting between number of
  • 00:28:14
    molecules and mass of a compound
  • 00:28:16
    remember now because we're talking about
  • 00:28:18
    a compound we're talking about molecules
  • 00:28:21
    okay so in the other example we're
  • 00:28:24
    talking about elements so we can
  • 00:28:25
    calculate number of atoms eventually
  • 00:28:27
    then we we can calculate also the number
  • 00:28:29
    of atoms if we want to determine from
  • 00:28:32
    from the molecule but for now we wants
  • 00:28:34
    to calculate the number of molecules
  • 00:28:36
    when we have a mass or data of the mass
  • 00:28:40
    of a compound so here is what is the
  • 00:28:43
    mass of 4.72 * 10 to the^
  • 00:28:47
    24 molecules of nitrogen
  • 00:28:51
    oxide so the data that we have is for
  • 00:28:55
    the the number of molecules is this one
  • 00:28:58
    or NO2 we wants to find the mass of NO2
  • 00:29:02
    that represent this amount of molecules
  • 00:29:05
    so we need to go from molecules to moles
  • 00:29:10
    remember that we don't have a specific
  • 00:29:12
    conversion factor that goes from
  • 00:29:13
    molecules to mass the only thing that we
  • 00:29:16
    know that has something related with
  • 00:29:17
    mass is the molar mass that is mole and
  • 00:29:21
    mass so molecules is particles okay this
  • 00:29:26
    in this example the mole is the number
  • 00:29:28
    of particles so if when we talk about
  • 00:29:31
    particles we that's that's that has to
  • 00:29:33
    make a click in our mind say number of
  • 00:29:36
    particles av's number okay so we need to
  • 00:29:40
    go from number of molecules to the mole
  • 00:29:43
    by using the avocados number and then
  • 00:29:46
    from moles to Mass using the second
  • 00:29:49
    coefficient factor that is the molar
  • 00:29:51
    mass okay so when we went to go from
  • 00:29:53
    particles to moles avocado number from
  • 00:29:56
    mole to Mass is going to be the marar
  • 00:29:59
    mass the same way is the opposite if we
  • 00:30:02
    wants to go from Mass to moles we need
  • 00:30:04
    to use the same kind of conversion
  • 00:30:06
    factor that we can obtain from the molar
  • 00:30:09
    mass and for moles to number of
  • 00:30:11
    particles we need to use a conversion
  • 00:30:14
    factor that can come from the avocados
  • 00:30:17
    number so here we need to use determine
  • 00:30:20
    the molar mass of NO2 and by doing that
  • 00:30:24
    we need to calculate it in this case is
  • 00:30:26
    46.01
  • 00:30:28
    G of uh nitrogen oxide divided by one Mo
  • 00:30:32
    of
  • 00:30:35
    NO2 so how can we do that here we have
  • 00:30:39
    our first uh relationship okay the ones
  • 00:30:41
    that we know fromage number and this one
  • 00:30:44
    is to obtain the molar mass of NO2 we
  • 00:30:46
    have one nitrogen and two oxygen from
  • 00:30:49
    our uh molecular chemical formula here
  • 00:30:53
    NO2 so we need to look for the atomic
  • 00:30:55
    mass of nitrogen and multiply by one the
  • 00:30:58
    atomic mass of oxygen and multiply by
  • 00:31:00
    two so the atomic mass of nitrogen is
  • 00:31:02
    14.1 * one is this one while the atomic
  • 00:31:06
    mass of oxygen is basically 16.0 * 2 so
  • 00:31:10
    that means that their molar mass of NO2
  • 00:31:13
    is
  • 00:31:14
    46.01
  • 00:31:15
    gram times mole so now we need to
  • 00:31:21
    go uh going back to the the problem need
  • 00:31:24
    we go from molecules to mass so first
  • 00:31:27
    first we change our molecules to moles
  • 00:31:31
    by using the av's number which a factor
  • 00:31:33
    we can cancel here the molecules and
  • 00:31:36
    then this moles can be cancell by the
  • 00:31:39
    units of mole that we can find in our
  • 00:31:41
    convergent factor from the molar mass so
  • 00:31:44
    we can cancel here the moles of NO2 and
  • 00:31:47
    we have the mass of NO2 that is
  • 00:31:51
    365 G of NO2 in
  • 00:31:55
    4.78 * 10 ^ 24
  • 00:31:59
    uh NO2
  • 00:32:00
    [Music]
  • 00:32:03
    molecules so chemical formulas as a
  • 00:32:05
    converstion factor we can also use those
  • 00:32:08
    chemical
  • 00:32:09
    formulas to go from number of molecules
  • 00:32:13
    to number of atoms that's I mentioned
  • 00:32:15
    before so here for example we have a
  • 00:32:17
    three left Clover an analogy how many
  • 00:32:21
    leaves of on 14 clovers so if each
  • 00:32:25
    Clover have three leaves okay that means
  • 00:32:28
    that we need to go from number of CLS
  • 00:32:31
    we're going to use the same u m solution
  • 00:32:34
    MTH number of clavers to number of
  • 00:32:36
    Leaves we have the conversion factor
  • 00:32:38
    that we can find by just looking okay
  • 00:32:40
    the FL so we then can use 14 CLS divided
  • 00:32:45
    by one cl multiply by three leaves this
  • 00:32:47
    is a conversion factor from here and now
  • 00:32:50
    we can have 42 leaves okay we can have
  • 00:32:53
    42 leaves from uh 14 CLS so the same way
  • 00:32:57
    we can use are chemical
  • 00:33:00
    formulas the formula for carbon
  • 00:33:03
    dioxide means there that there are two
  • 00:33:07
    oxym of atoms per one CO2 molecule okay
  • 00:33:12
    so we have another word two atoms of
  • 00:33:15
