Chapter 3 Stoichiometry: Calculations with Chemical Formulas and Equations

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Chapter 3Stoichiometry Calculations with
Chemical Formulas and Equations
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Chemical Equations

• We represent chemical reactions by writing
  chemical equations.
• Because atoms are neither created nor destroyed
  in chemical reactions we always balance chemical
  equations.
• A chemical eqn. Is balanced when equal numbers of
  each type of atom appear on both sides of the
  eqn. (some reactions have further requirements).
• We balance a chemical equation by inserting
  coefficients in front of the chemical formulas.
  The coefficients indicate the relative numbers of
  molecules (or formula units) of each species that
  are involved in the reaction.
• In contrast, the subscripts of chemical formula
  show the relative numbers of atoms of a
  particular element that are present within a
  formula unit.
• Never change a subscript within a chemical
  formula to balance the equation.
• The text outlines a method for balancing
  equations (by inspection) on pp. 76-79.

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Example 2
Example 1
Atoms on left Atoms on right K
K Cl Cl O O
Atoms on left Atoms on right N
N H H O O
Balanced eqn
Balanced eqn
Patterns of chemical Reactivity We omit this now
because we will cover it in Chpt 4
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• Atomic and Molecular Weights
• Mass and weight are not the same. Mass is
  independent of location and is the quantity used
  in scientific statements.
• Many scientists get sloppy and say atomic
  weight when they mean atomic mass.
• The Atomic Mass Scale
• Remember that almost all the mass of an atom is
  contributed by the protons and neutrons in the
  nucleus.
• The proton and neutron each have a mass of
  approximately 1 amu, so that the mass of an atom
  (in amu) of a particular isotope is nearly equal
  to its mass number.
• For example, the mass of an atom of
  is 36.9658 amu.
• This is close to 37, but not exactly equal to it
  because the protons and neutrons are not exactly
  1 amu in mass and because of other reasons that
  we will discuss when study nucleus in chpt 21.

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Average Atomic Masses

• While every atom has an integer number of protons
  and an integer number of neutrons, and although
  each of these contributes about 1 amu mass to the
  total mass of the atom, there is more to
  consider.
• The periodic table lists an atomic mass of 35.453
  amu for Cl. This is not close to an integer.
  The reason being is that atomic masses listed on
  the periodic table are average atomic masses for
  the mixture of isotopes found in a typical sample
  of the element.
• While each isotope has nearly integral mass, the
  average of these masses, weighted for the
  prevalence of each isotope might not be nearly an
  integer. Most elements turn out to have average
  atomic masses that are nearly an integer simply
  because most have one predominate isotope and
  that isotopic mass makes their major contribution
  to the average.
• The following example shows how average atomic
  mass can be calculated from the mass and
  prevalence of each isotope of an element.
• NOTE Average atomic mass is really determined
  through experiment and these calculations explain
  the measured values.

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• Example Calculate the average atomic mass of
  silicon, given the following data
• Isotope Abundance () Isotopic Mass (amu)
• 28Si 92.23 27.97693
• 29Si 4.67 28.97649
• 30Si 3.10 29.97276
• The isotopes are not equally represented in a
  typical sample of silicon. IN order to calculate
  the average mass, we must let the 28Si isotope
  make the largest contribution and count the 30Si
  isotope slightly less than the 29Si.
• We can weight the isotopic masses for their
  relative importance by multiplying each by its
  prevalence and then adding
• (27.97693 amu)(.9923)(28.97649
  amu)(.0467)(29.97376 amu)(.0310) 28.0855 amu
• 28.1 amu

