CHAPTER 3 | Chemical Reactions and Earth’s Composition
3.85. Collect and Organize
From the percent composition of a compound containing copper, chlorine, and oxygen, we are to determine the
formula. We have to convert the mass to moles, then find the lowest whole-number molar ratio for the
elements in the compound.
Analyze
If we assume 100 g of the compound, the percentage of each of the elements (24.2% Cu, 27.0% Cl, 48.8% O)
gives us the mass of the elements in that 100 g amount. Those masses can be converted into moles using the
molar masses of the elements from the periodic table. Then we compare the moles to find the molar ratio.
Solve
1 mol
 0.381 mol Cu
63.55 g
1 mol
27.0 g Cl 
 0.762 mol Cl
35.45 g
1 mol
48.8 g O 
 3.05 mol O
16.00 g
Dividing each of these mole amounts by the lowest mole amount (0.381) gives a ratio of 1 Cu : 2 Cl : 8 O. The
empirical formula for the compound therefore is CuCl2O8.
24.2 g Cu 
Think about It
In this case, the molar ratio, upon dividing the moles in 100 g of the substance, comes out to a whole-number
ratio. We did not have to multiply to obtain whole numbers for this compound.
3.92. Collect and Organize
From the data obtained from the combustion analysis for a compound containing carbon, hydrogen, and
oxygen, we are asked to determine the mass of carbon and hydrogen in the given mass of the compound and to
find the mass and moles of oxygen. From there we are able to determine the molar ratios of C, H, and O.
Analyze
The moles and masses of C and H can be calculated directly from the combustion analysis results. The oxygen
content will be the difference in the mass of the carbon plus hydrogen in the compound and the mass of the
0.100 g sample. The moles of oxygen can then be determined using the molar mass of oxygen from the
periodic table, and the ratio of carbon to hydrogen to oxygen can be found to determine the empirical formula.
Solve
0.1783 g CO 2 
1 mol CO 2
1 mol C

 4.051  10 –3 mol C
44.01 g
1 mol CO 2
12.011 g C
 4.866  10 2 g C
1 mol
1 mol H 2 O
2 mol H
0.0734 g H 2 O 

 8.147  10 –3 mol H
18.02 g
1 mol H 2 O
4.051  10 –3 mol C 
1.008 g H
 8.212  10 –3 g H
1 mol
Total mass of C and H = 4.866  10 2 g + 8.212  10 –3 g = 5.687  10 2 g
8.147  10 –3 mol H 
Mass of O present = 0.100 g – 0.05687 g = 0.043 g O
1 mol
Moles of O in compound = 0.043 g O 
 2.7  10 –3 mol O
15.999 g
80
Chemical Reactions and Earth’s Composition | 81
Dividing the moles of C, H, and O by the lowest molar amount (2.7  10–3 mol) gives a ratio of 1.5 C : 3 H : 1 O.
Multiplying this through by 2 to obtain whole-number ratios, we get an empirical formula of C3H6O2.
Think about It
This problem involves an additional step to determine the mass of carbon and hydrogen present so that we can
calculate the mass (and therefore the moles) of oxygen in the compound.
3.94. Collect and Organize
The compound from the bark contains oxygen as well as carbon and hydrogen. In this problem, therefore, we
have to determine the mass of carbon and hydrogen in the given mass of the compound to find the mass and
moles of oxygen. From there we are able to get the molar ratios of C, H, and O.
Analyze
The moles and masses of C and H can be calculated directly from the combustion analysis results. The oxygen
content is the difference between the mass of the carbon plus hydrogen in the compound and the mass of the
40.5 mg sample. The moles of oxygen can then be determined using the molar mass of oxygen from the
periodic table and the ratio of carbon to hydrogen to oxygen found to determine the empirical formula. We can
find the molecular formula from the given molar mass of 162 g/mol.
Solve
0.1100 g CO 2 
1 mol CO 2
1 mol C

