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AP Chemistry
AP Chemistry - Notes - Chapter 3 : Stoichiometry
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A. Stoichiometry - the study of quantities of substances consumed and produced in chemical reactions.
1. Atomic mass - the average mass of an atom of an element in atomic mass units (amu)
a. Atomic masses are based on 12C ("carbon twelve"), which is assigned a value of exactly 12 atomic
mass units.
b. The atomic masses of other elements are determined by comparison to 12C.
c. Mass spectrometer - currently best method for determination of atomic masses of atoms
- Procedure :
- atoms or molecules are placed in abeam of high-speed electrons (knocks electrons off the
atoms or molecules giving them a positive charge - cations)
- an electric field is then applied accelerating these cations which causes each cation to create
a magnetic field
- cations pass through an applied magnetic field
- cations with the least mass are deflected more than heavier cations and masses are
determined by comparing amounts of deflection
e.g. If , in a mass spectrometer, 13C is found to have a mass 1.0836 times that of 12C, then the
atomic mass of 13C = 1.0836 (12 amu) = 13.003 amu
d. The mass spectrometer can also be used to determine the isotopic composition of an element. When
a sample of an element is placed in a mass spectrometer, a mass spectrum can be obtained which
indicates the relative amounts of the various isotopes present in the sample.
e. The average mass of an atom (or mole) of an element is the sum of the fractions of each isotope
times their mass.
e.g. Naturally occurring carbon is composed of two isotopes, 12C (98.89%) and 13C (1.11%). The
average atomic mass for an atom of carbon = ( .9889 x 12 amu) + ( .0111 x 13.003) = 12.01
2. The Mole (abbrev. mol) - SI unit for the amount of a substance and is equal to the number of atoms in
exactly 12 grams of pure 12C.
- Mass spectrometry has determined this number to be 6.02214 x 1023 (Avogadro's number) (We
will use 4 sig.fig : 6.022 x 1023).
a. Calculations involving the mole
- calculate mass from atoms and vice versa
- calculate moles from mass and vice versa
3. Molar mass - the mass in grams of one mole of a substance( equals the atomic mass in grams)
- be able to calculate the molar mass of atoms, ions or molecules or numbers of atoms, ions or
molecules from moles or vice versa
4. Mass percent composition - the percent composition of a substance by mass
Sample Problem : What is the percent composition of glucose (C6H12O6)
mass percent of carbon =
mass of carbon
x 100% =
x 100% = 39.99%
mass of compound
mass percent of hydrogen =
mass percent of oxygen =
mass of hydrogen
x 100% =
x 100% = 6.727 %
mass of compound
mass of oxygen
x 100% =
x 100% = 53.28 %
mass of compound
5. Determining the Formula of a Compound
a. Empirical formula - lowest whole number ratio of the atoms in a compound
- Determination from percent composition data :
Step 1 : Determine mass composition (if not given) from percent composition (assume 100g
and then percents convert to grams)
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Step 2 : Determine moles of each element in the compound.
Step 3 : Determine the lowest whole number ratio of each element in the compound (empirical
- Determination of empirical formula from combustion data :
- Use stoichiometry to determine the grams of C and H from the balanced chemical equation
(if the combusted substance contains oxygen, determine the mass of oxygen from the mass
of the original compound)
Sample problem : A substance contains only carbon, hydrogen and oxygen. When 0.510 g of this
substance is burned, 1.122 g of CO2 and 0.612 g of H2O are produced. What is the empirical formula
of this compound?