    oxygen for each molecule of CO2 so this
  • 00:33:19
    is a conversion factor we can use this
  • 00:33:22
    the denominator and this the denominator
  • 00:33:25
    or vice versa it all depends of what is
  • 00:33:28
    the question and what we're looking for
  • 00:33:30
    okay the same way 12 dozens of oxygen
  • 00:33:34
    atoms are in one dozen of molecules of
  • 00:33:38
    CO2 because this basically twice the
  • 00:33:41
    number of CO2 of oxygen so the same
  • 00:33:45
    thing happens if we are talking about
  • 00:33:47
    moles two moles of oxygen are in one
  • 00:33:50
    mole of
  • 00:33:53
    CO2 so the conversion factor comes
  • 00:33:56
    directly from
  • 00:33:59
    formulas it Lex in one spider okay for
  • 00:34:02
    each spider we're going to find all
  • 00:34:04
    always it Lex in a chair we have four
  • 00:34:08
    legs and in one molecule of hydrogen we
  • 00:34:11
    have two atoms of hydrogen so all of
  • 00:34:14
    this are converstion factor using
  • 00:34:17
    basically this types of example so we
  • 00:34:20
    can understand a little bit more that
  • 00:34:23
    conversion factor from the
  • 00:34:26
    molecules so converting between moles of
  • 00:34:29
    a compound and moles of a constituent
  • 00:34:32
    element find the number of moles of
  • 00:34:35
    oxygen in 1.7 mole of calcium
  • 00:34:40
    carbonate so we start with this value
  • 00:34:42
    this is what we have in the problem we
  • 00:34:45
    have 1.7 mole of calcium carbonate and
  • 00:34:49
    we need to determine the number of moles
  • 00:34:51
    of
  • 00:34:52
    oxygen so we need to look for a
  • 00:34:55
    conversion factor that can relate the
  • 00:34:58
    number of moles of oxygen and the number
  • 00:35:00
    of moles of carbon uh calcium carbonate
  • 00:35:04
    and that will be from our uh
  • 00:35:08
    basically chemical formula in our
  • 00:35:11
    chemical formula we can see here that we
  • 00:35:13
    have three moles of oxygen per each mole
  • 00:35:16
    of calcium carbonate so the solution map
  • 00:35:19
    will be moles of calcium carbonate to
  • 00:35:21
    moles of oxygen we have three moles of
  • 00:35:25
    oxygen by each mole of calcium carbonate
  • 00:35:28
    this is our conviction conversion factor
  • 00:35:32
    okay so we have one mole 1.7 mole of car
  • 00:35:35
    calcium
  • 00:35:36
    carbonate times this will give us the MS
  • 00:35:39
    of
  • 00:35:42
    oxygen and here we have Point 5 point
  • 00:35:45
    sorry um 5.1 moles of oxygen so in 1.7
  • 00:35:51
    moles of carbonate we have 5.1 Mo
  • 00:35:55
    moles of oxygen so here we are using the
  • 00:36:00
    chemical formula okay the information
  • 00:36:03
    from the chemical formula to create a
  • 00:36:06
    conversion factor if we wants to
  • 00:36:09
    transform from moles of the compound to
  • 00:36:13
    mon to moles of a constituent element or
  • 00:36:16
    vice versa if he wants to start with Mo
  • 00:36:19
    of a constituent element to find how
  • 00:36:21
    many moles of the compound can be
  • 00:36:23
    produced but the same thing we need to
  • 00:36:25
    use a conversion factor that we can
  • 00:36:28
    create or obtain from the chemical
  • 00:36:32
    formula so let's do another example here
  • 00:36:35
    we need to convert from uh grams to of a
  • 00:36:40
    compound to a grams of a constituent
  • 00:36:43
    element so let's find the mass of sodium
  • 00:36:47
    in 50 gr of sodium
  • 00:36:49
    chloride so we start with the mass of
  • 00:36:53
    sodn chloride here okay and we to find
  • 00:36:56
    out the mass of sodium itself so that
  • 00:37:00
    means that maybe we need to go from the
  • 00:37:02
    mass of sodium chloride to the number of
  • 00:37:04
    moles of sodium chloride then from that
  • 00:37:08
    point we can determine the mo number of
  • 00:37:10
    moles of sodium and finally we can get
  • 00:37:14
    to the mass of sodium so here we have
  • 00:37:17
    basically our solution map we start with
  • 00:37:20
    the mass of the compound because that's
  • 00:37:23
    what we have in the problem 50 gram the
  • 00:37:26
    mass of sod chloride
  • 00:37:28
    so from the mass we can go to the moles
  • 00:37:30
    using the M mass of the compound sodium
  • 00:37:33
    chloride that is
  • 00:37:35
    5844 gram over one mole and we use this
  • 00:37:39
    we arrange of of that molar mass so we
  • 00:37:41
    can cancel the units of mass of sodium
  • 00:37:44
    chloride now we have the moles of sodium
  • 00:37:47
    chloride so we can go to from moles of
  • 00:37:49
    sodum chloride to moles of sodium by
  • 00:37:53
    using the conversion factor that we can
  • 00:37:55
    obtain from the chemical formula because
  • 00:37:59
    we can see here that in one mole of
  • 00:38:02
    sodium chloride how many moles of sodium
  • 00:38:04
    we have in the the chemical formula just
  • 00:38:06
    one so that's another conversion factor
  • 00:38:09
    that we can use to go from the M of
  • 00:38:11
    sodium chloride to the m of sodium and
  • 00:38:14
    finally you can go from the moles of
  • 00:38:15
    sodium to the mass of sodium using the
  • 00:38:19
    molar mass for sodium okay that is 22.9
  • 00:38:22
    G of sodium over one Mo we put here the
  • 00:38:26
