WORKSHEET #1 ATOMIC AND MOLECULAR MASS
What is the atomic mass of the following atoms? (Show units)
1. H
1.0 u
2. C
12.0 u
3. O
16.0 u
4. Cl
35.5 u
What is the molecular mass of the following compounds? (Show units)
5. KI
166.0 u
6. H2SO4
98.1 u
7. KNO3
101.1 u
8. Ca3(PO4)2 310.3 u
9. (NH4)2S  3(H2O) 122.1 u
10. C10H22
142.0 u
11. Na2C2O4 ● 7 H2O
260.0 u
WORKSHEET #2 THE MOLE CONCEPT
What is the mass of 1 mole of the following elements? (Show units)
1. H
1. 0 g
2. Na 23.0 g
3. O
16.0 g
What is the mass of 1 mole of the following substances? (Show units)
4. H2
2.0 g
5. O2
32.0 g
6. NaCl
58.5 g
7. ZnCl2
136.4 g
8. NaOH
40.0 g
9. H2SO4
98.1 g
10. Ba(OH)2  8(H2O) 315.3 g
WORKSHEET #3 MOLES AND GRAMS OF ELEMENTS
How many moles of atoms are in each of the following?
1. 10.3 g of sodium (Na)
10.3 g of Na
1.0 mol of Na
23.0 g of Na
= 0.4 moles of sodium atoms
2. 15.2 g of aluminum (Al)
15.2 g of Al
1.0 mol of Al
27.0 g of Al
= 0.6 moles of aluminum atoms
3. 10.3 g of cesium (Cs)
10.3 g of Cs
1.0 mol of Cs
132.9 g of Cs
= 0.08 moles of cesium atoms
4. 15.2 g of silicon (Si)
15.2 g of Si
1.0 mol of Si
28.1 g of Si
= 0.5 moles of silicon atoms
What is the mass of the following?
5. 18.0 moles of magnesium (Mg)
18.0 mol of Mg
24.3 g of Mg
1.0 mol of
Mg
= 437.4 g of Mg
6. 1.4 moles of argon (Ar)
1.4 mol of Ar
39.9 g of Ar
1.0 mol of Ar
= 55.9 g of argon
7. 18.0 moles of scandium (Sc)
18.0 mol of Sc
45.0 g of Sc
1.0 mol of
Sc
= 810.0 g of scandium
8. 1.4 moles of krypton (Kr)
1.4 mol of Kr
83.8 g of Kr
1.0 mol of Kr
= 117.3 g of krypton
WORKSHEET #4 MOLES AND GRAMS OF COMPOUNDS
How many moles of atoms are in each of the following?
1. 84.0 g of CaO
84.0 g of CaO
1.0 mol of CaO
56.1 g of CaO
= 1.5 moles of CaO
2. 21.7 g of LiBr
21.7 g of LiBr
1.0 mol of LiBr
86.8 g of LiBr
= 0.25 moles of LiBr
3. 44.0 g of CaS
44.0 g of CaS
1.0 mol of CaS
72.2 g of CaS
= 0.6 moles of CaS
4. 32.75 g of NaBr
32.75 g of NaBr
1.0 mol of NaBr
102.9 g of NaBr
= 0.3 moles of NaBr
What is the mass of the following?