Step 1 : Determine moles of C and H :
mol C = 1.122 g CO2 (1 mol CO2/44.01g) = 0.02549 mol CO2 (1mol C/1 mol CO2)= .02549 mol C
mol H = 0.612 g H2O (1 mol H2O/18.02g) = .0354 mol H2O(2 mol H/1mol H2O) = .0708 mol H
Step 2 : Determine mass of C, H and O
Mass C = .02549 mol C (12.01 g/mol C) = 0.3061 g C
Mass H = .0708 mol H(1.01g/mol H) = .0715 g H
Mass O = total mass of cpd - (mass C +H)
= 0.510 - (0.3061 + .0715) = .0784 g O
Step 3 - Determine empirical formula as in above problem
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b. Molecular formulas - the actual ratio of the atoms of elements in a compound and are whole number
multiples of their empirical formulas. For example if the molecular formula of a compound was
N2O4, the empirical formula would be NO2. In this case the molecular formula is 2 times the
empirical formula ( NO2 x 2 = N2O4 ). The mass of the molecular formula is necessarily also 2
times that of the mass of the empirical formula. This relationship can then be used to determine the
molecular formula of a compound if the empirical formula and molecular mass of the compound
are known. Here is a sample problem.
6. Chemical Equations - represent what happens in a chemical reaction
a. reactant Æ products (read "Reactants yield products")
b. conservation of atoms (mass) - atoms are neither created nor destroyed in chemical reactions, they
are recombined to form different substances
- mass is neither created nor destroyed chemical reactions (as opposed to nuclear reactions)
- chemical reactions must therefore be balanced - have same kinds and numbers of atoms on both
sides of the yields sign (Æ)
c. The physical states of the substances involved in a chemical reaction are represented by symbols :
(aq) - aqueous (dissolved in water)
(g) - gas
(l) - liquid
(s) - solid
7. Balancing chemical equations - usually by inspection
- it helps to do the most complex substance first
- never change the correct formulas of reactants or products (change the amounts of the substances
represented, do not change the substances)
- balance by placing coefficients in front of the reactant or product species
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8. Stoichiometric Calculations : Amounts of Reactants and Products
a. The following diagram represents the basic conversions used in stoichiometry :
- Calculations between moles and mass and moles and numbers of particles have already been
covered (moles and volume conversion will be covered later). The only new thing here is the use
of the mole ratio in a balanced equation to convert moles of one substance (known) to moles of
another (unknown)
Sample : From the balanced equation for the Haber process (N2 + 3H2 Æ 2 NH3) we see
that one mole of nitrogen is needed for every three moles of hydrogen present. How
many moles of nitrogen are needed if 1.2 moles of hydrogen are present?
moles nitrogen needed = 1.2 mol H 2 ⟨
1 mol N 2
⟩ = .40 mol N 2
3 mol H 2
9. Calculations Involving a Limiting Reactant.
a. Stoichiometric quantities - the exact amounts of reactants needed so that no amount of any reactant
will be left unused
b. Limiting Reactant - a reactant that there is proportionally less of so that it is used up first and limits
the amount of product that can be produced
- you must first determine which reactant is the limiting reactant when doing stoichiometry
problems involving limiting reactants
- to determine which reactant is limiting arbitrarily choose one reactant and use the mole ratio of
reactants to see if you have an excess or not of the other reactant
- e.g. Lets say you have 100.0 g of hydrogen and 200.0 grams of nitrogen and you are
going to combine them to form ammonia according to the Haber Process :
N2 + 3H2 Æ 2NH3
You can start with either the 100.0 g of H2, or the 200.0 g of N2 and convert to the other
reactant. Starting with grams of N2 and converting to moles of H2 :
200.0 g N 2 ⟨
1 mol N 2 3 mol H 2
⟩ = 21.41 mol H 2
28.02 g 1 mol N 2
Converting grams of H2 to moles H2 :
100.0 g H 2 ⟨
1 mol H 2
⟩ = 49.5 mol H 2
2.02 g
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Since the 200.0 g of N2 equates to a smaller quantity of H2 than what is available, N2 is
the limiting reactant. Or, to put it another way, since the 100.0g of H2 present is much
more than needed for the 200.0 g of N2, H2 is the excess reactant.
c. Excess reactant -opposite of limiting reactant - some will remain at the completion of a reaction
d. Theoretical yield - the amount of product a reaction should yield if 100% of the limiting reactant is
converted into the product
e. Actual yield - the amount of product that is actually produced when the reaction takes place
f. Percent yield - a measure of the efficiency of a chemical reaction or process
Percent yield =
actual yield
theoretical yield
x 100%