Worked Solutions
Chapter 9
Question 1
What is the empirical formula of each of the following compounds?
(a) ethene (C2H4)
Answer:
Ratio of carbon to hydrogen = 2 : 4.
Simplest whole number ratio = 1 : 2
Empirical formula = CH2
(b) sucrose (C12H22O11)
Answer:
Ratio of carbon to hydrogen to oxygen = 12 : 22 : 11.
Simplest whole number ratio = 12 : 22 : 11
Empirical formula = C12H22O11
(c) hydrogen peroxide (H2O2)
Answer:
Ratio of hydrogen to oxygen = 2 : 2.
Simplest whole number ratio = 1 : 1
Empirical formula = HO
(d) carbon dioxide (CO2)
Answer:
Ratio of carbon to oxygen = 1 : 2.
Simplest whole number ratio = 1 : 2
Empirical formula = CO2
(e) benzene (C6H6)
Answer:
Ratio of carbon to hydrogen = 6 : 6.
Simplest whole number ratio = 1 : 1
Empirical formula = CH
(f) ethanoic acid (CH3COOH)
Answer:
Ratio of carbon to hydrogen to oxygen = 2 : 4 : 2.
Simplest whole number ratio = 1 : 2 : 1
Empirical formula = CH2O
1
Question 2
A compound on analysis is found to contain 75% carbon and 25% hydrogen by mass. What is
the empirical formula of the compound?
Answer:
Element
Percentage Percentage / Ar Simplest Ratio
Carbon
75
Hydrogen 25
75/12 = 6.25
1
25/1 = 25
4
Empirical formula = CH4
Question 3
Analysis of a compound showed that it contained 60% carbon, 13.33% hydrogen and 26.66%
oxygen by mass. What is the empirical formula of the compound?
Answer:
Element
Percentage Percentage / Ar Simplest Ratio
Carbon
60
60/12 = 5
3
Hydrogen 13.33
13.33/1 = 13.33
8
Oxygen
26.66/16 = 1.66
1
26.66
Empirical formula = C3H8O
Question 4
Analysis of a compound showed that it contained 40% carbon, 6.66% hydrogen and 53.33%
oxygen by mass. What is the empirical formula of the compound?
Answer:
Element
Percentage Percentage / Ar Simplest Ratio
Carbon
40
40/12 = 3.33
1
Hydrogen 6.66
6.66/1 = 6.66
2
Oxygen
53.33/16 = 3.33
1
53.33
Empirical formula = CH2O
2
Question 5
Find the empirical formulas of compounds containing:
(a) 77.77 % Fe, 22.22 % O
Answer:
Element Percentage Percentage / Ar Simplest Ratio
Iron
77.77
77.77/56 = 1.39
1
Oxygen
22.22
22.22/16 = 1.39
1
Empirical formula = FeO
(b) 70% Fe, 30% O
Answer:
Element Percentage Percentage / Ar Simplest Ratio
Iron
70
70/56 = 1.25
2
Oxygen
30
30/16 = 1.875
3
Empirical formula = Fe2O3
(c) 72.41% Fe, 27.59% O
Answer:
Element Percentage Percentage / Ar Simplest Ratio
Iron
72.41
72.41/56 = 1.29
3
Oxygen
27.59
27.59/16 = 1.72
4
Empirical formula = Fe3O4
(d) 46.66% N, 53.33% O
Answer:
Element Percentage Percentage / Ar Simplest Ratio
Nitrogen 46.66
46.66/14 = 3.33
1
Oxygen
53.33/16 = 3.33
1
53.33
Empirical formula = NO
(e) 30.43% N, 69.57% O
3
Answer:
Element Percentage Percentage / Ar Simplest Ratio
Nitrogen 30.43
30.43/14 = 2.17
1
Oxygen
69.57/16 = 4.35
2
69.57
Empirical formula = NO2
(f) 39.81% Cu, 20.06% S, 40.13% O
Answer:
Element Percentage Percentage / Ar
Simplest Ratio
Copper
39.81
39.81/63.5 = 0.627 1
Sulfur
20.06
20.06/32 = 0.627
1
Oxygen
40.13
40.13/16 = 2.5
4
Empirical formula = CuSO4
(g) 56.95% Cu, 14.35% S, 28.70% O
Answer:
Element Percentage Percentage / Ar
Simplest Ratio
Copper
56.95
56.95/63.5 = 0.9 2
Sulfur
14.35
14.35/32 = 0.45
1
Oxygen
28.7
28.7/16 = 1.8
4
Empirical formula = Cu2SO4
(h) 2.13% H, 29.79% N, 68.09% O
Answer:
Element
Percentage Percentage / Ar Simplest Ratio
Hydrogen 2.13
2.13/1 = 2.13
1
Nitrogen
29.79
29.79/14 = 2.13
1
Oxygen
68.09
68.09/16 = 4.26
2
4
Empirical formula = HNO2
(i) 1.59% H, 22.22% N, 76.19% O
Answer:
Element
Percentage Percentage / Ar Simplest Ratio
Hydrogen 1.59
1.59/1 = 1.59
1
Nitrogen
22.22
22.22/14 = 1.59
1
Oxygen
76.19
76.19/16 = 4.76
3
Empirical formula = HNO3
Question 6
When 0.54 g of aluminium is heated with excess chlorine gas, 2.67 g of the metal chloride is
formed. What is the empirical formula of the chloride?
Answer:
Mass of aluminium consumed = 0.54 g
Mass of chlorine consumed = (2.67 – 0.54) g = 2.13 g
Moles of aluminium atoms consumed = 0.54 / 27 = 0.02
Moles of chlorine atoms consumed = 2.13 / 35.5 = 0.06
Ratio of aluminium atoms to chlorine atoms = 0.02 : 0.06 = 1 : 3
Empirical formula of aluminium chloride AlCl3
Question 7
When 1.53 g of vanadium is heated with excess oxygen, 2.49 g of the metal oxide is formed.
What is its empirical formula?
Answer:
Mass of vanadium consumed = 1.53 g
Mass of oxygen consumed = (2.49 – 1.53) g = 0.96 g
Moles of vanadium atoms consumed = 1.53 / 51 = 0.03
Moles of oxygen atoms consumed = 0.96 / 16 = 0.06
Ratio of vanadium atoms to oxygen atoms = 0.03 : 0.06 = 1 : 2
Empirical formula of magnesium oxide = VO2
Question 8
When 0.20 g of an oxide of copper is heated, and excess hydrogen gas is passed over it,
0.16 g of copper is formed. What is the empirical formula of the oxide?