5. 5.0 moles of water (H2O)
5.0 mol of H2O
18.0 g of H2O
1.0 mol of H2O
= 90.0 g of H2O
6. 3.4 moles of CuSO4
3.4 mol of CuSO4
159.6 g of CuSO4
1.0 mol of CuSO4
= 542.6 g of CuSO4
7. 15.0 moles of water
15.0 mol of H2O
18.0 g of H2O
1.0 mol of
H2O
= 270.0 g of H2O
8. 2.4 moles of CaSO4
2.4 mol of CaSO4
136.2 g of CaSO4
1.0 mol of CaSO4
= 326.9 g of CaSO4
WORKSHEET #5 MOLES AND NUMBERS
How many atoms are there in:
1. 2.3 moles
6.02 x 1023 atoms
2.3 mol
1.0 mol
= 1.4 x 10 atoms
2. 12.5 moles
24
12.5 mol
6.02 x 1023 atoms
1.0 mol
= 7.5 x 10 atoms
3. 3.3 moles
24
6.02 x 1023 atoms
3.3 mol
1.0 mol
= 2.0 x 1024 atoms
4. 10.5 moles
10.5 mol
6.02 x 1023 atoms
1.0 mol
= 6.3 x 10 atoms
How many moles equals:
5. 3.8 x 1025 atoms
24
3.8 x 1025
atoms
1.0 mol
6.02 x 1023 atoms
= 63.1 mol
6. 5.7 x 1028 molecules
5.7 x 1028 cules
1.0 mol
6.02 x 1023 cules
= 94684 mol = 9.5 x 104 mol
7. 4.8 x 1025 atoms
4.8 x 1025
atoms
1.0 mol
6.02 x 1023 atoms
= 79.7 mol
8. 3.7 x 1028 molecules
3.7 x 1028 cules
1.0 mol
6.02 x 1023 cules
= 6.1 x 104 mol
WORKSHEET #6 MOLE-MOLE PROBLEMS
The following questions refer to this equation:
4Fe(s) + 3O2(g) → 2Fe2O3(s)
1. How many moles of iron (III) oxide are produced when 2.4 moles of iron is oxidized?
2.4 mol Fe
2 mol Fe2O3
4 mol Fe
= 1.2 mol Fe2O3
2. How many moles of oxygen gas are consumed when 5.7 moles of ferric oxide are produced?
5.7 mol Fe2O3
3 mol O2
2 mol Fe2O3
= 8.6 mol O2
The following questions refer to this equation:
Cu(s) + 2H2SO4(aq) → CuSO4(aq) + 2H2O(l) + SO2(g)
3. How many moles of sulfuric acid are needed to react with 7.2 moles of copper metal?
7.2 mol Cu
2 mol H2SO4
1 mol Cu
= 14.4 mol H2SO4
4. How many moles of sulfer dioxide are produced when 3.9 moles of water is produced?
3.9 mol H2O
1 mol SO2
2 mol H2O
= 2.0 mol SO2
The following questions refer to this equation:
2Al(s)
+
3ZnCl2(aq)
→
2AlCl3(aq)
+
3Zn(s)
5. How many moles of aluminum chloride are produced when 3.5 moles of zinc chloride are
consumed?
3.5 mol ZnCl2
2 mol AlCl3
3 mol ZnCl2
= 2.3 mol AlCl3
6. How many moles of aluminum are needed to react with 1.28 moles of zinc chloride?
1.28 mol ZnCl2
2 mol Al
3 mol ZnCl2
= 0.85 mol Al
WORKSHEET #7 MASS-MASS PROBLEMS
The following questions refer to this equation:
2Al(s) + 3Br2(l) → 2AlBr3(s)
1. What mass of aluminum bromide is produced when 2.0 g of aluminum is consumed?
2.0 g Al
1 mol Al
2 mol AlBr3
266.7 g AlBr3
27.0 g Al
2 mol Al
1 mol AlBr3
= 19.8 g AlBr3
2. What mass of bromine is required to produce 10.0 g of aluminum bromide?
10.0 g AlBr3
1 mol AlBr3
3 mol Br2
159.8 g Br2
266.7 g AlBr3
2 mol AlBr3
1 mol Br2
= 9.0 g Br2
The following questions refer to this equation:
Mg(s) + 2HCl(aq) → MgCl2(s) + H2(g)