    moles in the denominator so we can
  • 00:38:27
    cancel the moles of sodium so in other
  • 00:38:31
    words the solution will be 50 g okay of
  • 00:38:34
    sodium chloride and we can uh substitute
  • 00:38:37
    all this value here in this equation
  • 00:38:39
    multiply 15 by 22.9 and then divide it
  • 00:38:42
    by
  • 00:38:43
    5844 and that will be 5.9 gr of sodium
  • 00:38:47
    okay so because 1 * 1 * 21 this will be
  • 00:38:50
    22.9 and then the denominator 58.4 * 1 *
  • 00:38:54
    1 will be 58.4 so we can multiply 15 *
  • 00:38:59
    22.9 ided by
  • 00:39:02
    5844 and that will be the mass of sodium
  • 00:39:06
    that we can obtain from 15 grams of
  • 00:39:10
    sodium floride so for each 15 grams that
  • 00:39:12
    we ate eat every day of sodum fluoride
  • 00:39:16
    from those 50 15 grams 5.9 grams are
  • 00:39:20
    from
  • 00:39:22
    sodium so as we saw before there is a
  • 00:39:25
    more relationship from the chemic
  • 00:39:27
    formula here we have one Mo of C4 Tetra
  • 00:39:31
    chloride of carbon Tetra chloride carbon
  • 00:39:34
    okay um and here we have uh four moles
  • 00:39:37
    okay that we can obtain from the one
  • 00:39:40
    mole of ccl4 so the relationship
  • 00:39:43
    inherent in a chemical formula allow us
  • 00:39:46
    to convert between moles of the compound
  • 00:39:49
    and moles of a constituent element and
  • 00:39:52
    vice versa so that's how we can use the
  • 00:39:56
    chemical formula as a conversion factor
  • 00:39:59
    as you can see here one mole of
  • 00:40:00
    cl4 can produce four moles of CCL of Cl
  • 00:40:04
    all right and what about if we need to
  • 00:40:07
    look for the relationship using the C4
  • 00:40:11
    of moles of
  • 00:40:14
    carbon so if we need to determine the
  • 00:40:16
    moles of carbon the conversion factor
  • 00:40:19
    from here would be one mole of
  • 00:40:22
    ccl4 okay will produce one mole of
  • 00:40:26
    carbon because here in the chemical
  • 00:40:29
    formula we see that for each ccl4 we
  • 00:40:31
    have just one carbon and for each ccl4
  • 00:40:34
    we have four moles of
  • 00:40:38
    choride now talking about um compounds
  • 00:40:42
    that have chlorine synthetic compounds
  • 00:40:46
    known as chlorofluorocarbon CFCs CFCs
  • 00:40:50
    are destroying a vital compound called
  • 00:40:53
    ozone in Earth upper atmosphere CFC are
  • 00:40:58
    chemically iner molecules used primarily
  • 00:41:01
    as refrigerants and Industrial solvents
  • 00:41:04
    the upper
  • 00:41:06
    atmosphere sunlight break bonds within
  • 00:41:09
    cfc's resulted in the release of
  • 00:41:12
    chlorine atoms and there's the situation
  • 00:41:15
    chlorine chlorine atoms react with ozone
  • 00:41:18
    and destroyed it by converting from
  • 00:41:20
    ozone to O2 ozone is O3 and O2 is oxygen
  • 00:41:26
    the finding of o overpopulated areas is
  • 00:41:29
    dangerous because ultraviolet light can
  • 00:41:32
    harm living things and induce skin
  • 00:41:34
    cancer in humans because those o the the
  • 00:41:37
    ozone protect us against the ultraviolet
  • 00:41:40
    light from the
  • 00:41:42
    sun most developed Nations banned
  • 00:41:45
    production of cfc's on January 1st in
  • 00:41:49
    1996 the cfc's still lur in older
  • 00:41:52
    refrigerators and air conditions units
  • 00:41:55
    and can leak into the atmosphere here
  • 00:41:57
    and destroy the
  • 00:42:00
    ozone now upper atmospheric ozone is
  • 00:42:03
    important as I mentioned before because
  • 00:42:05
    it acts as a shield to protect life and
  • 00:42:08
    on Earth from harmful ultraviolet life
  • 00:42:11
    so the ozone is basically around the
  • 00:42:14
    earth okay those molecules of O3 and the
  • 00:42:17
    UV is absorbed by those molecules and
  • 00:42:20
    avoid to to impact directly to the Earth
  • 00:42:23
    and also affect us now the Antarctic
  • 00:42:27
    ozone hole area from 1980 to uh 2012 is
  • 00:42:31
    right here in this graph this is
  • 00:42:33
    basically the years and this is the size
  • 00:42:36
    of the hole of the hole in the ozone
  • 00:42:38
    goes to the arc the darkest blue colar
  • 00:42:41
    indicate the lowest ozone level so that
  • 00:42:43
    means that this area is less protected
  • 00:42:46
    than the rest of the world now in in the
  • 00:42:48
    previous slide we mentioned that in 1996
  • 00:42:52
    a lot of nation banned the production of
  • 00:42:54
    CFCs and we can see here that in uh from
  • 00:42:57
    1995 on there is like like a flat up
  • 00:43:01
    there because um there was no production
  • 00:43:03
    of CFCs so that's why it's important to
  • 00:43:06
    understand and learn about the different
  • 00:43:08
    compounds with
  • 00:43:10
    Florine and the impact in our
  • 00:43:15
    lives let's talk now about the mass
  • 00:43:17
    percent composition of
  • 00:43:20
    compounds this also known as the mass
  • 00:43:23
    percent of an element is the element's
  • 00:43:25
    percentage of the total mass of the
  • 00:43:28
    compound so for example here we have the
  • 00:43:30
    mass percent of element X is the mass of
  • 00:43:33
    X in a sample of the compound divided by
  • 00:43:36
    the mass of the sample of the compound