5
Answer:
Mass of copper in the compound = 0.16 g
Mass of oxygen in the compound = (0.2 – 0.16) g = 0.04 g
Moles of copper atoms in the compound = 0.16 / 63.5 = 0.0025
Moles of oxygen atoms in the compound = 0.04 / 16 = 0.0025
Ratio of copper atoms to oxygen atoms = 0.0025 : 0.0025 = 1 : 1
Empirical formula of copper oxide = CuO
Question 9
When 2.31 g of lead iodide is decomposed completely by heating, 1.04 g of lead is formed.
What is the empirical formula of lead iodide?
Answer:
Mass of lead in the compound = 1.04 g
Mass of iodine in the compound = (2.31 – 1.04) g = 1.27 g
Moles of lead atoms in the compound = 1.04 / 207 = 0.005
Moles of iodine atoms in the compound = 1.27 / 127 = 0.01
Ratio of lead atoms to iodine atoms = 0.005 : 0.01 = 1 : 2
Empirical formula of lead iodide = PbI2
Question 10
Urea has an empirical formula of CON2H4 and a relative molecular mass of 60. Find its
molecular formula.
Answer
The formula mass of CON2H4= 12 + 16 + 2(14) + 4(1) = 60
The relative molecular mass of urea = 60
Number of CON2H4 units in a urea molecule = 60 / 60 = 1
Molecular formula of urea = CON2H4
Question 11
Glucose has an empirical formula of CH2O and a relative molecular mass of 180. Find its
molecular formula.
Answer:
The formula mass of CH2O = 12 + 2(1) + 16= 30
The relative molecular mass of glucose = 180
Number of CH2O units in a glucose molecule = 180 / 30 = 6
Molecular formula of glucose = C6H12O6
6
Question 12
Heptane has a relative molecular mass of 100, and contains 84% carbon and 16% hydrogen
by mass. Find its molecular formula.
Answer:
Element
Percentage Percentage / Ar Simplest Ratio
Carbon
84
84/12 = 7
7
Hydrogen 16
16/1 = 16
16
Empirical formula = C7H16
The formula mass of C7H16 = 7(12) + 16(1) = 100
The relative molecular mass of heptane = 100
Number of C7H16 units in a heptane molecule = 100 / 100 = 1
Molecular formula of heptane = C7H16
Question 13
Butanoic acid has a relative molecular mass of 88. On analysis, it is found to contain 54.54%
carbon, 9.09% hydrogen and 36.36% oxygen by mass. What is the molecular formula of
butanoic acid?
Answer:
Element
Percentage Percentage / Ar
Simplest Ratio
Carbon
54.54
54.54/12 = 4.54
2
Hydrogen 9.09
9.09/1 = 9.09
4
Oxygen
36.36/16 = 2.273 1
36.36
Empirical formula = C2H4O
The formula mass of C2H4O = 2(12) + 4(1) + 16 = 44
The relative molecular mass of butanoic acid = 88
Number of C2H4O units in a butanoic acid molecule = 88 / 44 = 2
Molecular formula of butanoic acid = C4H8O2
Question 14
On analysis, a sweet smelling compound made from butanoic acid is found to contain 62%
carbon, 10.4% hydrogen and 27.6% oxygen by mass. If the relative molecular mass of the
compound is 116, find its molecular formula.
Answer:
7
Element
Percentage Percentage / Ar Simplest Ratio
Carbon
62
62/12 = 5.17
3
Hydrogen 10.4
10.4/1 = 10.4
6
Oxygen
27.6/16 = 1.725
1
27.6
Empirical formula = C3H6O
The formula mass of C3H6O = 3(12) + 6(1) + 16 = 58
The relative molecular mass of the compound = 116
Number of C3H6O units in the compound molecule = 116 / 58 = 2
Molecular formula of the compound = C6H12O2
Question 15
Fructose, a sugar that occurs in honey, has the following composition by mass: 40% carbon,
6.66% hydrogen and 53.33% oxygen. If the relative molecular mass of fructose is 180, find its
molecular formula.
Answer:
Element
Percentage Percentage / Ar Simplest Ratio
Carbon
40
40/12 = 3.33
1
Hydrogen 6.66
6.66/1 = 6.66
2
Oxygen
53.33/16 = 3.33
1
53.33
Empirical formula = CH2O
The formula mass of CH2O =12 + 2(1) + 16 = 30
The relative molecular mass of fructose = 180
Number of CH2O units in the compound molecule = 180 / 30 = 6
Molecular formula of the compound = C6H12O6
Question 16
Calculate the percentage by mass of nitrogen present in ammonium chloride (NH4Cl).
Answer:
Moles of nitrogen per mole of ammonium chloride = 1
Mass of nitrogen per mole of ammonium chloride = 14 g
Molar mass of ammonium chloride = 53.5 g mol-1
Percentage of nitrogen in ammonium sulfate = 14 X 100 / 53.5%
= 26.17%
8
Question 17
Calculate the percentage by mass of nitrogen present in urea (CO(NH2)2).
Answer:
Moles of nitrogen per mole of urea = 2
Mass of nitrogen per mole of urea = 28 g
Molar mass of urea = 60 g mol-1
Percentage of nitrogen in urea = 28 X 100 / 60%
= 46.67%
Question 18
Calculate the percentage by mass of nitrogen present in ammonium nitrate (NH4NO3).
Answer:
Moles of nitrogen per mole of ammonium nitrate = 2
Mass of nitrogen per mole of ammonium nitrate = 28 g
Molar mass of ammonium nitrate = 80 g mol-1
Percentage of nitrogen in ammonium nitrate = 28 X 100 / 80%
= 35%
Question 19
Calculate the percentage by mass of carbon in ethanol (C2H5OH).
Answer:
Moles of carbon per mole of ethanol = 2
Mass of carbon per mole of ethanol = 24 g
Molar mass of ethanol = 46 g mol-1
Percentage of carbon in ethanol = 24 X 100 / 46%
= 52.17%
Question 20
Calculate the percentage by mass of (a) nitrogen (b) phosphorus present in (NH4)2HPO4.
Answer:
(a) Moles of nitrogen per mole of (NH4)2HPO4= 2
Mass of nitrogen per mole of (NH4)2HPO4 = 28 g
Molar mass of (NH4)2HPO4 = 132 g mol-1
Percentage of nitrogen in (NH4)2HPO4 = 28 X 100 / 132%
= 21.21%
9
(b) Moles of phosphorus per mole of (NH4)2HPO4= 1
Mass of phosphorus per mole of (NH4)2HPO4 = 31 g
Molar mass of (NH4)2HPO4 = 132 g mol-1
Percentage of phosphorus in (NH4)2HPO4 = 31 X 100 / 132%
= 23.48%
Question 21
Calculate the percentage by mass of each element in (a) sulfuric acid (H2SO4) (b) sodium
hydroxide (NaOH) (c) sodium chloride (NaCl) (d) anhydrous sodium carbonate (Na2CO3)
(e) potassium manganate(VII) (KMnO4) (f) sodium hydrogencarbonate (NaHCO3) (g) silver
nitrate (AgNO3) (h) ammonia (NH3) (i) calcium hydroxide (Ca(OH)2) (j) potassium nitrate
(KNO3).