3. How many grams of magnesium are needed to produce 100 g of hydrogen gas?
100 g H2
1 mol H2
1 mol Mg
24.3 g Mg
2.0 g H2
1 mol H2
1 mol Mg
= 1215.0 g Mg
4. How much hydrogen gas is produced if 500 g of magnesium chloride is produced?
500 g MgCl2
1 mol MgCl2
1 mol H2
2.0 g H2
95.3 g MgCl2
1 mol MgCl2
1 mol H2
= 10.5 g H2
The following questions refer to this equation:
Cu(s) + 2H2SO4(aq) → CuSO4(aq) + 2H2O(l) + SO2(g)
5. What mass of sulfur dioxide is produced when 10.4 g of copper is reacted?
10.4 g Cu
1 mol Cu
1 mol SO2
64.1 g SO2
63.5 g Cu
1 mol Cu
1 mol SO2
= 10.5 g SO2
6. What mass of copper is needed to react with 15 g of sulfuric acid?
15 g H2SO4(aq)
= 4.9 g Cu
1 mol H2SO4(aq)
1 mol Cu
63.5 g Cu
98.1 g H2SO4(aq)
2 mol H2SO4(aq)
1 mol Cu
WORKSHEET #8 LIMITING REACTANT MASS-MASS PROBLEMS
1. If you have 25.0 g of sodium and 30.0 g of sulfuric acid, how many grams of hydrogen gas be
produced? What is the limiting reactant?
2Na(s) + H2SO4(aq) → Na2SO4(aq) + H2(g)
25.0 g Na
1 mol Na
1 mol H2
2.0 g H2
23.0 g Na
2 mol Na
1 mol H2
= 1.1 g H2
30.g H2SO4
1 mol H2SO4
1 mol H2
2.0 g H2
98.1 g H2SO4
1 mol H2SO4
1 mol H2
= 0.6 g H2
Therefore, sulfuric acid is the limiting reactant and sodium is in excess and 0.6 g of H2 is
produced.
2. If you have 13.0 g of copper and 20.0 g of nitric acid, how many grams of water is produced?
What is the limiting reactant?
3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
13.0 g Cu
1 mol Cu
4 mol H2O
18.0 g H2O
63.5 g Cu
3 mol Cu
1 mol H2O
= 4.9 g H2O
20.g HNO3
1 mol HNO3
4 mol H2O
18.0 g H2O
63.0 g HNO3
8 mol HNO3
1 mol H2O
= 2.9 g H2O
Therefore, nitric acid is the limiting reactant and copper is in excess and 2.9 g of H2O is
produced.
3. How much zinc chloride is produced when 10 g of zinc is reacted with 20 grams of
hydrochloric acid? What is the limiting reactant?
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
10.0 g Zn
1 mol Zn
1 mol ZnCl2
136.4 g ZnCl2
65.4 g Zn
1 mol Zn
1 mol ZnCl2
1 mol HCl
1 mol ZnCl2
136.4 g ZnCl2
36.5 g HCl
2 mol HCl
1 mol ZnCl2
= 20.9 g ZnCl2
20.0 g HCl
= 37.4 g ZnCl2
Zinc is the limiting reactant and hydrochloric acid is in excess and 20.9 g of ZnCl2 is produced.
WORKSHEET #9 MASS-VOLUME PROBLEMS AT STP
The following questions refer to this equation:
Zn(s) + 2HCl(aq)  ZnCl2(aq) + H2(g)
1. What volume of hydrogen gas is produced at STP when 10.0 g of hydrochloric acid are
consumed?
10.0 g HCl
1 mol HCl
1 mol H2
22.4 L H2
36.5 g HCl
2 mol HCl
1 mol H2
= 3.1 L H2
2. What mass of zinc was consumed if 10.0 L of hydrogen gas were produced at STP?
10.0 L H2
1 mol H2
1 mol Zn
65.4 g Zn
22.4 L H2
1 mol H2
1 mol Zn
= 29.2 g Zn
The following questions refer to this equation:
2CH3OH(l) + 3O2(g)  2CO2(g) + 4H2O(g)