  • 00:43:38
    times 100 so basically the mass percent
  • 00:43:41
    of the element X is the mass of x
  • 00:43:43
    divided by the total mass of the
  • 00:43:48
    compound so 358 G sample of chromium
  • 00:43:53
    reacts with oxygen to form 523 G of
  • 00:43:57
    metal
  • 00:43:58
    oxide the mass percent of chromium will
  • 00:44:02
    be when all react together and produce
  • 00:44:04
    this will be point will be basically the
  • 00:44:07
    mass of chromium divided by the mass of
  • 00:44:09
    the metal
  • 00:44:10
    oxide so it will be 358 G divided by 523
  • 00:44:16
    G this is the mass of the chromide as we
  • 00:44:18
    can see
  • 00:44:20
    here and this is the total mass of the
  • 00:44:24
    metal oxide time 100 this will be equal
  • 00:44:26
    to
  • 00:44:28
    68.5% of
  • 00:44:30
    chromium in that
  • 00:44:32
    oxide so that's how you can determine
  • 00:44:34
    the mass percent of an
  • 00:44:40
    element so use mass percent composition
  • 00:44:42
    as conversion factor we can use mass
  • 00:44:44
    percent composition as a conversion
  • 00:44:46
    factor between grams of a an element and
  • 00:44:50
    the grams of the
  • 00:44:51
    compound so if we have for example uh a
  • 00:44:55
    sample a sample of s chloride 39% of
  • 00:44:59
    sodium that means that we have 39 G of
  • 00:45:03
    sodium in 100 grams of sodium chloride
  • 00:45:07
    because the composition of sodium in
  • 00:45:09
    that sample of sodium chloride is
  • 00:45:12
    39% so that means that 39 of of the 100
  • 00:45:16
    part are from sodium while the
  • 00:45:19
    61% is from chloride so by using the
  • 00:45:23
    sodium uh relation we can find out a
  • 00:45:27
    conversion factor okay about 39 grams of
  • 00:45:30
    sodium for for each 100 gram of sodium
  • 00:45:34
    chloride and that will be the conation
  • 00:45:37
    factor obtained from the mass percent
  • 00:45:41
    composition so here we have the mass
  • 00:45:44
    percent composition of Florine 39% so we
  • 00:45:47
    can have two conversion factor obtained
  • 00:45:49
    from that uh percent because we can say
  • 00:45:52
    that we have 39 gram of sodium in 100
  • 00:45:55
    gram of sodium chloride
  • 00:45:57
    or 100 G of fum
  • 00:45:59
    chloride over 39 G of sodium so this
  • 00:46:04
    fractions are conversion factor factors
  • 00:46:07
    between the grams of sodium and the
  • 00:46:10
    grams of sodium chloride in a sample of
  • 00:46:13
    39% sodium in sodium
  • 00:46:18
    chloride so as me we mentioned at the
  • 00:46:20
    beginning of this video the FDA
  • 00:46:22
    recommends uh to adults to consume 2.4
  • 00:46:25
    grams of sodium per day
  • 00:46:27
    so how many grams of sod chloride can
  • 00:46:29
    you consume and still be within the FDA
  • 00:46:33
    guidelines so sodum chloride is
  • 00:46:36
    39% of sodium by mass as we mentioned
  • 00:46:41
    before so if we have for example if we
  • 00:46:44
    need for example 2.4 grams of sodium
  • 00:46:47
    okay and we have this conversion factor
  • 00:46:49
    that says that for each 100 gram of
  • 00:46:52
    sodium chloride we have 39 grams of
  • 00:46:54
    sodium because the sample is 39% sodium
  • 00:46:58
    and sodium chloride and once we
  • 00:47:00
    determine the grams or mass of sodium
  • 00:47:02
    chloride we can use this conversion
  • 00:47:04
    factor okay and we need to and we know
  • 00:47:07
    that we need to go from the mass of
  • 00:47:09
    sodium to the mass of sodium chloride by
  • 00:47:12
    using the converstion factor of
  • 00:47:15
    the um mass percent
  • 00:47:18
    composition so we can multiply 2.4 * 100
  • 00:47:24
    and divide it by 39
  • 00:47:28
    and this will give us the mass of
  • 00:47:30
    Southern chlorine that we need to eat
  • 00:47:32
    every day okay to obtain the uh amount
  • 00:47:37
    recommended of sodium per day by FDA so
  • 00:47:41
    if we wants to eat 2.4 grams of sodium
  • 00:47:44
    we need to really eat 6.2 grams of
  • 00:47:47
    sodium chloride okay in in our diet in a
  • 00:47:50
    way that we can consume the uh levels
  • 00:47:53
    that FDA can for adults
  • 00:48:00
    So based on the chemical formula the
  • 00:48:02
    mass percent of element Chlorine in
  • 00:48:05
    compound C cl2 F2 is as follow mass
  • 00:48:08
    percent of Cl would be two times the
  • 00:48:11
    molar mass of chlorine because we have
  • 00:48:13
    two moles of chlorine there divided by
  • 00:48:17
    the mol mass of the compound C2 F2 * 100
  • 00:48:22
    okay so because we have two moles of
  • 00:48:25
    that element
  • 00:48:27
    then we need to multiply twice the M
  • 00:48:29
    mass of that element and that will give
  • 00:48:32
    gave us the mass of that element in this
  • 00:48:35
    in this molecule this compound and we
  • 00:48:38
    then divide it by the M mass of the
  • 00:48:39
    whole compound times 100 that will give
  • 00:48:42
    us the mass percent for chlorine the
  • 00:48:44
    same thing will be with Florine because
  • 00:48:46
    we have two if we need to calculate the
  • 00:48:48
    mass percent for Florine will be two
  • 00:48:51
    times the mar mass of Florine divided by
  • 00:48:54
    the mar mass of CCL
  • 00:48:57