Answer:
(a) Moles of hydrogen per mole of sulfuric acid = 2
Mass of hydrogen per mole of sulfuric acid = 2 g
Molar mass of sulfuric acid = 98 g mol-1
Percentage of hydrogen in sulfuric acid = 2 X 100 / 98%
= 2.04%
Moles of sulfur per mole of sulfuric acid = 1
Mass of sulfur per mole of sulfuric acid = 32 g
Molar mass of sulfuric acid = 98 g mol-1
Percentage of sulfur in sulfuric acid = 32 X 100 / 98%
= 32.65%
Moles of oxygen per mole of sulfuric acid = 4
Mass of oxygen per mole of sulfuric acid = 64 g
Molar mass of sulfuric acid = 98 g mol-1
Percentage of oxygen in sulfuric acid = 64 X 100 / 98%
= 65.31%
(b) Moles of sodium per mole of sodium hydroxide = 1
Mass of sodium per mole of sodium hydroxide = 23 g
Molar mass of sodium hydroxide = 40 g mol-1
Percentage of sodium in sodium hydroxide = 23 X 100 / 40%
= 57.5%
Moles of oxygen per mole of sodium hydroxide = 1
Mass of oxygen per mole of sodium hydroxide = 16 g
10
Molar mass of sodium hydroxide = 40 g mol-1
Percentage of oxygen in sodium hydroxide = 16 X 100 / 40%
= 40%
Moles of hydrogen per mole of sodium hydroxide = 1
Mass of hydrogen per mole of sodium hydroxide = 1 g
Molar mass of sodium hydroxide = 40 g mol-1
Percentage of hydrogen in sodium hydroxide = 1 X 100 / 40%
= 2.5%
(c) Moles of sodium per mole of sodium chloride = 1
Mass of sodium per mole of sodium chloride = 23 g
Molar mass of sodium chloride = 58.5 g mol-1
Percentage of sodium in sodium chloride = 23 X 100 / 58.5%
= 39.32%
Moles of chlorine per mole of sodium chloride = 1
Mass of chlorine per mole of sodium chloride = 35.5 g
Molar mass of sodium chloride = 58.5 g mol-1
Percentage of chlorine in sodium chloride = 35.5 X 100 / 58.5%
= 60.68%
(d) Moles of sodium per mole of anhydrous sodium carbonate = 2
Mass of sodium per mole of anhydrous sodium carbonate = 46 g
Molar mass of anhydrous sodium carbonate = 106 g mol-1
Percentage of sodium in anhydrous sodium carbonate = 46 X 100 / 106%
= 43.4%
Moles of carbon per mole of anhydrous sodium carbonate = 1
Mass of carbon per mole of anhydrous sodium carbonate = 12 g
Molar mass of anhydrous sodium carbonate = 106 g mol-1
Percentage of carbon in anhydrous sodium carbonate = 12 X 100 / 106%
= 11.32%
Moles of oxygen per mole of anhydrous sodium carbonate = 3
Mass of oxygen per mole of anhydrous sodium carbonate = 48 g
Molar mass of anhydrous sodium carbonate = 106 g mol-1
Percentage of oxygen in anhydrous sodium carbonate = 48 X 100 / 106%
= 45.28%
(e) Moles of potassium per mole of potassium manganate(VII) = 1
Mass of potassium per mole of potassium manganate(VII) = 39 g
Molar mass of potassium manganate(VII) = 158 g mol-1
Percentage of potassium in potassium manganate(VII) = 39 X 100 / 158%
11
= 24.68%
Moles of manganese per mole of potassium manganate(VII) = 1
Mass of manganese per mole of potassium manganate(VII) = 55 g
Molar mass of potassium manganate(VII) = 158 g mol-1
Percentage of manganese in potassium manganate(VII) = 55 X 100 / 158%
= 34.8%
Moles of oxygen per mole of potassium manganate(VII) = 4
Mass of oxygen per mole of potassium manganate(VII) = 64 g
Molar mass of potassium manganate(VII) = 158 g mol-1
Percentage of oxygen in potassium manganate(VII) = 64 X 100 / 158%
= 40.5%
(f) Moles of sodium per mole of sodium hydrogencarbonate = 1
Mass of sodium per mole of sodium hydrogencarbonate = 23 g
Molar mass of sodium hydrogencarbonate = 84 g mol-1
Percentage of sodium in sodium hydrogencarbonate = 23 X 100 / 84%
= 27.38%
Moles of hydrogen per mole of sodium hydrogencarbonate = 1
Mass of hydrogen per mole of sodium hydrogencarbonate = 1 g
Molar mass of sodium hydrogencarbonate = 84 g mol-1
Percentage of hydrogen in sodium hydrogencarbonate = 1 X 100 / 84%
= 1.19%
Moles of carbon per mole of sodium hydrogencarbonate = 1
Mass of carbon per mole of sodium hydrogencarbonate = 12 g
Molar mass of sodium hydrogencarbonate = 84 g mol-1
Percentage of carbon in sodium hydrogencarbonate = 12 X 100 / 84%
= 14.28%
Moles of oxygen per mole of sodium hydrogencarbonate = 3
Mass of oxygen per mole of sodium hydrogencarbonate = 48 g
Molar mass of sodium hydrogencarbonate = 84 g mol-1
Percentage of oxygen in sodium hydrogencarbonate = 48 X 100 / 84%
= 57.14%
(g) Moles of silver per mole of silver nitrate = 1
Mass of silver per mole of silver nitrate = 108 g
Molar mass of silver nitrate = 170 g mol-1
12
Percentage of silver in silver nitrate = 108 X 100 / 170%
= 63.53%
Moles of nitrogen per mole of silver nitrate = 1
Mass of nitrogen per mole of silver nitrate = 14 g
Molar mass of silver nitrate = 170 g mol-1
Percentage of nitrogen in silver nitrate = 14 X 100 / 170%
= 8.24%
Moles of oxygen per mole of silver nitrate = 3
Mass of oxygen per mole of silver nitrate = 48 g
Molar mass of silver nitrate = 170 g mol-1
Percentage of oxygen in silver nitrate = 48 X 100 / 170%
= 28.24%
(h) Moles of nitrogen per mole of ammonia = 1
Mass of nitrogen per mole of ammonia = 14 g
Molar mass of ammonia = 17 g mol-1
Percentage of nitrogen in ammonia = 14 X 100 / 17%
=82.35%
Moles of hydrogen per mole of ammonia = 3
Mass of hydrogen per mole of ammonia = 3 g
Molar mass of ammonia = 17 g mol-1
Percentage of hydrogen in ammonia = 3 X 100 / 17%
= 17.65%
(i) Moles of calcium per mole of calcium hydroxide = 1
Mass of calcium per mole of calcium hydroxide = 40 g
Molar mass of calcium hydroxide = 74 g mol-1
Percentage of calcium in calcium hydroxide = 40 X 100 / 74%
= 54.05%
Moles of oxygen per mole of calcium hydroxide = 2
Mass of oxygen per mole of calcium hydroxide = 32 g
Molar mass of calcium hydroxide = 74 g mol-1
Percentage of oxygen in calcium hydroxide = 32 X 100 / 74%
= 43.24%
Moles of hydrogen per mole of calcium hydroxide = 2
Mass of hydrogen per mole of calcium hydroxide = 2 g
Molar mass of calcium hydroxide = 74 g mol-1
13
Percentage of hydrogen in calcium hydroxide = 2 X 100 / 74%
= 2.7%
(j) Moles of potassium per mole of potassium nitrate = 1
Mass of potassium per mole of potassium nitrate = 39 g
Molar mass of potassium nitrate = 101 g mol-1
Percentage of potassium in potassium nitrate = 39 X 100 / 101%
= 38.61%
Moles of nitrogen per mole of potassium nitrate = 1
Mass of nitrogen per mole of potassium nitrate = 14 g
Molar mass of potassium nitrate = 101 g mol-1
Percentage of nitrogen in potassium nitrate = 14 X 100 / 101%
= 13.86%
Moles of oxygen per mole of potassium nitrate = 3
Mass of oxygen per mole of potassium nitrate = 48 g
Molar mass of potassium nitrate = 101 g mol-1
Percentage of oxygen in potassium nitrate = 48 X 100 / 101%
= 47.52%
Question 23
The structural formula of propane is
H
H
H



H C  CCH



H
H
H
Find (a) its molecular formula and (b) its empirical formula.