3. What volume of carbon dioxide at STP will be produced if 7.5 g of methanol are consumed?
7.5 g CH3OH
1 mol CH3OH
2 mol CO2
22.4 L CO2
32.0 g CH3OH
2 mol CH3OH
1mol CO2
= 5.3 L CO2
4. What mass of methanol was combusted if 80 L of water vapor was produced at STP?
80 L H2O
1 mol H2O
22.4 L H2O
= 57.1 g CH3OH
2 mol CH3OH
32.0 g CH3OH
4 mol H2O
1 mol CH3OH
WORKSHEET #10 MASS-VOLUME PROBLEMS NOT AT STP
The following questions refer to this equation:
2Al(s) + 3Cl2(g)  2AlCl3(s)
1. What volume of chlorine gas at 1.0 atm and 22°C is needed to react with 5 g of aluminum?
5 g Al
1 mol Al
3 mol Cl2
27.0 g Al
= 0.28 mol Cl2
2 mol Al
V = (0.28 mol)(0.082 atm L / mol K)(295 K) / (1.0 atm) = 6.7 L
2. What mass of aluminum chloride is produced from 300 mL of chlorine gas at 98 kPa and
25°C?
n = (98 kPa)(0.300 L) / (8.31 kPa L / mol K)( 295 K) = 0.012 mol Cl2
0.012 mol Cl2
2 mol AlCl3
133.5 g AlCl3
3 mol Cl2
1 mol AlCl3
= 1.07 g AlCl3
The following questions refer to butane burned to produce carbon dioxide and water vapor.
2C4H10(g) + 13O2(g)  8CO2(g) + 10H2O(g)
3. What volume of carbon dioxide is produced when 0.025 g of butane is burned at 95 kPa and
1800°C?
0.025 g C4H10
1 mol C4H10
8 mol CO2
58.0 g C4H10
2 mol C4H10
= 0.0.002 mol C4H10
V = (0.002 mol C4H10)(8.31 kPa L / mol K)(2073 K) / (95 kPa) = 0.36 L
4. What mass of water vapor is produced when 0.5 L of oxygen is consumed at 743 mmHg and
1900°C?
n = (743 mmHg)(0.5 L)/ (62.3 mmHg L / mol K)(2173 K) = 0.003 mol O2
0.003 mol O2
= 0.04 g
10 mol H2O
13 mol O2
18.0 g H2O
1 mol H2O
WORKSHEET #11 VOLUME-VOLUME PROBLEMS AT STP
The following questions refer to burning methane to produce carbon dioxide and water vapor.
1. What volume of carbon dioxide is produced when 100 L of methane are burned at STP?
100 L CH4
1 L CO2
1 L CH4
= 100 L CO2
2. What volume of water vapor is produced when 250 L of carbon dioxide is produced at STP?
250 L CO2
2 L H2O
1 L CO2
= 500 L H2O
The questions refer to the following reaction which occurs at STP:
2C2H6(g) + 7O2(g)  4CO2(g) + 6H2O(g)
3. What volume of oxygen is required to burn 2000 L of ethane (C2H6) at STP?
2000 L C2H6
7 L O2
2 L C2H6
= 7000 L O2
4. What volume of carbon dioxide is produced when 2000 L of ethane is burned at STP?
2000 L C2H6
4 L CO2
2 L C2H6
= 4000 L CO2
WORKSHEET #12 PERCENT COMPOSITION FROM A FORMULA
What is the percent composition of the following?
