Chem 1
Dr. Casagrande / Mrs. Rinaldi
Chapter 11 Review Sheet Answer Key
Problem Sets
74. What is molar mass?
Molar mass is the mass in grams of one mole of any element or compound. For an element, it is the average
atomic mass in grams. For a compound it is the sum of the molar masses of all of the elements (each
multiplied by its subscript).
75. Which contains more atoms, a mole of silver atoms or a mole of gold atoms Explain your answer.
They both contain the same number of atoms because a mole of anything contains 6.022×1023 particles.
76. Discuss the relationships that exist between the mole, molar mass, and Avogadro’s number.
Molar mass is the mass in grams of one mole of any substance. Avogadro’s number is the number of
representative particles in one mole. The mass of 6.022×1023 representative particles of a substance is the
molar mass of the substance.
77. Which has a greater mass, a mole of silver atoms or a mole of gold atoms? Explain your answer.
A mole of gold atoms has more mass since Au has a higher molar mass than silver.
78. Explain the difference between atomic mass (amu) and molar mass (gram).
Atomic mass (amu) is the mass of an individual particle, while molar mass is the mass of one mole of
particles.
82. Which of the following molecules contains the most moles of carbon atoms per mole of the compound:
ascorbic acid (C6H8O6), glycerin (C3H8O3), or vanillin (C8H8O3)? Explain.
The formula for vanillin shows that it has 8 moles of carbon per mole of vanillin while ascorbic acid has 6
mole of carbon and glycerin has 3, so 1 mole of vanillin contains the most moles of carbon.
89. Determine the number of representative particles in each of the following:
6.022 ×10 23 atoms Ag
23
a. ? atoms Ag = 0.250 mol Ag ×
= 1.51×10 atoms Ag
1 mol Ag
6.022 ×10 23 f.un. NaCl
= 5.15 ×10 21 f.un. NaCl
1 mol NaCl
23
6.022 ×10 molec CO2
= 2.13 ×10 25 molec CO2
c. ? molec CO2 = 35.3 mol CO2 ×
1 mol CO2
–3
b. ? f.un. NaCl = 8.56 ×10 mol NaCl ×
23
6.022 ×10 molec N 2
= 2.56 ×10 23 molec N 2
1 mol N 2
90. Determine the number of moles in each of the following:
1 mol Pb
a. ? mol Pb = 3.25 ×10 20 atoms Pb ×
= 5.40 ×10−4 mol Pb
23
6.022 ×10 atoms Pb
1 mol glucose
24
= 8.24 mol glucose
b. ? mol glucose = 4.96 ×10 molec glucose ×
6.022 ×10 23 molec glucose
d. ? molec N 2 = 0.425 mol N 2 ×
1 mol NaOH
= 0.259 mol NaOH
6.022 ×10 23 f.un. NaOH
1 mol Cu 2+
2+
25
2+
= 20.8 mol Cu 2+
d. ? mol Cu = 1.25 ×10 ions Cu ×
23
2+
6.022 ×10 ions Cu
91. Make the following conversions.
a. 1.51×1015 atoms Si to mol Si
1 mol Si
? mol Si = 1.51×1015 atoms Si ×
= 2.51×10 –9 mol Si
23
6.022 ×10 atoms Si
–2
b. 4.25×10 mol H2SO4 to molecules H2SO4
6.022 ×10 23 molec H 2SO4
? molec H 2SO4 = 4.25 ×10 –2 mol H 2SO4 ×
= 2.56 ×10 22 molec H 2SO4
1 mol H 2SO4
c. ? mol NaOH = 1.56 ×10 23 f.un. NaOH ×
92. How many molecules are contained in each of the following?
23
a. ? molec CS2 = 1.35 mol CS2 × 6.022 ×10 molec CS2 = 8.13 ×10 23 molec CS2
1 mol CS2
23
b. ? molec As2O3 = 0.254 mol As2O3 × 6.022 ×10 molec As2O3 = 1.53 ×10 23 molec As2O3