Answer:
(a) The molecular formula is found by counting the atoms of each element in the
structural formula.
Molecular formula of propane = C3H8
(b) Empirical formula of propane = C3H8
Question 24
Balance each of the following chemical equations:
(a) MgCl2 + AgNO3 → Mg(NO3)2 + AgCl
Answer:
Magnesium: Left hand side 1, right hand side 1, no change
14
Chlorine: Left hand side 2, right hand side 1, so AgCl X 2
MgCl2 + AgNO3 → Mg(NO3)2 + 2AgCl
Silver: Left hand side 1, right hand side 2, so AgNO3 X 2
MgCl2 + 2AgNO3 → Mg(NO3)2 +2AgCl
Nitrogen: Left hand side 2, right hand side 2, no change
Oxygen: Left hand side 6, right hand side 6, no change
MgCl2 + 2AgNO3 → Mg(NO3)2 +2AgCl
(b) N2 + H2 → NH3
Answer:
Nitrogen: Left hand side 2, right hand side 1, so NH3 X 2
N2 + H2 →2NH3
Hydrogen: Left hand side 2, right hand side 6, so H2 X 3
N2 + 3H2 → 2NH3
(c) Fe + HCl →FeCl2 + H2
Answer:
Iron: Left hand side 1, right hand side 1, no change
Hydrogen: Left hand side 1, right hand side 2, so HCl X 2
Fe + 2HCl →FeCl2 + H2
Chlorine: Left hand side 2, right hand side 2, no change
Fe + 2HCl →FeCl2 + H2
(d) Al + O2 → Al2O3
Answer:
Aluminium: Left hand side 1, right hand side 2, so Al X 2
2Al + O2 → Al2O3
Oxygen: Left hand side 2, right hand side 3, so O2 X 3 and Al2O3 X 2
2Al + 3O2 → 2Al2O3
There are now 4 Al on the right hand side so Al X 4
4Al + 3O2 → 2Al2O3
(e) H2SO4 + NaOH → Na2SO4 + H2O
Answer:
Hydrogen: Left hand side 3, right hand side 2, so H2O X 2 and NaOH X 2
H2SO4 +2NaOH → Na2SO4 +2H2O
15
Sulfur: Left hand side 1, right hand side 1, no change
Oxygen: Left hand side 6, right hand side 6, no change
H2SO4 +2NaOH → Na2SO4 +2H2O
(f) C4H10 + O2 → CO2 + H2O
Answer:
Carbon: Left hand side 4, right hand side 1, so CO2 X 4
C4H10 + O2 → 4CO2 + H2O
Hydrogen: Left hand side 10, right hand side 2, so H2O X 5
C4H10 + O2 → 4CO2 + 5H2O
Oxygen: Left hand side 2, right hand side 13, so O2 X 6 12
C4H10 + 6 12O2 → 4CO2 + 5H2O
(g) Al + Cl2 → AlCl3
Answer:
Aluminium: Left hand side 1, right hand side 1, no change
Chlorine: Left hand side 2, right hand side 3, so Cl2 X 3 and AlCl 3 X 2
Al + 3Cl2 → 2AlCl3
There are now 2 Al on the right hand side so Al X 2
2Al + 3Cl2 → 2AlCl3
(h) H2 + Cl2 → HCl
Answer:
Hydrogen: Left hand side 2, right hand side 1, so HCl X 2
H2 + Cl2 → 2HCl
Chlorine: Left hand side 2, right hand side 2, no change
H2 + Cl2 → 2HCl
(i) Ca(OH)2 + CO2 → CaCO3 + H2O
Answer:
Calcium: Left hand side 1, right hand side 1, no change
Oxygen: Left hand side 4, right hand side 4, no change
Hydrogen: Left hand side 2, right hand side 2, no change
Carbon: Left hand side 1, right hand side 1, no change
Ca(OH)2 + CO2 → CaCO3 + H2O
(j) ZnS + O2 → ZnO + SO2
Answer:
Zinc: Left hand side 1, right hand side 1, no change
16
Sulfur: Left hand side 1, right hand side 1, no change
Oxygen: Left hand side 2, right hand side 3, so O2 X 112
ZnS + 112O2 → ZnO + SO2
(k) (NH4)2Cr2O7 → Cr2O3 + H2O + N2
Answer:
Nitrogen: Left hand side 2, right hand side 2, no change
Hydrogen: Left hand side 8, right hand side 2, so H2O X 4
(NH4)2Cr2O7 → Cr2O3 + 4H2O + N2
Chromium: Left hand side 2, right hand side 2, no change
Oxygen: Left hand side 7, right hand side 7, no change
(NH4)2Cr2O7 → Cr2O3 + 4H2O + N2
Question 25
Oxygen gas is prepared by decomposing hydrogen peroxide, according to the equation
2H2O2(aq) → O2(g) + 2H2O(l)
How many moles of oxygen are produced when 5 moles of hydrogen peroxide are
decomposed?