1. NaBr
1 mol Na = 23.0 g
1 mol Br = 79.9 g
1 mol NaBr = 102.9 g
% Na = (23.0 g / 102.9 g) x 100% = 22.4%
% Br = (79.9 g / 102.9 g) x 100% = 77.6%
2. H2O2
2 mol H = 2.0 g
2 mol O = 32.0 g
1 mol H2O2 = 34.0 g
% H = (2.0 g / 34.0 g) x 100% = 5.9 %
% O = ( 32.0 g / 34.0 g) x 100% = 94.1%
3. Al2(SO4)3
2 mol Al = 54.0 g
3 mol S = 96.3 g
12 mol O = 192.0 g
1 mol Al2(SO4)3 = 342.3 g
% Al = (54.0 g / 342.3 g) x 100% = 15.8%
% S = (96.3 g / 342.3 g) x 100% = 28.1%
% O = (192.0 g / 342.3 g) x 100% = 56.1%
4. Fe2O3
2 mol Fe = 111.6 g
3 mol O = 48.0 g
1 mol Fe2O3 = 159.6 g
% Fe = (111.6 g / 159.6 g) x 100% =69.9%
% O = (48.0 g / 159.6 g) x 100% = 30.1%
5. Na2O2
2 mol Na = 46.0 g
2 mol O = 32.0 g
1 mol Na2O2 = 78.0 g
% Na = (46.0 g / 78.0 g) x 100% = 59.0%
% O = (32.0 g / 78.0 g) x 100% = 41.0%
6. Al2(SO3)3
2 mol Al = 54.0 g
3 mol S = 96.3 g
9 mol O = 144.0 g
% Al = (54.0 g / 294.3 g) x 100% = 18.3%
% S = (96.3 g / 294.3 g) x 100% = 32.7%
% O = (144.0 g / 294.3 g) x 100% = 48.9% or about 49%
1 mol Al2(SO3)3 = 294.3 g
WORKSHEET #13 EMPIRICAL FORMULA FROM PERCENT COMPOSITION
1. A compound that consists of 88.8% copper and 11.2% oxygen.
88.8 g of Cu
1.0 mol of Cu
63.5 g of Cu
= 1.4 mol Cu
11.2 g of O
1.0 mol of O
16.0 g of O
= 0.7 mol O
1.4 mol Cu / 0.7 mol O = 2 Cu / 1 O, therefore the empirical formula is Cu 2O
2. A compound that consists of 36.7% potassium, 33.3% chlorine, and 30.0% oxygen.
36.7 g of K
1.0 mol of K
39.1 g of K
= 0.94 mol K (sometimes you need to keep more digits to get nice ratios later)
33.3 g of Cl
1.0 mol of Cl
35.5 g of Cl
= 0.94 mol Cl
30.0 g of O
1.0 mol of O
16.0 g of O
= 1.88 mol O
1.88 mol O / 0.94 mol Cl = 2 O / 1 Cl 0.94 mol K / 0.94 mol Cl = 1 K / 1 Cl
Therefore, the empirical formula is KClO2
3. A compound that contains 46.2% carbon and 53.8% nitrogen?
46.2 g C
1.0 mol C
12.0 g C
= 3.8 mol C
53.8 g N
1.0 mol N
14.0 g N
= 3.8 mol N
3.8 mol C / 3.8 mol N = 1 C / 1 N, therefore the empirical formula is CN
4. A compound that consists of 21.6%sodium, 33.3% chlorine, and 45.1% oxygen.
21.6 g of Na
1.0 mol of Na
23.0 g of Na
= 0.94 mol Na
33.3 g of Cl
1.0 mol of Cl
35.5 g of Cl
= 0.94 mol Cl
45.1 g of O
1.0 mol of O
16.0 g of O
= 2.82 mol O
2.82 mol O / 0.94 mol Cl = 3 O / 1 Cl 0.94 mol Na / 0.9 mol Cl = 1 Na / 1 Cl
Therefore the empirical formula is NaClO3
WORKSHEET #14 DETERMING EMPIRICAL FORMULA FROM MASS DATA
1. A 13.20 g sample of a gas consists of 3.60 g of carbon and 9.60 g of oxygen.
3.60 g of C
1.0 mol of C
12.0 g of C
= 0.3 moles of C
9.60 g of O
1.0 mol of O
16.0 g of O
= 0.6 moles of O
0.6 mol O / 0.3 mol C = 2 O / 1 C, therefore the empirical formula is CO2
2. A hydrocarbon is a compound consisting of hydrogen and carbon. A sample consists of 2.50 g of carbon and 0.83 g of hydrogen.