1 mol As2O3
93. How many moles contain each of the following?
1 mol CO2
15
= 2.08 ×10−9 mol CO2
a. ? mol CO2 = 1.25 ×10 molec CO2 ×
23
6.022 ×10 molec CO2
1 mol NaNO3
21
= 5.96 ×10−3 mol NaNO3
b. ? mol NaNO3 = 3.59 ×10 f.un. NaNO3 ×
23
6.022 ×10 molec NaNO3
18
96. If a snowflake contains 1.9×10 molecules of water, how many moles of water does it contain?
1 mol H 2O
? mol H 2O = 1.9 ×1018 molec H 2O ×
= 3.2 ×10 –6 mol H 2O
23
6.022 ×10 molec H 2O
99. Make the following conversions:
9.65 g Li
a. ? g Li = 3.50 mol Li ×
= 24.3 g Li
1 mol Li
1 mol Co
b. ? mol Co = 7.65 g Co ×
= 0.130 mol Co
58.93 g Co
101. Convert the following to mass in grams.
1 mol H
1.008 g H
25
×
= 145 g H
b. ? g H = 8.65 ×10 atoms H ×
23
6.022 ×10 atoms H 1 mol H
!###########################"
143.6 mol H
1 mol O
16.00 g O
c. ? g O = 1.25 ×10 22 atoms O ×
×
= 0.332 g O
23
1 mol O
6.022 ×10 atoms O
!###########################
"
0.02076 mol O
103. Calculate the number of atoms in each of the following:
1 mol Hg 6.02 ×10 23 atoms Hg
a. ? atoms Hg = 25.8 g Hg ×
×
= 7.75 ×10 22 atoms Hg
200.6 g Hg
1 mol Hg
!##############"
0.1286 mol Hg
1 mol Ar 6.022 ×10 23 atoms Ar
c. ? atoms Ar = 150 g Ar ×
×
= 2.3 ×10 24 atoms Ar
39.95 g Ar
1 mol Ar
!#############"
3.76 mol Ar
108. How many moles of oxygen atoms are contained in the following?
4 mol O
= 10.0 mol O
1 mol KMnO4
b. ? mol O = 45.9 mol CO2 × 2 mol O = 91.8 mol O
1 mol CO2
112. Calculate the molar mass of each of the following:
a. ascorbic acid (C6H8O6)
6 C × 12.01 g +8 H × 1.008 g +6 O × 16.00 g =176.12 g/mol C6H8O6
b. sulfuric acid (H2SO4)
2 H × 1.008 g +1 S × 32.07 g +4 O × 16.00g =98.09 g/mol H2SO4
a. ? mol O = 2.50 mol KMnO4 ×
pg. 2
Chapter 11 Review Key
114. How many moles are in 100.0 g of each of the following compounds?
a. dinitrogen monoxide (N2O)
MM = 2 N × 14.01 g + 1 O × 16.00 g = 44.02 g/mol N2O
1 mol N 2O
? mol N 2O = 100.0 g N 2O ×
= 2.272 g N 2O
44.02 g N 2O
b. methanol (CH3OH)