Answer:
From the equation
2H2O2(aq) → O2(g) + 2H2O(l)
2 moles
Therefore
1 mole
2 moles
1 mole
0.5 moles
1 mole
Therefore 5 X ( 1 mole
0.5 moles
1 mole )
= 5 moles
2.5 moles
5 moles
2.5 moles of oxygen are produced.
Question 26
Ammonia is prepared from nitrogen and hydrogen:
N2(g) + 3H2(g) → 2NH3(g)
(a) How many moles of hydrogen are needed to react fully with 0.5 moles of nitrogen?
(b) How many moles of ammonia will be formed?
Answer:
From the equation
N2(g)
1 mole
+
3H2(g)
3 moles
→
2NH3(g)
2 moles
17
Therefore 0.5 X (1 mole
3 moles
2 moles)
1.5 moles
1 mole
= 0.5 moles
Answers: (a) 1.5 moles of hydrogen (b) 1 mole of ammonia
Question 27
Butane gas burns in air according to the equation
C4H10(g) + 61/2O2(g) → 4CO2(g) + 5H2O(g)
How many moles of oxygen are needed for the complete combustion of 4 moles of butane?
Answer:
From the equation
+ 61/2O2(g) → 4CO2(g) + 5H2O(g)
C4H10(g)
1 mole
6.5 moles
4 moles
5 moles
Therefore
4 X (1 mole
6.5 moles
4 moles
5 moles)
=
26 moles
16 moles
20 moles
2 moles
25 moles O2 are needed.
Question 28
Hydrogen and chlorine react to form hydrogen chloride:
H2(g) + Cl2(g) → 2HCl(g)
How many moles of hydrogen are needed to make 10 moles of hydrogen chloride?
Answer:
From the equation
H2(g)
+
Cl2(g)
→
2HCl(g)
1 mole
1 mole
2 moles
0.5 moles
0.5 moles
1 mole
0.5 moles
1 mole)
5 moles
10 moles
Therefore
10 X (0.5 moles
=
5 moles
5 moles of hydrogen are needed.
Question 29
A solution containing 7.4 g of calcium hydroxide reacts fully with carbon dioxide according
to the equation
Ca(OH)2(aq) + CO2(g) → CaCO3(s) + H2O(l)
What mass of calcium carbonate is formed?
18
Answer:
7.4 g Ca(OH)2 = 7.4 / 74 moles Ca(OH)2 = 0.1 moles Ca(OH)2
From the equation
Ca(OH)2(aq) + CO2(g) →
1 mole
Therefore: 0.1 moles
1 mole
0.1 moles
CaCO3(s) +
H2O(l)
1 mole
0.1 moles
1 mole
0.1 moles
1 mole CaCO3= 100 g CaCO3
0.1 moles CaCO3 = 0.1 X 100 g = 10 g
10 g calcium carbonate are formed.
Question 30
Ethane burns in air according to the equation
C2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l)
If 8.4 litres of ethane (measured at s.t.p.) are used to react with excess oxygen, what mass of
water is formed?
Answer:
8.4 litres C2H6 = 8.4/22.4 moles C2H6 = 0.375 moles C2H6
C2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l)
From the equation
1 mole
Therefore:
3.5 moles
2 moles
3 moles
3 X 0.375 = 1.125 moles
0.375 moles
1 mole H2O = 18 g H2O
1.125 moles H2O = 1.125 X 18 g = 20.25 g
Question 31
If 112 g of iron are reacted fully with hydrochloric acid:
Fe(s) + 2HCl(aq) ) → FeCl2(aq) + H2(g)
what volume of hydrogen (measured at s.t.p.) is produced?
Answer:
112 g iron = 112 / 56 moles Fe = 2 moles Fe
Fe(s) + 2HCl(aq) ) → FeCl2(aq) + H2(g)
Therefore
2 X
1 mole 2 moles
1 mole
1 mole
(1 mole 2 moles
1 moles
1 mole)
= 2 moles 4 moles
2 moles
2 moles
1 mole of hydrogen at s.t.p. = 22.4 l H2
2 moles at s.t.p. = 2 X 22.4 l = 44.8 l
19
Question 32
Silver oxide breaks down on heating to give silver and oxygen:
2Ag2O(s) → 4 Ag(s) + O2(g)
If 92.8 g of silver oxide are used, what mass of silver is formed, and what volume of oxygen
(measured at s.t.p.) is produced?
Answer:
92.8 g silver oxide = 92.8 / 232 moles Ag2O = 0.4 moles Ag2O
2Ag2O(s) → 4 Ag(s) + O2(g)
Therefore
2 moles
4 moles 1 mole
0.4 moles
0.8 moles 0.2 moles
1 mole Ag = 108 g
0.8 moles Ag = 0.8 X 108 = 86.4 g.
1 mole of oxygen at s.t.p. = 22.4 l O2
0.2 moles at s.t.p. = 0.2 X 22.4 l = 4.48 l
Question 33
Phosphorus burns in oxygen, forming phosphorus pentoxide, according to the equation:
4P(s) ) + 5O2(g) → 2P2O5(s)
What volume of oxygen (measured at s.t.p.) is consumed, and what mass of phosphorus is
consumed, when 35.5 g of phosphorus pentoxide is formed?
Answer:
35.5 g phosphorus pentoxide = 35.5 / 142 moles P2O5 = 0.25 moles P2O5
4P(s) + 5O2(g) → 2P2O5(s)
4 moles
Therefore
5 moles
2 moles
0.5 moles 0.625 moles 0.25 moles
1 mole of oxygen at s.t.p. = 22.4 l O2
0.625 moles at s.t.p. = 0.625 X 22.4 l = 14 l O2
1 mole P = 31 g
0.5 moles P = 0.5 X 31 g = 15.5 g P.
Question 34
Ammonium dichromate (31.5 g) decomposes fully on heating:
(NH4)2Cr2O7(s) → Cr2O3(s) + 4H2O(l) + N2(g)
What mass of chromium(III) oxide is formed, and what volume of nitrogen (measured at
s.t.p.) is produced? How many water molecules are formed?
20
Answer:
31.5 g ammonium dichromate = 31.5 / 252 moles (NH4)2Cr2O7 = 0.125 moles (NH4)2Cr2O7
(NH4)2Cr2O7(s) → Cr2O3(s) + 4H2O(l) + N2(g)
1 mole
1 mole
4 moles 1 mole
Therefore
0.125 X (1 mole
= 0.125 moles
1 mole 4 moles
1 mole)
0.125 moles 0.5 moles 0.125 moles
1 mole chromium (III) oxide = 152 g
0.125 moles Cr2O3 = 0.125 X 152 = 19 g
1 mole of nitrogen at s.t.p. = 22.4 l N2
0.125 moles at s.t.p. = 0.125 X 22.4 l = 2.8 l
1 mole of water molecules = 6 X 1023 molecules
0.5 moles of water molecules = 0.5 X 6 X 1023 molecules = 3 X 1023 molecules
Question 35
Ammonia gas can be prepared from ammonium chloride and sodium hydroxide:
NH4Cl(s) + NaOH(s) → NH3(g) + NaCl(s) + H2O(l)
If 10.7 g ammonium chloride and 10 g sodium hydroxide are used, show that the sodium
hydroxide is present in excess. Calculate the mass of sodium chloride and the volume of
ammonia (measured at s.t.p.) formed.
