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CHAPTER THREE
STOICHIOMETRY
For Review
1.
Counting by weighing utilizes the average mass of a particular unit of substance. For marbles,
a large sample size will contain many different individual masses for the various marbles.
However, the large sample size will have an average mass so that the marbles behave as if
each individual marble has that average mass. This assumption is valid as long as the sample
size is large. When a large sample of marbles is weighed, one divides the total mass of
marbles by the average mass of a marble and this will equal the number of marbles present.
For atoms, because we can’t count individual atoms, we “count” the atoms by weighing them
and convert the mass in grams to the quantity of atoms in the sample. The mole scale of
atoms is a huge number (6.022 × 1023 atoms = 1 mol), so the assumption that a weighable
sample size behaves as a bunch of atoms each with an average mass is valid and very useful.
2.
The masses of all the isotopes are relative to a specific standard. The standard is one atom of
the carbon-12 isotope weighing exactly 12.0000 amu. One can determine from experiment
how much heavier or lighter any specific isotope is than 12C. From this information, we
assign an atomic mass value to that isotope. For example, experiment tells one that 16O is
about 4/3 heavier than 12C, so a mass of 4/3(12.00) = 16.00 amu is assigned to 16O.
3.
The two major isotopes of boron are 10B and 11B. The listed mass of 10.81 is the average
mass of a very large number of boron atoms.
4.
There are several ways to do this. The three conversion factors to use are Avogadro’s
number, the molar mass, and the chemical formula. Two ways to use these conversions to
convert grams of aspirin to number of H atoms are given below:
molar mass aspirin = 9(12.01) + 8(1.008) + 4(16.00) = 180.15 g/mol
1.00 g C9H8O4 ×
1 mol C 9 H 8 O 4
8 mol H
6.022 × 10 23 atoms H
×
×
180.15 g C 9 H 8 O 4
mol C 9 H 8 O 4
mol H
= 2.67 × 1022 H atoms
or
1.00 g C9H8O4 ×
1 mol C 9 H 8 O 4
6.022 × 10 23 molecules C 9 H 8 O 4
×
×
180.15 g C 9 H 8 O 4
mol C 9 H 8 O 4
8 atoms H
= 2.67 × 1022 H atoms
molecule C 9 H 8 O 4
Of course, the answer is the same no matter the order of the conversion factors.
2
5.
CHAPTER 3
STOICHIOMETRY
CxHyOz + oxygen → x CO2 + y/2 H2O
From the equation above, the only reactant that contains carbon is the unknown compound
and the only product that contains carbon is CO2. From the mass of CO2 produced, one can
calculate the mass of C present which is also the mass of C in CxHyOz. Similarly, all the
hydrogen in the unknown compound ends up as hydrogen in water. From the mass of H2O
produced, one can calculate the mass of H in CxHyOz. Once the mass of C and H are known,
the remainder of the compound is oxygen. From the mass of C, H, and O in the compound,
one can then go on to determine the empirical formula.
6.
The molecular formula tells us the actual number of atoms of each element in a molecule (or
formula unit) of a compound. The empirical formula tells only the simplest whole number
ratio of atoms of each element in a molecule. The molecular formula is a whole number
multiple of the empirical formula. If that multiplier is one, the molecular and empirical
formulas are the same. For example, both the molecular and empirical formulas of water are
H2O. For hydrogen peroxide, the empirical formula is OH; the molecular formula is H2O2.
7.
The product of the reaction has two A atoms bonded to a B atom for a formula of A2B. The
initial reaction mixture contains 4 A2 and 8 AB molecules and the final reaction mixture
contains 8 A2B molecules. The reaction is:
8 AB(g) + 4 A2(g) → 8 A2B(g)
Using the smallest whole numbers, the balanced reaction is:
2 AB(g) + A2(g) → 2 A2B(g)
2.50 mol A2 ×
2 mol A 2 B
= 5.00 mol A2B
mol A 2
The atomic mass of each A atom is 40.0/2 = 20.0 amu and the atomic mass of each B atom is
30.0 ! 20.0 = 10.0 amu. The mass of A2B = 2(20.0) + 10.0 = 50.0 amu.
15.0 g AB ×
1 mol A 2
40.0 g A 2
1 mol AB
×
×
= 10.0 g A2
30.0 g AB 2 mol AB
mol A 2
From the law of conservation of mass, the mass of product is:
10.0 g A2 + 15.0 g AB = 25.0 g A2B
or by stoichiometric calculation:
15.0 g AB ×
1 mol AB 1 mol A 2 B 50.0 g A 2 B
×
×
= 25.0 g A2B
30.0 g AB
mol AB
mol A 2 B
or
10.0 g A2 ×
1 mol A 2
2 mol A 2 B 50.0 g A 2 B
×
×
= 25.0 g A2B
40.0 g A 2
mol A 2
mol A 2 B
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CHAPTER 3
STOICHIOMETRY
Generally, there are several ways to correctly do a stoichiometry problem. You should choose
the method you like best.
8.
A limiting reactant problem gives you initial masses of at least two of the reactants and then
asks for the amount of product that can form. Because one doesn’t know which reactant runs
out first and hence determines the mass of product formed, this is a more difficult problem.
The first step in solving the problem is to figure which reactant runs out first (is limiting).
The strategy outlined in the text is to calculate the mole ratio of reactants actually present and
compare this mole ratio to that required from the balanced reaction. Whichever ratio is larger
allows one to deduce the identity of the limiting reactant and can, in turn, be used to calculate
the amount of product formed. Another strategy is to pick one of the reactants and then
calculate the mass of the other reactant required to react with it. By comparing the calculated
mass to the actual mass present in the problem, one can deduce the identity of the limiting
reactant and go on to solve the problem. A third common strategy is to assume each reactant
is limiting and then calculate for each reactant the amount of product that could form. This
gives two or more possible answers. The correct answer is the mass of product that is
smallest. Even though there is enough of the other reactant to form more product, once the
smaller amount of product is formed, the limiting reactant has run out.
9.
Balanced reaction: 2 SO2(g) + O2(g) → 2 SO3(g)
We have 6 SO2 and 6 O2 molecules present. If all six of the SO2 molecules react, then 3
molecules of O2 will react producing 6 molecules of SO3. These numbers were determined
using the balanced reaction. Since 6 molecules of O2 are present, and only 3 react when the
SO2 reacts completely, SO2 is limiting. The product mixture will contain 6 – 6 = 0 SO2
molecules, 6 – 3 = 3 O2 molecules in excess, and 6 molecules of SO3 formed.
96.0 g SO2 ×
1 mol SO 2
1 mol O 2
32.00 g O 2
×
×
= 24.0 g O2
64.07 g SO 2
2 mol SO 2
mol O 2
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CHAPTER 3
STOICHIOMETRY
Because 32.0 g O2 are actually present, O2 is in excess and SO2 is the limiting reactant. Note
that if the calculated amount of O2 was greater than 32.0 g, then we would have deduced that
O2 is limiting. Solving the rest of the problem:
96.0 g SO2 ×
10.
2 mol SO 3 80.07 g SO 3
1 mol SO 2
×
×
= 120. g SO3
64.07 g SO 2
2 mol SO 2
mol SO 3
Side reactions may occur. For example, in the combustion of CH4 (methane) to CO2 and
H2O, some CO may also form. Also, reactions only go part way to completion, instead
reaching a state of equilibrium where both reactants and products are present (see Ch. 13).