 2.50  10 –3 mol C
44.01 g
1 mol CO 2
12.01 g C
 0.0300 g C
1 mol
1 mol H 2 O
2 mol H
0.0225 g H 2 O 

 2.50  10 –3 mol H
18.02 g
1 mol H 2 O
2.50  10 –3 mol C 
1.01 g H
 0.00252 g H
1 mol
Total mass of C and H = 0.0300 g + 0.00252 g = 0.0325 g
Mass of O present = 0.0405 g – 0.0325 g = 0.0080 g O
1 mol
Moles of O in compound = 0.0080 g O 
 5.0  10 –4 mol O
16.00 g
Dividing the moles of C, H, and O by the lowest molar amount (5.0  10–4 mol) gives a ratio of 5 C : 5 H :1 O.
This gives an empirical formula of C 5H5O. The molar mass of this empirical formula is 81.09 g/mol. This is
half that of the molar mass for the molecular formula (given in the problem as 162 g/mol), so the molecular
formula is C10H10O2.
2.50  10 –3 mol H 
Think about It
This problem involves an additional step to determine the mass of carbon and hydrogen present so that we can
calculate the mass (and therefore the moles) of oxygen in the compound.
3.103. Collect and Organize
We have to determine first the moles of each reactant we have, then calculate the moles and mass of oxygen
each reactant would theoretically give. The reactant that gives the least amount of O2 is the limiting reactant
and that amount of O2 produced is the theoretical yield for the reaction.
Analyze
We are given the balanced chemical equation for the reaction. For every 4 mol of KO 2 used, 3 mol of O2 are
produced, and for every 2 mol of CO2 used, 3 mol of O2 are produced. The molar mass of KO2 computed from
the molar masses of the elements in the periodic table is 71.10 g/mol. For CO 2, the molar mass is 44.01 g/mol,
and for O2, the molar mass is 32.00 g/mol.
82 | Chapter 3
Solve
The theoretical yield of oxygen from the KO2 is
1 mol KO2
3 mol O2
32.00 g O2
2.50 g KO2 


 0.844 g O2
71.10 g
4 mol KO2 1 mol O2
The theoretical yield of oxygen from the CO2 is
1 mol CO2
3 mol O2
32.00 g O2
4.50 g CO2 