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Try this Approximate the average atomic mass of
neon given the following data Isotope Abundance
() Answer is 2.0 x 101 amu 20Ne 90.51 21Ne 0.
27 22Ne 9.22
Formula and Molecular Weights It should come as
no surprise that the mass of several atoms is
equal to the sum of the masses of the atoms. A
molecule is a group of atoms (as is a formula
unit for non-molecular compounds) and so we can
easily calculate molecular weights (in
amu). Given a chemical formula, we use the
subscripts to figure out how many atoms of each
element are present and then we add up the
average atomic mass of each atom. Example
calculate the molecular weight of vitamin C
(C6H8O6). Atomic masses of elements Carbon
12.011 amu hydrogen 1.00794 amu oxygen 15.9994
amu 6(12.011 amu)8(1.00794 amu)6(15.9994 amu)
176.12592 amu 176.13 amu
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Percent Composition A compound is made up of two
or more elements. Each element present will
contribute to the mass of the compound. The
present composition (by mass) expresses the
relative contribution to the molecular weight of
the compound by each element in the
compound. For element X,
Example calculate the percent composition (by
mass) of each element in phosphoric acid
(H3PO4). Molecular weight 3(1.00794
amu)(30.97376 amu)4(15.9994 amu) 97.99518
amu For a single molecule
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CHECK 3.08568 31.6074 65.3069
100.0000
The Mole Chemists work in mass units of grams.
We cannot weight out a small number of
amus. However, if we took a a large enough
sample of a substance, then the mass of the
sample would be several grams (and many, many
amus). If we took just the right number of
atoms, we could manage to have X grams of an
element whose atoms each has a mass of X amus.
For example, the most common isotope of carbon
has an atomic weight of 12.00 amu. If we weight
out 12.00 g of carbon, we have some particular
number (very large number) of carbon atoms. What
is the magic number? The convention is to define
the masses of all atoms by comparing them to the
12C standard. The carbon-12 atom is assigned a
mass of exactly 12 atomic mass units (amu). In
other words, one amu is a mass equal to 1/12th of
the mass of one carbon-12 standard. If an atom
was half as massive as the 12C atom, it would
have a mass of (0.5)(12 amu) 6 amu. This same
calculation can be done for something twice as
massive or any other multiple value.
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Since we are taking some number of (12 amu) 12C
atoms that have a total mass of 12 g, this is the
number of amus in a gram. All other atomic
masses are based upon the 12C standard. Thus,
taking this number of atoms for any element means
we have a sample whose mass in grams is
equivalent in magnitude to the mass of an atom in
amu. The number is 6.0221367 x 1023, and it is
called Avogadros number. We call Avogadro's
number of things (no matter what they are) a mole
of those things. We can now say that 1 mol of
12C has a mass of 12.00g, 1 mol of carbon sample
has a mass of 12.011 g, 1 mol of iron has a mass
of 55.847 g, etc. This can be extended to
molecules since they are just collections of
atoms. A water molecule has a mass of about
18.0153 amu, and a mole of water has a mass of
about 18.0153 g. Alternatively, we say that
18.0153 g of water must contain 6.022 x 1023
molecules. Since a mole of carbon has a mass of
12.011 g, we say that the molar mass of carbon is
12.011 g. Molar mass of water is 18.0153 g. We
can now interconvert between all these quantities.
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Example What is the mass in grams of 1.00 x 1012
Pb atoms?
Answer 3.44 x 10-10 g
Example Calculate the number of N,C,O, and H
atoms in 1.68 x 104g of urea (NH2)2CO.
Answer 6.74 x 1026 H atoms
Since percent composition and empirical formula
both express the relative amounts of elements
present in a compound, it should come as no
surprise that one can be determined from the
other. Example What is the empirical formula
of vitamin C, if it contains 40.92 C, 4.58 H,
and 54.5 O by mass? Next slide
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We are given only relative amounts. We can
assume any convenient size sample. Assume the
sample is 100.00 g of vitamin C. Outline of
Empirical Formula Method Step 1 Find the mass
of each element present. (This might involve an
assumption.) Step 2 Divide each elements mass
by its molar mass to obtain moles. Step 3
Write a chemical formula using the moles as
tentative subscripts. Step 4 Divide each
subscript by the smallest subscript. If all
subscripts are integers, you are done! Step 5
Multiply all subscripts by the smallest whole
number that converts them all to integers. If
they arent integers by now, youve made a
mistake!
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In a real lab you are unlikely to have complete
percent composition data. Instead, you take a
certain sample of an unknown compound and you
attempt to determine the mass of each element
present.
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Example When 11.5 g of ethanol is burned in
oxygen, 22.0 g of CO2 and 13.5 g of H2O is
produced. Find the empirical formula of
ethanol? The empirical formula is a subunit of
the molecular formula. Empirical molecular CH
C6H6 NO2 N2O4 H2O H2O Since the molecular
formula contains one or more copies of the
empirical formula, the molecular mass is always
a multiple of the empirical mass. Solve them by
finding the empirical formula and its mass,
determining how many times heavier the molecule
is than the empirical subunit , and multiplying
the empirical formula accordingly. Example
The empirical formula of acetic acid is found to
be CH2O. If the molar Mass of acetic acid is
60g, what is the molecular formula? The
empirical mass is (12.011g)2(1.00794 g)
15.9994 g) 30.026 g The molar mass of the
compound must be some multiple of this. 60g/
30.026 2 The molecular formula is C2H4O2.
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Quantitative Information from Balanced Equations

• Stoichiometry is the analysis of the mass
  relationships between the reactants and products
  of a chemical reaction. To perform
  stoichiometric calculations we always work in
  moles.
• 1. Obtain the balanced equation for the
  reaction.
• 2. Identify which quantities are known and
  convert them into moles.
• 3. use the coefficients of the balanced
  equation to relate the moles of the unknown
  quantities to the known quantities.
• 4. Convert the calculated numbers of moles (of
  unknown quantities) into the desired units
  (e.g. mass, volume, etc.).
• 5. Check that the answers are reasonable.
• (See figure 3.13)
• Details of Step 3 (textbook method)
• The text recommends that the moles of
  unknown quantities be calculated from the
  moles of known quantities through the factor
  label method. The conversion factors are
  obtained through equivalence relations derived
  from the coefficients in the balanced
  equations.