    two and if this will be the mass percent
  • 00:49:00
    of carbon will be just one time the M
  • 00:49:02
    mass of carbon divided by the molar mass
  • 00:49:06
    of the whole um
  • 00:49:09
    compound so let's do an example
  • 00:49:12
    calculate the mass percent of Chlorine
  • 00:49:14
    in C in C2 cl4 F2 that is fre on
  • 00:49:19
    114 so we have the compound we need to
  • 00:49:24
    calculate the percent mass of chlorine
  • 00:49:27
    so we have how many chlorines per
  • 00:49:30
    molecule of C2 cl4 F2 we have four so
  • 00:49:34
    that means that we need to multiply the
  • 00:49:36
    mar mass of chlorines Time 4 and then
  • 00:49:38
    divide it by the mar mass of
  • 00:49:42
    c24
  • 00:49:43
    F2 so that's we have here the percent of
  • 00:49:47
    Chlorine okay will be four times the mar
  • 00:49:51
    mass of chlorine divided by the mar mass
  • 00:49:55
    of the compound
  • 00:49:57
    that will be the equation that we need
  • 00:49:59
    to
  • 00:50:05
    solve so in other words basically we
  • 00:50:07
    have the mass percent of an element X as
  • 00:50:09
    mentioned before it will be the mass of
  • 00:50:11
    element X in one mole okay divided by
  • 00:50:14
    one mole of the
  • 00:50:18
    compound so here we have in the in the
  • 00:50:22
    solution of 3 114 the M mass of chlorine
  • 00:50:26
    is
  • 00:50:27
    35.45 * 4 okay will be 41.8 this is the
  • 00:50:32
    marar mass for chlorine and the amount
  • 00:50:36
    of mass for chlorine in the compound and
  • 00:50:39
    the molar mass of the compound would be
  • 00:50:41
    two times the atomic Mar mass of carbon
  • 00:50:46
    four times the mar mass of chlorine and
  • 00:50:49
    two times the M mass of Florine so this
  • 00:50:51
    is the amount of mass due to Florine
  • 00:50:55
    this one due to Florine and this this is
  • 00:50:57
    to this due to carbon okay that all of
  • 00:51:01
    them will give us the molar mass of the
  • 00:51:04
    compound that is
  • 00:51:07
    23.8 gram okay so this is the mar mass
  • 00:51:10
    for the compound and this is the the
  • 00:51:12
    mass basically that we have from
  • 00:51:15
    chlorine in this compound so this will
  • 00:51:18
    be 141.5 that is this divided by 203
  • 00:51:23
    that is the mass for the compound time
  • 00:51:27
    100 so we have a total the mass percent
  • 00:51:31
    for chlorine of
  • 00:51:34
    6958 percent so that's the percent of
  • 00:51:37
    Florine I mean of chlorine in this uh
  • 00:51:41
    compound if we would like to determine
  • 00:51:44
    the percent of Florine it will be the M
  • 00:51:47
    mass of Florine times two and then that
  • 00:51:50
    that product you put that product here
  • 00:51:52
    divided by this amount because it's the
  • 00:51:53
    same compound so you need to use the
  • 00:51:55
    same Mar mass of the compound times 100
  • 00:51:59
    that will give you the M mass for I mean
  • 00:52:02
    the mass percent of the
  • 00:52:07
    floride so now that's we are talking
  • 00:52:09
    about fluoride fline fluoride
  • 00:52:12
    strengthens tool enemo which prevents
  • 00:52:16
    tooth
  • 00:52:17
    decay too much fluoride can cause teeth
  • 00:52:20
    to become Brown and spot a condition
  • 00:52:23
    known as Dental
  • 00:52:25
    fosis EX extremely high levels can led
  • 00:52:28
    to skeletal
  • 00:52:30
    fosis the scientific consensus is that
  • 00:52:34
    like many minerals fluoride shows some
  • 00:52:37
    health benefits at certain levels about
  • 00:52:40
    1 to four milligrams a day for adults
  • 00:52:42
    but can have tental effects at higher
  • 00:52:45
    levels so basically this is what is
  • 00:52:47
    recommended to have every single day
  • 00:52:50
    adults who drinks between one and two
  • 00:52:52
    liters of water per day would receive
  • 00:52:55
    the beneficial amount of fluoride from
  • 00:52:58
    the water okay so one of the principal
  • 00:53:01
    sources of fluoride is the water
  • 00:53:05
    fluoride is often added to water as
  • 00:53:08
    sodium
  • 00:53:09
    fluoride now what is the mass percent
  • 00:53:12
    composition of fline in sodium
  • 00:53:16
    fluoride how many grams of sodium
  • 00:53:19
    fluoride should be added to one uh to
  • 00:53:22
    1500 liters of water to fluoridate it at
  • 00:53:26
    levels of 1 milligram of Florine per
  • 00:53:29
    liter okay so this question must be
  • 00:53:32
    answered and we're going to work with
  • 00:53:33
    them by using all this conversion factor
  • 00:53:37
    uh we can do that by using conversion
  • 00:53:40
    factors so let's talk now about
  • 00:53:43
    empirical formulas from mass percent
  • 00:53:48
    composition here we have chemical
  • 00:53:50
    formula okay and we have the mass
  • 00:53:53
    percent composition an empirical formula
  • 00:53:55
    give us give us only the smallest whole
  • 00:53:58
    number ratio of each type of atom in a
  • 00:54:01
    compound not the specific number of each
  • 00:54:04
    type of atom in a molecule so this give
  • 00:54:07
    it just the ratio of a compound okay by
  • 00:54:11
    by looking at them at the element level
  • 00:54:15
    the molecular formula is always a whole
  • 00:54:18
    number multiple of the emperical form
  • 00:54:22
