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
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 RatioCarbon
/ 75 / 75/12 = 6.25 / 1Hydrogen / 25 / 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 RatioCarbon
/ 60 / 60/12 = 5 / 3Hydrogen / 13.33 / 13.33/1 = 13.33 / 8
Oxygen / 26.66 / 26.66/16 = 1.66 / 1
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 RatioCarbon
/ 40 / 40/12 = 3.33 / 1Hydrogen / 6.66 / 6.66/1 = 6.66 / 2
Oxygen / 53.33 / 53.33/16 = 3.33 / 1
Empirical formula = CH2O
Question 5
Find the empirical formulas of compounds containing:
(a) 77.77 % Fe, 22.22 % O
Answer:
Element
/ Percentage / Percentage / Ar / Simplest RatioIron
/ 77.77 / 77.77/56 = 1.39 / 1Oxygen / 22.22 / 22.22/16 = 1.39 / 1
Empirical formula = FeO
(b) 70% Fe, 30% O
Answer:
Element
/ Percentage / Percentage / Ar / Simplest RatioIron
/ 70 / 70/56 = 1.25 / 2Oxygen / 30 / 30/16 = 1.875 / 3
Empirical formula = Fe2O3
(c) 72.41% Fe, 27.59% O
Answer:
Element
/ Percentage / Percentage / Ar / Simplest RatioIron
/ 72.41 / 72.41/56 = 1.29 / 3Oxygen / 27.59 / 27.59/16 = 1.72 / 4
Empirical formula = Fe3O4
(d) 46.66% N, 53.33% O
Answer:
Element
/ Percentage / Percentage / Ar / Simplest RatioNitrogen
/ 46.66 / 46.66/14 = 3.33 / 1Oxygen / 53.33 / 53.33/16 = 3.33 / 1
Empirical formula = NO
(e) 30.43% N, 69.57% O
Answer:
Element
/ Percentage / Percentage / Ar / Simplest RatioNitrogen
/ 30.43 / 30.43/14 = 2.17 / 1Oxygen / 69.57 / 69.57/16 = 4.35 / 2
Empirical formula = NO2
(f) 39.81% Cu, 20.06% S, 40.13% O
Answer:
Element
/ Percentage / Percentage / Ar / Simplest RatioCopper
/ 39.81 / 39.81/63.5 = 0.627 / 1Sulfur
/ 20.06 / 20.06/32 = 0.627 / 1Oxygen / 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 RatioCopper
/ 56.95 / 56.95/63.5 = 0.9 / 2Sulfur
/ 14.35 / 14.35/32 = 0.45 / 1Oxygen / 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 RatioHydrogen
/ 2.13 / 2.13/1 = 2.13 / 1Nitrogen
/ 29.79 / 29.79/14 = 2.13 / 1Oxygen / 68.09 / 68.09/16 = 4.26 / 2
Empirical formula = HNO2
(i) 1.59% H, 22.22% N, 76.19% O
Answer:
Element
/ Percentage / Percentage / Ar / Simplest RatioHydrogen
/ 1.59 / 1.59/1 = 1.59 / 1Nitrogen
/ 22.22 / 22.22/14 = 1.59 / 1Oxygen / 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?
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
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 RatioCarbon
/ 84 / 84/12 = 7 / 7Hydrogen / 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 RatioCarbon
/ 54.54 / 54.54/12 = 4.54 / 2Hydrogen / 9.09 / 9.09/1 = 9.09 / 4
Oxygen / 36.36 / 36.36/16 = 2.273 / 1
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:
Element
/ Percentage / Percentage / Ar / Simplest RatioCarbon
/ 62 / 62/12 = 5.17 / 3Hydrogen / 10.4 / 10.4/1 = 10.4 / 6
Oxygen / 27.6 / 27.6/16 = 1.725 / 1
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 RatioCarbon
/ 40 / 40/12 = 3.33 / 1Hydrogen / 6.66 / 6.66/1 = 6.66 / 2
Oxygen / 53.33 / 53.33/16 = 3.33 / 1
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%
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%
(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
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%
= 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
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
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 HH
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
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 AgNO3X 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 H2X 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 O2X 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
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 CO2X 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 Cl2X 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
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 1 mole 2 moles
Therefore 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) + 3H2(g) → 2NH3(g)
1 mole 3 moles 2 moles
Therefore 0.5 X (1 mole 3 moles 2 moles)
= 0.5 moles 1.5 moles 1 mole
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
C4H10(g) + 61/2O2(g) → 4CO2(g) + 5H2O(g)
1 mole 6.5 moles 4 moles 5 moles
Therefore
4 X (1 mole 6.5 moles 4 moles 5 moles)
= 2 moles 26 moles 16 moles 20 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
Therefore
10 X (0.5 moles 0.5 moles 1 mole)
= 5 moles 5 moles 10 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?
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) → CaCO3(s) + H2O(l)
1 mole 1 mole 1 mole 1 mole
Therefore: 0.1 moles 0.1 moles 0.1 moles 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
From the equationC2H6(g) + 3 1/2O2(g) → 2CO2(g) + 3H2O(l)
1 mole 3.5 moles 2 moles 3 moles
Therefore:0.375 moles3 X 0.375 = 1.125 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)
1 mole 2 moles 1 mole 1 mole
Therefore 2 X (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
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)
2 moles 4 moles 1 mole
Therefore 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 5 moles 2 moles
Therefore0.5 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?
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 mole1 mole 4 moles 1 mole)
= 0.125 moles0.125 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