Answer:
Moles of ammonium chloride present initially = 10.7 / 53.5 = 0.2
Moles of sodium hydroxide present initially = 10 / 40 = 0.25
NH4Cl(s) + NaOH(s) → NH3(g) + NaCl(s) + H2O(l)
1 mole
1 mole
0.2 moles 0.2 moles
1 mole 1 mole 1 mole
0.2 moles 0.2 moles 0.2 moles
Since the 0.2 moles of ammonium chloride that are present initially will react fully with 0.2
moles of sodium hydroxide, and there are 0.25 moles of sodium hydroxide present initially, it
is clear that it is the sodium hydroxide that is present in excess.
(ii) Ammonium chloride is the limiting reagent.
NH4Cl(s) + NaOH(s) → NH3(g) + NaCl(s) + H2O(l)
0.2 moles 0.2 moles
0.2 moles 0.2 moles 0.2 moles
Amount of sodium chloride formed = 0.2 moles
Mass of sodium chloride formed = 0.2 X 58.5 g = 11.7 g
Amount of ammonia formed = 0.2 moles
Volume (at s.t.p.) of ammonia formed = 0.2 X 22.4 l = 4.48 l
21
22
Question 36
Sodium sulfite reacts with sulfuric acid as follows:
Na2SO3(s) + H2SO4(aq) → Na2SO4(aq) + SO2(g) + H2O(l).
If a solution containing 4.9 g sulfuric acid is added to 50.4 g sodium sulfite, show that the
sodium sulfite is present in excess. Calculate the mass of sodium sulfate and the volume of
sulfur dioxide (measured at s.t.p.) formed.
Answer:
Moles of sulfuric acid present initially = 4.9 / 98 = 0.05
Moles of sodium sulfite present initiallyl = 50.4 / 126 = 0.4
Na2SO3(s) + H2SO4(aq) → Na2SO4(aq) + SO2(g) + H2O(l).
1 mole
1 mole
0.05 moles 0.05 moles
1 mole
1 mole 1 mole
0.05 moles 0.05 moles 0.05 moles
Since the 0.05 moles of sulfuric acid that are present initially will react fully with 0.05 moles
of sodium sulfite, and there are 0.4 moles of sodium sulfite present initially, it is clear that it is
the sodium sulfite that is present in excess.
Sulfuric acid is the limiting reagent.
Na2SO3(s) + H2SO4(aq) → Na2SO4(aq) + SO2(g) + H2O(l).
0.05 moles 0.05 moles 0.05 moles 0.05 moles 0.05 moles
Amount of sodium sulfate formed = 0.05 moles
Mass of sodium sulfate formed = 0.05 X 142 g = 7.1 g
Amount of sulfur dioxide formed = 0.05 moles
Volume (at s.t.p.) of sulfur dioxide formed = 0.05 X 22.4 l = 1.12 l
Question 37
Ethanal (CH3CHO) is prepared in the laboratory by reacting ethanol
(C2H5OH) with acidified sodium dichromate solution:
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(aq) + 7H2O(l)
If 29.8 g of sodium dichromate (Na2Cr2O7.2 H2O) and 18.4 g of ethanol are used in an
experiment to prepare ethanal, which of the two reagents is present in excess? Calculate the
maximum amount of ethanal that could be obtained in this experiment.
Answer:
Moles of sodium dichromate present initially = 29.8 / 298 = 0.1
Moles of ethanol present initially = 18.4 / 46 = 0.4
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(aq) + 7H2O(l)
3 moles
1 mole
0.3 moles 0.1 moles
3 moles
2 moles
0.3 moles 0.2 moles
7 moles
0.7 moles
23
Since the 0.3 moles of ethanol that are present initially will react fully with 0.1 moles of
sodium dichromate, and there are 0.4 moles of ethanol present initially, it is clear that it is the
ethanol that is present in excess.
Sodium dichromate is the limiting reagent.
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(aq) + 7H2O(l)
0.3 moles 0.1 moles
0.3 moles
0.2 moles 0.7 moles
Amount of ethanal formed = 0.3 moles
Question 38
Ethanoic acid (CH3COOH) is prepared in the laboratory by reacting ethanol (C2H5OH) with
acidified sodium dichromate solution:
3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH(l) + 4Cr3+(aq) + 11H2O(l)
If 14.9 g of sodium dichromate (Na2Cr2O7.2 H2O) and 4 g of ethanol are used in an
experiment to prepare ethanoic acid, which of the two reagents is present in excess? Calculate
the maximum amount of ethanoic acid that could be obtained in this experiment.
Answer:
Moles of sodium dichromate present initially = 14.9/ 298 = 0.05
Moles of ethanol present initially = 4 / 46 = 0.0875
3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH(l) + 4Cr3+(aq) + 11H2O(l)
0.05 X
3 moles
2 moles
3 moles
4 moles 11 moles
1.5 moles
1 mole
1.5moles
2 moles 5.5 moles
(1.5 moles
1 mole
1.5moles
= 0.075 moles 0.05moles
2 moles 5.5 moles)
0.075 moles 0.1 moles 0.275 moles
Since the 0.05 moles of ethanol that are present initially will react fully with 0.08 moles of
sodium dichromate, and there are 0.0875 moles of ethanol present initially, it is clear that it is
the ethanol that is present in excess.
Sodium dichromate is the limiting reagent.
3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH(l) + 4Cr3+(aq) + 11H2O(l)
0.075moles 0.05moles
0.075 moles 0.1 moles 0.275 moles
Amount of ethanoic acid formed = 0.075 moles
Question 39
Oxygen can be prepared from potassium chlorate using the reaction
2KClO3(s) → 2KCl(s) + 3O2(g)
If 30 g of KClO3 are used, and 8 g O2 are collected, calculate the percentage yield of oxygen.
24
Answer:
30 g KClO3 = 30 / 122.5 moles KClO3 = 0.245 moles
2KClO3(s) →
2KCl(s) +
3O2(g)
2 mole
2 mole
3 mole
0.245 moles
0.245 moles 0.3675 moles
Theoretical yield = 0.3675 moles O2 = 0.3675 X 32 g O2 = 11.76 g
Actual yield = 8 g
Actual yield X 100
8 X 100
Percentage yield =  % = 
Theoretical yield
11.76
= 68.03 %
Question 40
Nitrogen monoxide is prepared using the reaction
3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
If 12.7 g of copper are used to produce 1.12 l NO (measured at s.t.p.), calculate the percentage
yield of NO.