MM = 1 C × 12.01 g + 4 H × 1.008 g + 1 O × 16.00 g = 32.04 g/mol CH3OH
1 mol CH 3OH
? mol CH 3OH = 100.0 g CO2 ×
= 3.121 mol CH 3OH
32.04 g CH 3OH
115. What is the mass of each of the following?
a. MM = 1 Cu × 63.55 g + 2 Cl × 35.45 g = 135.45 g/mol CuCl2
135.45 g CuCl 2
= 6.05 g CuCl 2
1 mol CuCl 2
b. MM = 1 Ca × 40.08 g + 2 O × 16.00 g + 2 H × 1.008 g = 74.10 g/mol Ca(OH)2
74.10 g Ca(OH)2
? g Ca(OH)2 = 1.25 × 10 2 mol Ca(OH)2 ×
= 9.26 × 103 g Ca(OH)2
1 mol Ca(OH)2
116. Determine the number of moles in each of the following.
a. MM = 2 Na × 22.99 g + 1 S × 32.07 g = 78.05 g/mol Na2S
1 mol Na 2S
? mol Na 2S = 1.25 × 10 2 g Na 2S ×
= 1.60 mol Na 2S
78.05 g Na 2S
b. MM = 2 H × 1.008 g + 1 S × 32.07 g = 34.09 g/mol H2S
1 mol H 2S
? mol H 2S = 0.145 g H 2S ×
= 4.25 × 10 –3 mol H 2S
34.09 g H 2S
? g CuCl 2 = 4.50 × 10 –2 mol CuCl 2 ×
124. Calculate the mass of 3.62×1024 molecules of glucose (C6H12O6)
MM = 6 C × 12.01 g + 12 H × 1.008 g + 6 O × 16.00 g = 180.2 g/mol C6H12O6
1 mol C6H12O6
180.2 g C6H12O6
? g C6H12O6 = 3.62 ×10 24 mlcl C6H12O6 ×
×
= 1080 g C6H12O6
23
6.022 ×10 mlcl C6H12O6 1 mol C6H12O6
!##################################"
6.011 mol C6H12O6
125. Determine the number of molecules of ethanol (C2H5OH) in 47.0 g.
MM = 2 C × 12.01 g + 6 H × 1.008 g + 1 O × 16.00 g = 46.07 g/mol C2H5OH
1 mol C 2 H 5OH 6.02 ×10 23 mlcl C 2 H 5OH
? molec C 2 H 5OH = 47.0 g C 2 H 5OH ×
×
= 6.14 ×10 23 mlcl C 2 H 5OH
46.07 g C 2 H 5OH
1 mol C 2 H 5OH
!#######################"
1.020 mol C 2 H 5OH
132. Determine the number of chloride ions in 10.75 g of magnesium chloride.
MM = 1 Mg × 24.31 g + 2 Cl × 35.45 g = 95.21 g/mol MgCl2
1 mol MgCl 2
2 mol Cl –
6.022 ×10 23 ions Cl −
−
23
–
? ions Cl = 10.75 g MgCl 2 ×
×
×
= 1.360 ×10 ions Cl
−
95.21 g MgCl 2 1 mol MgCl 2
1 mol Cl
!###################" !########
"
−
0.1129 mol MgCl 2
0.2258 mol Cl
136. Express the composition of each of the following as the mass percent of its elements (percent
composition).
a. sucrose (C12H22O11)
MM=12 C × 12.01 g + 22 H × 1.008 g + 11 O × 16.00 g = 342.3 g/mol C12H22O11
Chapter 11 Review Key
pg. 3
12 C × 12.01 g/mol
22 H × 1.008 g/mol
× 100% = 42.10% C ; %H =
× 100% = 6.472% H ;
342.3 g/mol
342.3 g/mol
11 O × 16.00 g/mol
%O =
× 100% = 51.42% O
342.3 g/mol
b. magnetite (Fe3O4)
MM=3 Fe × 55.85 g + 4 O × 16.00 g = 231.55 g
3 Fe × 55.85 g/mol
4 O × 16.00 g/mol
%Fe =
× 100 = 72.36% Fe ; %O =
× 100 = 27.64% O
231.54 g/mol
231.55 g/mol
c. aluminum sulfate (Al2(SO4)3)
MM = 2 Al × 26.98 g + 3 S × 32.07 g + 12 O × 16.00 g = 342.2 g/mol
2 Al × 26.98 g/mol
3 S × 32.07 g/mol
%Al =
× 100% = 15.77% Al; %S =
× 100% = 28.12% S;
342.2 g/mol
342.2 g/mol
12 O × 16.00 g/mol
%O =
× 100% = 51.11% O
342.2 g/mol
138. Determine the empirical formula for each of the following compounds:
a. ethylene (C2H4) CH2
b. ascorbic acid (C6H8O6) C3H4O3
c. naphthalene (C10H8) C5H4
142. Aspartame, an artificial sweetener, has the formula C14H18N2O5. Determine the percent composition of
aspartame.