    for example the molecular formula for
  • 00:54:24
    hydrogen peroxide is H2 2 and its
  • 00:54:27
    empirical formula is ho okay so here we
  • 00:54:30
    have the molecular formula and this is
  • 00:54:33
    the empirical formula from this
  • 00:54:35
    molecular formula and sometimes we
  • 00:54:39
    have situation where the molecular
  • 00:54:42
    formula is the same as the peral formula
  • 00:54:44
    as for example in water H2O H2O is the
  • 00:54:47
    molecular formula but also is the
  • 00:54:50
    empirical
  • 00:54:51
    formula so the molecular formula is
  • 00:54:54
    equal to the empirical formula time n
  • 00:54:57
    where n is one 2 3
  • 00:55:00
    Etc n is equal to two four hydren
  • 00:55:04
    peroxide because here we see that hydren
  • 00:55:07
    peroxide is twice okay the empirical
  • 00:55:10
    formula of ho so that's why the N is
  • 00:55:14
    going to be equal to two for hydrogen
  • 00:55:20
    peroxide now we can calculate empirical
  • 00:55:22
    formula from experimental data we use
  • 00:55:25
    the example of water to give us uh to
  • 00:55:28
    tell you how how can you do that with
  • 00:55:31
    the compost a sample of water in the lab
  • 00:55:34
    and find that it produced three gram of
  • 00:55:37
    hydrogen and 24 gram of
  • 00:55:40
    oxygen by using this data how can we
  • 00:55:44
    determine the empirical formula okay so
  • 00:55:48
    we create a de composition of water here
  • 00:55:50
    and we observe that we have 30 I mean
  • 00:55:53
    three grams of hydrogens and 2 4 G of
  • 00:55:59
    oxy so how many moles of each element
  • 00:56:02
    are formed during the decomposition of
  • 00:56:04
    water the first step to determine the
  • 00:56:08
    empirical formula if we have as initial
  • 00:56:11
    data a mass we need to transform that
  • 00:56:14
    mass okay that data with unit of mass to
  • 00:56:17
    moles okay so if we have three grams of
  • 00:56:20
    hydrogen how can we change from grams to
  • 00:56:24
    moles using what
  • 00:56:27
    we can use the M Mass the units of M
  • 00:56:30
    Mass remember that is g per mole so we
  • 00:56:34
    can go
  • 00:56:35
    from the three G of hydrogen and divide
  • 00:56:39
    it here by the M mass of hydrogen 1.01
  • 00:56:43
    gram of hydrogen is equal to one mole
  • 00:56:45
    and by this we have the M of hydrogen
  • 00:56:47
    the same thing we can do with oxygen we
  • 00:56:49
    can multiply the mass of oxygen times
  • 00:56:52
    one mole and then divided by 16.0 gr of
  • 00:56:55
    oxygen this is the molar mass of oxygen
  • 00:56:59
    this is the molar mass of hydrogen and
  • 00:57:01
    it's flipped because we need to to have
  • 00:57:04
    those units of gram in the denominator
  • 00:57:07
    so that we can cancel the units of gram
  • 00:57:10
    here from hydrogen and also from oxygen
  • 00:57:13
    so here we have that we can produce
  • 00:57:15
    three moles of hydrogen okay and we can
  • 00:57:18
    produce 1.5 moles of oxygen so that
  • 00:57:21
    means that in this sample the ratio
  • 00:57:24
    basically goes from 3 moles of hydrogen
  • 00:57:27
    for every 1.5 moles of oxygen so we can
  • 00:57:32
    create and we can write what is known as
  • 00:57:34
    a pseudo formula for water it will be
  • 00:57:37
    h3o
  • 00:57:39
    1.5 okay but as we have been said before
  • 00:57:43
    we need to have basically um whole
  • 00:57:46
    number subscri okay we can't have
  • 00:57:49
    decimals
  • 00:57:51
    so but if we can look here 1.5 we
  • 00:57:54
    multiply by two okay B basically sorry
  • 00:57:57
    we need to divide this by the smallest
  • 00:58:00
    subscrip okay so we divide this by the
  • 00:58:02
    smallest subscrip and we have 1.5 ID 1.5
  • 00:58:07
    3id 1.5 and from here from this division
  • 00:58:10
    we have the subscrip okay for each
  • 00:58:13
    element and we can find out that the
  • 00:58:16
    molecular formula is H2O so our
  • 00:58:19
    empirical formula for water which in
  • 00:58:21
    this case also happens to be the
  • 00:58:22
    molecular formula is H2O so that's the
  • 00:58:25
    way that how how we can determine or we
  • 00:58:27
    can calculate the empirical formula by
  • 00:58:31
    using data okay of the mass of each of
  • 00:58:34
    the element that create that compound we
  • 00:58:37
    have before as we mentioned here the
  • 00:58:40
    composition of water producing three
  • 00:58:41
    Gams of hydrogen and 24 grams of oxygen
  • 00:58:44
    so by using this we can calculate we can
  • 00:58:48
    find out the empirical formula okay once
  • 00:58:51
    again we have the mass of the element we
  • 00:58:54
    divide that mass by the m mass of each
  • 00:58:56
    element and that will give us the um the
  • 00:59:01
    moles so we can write a sudo formula
  • 00:59:04
    okay as we WR WR here H3 1.5 but we need
  • 00:59:10
    to have whole numbers so to do that we
  • 00:59:12
    divide this two by the smallest one in
  • 00:59:16
    this case is 1.5 so 1.5 divided by 1.5
  • 00:59:19
    is 1 3id 1.5 is 2 so H2O is basically in
  • 00:59:24
    our case the emper formula as well as
  • 00:59:28
    the molecular
  • 00:59:31
    formula
  • 00:59:33
    so these are some rules or some steps to
  • 00:59:38
    uh determine the brdal formula from