Answer:
12.7 g Cu = 12.7 / 63.5 moles Cu = 0.2 moles
3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)
3 moles
8 moles
1 mole 2.667 moles
3 moles
1 mole
0.2 moles
2 moles
4 moles
0 .667 moles 1.33 moles
0.133 moles
Theoretical yield = 0.133 moles NO = 0.133 X 22.4 l NO = 2.979 l
Actual yield = 1.12 l
Actual yield X 100
1.12 X 100
Percentage yield =  % = 
Theoretical yield
2.979
= 37.60 %
Question 41
Ethanoic acid (CH3COOH) is prepared in the laboratory by reacting ethanol (C2H5OH) with
acidified sodium dichromate solution:
3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH (l) + 4Cr3+(aq) + 11H2O(l)
If excess sodium dichromate and 16.5 g of ethanol are used in an experiment to prepare
ethanoic acid, and 17.2 g of ethanoic acid is formed, calculate the percentage yield of
ethanoic acid.
Answer:
16.5 g of ethanol = 16.5 / 46 moles
= 0.3587 moles
25
3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH (l) + 4Cr3+(aq) + 11H2O(l)
3 moles
3 moles
1 mole
1 mole
0.3587 moles
0.3587 moles
Theoretical yield of ethanoic acid = 0.3587 moles = 0.3587 moles X 60 g
= 21.522 g
Actual yield of ethanoic acid = 17.2 g
Actual yield X 100
17.2 X 100
Percentage yield =  % =  %
Theoretical yield
21.522
= 79.92%
Question 42
Ethanal (CH3CHO) is prepared in the laboratory by reacting ethanol
(C2H5OH) with acidified sodium dichromate solution:
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(aq) + 7H2O(l)
If 7.4 g of sodium dichromate (Na2Cr2O7.2 H2O) and 5.6 g of ethanol were used in an
experiment to prepare ethanal, which of the two reagents was present in excess? After
purification, 1.2 g of ethanal was obtained. Calculate the percentage yield of ethanal.
Answer:
Moles of sodium dichromate present initially = 7.4 / 298 = 0.0248
Moles of ethanol present initially = 5.6 / 46 = 0.12
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(aq) + 7H2O(l)
3 moles
1 mole
0.075 moles 0.025 moles
3 moles
2 moles
7 moles
0.075 moles
Since the 0.025 moles of sodium dichromate that are present initially will react fully with
0.075 moles of ethanol, and there are 0.12 moles of ethanol present initially, it is clear that it
is the ethanol that is present in excess.
Sodium dichromate is the limiting reagent.
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(aq) + 7H2O(l)
0.075 moles 0.025 moles
0.075 moles
Amount of ethanal formed = 0.075 moles
Theoretical yield of ethanal = 0.075 moles = 0.075 moles X 44 g
= 3.3 g
Actual yield of ethanal = 1.2 g
26
Actual yield X 100
1.2 X 100
Percentage yield =  % =  %
Theoretical yield
3.3
= 36.37%
Question 43
A compound contains 43.4% sodium, 11.3% carbon and 45.3% oxygen by mass. Find its
empirical formula.
Answer:
Element Percentage Percentage / Ar
Simplest Ratio
Sodium
43.4
43.4 / 23 = 1.886
2
Carbon
11.3
11.3 / 12 = 0.942
1
Oxygen
45.3
45.3 / 16 = 2.831 3
Ratio of sodium atoms to carbon atoms to oxygen atoms
=2:1:3
Empirical formula = Na2CO3
Question 44
A compound of relative molecular mass 74 was analysed and found to contain 43.24%
oxygen, 48.65% carbon and 8.11% hydrogen by mass. Calculate the molecular formula of the
compound.
Answer:
Element
Percentage Percentage / Ar
Oxygen
43.24
43.24 / 16= 2.70 2
Carbon
48.65
48.65 / 12= 4.05 3
Hydrogen 8.11
8.11 / 1= 8.11
Simplest Ratio
6
Ratio of oxygen atoms to carbon atoms to hydrogen atoms
=2:3:6
Empirical formula = C3H6O2
Formula mass of C3H6O2
= 74
Molecular formula = C3H6O2
Question 45
Caffeine contains 49.48% carbon, 28.87% nitrogen, 16.50% oxygen and 5.15% hydrogen by
mass. Find the molecular formula of caffeine, given that its relative molecular mass is 194.
27
Answer:
Element
Percentage Percentage / Ar
Carbon
49.48
49.48 / 12 = 4.12 4
Nitrogen
28.87
28.87 / 14 = 2.06 2
Oxygen
16.50
16.50 / 16 = 1.03 1
Hydrogen 5.15
5.15 / 1 = 5.15
Simplest Ratio
5
Empirical formula of caffeine = C4N2OH5
Formula mass of C4N2OH5 = 97
Relative molecular mass of caffeine = 194
Number of C4N2OH5 units in a caffeine molecule = 194 / 97 = 2
Molecular formula of caffeine = C8N4O2H10
Question 46
Calculate the percentage by mass of (a) carbon (b) hydrogen and (c) oxygen in aspirin, given
that the empirical formula of aspirin is C9H8O4.
Answer:
Moles of carbon per mole of aspirin = 9
Mass of carbon per mole of aspirin = 108
Molar mass of aspirin = 180 g mol-1
Percentage of carbon in aspirin = 108 X 100 / 180%
= 60%
Moles of hydrogen per mole of aspirin = 8
Mass of hydrogen per mole of aspirin = 8
Molar mass of aspirin = 180 g mol-1
Percentage of hydrogen in aspirin = 8 X 100 / 180%
= 4.44%
Moles of oxygen per mole of aspirin = 4
Mass of oxygen per mole of aspirin = 64
Molar mass of aspirin = 180 g mol-1
Percentage of oxygen in aspirin = 64 X 100 / 180%
= 35.55%
Question 47
The reaction of 12.6 g sodium sulfite with sulfuric acid occurs according to the equation:
28
Na2SO3(s) + H2SO4(aq) → Na2SO4(aq) + SO2(g) + H2O(l)
(a) How many moles of sodium sulfite does 12.6 g represent?
(b) What mass of sodium sulfate is formed?
(c) What volume of sulfur dioxide gas (measured at s.t.p.) is produced?
(d) How many molecules of water are formed?
(e) What mass of sulfuric acid is used up in the reaction?