MM = 14 C × 12.01 g + 18 H × 1.008 g + 2 N × 14.01 g + 5 O × 16.00 g = 294.3 g/mol
14 × 12.01 g/mol
18 × 1.008 g/mol
%C =
× 100% = 57.12% C; %H =
× 100% = 6.165% H;
294.3 g/mol
294.3 g/mol
2 × 14.01 g/mol
5 × 16.00 g/mol
%N =
× 100% = 9.521% N; %O =
× 100% = 27.18% O
294.3 g/mol
294.3 g/mol
144. Monosodium glutamate (MSG) is sometimes added to food to enhance flavor. Analysis determined this
compound to be 35.5% C, 4.77% H, 8.29% N, 13.6% Na, and 37.9% O. What is the empirical formula for
MSG?
1 mol
1 mol
35.5 g C ×
= 2.96 mol C ÷ 0.592 = 5 mol C; 4.77 g H ×
= 4.73 mol H ÷ 0.592 = 8 mol H
12.01 g
1.008 g
1 mol
1 mol
8.29 g N ×
= 0.592 mol N ÷ 0.592 = 1 mol N; 13.6 g Na ×
= 0.592 mol Na ÷ 0.592 = 1 mol Na
14.01 g
22.99 g
1 mol
37.9 g O ×
= 2.37 mol O ÷ 0.592 = 4 mol O: C5H 8 NO4 Na
16.00 g
146. Vanadium oxide is used as an industrial catalyst. The percent composition of this oxide is 56.0% vanadium
and 44.0% oxygen. Determine the empirical formula for vanadium oxide.
1 mol
⎫
56.0 g V ×
= 1.10 mol V ÷ 1.10 = 1 × 2 = 2 mol V ⎪
50.94 g
⎪
⎬ V2 O5
1 mol
44.0 g O ×
= 2.75 mol O ÷ 1.10 = 2.5 × 2 = 5 mol O ⎪
⎪⎭
16.00 g
%C =
pg. 4
Chapter 11 Review Key
149. Analysis of a compound containing chlorine and lead reveals that the compound is 59.37% lead. The molar
mass of the compound is 349.0 g/mol. What is the empirical formula for the chloride? What is the
molecular formula?
1 mol
⎫
59.37 g Pb ×
= 0.2865 mol Pb ÷ 0.2865 = 1 mol Pb ⎪
207.2 g
⎪
⎬ PbCl4
1 mol
40.63 g Cl ×
= 1.146 mol Cl ÷ 0.2865 = 4 mol Cl ⎪
⎪⎭
35.45 g
1. Which has more atoms, 1 g of silver (Ag) or 1 g of gold (Au)? Explain your answer.
1 g of silver has more atoms because Ag has a smaller molar mass; in order to have the same mass, there
must be more of the lighter atoms.
2. What is the percent composition of the compound 2-propanol, C3H6O (rubbing alcohol)?
MM = 3 C × 12.01 g + 6 H × 1.008 g + 1 O × 16.00 g = 58.08 g/mol
36.03 g C
6.048 g H
%C =
× 100% = 62.04% C; %H =
× 100% = 10.41% H;
58.08 g C3H 6O
58.08 g C3H 6O
16.00 g O
× 100% = 27.55% O
58.08 g C3H 6O
3. If you decomposed 45.0 g of 2-propanol, how many grams of carbon would you obtain? [Use the %C from
the previous question.]
62.04 % C
? g C = 45.0 g C3H 6O ×
= 27.9 g C
100 % C3H 6O
4. One of the most deadly poisons, strychnine, has the composition 75.42% C, 6.63% H, 8.38% N; the rest is
oxygen. Calculate the empirical formula of strychnine, arranging the elements in the order given.
⎫
1 mol
75.42 g C ×
= 6.280 mol C ÷ 0.598 = 10.5 mol C × 2 = 21 ⎪
12.01 g
⎪
⎪
1 mol
6.63 g H ×
= 6.58 mol H ÷ 0.598 ≈ 11 mol H × 2 = 2 ⎪
⎪
1.008 g
⎬ C 21H 22 N 2O2
1 mol
8.39 g N ×
= 0.599 mol N ÷ 0.598 = 1 mol N × 2 = 2 ⎪
⎪
14.01 g
⎪
1 mol
9.57 g O ×
= 0.598 mol O ÷ 0.598 = 1 mol O × 2 = 2 ⎪
⎪
16.00 g
⎭
%O =
Chapter 11 Review Key
pg. 5