 4.91 g O2
44.01 g
2 mol CO2
1 mol O2
Because KO2 is limiting in this reaction, it is depleted first, producing only 0.844 g O2 in the reaction.
Think about It
Not only was KO2 present in lower gram amounts in the reaction it also has a significantly larger molar mass.
This means that it is present in the least amount for this reaction. Be careful, though; you must also consider
the molar ratio of the reactants to products in the balanced equation.
3.105. Collect and Organize
To solve this problem we first write a balanced chemical equation. To find the amount of excess reactant left
over at the end of the reaction, we have to determine the limiting reactant. Knowing that, we can compute the
number of moles (and then mass) of the excess reactant used in the reaction to subtract from what was initially
present before the reaction.
Analyze
The molecular formulas for the compounds are needed to write the balanced chemical equation. Ammonia is
NH3, hydrogen chloride is HCl, and ammonium chloride is NH 4Cl. We need the molar masses of the reactants
to compute the moles of each present. The molar mass of NH 3 is 17.03 g/mol, and for HCl the molar mass is
36.46 g/mol.
Solve
The balanced chemical equation is
NH3( g) + HCl( g)  NH4Cl(s)
The moles of NH3 present at the start of the reaction:
1 mol
3.0 g NH3 
 0.18 mol NH3
17.03 g
The moles of HCl present at the start of the reaction:
1 mol
5.0 g HCl 
 0.14 mol HCl
36.46 g
Comparing these two mole amounts and considering that 1 mol of NH 3 reacts with 1 mol of HCl in the
balanced equation, we see that HCl is the limiting reactant. Also, we know that, in the reaction 0.14 mol of
NH3 is used up in the reaction and therefore
0.18 mol – 0.14 mol = 0.04 mol of NH3
will be leftover. We can convert this excess of NH3 in moles to grams with the molar mass of NH3.
17.03 g
0.04 mol NH3 
 0.7 g NH3
1 mol NH3
Thus, 0.7 g of NH3 remains in excess at the end of the reaction.
Think about It
Although there was a higher gram amount of HCl in the reaction, it proved to be the limiting reactant. Notice,
too, for this problem we did not even need to consider the theoretical yield of the product. We focused only on
the amounts and molar relationships of the reactants.
Chemical Reactions and Earth’s Composition | 83
3.110. Collect and Organize
Given the mass of ions present in seawater, we are asked first to determine the moles of each ion present.
From that information, we can then determine whether there is sufficient chloride ion present to form
magnesium chloride and sodium chloride.
Analyze
To calculate the moles of each ion, we need the molar mass of each species. Since the mass of the electron is
negligible, the molar masses of the ions are the masses of the elements from the periodic table. From that
information, we can compute the moles for part a. For part b the moles of the chloride ion present must be
equal to or greater than the combined moles of chloride needed to form sodium chloride (NaCl) and
magnesium chloride (MgCl2).
Solve
(a) The molar amount of each ion present in seawater is
1 mol Cl –
19.4 g Cl – 
 0.547 mol Cl –
35.453 g
10.8 g Na + 
1 mol Na +
 0.470 mol Na +
22.990 g
1 mol Mg 2+
 0.0531 mol Mg 2+
24.305 g
(b) To form sodium chloride (NaCl), we would need 0.470 mol of chloride ions. To form 0.0531 mol of
magnesium chloride (MgCl2), we would need 0.0531  2 = 0.106 mol Cl–. Adding these gives 0.576 mol of
Cl– needed to form the chloride salts. We have only 0.547 mol of Cl –, so we do not have sufficient chloride
ion to form the salts.
1.29 g Mg 2+ 
Think about It
There must be charge balance in the seawater. Because there is insufficient chloride to form the salts, there
must be another anion present to maintain charge balance.
3.116. Collect and Organize
We first need a balanced chemical equation. We then consider how much phosgenite can be produced from
10.0 g of PbO and NaCl; we have to determine which of the two reactants is limiting. Finally, we are asked to
look carefully at the formula of the compound to determine the formulas of two lead compounds in
phosgenite.
Analyze
To determine the limiting reactant and theoretical yield, we use the balanced equation in part a and the molar
masses of PbO (223.2 g/mol), NaCl (58.44 g/mol), and phosgenite (Pb 2Cl2CO3, 545.3 g/mol).
Solve
(a) The balanced equation is
2 PbO (s) + 2 NaCl(aq) + H2O ( ) + CO2(g)  Pb2Cl2CO3(s) + 2 NaOH(aq)
(b) From 10.0 g of PbO, we can theoretically obtain
10.0 g PbO 
1 mol 1 mol Pb2 Cl2 CO3 545.3 g Pb2 Cl2 CO3


 12.2 g Pb2 Cl2 CO3
223.2 g
2 mol PbO
1 mol
From 10.0 g of NaCl, we can theoretically obtain
1 mol 1 mol Pb2 Cl2 CO3 545.3 g Pb2 Cl2 CO3


 46.7 g Pb 2Cl2CO3
58.44 g
2 mol NaCl
1 mol
PbO is the limiting reactant, and 12 g of the phosgenite will be produced.
(c) Because chloride has a 1– charge and carbonate has a 2– charge as a polyatomic anion, in the formula
Pb2Cl2CO3 we can see PbCl2 and PbCO3.
10.0 g NaCl 
84 | Chapter 3
Think about It
Because the molar mass of sodium chloride is low, the number of moles of NaCl in 10 g is much greater than the
moles of PbO present in 10.0 g. It is not surprising, therefore, that PbO is the limiting reactant in this problem.