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Example How many moles of H2 can be formed by
complete reaction of 6.23 mole Li with water.
Example How many moles of H2 can be
formed by complete reaction of 6.23 mole Li with
water.
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Limiting Reagents

• If we combine two or more reactants and a
  chemical reaction occurs, it is unlikely that we
  would have mixed exactly the proper amounts
  together so that all the reactants are consumed
  without any excess remaining.
• Example 2H2 O2 ? 2H2O
• The stoichiometric amounts would be 2 parts H2
  per 1 part O2 such as
• 2 mol H2 and 1 mol O2
• or
• 4 mol H2 and 2 mol O2
• or
• 0.5 mol H2 and 0.25 mol O2
• Usually, the reactants are not present in the
  proportions suggested by the coefficients in the
  chemical equation. In fact, we usually do not
  even attempt to mix the reactants in the
  stoichiometric amounts. If we combine the
  reactants in other than the stoichiometric
  amounts, one reactant will be used up before
  the other(s) and the reaction will stop.
• The reactant which is completely consumed first
  LIMITS the amount of reaction that can occur and
  it is the limiting reagent. Any reactant that
  remains is Excess and we call that reactant an
  excess reagent. See figure 3.15

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Determination of limiting reagent (textbook
method 1)

• Calculate how much of each reactant is
  available, in moles.
• Choose one reactant, and then evaluate the number
  of moles of the other reactant required to
  react with the chosen substance.
• Compare the amount calculated in step 2 to the
  amount available.
• If the available amount of a given reactant is
  larger than the required amount it is an excess
  reagent.
• If the available amount of a given reactant is
  less than the required amount it is the
  limiting reagent.
• Example Urea is prepared through the reaction
  2NH3 CO2?(NH2)2CO H2O
• In one process, 637.2 g of NH3 are allowed to
  react with 1142 g of CO2. Which is the limiting
  reagent?
• Answer NH3 is the limiting reagent.

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Determination of Limiting Reagent (textbook
method 2 EZ method)

• Calculate how much of each reactant is
  available, in moles.
• For each reactant, evaluate how much of a given
  product can be produced.
• Whichever reactant would produce the smallest
  amount of a given product is the limiting
  reagent.
• Example Urea is prepared through the reaction
• 2NH3 CO2? (NH2)2CO H2O
• In one process, 637.2 g of NH3 are allowed to
  react with 1142 g of CO2. Which is the limiting
  reagent?
• Answer NH3 is the limiting reagent.
• Once the limiting reagent has been identified, it
  can be used to predict a) How much product will
  form. b) How much of the excess reagents will be
  consumed.
• Example Fe(s) S(l)? FeS(s)
• 7.62 g Fe are allowed to react with 8.67 g S.
  Which reagent is limiting? What mass of FeS will
  form? How much excess reagent will be left?

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Yields of Reaction

• Anytime that you calculate the amount of product
  from the amount of (limiting) reactant, you are
  assuming that the reaction will run perfectly,
  will go to completion, and that you will be able
  to extract all the product from the final
  reaction mixture.
• You are calculating the maximum possible amount
  of product. This is the theoretical yield. In
  real life, which is less than perfect, you never
  get the theoretical yield.
• The reaction will have by-products, you may not
  be able to extract all the product, and it may
  reach equilibrium relatively early. The amount
  of product that is obtained when the reaction is
  carried out is the actual yield. The actual
  yield is always less than the theoretical yield.
  The actual yield is an experimental result.
• We are often interested in how close the actual
  yield is to the theoretical yield. This is a
  measure of how well we did.
• To calculate percent yield

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Example

• What is the theor. amount of O2 that this
  reaction will give from 2.00 x 102 g
• nitroglycerin?
• b) If the actual yield is 6.55 g O2, what is the
  yield?

Answers a) Theo. Yield of oxygen is 7.05 g b)
Percent yield is 93.0

• Example From the combustion of 4.66 g butane,
  C4H10, and 11.1L of O2
• (density, 1.43 g/L)
• C4H10(g) O2(g)?CO2(g) 5H2O(g)
• 12.7 g of CO2 was collected.
• What is the percent yield?
• Calculate the mass of excess reactants?