  • 00:59:39
    experimental data first write down as
  • 00:59:42
    given the masses of each element present
  • 00:59:45
    in a sample of the compound if you are
  • 00:59:47
    given mass percent composition assume a
  • 00:59:49
    100 G sample and calculate the masses of
  • 00:59:52
    each element from the given percentage
  • 00:59:55
    then then convert each of the masses
  • 00:59:58
    this in step one two moles as I
  • 01:00:00
    mentioned before by using the
  • 01:00:02
    appropriate molar mass for each element
  • 01:00:04
    as a conversion factor so that way when
  • 01:00:07
    you use the M mass and multiply by the
  • 01:00:09
    mass you will have the
  • 01:00:11
    moles then you can write a pseudo
  • 01:00:14
    formula for the compound using the moles
  • 01:00:17
    of each element as a
  • 01:00:19
    subsrate and then you need to divide all
  • 01:00:22
    the subsrate in the formula by the
  • 01:00:24
    smallest subsrate
  • 01:00:28
    if the subss are not whole numbers
  • 01:00:31
    multiply all by all the subscrip by a
  • 01:00:34
    small whole number as I'm going to show
  • 01:00:36
    in the next slide uh table to arrive at
  • 01:00:39
    a whole number
  • 01:00:42
    subscript so see for
  • 01:00:44
    example uh there's a fractional
  • 01:00:46
    subscript that you're going to have is
  • 01:00:47
    0.1 you multiply by 10 is is2 you multip
  • 01:00:51
    5 25 multip by 4 23 multiply 3 so if
  • 01:00:56
    your subs is any of this this is the
  • 01:00:58
    factor that you need to multiply to have
  • 01:01:01
    eventually a whole
  • 01:01:04
    number following
  • 01:01:07
    example a 3.24 gram sample of titanium
  • 01:01:11
    reacts with oxygen to form 5.40 gram of
  • 01:01:15
    the metal oxide What is the empirical
  • 01:01:18
    formula of the metal oxide so we need to
  • 01:01:21
    know the emperical formula for the metal
  • 01:01:25
    oxide that is produced by Titanium with
  • 01:01:29
    oxygen so what we have initial is the
  • 01:01:32
    mass of titanium okay the sample of
  • 01:01:35
    titanium and the mass of the metal oxide
  • 01:01:39
    so we need to find out the empirical
  • 01:01:42
    formula you cannot convert mass of metal
  • 01:01:46
    oxide into moles because you don't know
  • 01:01:49
    the uh chemical formula for this metal
  • 01:01:52
    oxide yet so and that's the what you
  • 01:01:56
    need eventually that's what you need to
  • 01:01:57
    to find out okay
  • 01:01:59
    so what we can do first of all you're
  • 01:02:02
    given the mass of the initial titanium
  • 01:02:05
    and the mass of the oxide but we need to
  • 01:02:08
    determine the mass of oxygen how can you
  • 01:02:12
    think that we can determine the mass of
  • 01:02:14
    oxygen if this is what we
  • 01:02:17
    have so remember the law of conservation
  • 01:02:20
    of mass that can be destroyed they're
  • 01:02:22
    created so we have here that the total
  • 01:02:26
    mass of the com of the the reaction of
  • 01:02:29
    tanum with oxygen is 5.40 so if we
  • 01:02:33
    subtract
  • 01:02:34
    3.24 or 540 5.40 what we're going to
  • 01:02:39
    have is the mass of oxygen because this
  • 01:02:42
    metal oxide is the com is the
  • 01:02:43
    combination of the mass of oxygen plus
  • 01:02:47
    the titanium so if we subtract the mass
  • 01:02:49
    of the titanium we're going to have the
  • 01:02:51
    mass of the oxygen so the difference of
  • 01:02:54
    this will be the mass of
  • 01:02:56
    oxygen that combined with the titanium
  • 01:02:58
    will produce the metal
  • 01:03:01
    oxide so to find the mass of oxygen we
  • 01:03:04
    need to subtract the mass of titanium
  • 01:03:06
    from the mass of the metal oxide that
  • 01:03:08
    would be 5.40 minus 3.24 and this is the
  • 01:03:13
    mass of oxygen so now that we have the
  • 01:03:16
    mass of oxygen what we need to
  • 01:03:19
    do eventually we need
  • 01:03:21
    to convert that mass in moles and then
  • 01:03:25
    divide
  • 01:03:27
    divided by the smallest number okay once
  • 01:03:30
    we created the pseudo um empirical
  • 01:03:33
    formula okay we divided by the smallest
  • 01:03:35
    and then we can have the empirical
  • 01:03:41
    formula so now we can calculate
  • 01:03:43
    molecular formula four compounds from
  • 01:03:45
    emperical formula and form from Mar
  • 01:03:48
    masses so once we have the empirical
  • 01:03:51
    formula we can um determine the M uh the
  • 01:03:55
    molecular formul for that compound but
  • 01:03:58
    we need to have the M Mass to do that so
  • 01:04:01
    the molecular formula is always a whole
  • 01:04:03
    number multiple of the empirical
  • 01:04:05
    formula we need to find the n in the
  • 01:04:09
    expression okay remember that the
  • 01:04:11
    molecular formula is equal to the
  • 01:04:13
    empirical formulas time n so we need
  • 01:04:16
    this value of
  • 01:04:18
    n we can find n in the expression of
  • 01:04:23
    mass equal to the empirical formula mass
  • 01:04:25
    time n n will be equal to the m mass
  • 01:04:29
    divided by the empirical formula Mar
  • 01:04:31
    Mass okay so if we have the empirical
  • 01:04:34
    formula okay we can find out the molar
  • 01:04:37
    mass of that empirical formula and