2.50 g of C
1.0 mol of C
12.0 g of C
= 0.2 moles of C
0.83 g of H
1.0 mol of H
1.0 g of H
= 0.8 moles of H
0.8 mol H / 0.2 mol C = 4 H / 1 C, therefore the empirical formula is CH4
3. A compound consisting of 23.00 g of sodium, 6.00 g of carbon, and 24.00 g of oxygen.
23.00 g of Na
1.0 mol of Na
23.0 g of Na
= 1.0 moles of Na
6.00 g of C
1.0 mol of C
12.0 g of C
= 0.5 moles of C
24.00 g of O
1.0 mol of O
16.0 g of O
= 1.5 moles of O
1.5 mol O / 0.5 mol C = 3 O / 1 C
1.0 mol Na / 0.5 mol C = 2 Na / 1 C
Therefore, the empirical formula is Na2CO3
4. A 15.0 g sample of iron chloride consisting of 5.17 g iron and 9.83 g of chlorine.
5.17 g of Fe
1.0 mol of Fe
55.8 g of Fe
= 0.093 moles of Fe
9.83 g of Cl
1.0 mol of Cl
35.5 g of Cl
= 0.28 moles of Cl
0.28 mol Cl / 0.93 mol Fe = 3 Cl / 1 Fe, therefore the empirical formula is FeCl3
5. A hydrocarbon is a compound consisting of hydrogen and carbon. A sample consists of 2.500 g of carbon and 0.625 g of hydrogen.
2.500 g of C
1.0 mol of C
12.0 g of C
= 0.208 moles of C
0.625 g of H
1.0 mol of H
1.0 g of H
= 0.625 moles of H
0.625 mol H / 0.208 mol C = 3 H / 1 C, therefore the empirical formula is CH3
WORKSHEET #15 MOLECULAR FORMULA
1. What is the molecular formula of a hydrocarbon consisting of 92.3 % carbon and 7.7% hydrogen? The molecular mass is 26.0 u.
92.3 g of C
1.0 mol of C
12.0 g of C
= 7.7 mol C
7.7 g of H
1.0 mol of H
1.0 g of H
= 7.7 mol H
7.7 mol H / 7.7 mol C = 1 H / 1 C, therefore the empirical formula is CH
The molecular mass based on the empirical formula is 12.0 u + 1.0 u = 13.0 u
26.0 u / 13.0 u = 2, multiply empirical formula subscripts by this, so molecular formula is C2H2
2. What is the molecular formula of an organic compound consisting of 74.05% bromine, 22.24% carbon, and 3.71% hydrogen? The molecular
mass is 215.8 u.
74.05 g of Br
1.0 mol of Br
79.9 g of Br
= 0.92 mol Br
22.24 g of C
1.0 mol of C
12.0 g of C
= 1.85 mol C
3.71 g of H
1.0 mol of H
1.0 g of H
= 3.71 mol H
3.71 mol H / 0.92 mol Br = 4 H / 1 Br
1.85 mol C / 0.92 mol Br = 2 C / 1 Br
The molecular mass based on the empirical formula is 107.9 u
215.8 u / 107.9 u = 2, multiply subscripts of empirical formula by this to get C4H8Br2
Therefore the empirical formula is C2H4Br
3. What is the molecular formula of a consisting of 46.2% carbon and 53.8% nitrogen? The molecular mass is 52.0 u.
46.2 g C
1.0 mol C
12.0 g C
= 3.8 mol C
53.8 g N
1.0 mol N
14.0 g N
= 3.8 mol N
3.8 mol C / 3.8 mol N = 1 C / 1 N, therefore the empirical formula is CN
The molecular mass based on the empirical formula is 26.0 u
52.0 u / 26.0 u = 2, so the molecular formula is C2N2
4. What is the molecular formula of an compound consisting of 28.2% potassium, 25.6% chlorine, and 46.2% oxygen? The molecular mass is 415.5
u.
28.2 g of K
1.0 mol of K
39.1 g of K
= 0.72 mol K
25.6 g of Cl
1.0 mol of Cl
35.5 g of Cl
= 0.72 mol Cl
46.2 g of O
1.0 mol of O
16.0 g of O
= 2.88 mol O
2.88 mol O / 0.72 mol Cl = 4 O / 1 Cl
0.72 mol K / 0.72 mol Cl = 1 K / 1 Cl Therefore, the empirical formula is KClO4
The molecular mass based on the empirical formula is 138.5 u 415.5 u / 138.5 u = 3, so the molecular formula is K3(ClO4)3