Answer:
(a) 12.6 g sodium sulfite = 12.6 / 126 = 0.1 moles
(b) From the equation
Na2SO3(s) + H2SO4(aq) → Na2SO4(aq) + SO2(g) + H2O(l)
1 mole
1 moles
1 mole
1 mole
1 mole
Therefore:
0.1 moles 0.1 moles
0.1 moles 0.1 moles 0.1 moles
1 mole Na2SO4= 142 g Na2SO40.1 moles Na2SO4 = 0.1 X 142 g = 14.2 g
(c) 1 mole SO2 gas = 22.4 l 0.1 moles SO2 gas = 0.1 X 22.4 = 2.24 l
(d) 1 mole of water molecules = 6 X 10 23 molecules
0.1 moles of water molecules = 0.1 X 6 X 1023 molecules = 6 X 1022 molecules
(e) 1 mole of sulfuric acid = 98 g
0.1 moles of sulfuric acid = 0.1 X 98 g = 9.8 g
Question 48
Ethanol burns in air according to the equation:
C2H5OH(l) + 3O2(g) →2CO2(g) + 3H2O(g)
If 23 g of ethanol is burned completely, answer the following
Questions:
(a) What volume of oxygen (measured at s.t.p.) is used up in the reaction?
(b) What mass of water vapour is produced?
(c) How many molecules of carbon dioxide are formed?
Answer:
23 g ethanol = 23 / 46 = 0.5 moles ethanol
From the equation
C2H5OH(l) + 3O2(g)
1 mole
3 moles
→
2CO2(g) + 3H2O(g)
2 moles 3 moles
29
Therefore: 0.5 moles 1.5 moles
(a)
1 mole 1.5 moles
1 mole oxygen = 22.4 l
1.5 moles oxygen = 1.5 X 22.4 = 33.6 l
(b)
1 mole H2O = 18 g H2O
1.5 moles H2O = 1.5 X 18 g = 27 g
(c)
1 mole carbon dioxide molecules = 6 X 1023 molecules
Question 49
In an experiment, it as found that 5 g methylbenzene (C6H5CH3) reacted with potassium
manganate(VII) solution, forming 1.22 g benzoic acid (C6H5COOH), according to the
equation:
C6H5CH3(l) + 2MnO4-(aq) → 2MnO2(s) + C6H5COOH(aq) + OH-(aq)
(a) How many moles of methylbenzene were present initially?
(b) Calculate the percentage yield of benzoic acid.
Answer:
(a)5 g of methyl benzene = 5 / 92 moles
= 0.054 moles
(b) C6H5CH3(l) + 2MnO4-(aq) → 2MnO2(s) + C6H5COOH(aq) + OH-(aq)
1 mole
1 mole
0.054 moles
0.054 moles
Theoretical yield of benzoic acid = 0.054 moles = 0.054 X 122 g
= 6.59 g
Actual yield of ethanoic acid = 1.22 g
Actual yield X 100
1.22 X 100
Percentage yield =  % =  %
Theoretical yield
6.59
= 18.51%
Question 50
A sample of ethanoic acid was prepared by reacting 6.8 cm3 of ethanol (density 0.8 g cm-3) with an
acidified solution containing 32 g of sodium dichromate (Na2Cr2O7.2 H2O). The reaction may be
represented by the equation
3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH (l) + 4Cr3+(aq) + 11H2O(l)
30
After purification, it was found that 3 cm3 of ethanoic acid (density
1.06 g cm-3) were formed.
(b) Determine the limiting reactant.
(c) Calculate the percentage yield.
Answer:
(b) Density = mass/volume
Mass = volume X density
6.8 cm3 of ethanol = 6.8 X 0.8 g ethanol = 5.44 g ethanol = 5.44 / 46 moles
= 0.1183 moles
32 g sodium dichromate = 32 / 298 = 0.1074 moles
3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH(l) + 4Cr3+(aq) + 11H2O(l)
3 moles
2 moles
3 moles
4 moles
11 moles
1.5 moles
1 moles
1.5 moles
2 moles
5.5 moles
0.1074 X (1.5 moles
1 moles
1.5moles
2 moles
5.5 moles)
0.1611
0.1074
Since the 0.1074 moles of sodium dichromate that are present will react fully with 0.1611
moles of ethanol, and there are 0.1183 moles of ethanol present initially, it is clear that it is
the sodium dichromate that is present in excess.
Ethanol is the limiting reagent.
(c) Density = mass/volume
Mass = volume X density
6.8 cm3 of ethanol = 6.8 X 0.8 g ethanol = 5.44 g ethanol = 5.44 / 46 moles
= 0.1183 moles
3C2H5OH(l) + 2Cr2O72-(aq) + 16H+(aq) → 3CH3COOH (l) + 4Cr3+(aq) + 11H2O(l)
3 moles
3 moles
1 mole
1 mole
0.1183 moles
0.1183 moles
Theoretical yield of ethanoic acid = 0.1183 mole = 0.1183 moles X 60 g = 7.098 g
Actual yield of ethanoic acid = 3 cm3 = 3 X 1.06 g = 3.18 g
Actual yield X 100
3.18 X 100
Percentage yield =  % =  %
Theoretical yield
7.098
= 44.80%
31
Question 51
A sample of ethanal was prepared by reacting 9.2 cm3 of ethanol (density 0.8 g cm-3) with an
acidified solution containing 11 g of sodium dichromate (Na2Cr2O7.2 H2O). The reaction may
be represented by the equation
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(aq) + 7H2O(l)
After purification, it was found that 1.2 g of ethanal were formed.
(a) Determine the limiting reactant.
(b) Calculate the percentage yield of ethanal.
Answer:
(a) Density = mass/volume
Mass = volume X density
9.2 cm3 of ethanol = 9.2 X 0.8 g ethanol = 7.36 g ethanol = 7.36 / 46 moles
= 0.16 moles
11 g sodium dichromate = 11 / 298 = 0.0369 moles
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(
3 moles
1 mole
3 moles
aq)
2 moles
+ 7H2O(l)
7 moles
0.0369 x
(3 moles
0.1107
1 mole
3 moles
2 moles
7 moles)
0.0369
Since the 0.0369 moles of sodium dichromate that are present will react fully with 0.1107
moles of ethanol, and there are 0.16 moles of ethanol present initially, it is clear that it is the
ethanol that is present in excess.
Sodium dichromate is the limiting reagent.
(b) Actual yield =1.2 g ethanal
3C2H5OH(l) + Cr2O72-(aq) + 8H+(aq) → 3CH3CHO(l) + 2Cr3+(
3 moles
aq)
+ 7H2O(l)
1 mole
3 moles
2 moles
7 moles
1 mole
3 moles
2 moles
7 moles)
0.0369 X
(3 moles
Theoretical yield of ethanal = 0.0369 moles x 3 = 0.1107 moles =
0.1107 X 44 g = 4.871 g
Actual yield X 100
1.2
X 100
Percentage yield =  % =  %
Theoretical yield
4.871
= 24.64%
32