  • 01:04:39
    somehow in the problem they must provide
  • 01:04:42
    information about the marolar mass so we
  • 01:04:44
    can divide the molar mass of the
  • 01:04:47
    compound divided by the marar mass of
  • 01:04:49
    the empirical formula and that will give
  • 01:04:51
    us the N so we can multiply to the
  • 01:04:54
    empirical formula and eventually have
  • 01:04:56
    there the molecular formula for that
  • 01:05:00
    compound so calculating the molecular
  • 01:05:04
    formula for compounds in this going to
  • 01:05:06
    do the example of
  • 01:05:08
    fructose so find out the molecular
  • 01:05:10
    formula for fructose with an with
  • 01:05:12
    empirical formula ch2o and the molar
  • 01:05:15
    mass is
  • 01:05:17
    18.2 gram per mole okay so the molecular
  • 01:05:21
    formula is a whole number multiple of
  • 01:05:24
    the ch2o this is the empirical formula
  • 01:05:27
    we need to calculate the N value okay so
  • 01:05:30
    we can multiply that n value for each of
  • 01:05:33
    the subsrate okay of this empirical
  • 01:05:35
    formula to create the molecular formula
  • 01:05:40
    for
  • 01:05:43
    frose so for fructose the empirical
  • 01:05:46
    formula mass will be 1 *
  • 01:05:49
    12.01 + 2 * 1.01 + 16 because we have
  • 01:05:54
    one carbon two hydrogen and one oxygen
  • 01:05:57
    so we can see here see one carbon two
  • 01:06:00
    hydrogen one oxygen so this is the we
  • 01:06:03
    need to multiply one by the uh Mar mass
  • 01:06:06
    of carbon two mol mass of hydrogen one M
  • 01:06:09
    mass of oxygen so that's what we have
  • 01:06:11
    here one M mass of carbon two M mass of
  • 01:06:15
    hydrogen one the M mass of oxygen and
  • 01:06:18
    this will be equal to
  • 01:06:20
    3.03 this is the m mass for the
  • 01:06:23
    empirical formula and they give
  • 01:06:26
    the um Mar mass for the compound so we
  • 01:06:30
    will now we can divide the mar molecular
  • 01:06:33
    mass Mar mass for the for fructose
  • 01:06:36
    divided by the M mass for the veral
  • 01:06:38
    formula and this will be equals to six
  • 01:06:40
    so the N is equal to six so that means
  • 01:06:44
    that we need to multiply each of the
  • 01:06:45
    subsquare of the empirical formula for
  • 01:06:48
    fructose times
  • 01:06:50
    six so the empirical is CH H2O so will
  • 01:06:54
    be 6 * 1 C6 C * 2 h12 6 * 1 o6 this now
  • 01:07:02
    is the molecular formula of so from the
  • 01:07:05
    empirical formula we can calculate the
  • 01:07:08
    and or evaluate or find out the
  • 01:07:12
    molecular formula of a
  • 01:07:15
    compound so calculating molecular
  • 01:07:17
    formulas for a compound use the marar
  • 01:07:20
    mass which is given and the emperical
  • 01:07:22
    formula Mar Mass which you can calculate
  • 01:07:25
    okay by the empirical
  • 01:07:28
    formula to determine the N that is the
  • 01:07:31
    integer by which you must multiply the
  • 01:07:33
    empirical formula to get the molecular
  • 01:07:36
    formula then multiply the subsrate in
  • 01:07:39
    the empirical formula by n to arrive to
  • 01:07:43
    the molecular formula so this is how you
  • 01:07:46
    can find out the molecular formula by
  • 01:07:49
    using the empirical
  • 01:07:52
    form so let's review our chapter six
  • 01:07:56
    the mole concept the mole is a specific
  • 01:07:58
    number 6.02 * 10 the^ 23r particles okay
  • 01:08:03
    that allows us to easily count atoms or
  • 01:08:06
    molecules by weighing them one mole of
  • 01:08:09
    any element has a mass equivalent to its
  • 01:08:11
    atomic mass in
  • 01:08:14
    grams one mole of any compound has a
  • 01:08:17
    mass equivalent to its formula mass in
  • 01:08:20
    grams and the mass of one mole of an
  • 01:08:23
    element or compound is its molar
  • 01:08:27
    mass about the chemical formulas and
  • 01:08:29
    chemical composition the chemical
  • 01:08:31
    formulas indicate the relative number of
  • 01:08:34
    each kind of element in a compound these
  • 01:08:37
    numbers are based on atoms or
  • 01:08:40
    moles by using the molar masses the
  • 01:08:43
    information in the chemical formula can
  • 01:08:44
    be used to determine determine the
  • 01:08:48
    relative masses of each kind of element
  • 01:08:50
    in a
  • 01:08:51
    compound and total mass of a sample
  • 01:08:56
    of a compound can be related to the
  • 01:08:59
    masses of the constituent elements
  • 01:09:01
    contained in the
  • 01:09:02
    compound and finally the empirical and
  • 01:09:05
    molecular formulas from Lab
  • 01:09:07
    data we can refer to the relative masses
  • 01:09:10
    of each kind of element within a
  • 01:09:12
    compound to determine the empirical
  • 01:09:15
    formula of the compound if the chemist
  • 01:09:18
    also knows the molar mass of the
  • 01:09:21
    compound he or she can also determine
  • 01:09:24
    its molecular
  • 01:09:27
    for and this will be all with chapter
  • 01:09:30
    six chemical composition from the
  • 01:09:33
    introductory Chemistry by Dr tro
  • 01:09:41
    [Music]
Tags
  • Chemistry
  • Moles
  • Chemical Composition
  • Sodium
  • CFCs
  • Molar Mass
  • Empirical Formula
  • Molecular Formula
  • Avogadro's Number
  • Conversions