Mr Cole Chemistry

Percentage Yield and Atom Economy Questions

Percentage Yield & Atom Economy

Amount of Substance Worksheet

Easy Questions

1. 4.00g of magnesium reacts with excess hydrochloric acid producing magnesium chloride and hydrogen. When the magnesium chloride was crystalised, it weighed 12.2g.
a) What was the atom economy of the reaction?
Mg + 2HCl → MgCl2 + H2
Desired Product: MgCl2 Mr(MgCl2) = 24.3 + 2(35.5) = 95.3
Total Mr(Reactants/Products) = 95.3 + 2.0 = 97.3
Atom Economy = (95.3 / 97.3) × 100
Answer = 97.9%
b) What was the percentage yield of magnesium chloride?
Theoretical Yield n(Mg) = 4.00 / 24.3 = 0.1646 mol
Reaction ratio is 1:1, so n(MgCl2) = 0.1646 mol
Theoretical Mass = 0.1646 × 95.3 = 15.69 g
Percentage Yield Yield = (Actual / Theoretical) × 100
Yield = (12.2 / 15.69) × 100
Answer = 77.8%
2. 100g of barium chloride were reacted with excess sulfuric acid, producing barium sulfate and hydrochloric acid. The barium sulfate was filtered and dried, producing 90.1g of solid.
a) What was the atom economy of the reaction?
BaCl2 + H2SO4 → BaSO4 + 2HCl
Desired Product: BaSO4 Mr(BaSO4) = 137.3 + 32.1 + 64.0 = 233.4
Total Mr = 233.4 + 2(36.5) = 306.4
Atom Economy = (233.4 / 306.4) × 100
Answer = 76.2%
b) What was the percentage yield of barium sulfate?
Theoretical Yield Mr(BaCl2) = 137.3 + 71.0 = 208.3
n(BaCl2) = 100 / 208.3 = 0.480 mol
Theoretical Mass BaSO4 = 0.480 × 233.4 = 112.0 g
Percentage Yield Yield = (90.1 / 112.0) × 100
Answer = 80.4%
3. 150kg of iron (III) oxide are reduced to iron using carbon monoxide. 101kg of iron is formed.
a) What was the atom economy of the reaction?
Fe2O3 + 3CO → 2Fe + 3CO2
Desired Product: 2Fe Mass of Desired (2Fe) = 2 × 55.8 = 111.6
Total Mass (Reactants) = 159.6 + 3(28.0) = 243.6
Atom Economy = (111.6 / 243.6) × 100
Answer = 45.8%
b) What was the percentage yield of iron?
Theoretical Yield Mr(Fe2O3) = 159.6
n(Fe2O3) = 150,000 / 159.6 = 939.85 mol
n(Fe) = 939.85 × 2 = 1879.7 mol
Mass(Fe) = 1879.7 × 55.8 = 104,887 g = 104.9 kg
Percentage Yield Yield = (101 / 104.9) × 100
Answer = 96.3%
4. 1.40kg of phosphorous was reacted with excess chlorine. Given the percentage yield was 92.0% state the mass of phosphorous trichloride produced.
Working Out
2P + 3Cl2 → 2PCl3
Theoretical Mass n(P) = 1400 / 31.0 = 45.16 mol
Reaction is 1:1 ratio for P:PCl3
Mr(PCl3) = 31.0 + 3(35.5) = 137.5
Theoretical Mass = 45.16 × 137.5 = 6209.7 g = 6.21 kg
Actual Mass Mass = 6.21 × 0.92
Answer = 5.71 kg
5. 120g of barium carbonate was thermally decomposed, producing 100g of solid product.
a) What was the atom economy of the reaction?
BaCO3 → BaO + CO2
Mr(BaO) = 153.3, Mr(BaCO3) = 197.3
Atom Economy = (153.3 / 197.3) × 100
Answer = 77.7%
b) What was the percentage yield of barium oxide?

Note: The question states 100g of solid product, but this implies a mix of unreacted Carbonate and Oxide. The calculation below derives the yield from the mass lost as CO2.

Actual Yield Calculation Mass lost (CO2) = 120g – 100g = 20g
n(CO2) = 20 / 44.0 = 0.455 mol
n(BaO produced) = 0.455 mol
Actual Mass BaO = 0.455 × 153.3 = 69.7 g
Theoretical Yield Calculation n(BaCO3 start) = 120 / 197.3 = 0.608 mol
Theoretical Mass BaO = 0.608 × 153.3 = 93.2 g
Percentage Yield Yield = (69.7 / 93.2) × 100
Answer = 74.8%

Medium Difficulty Questions

1. 50.0g of sodium reacted with hydrochloric acid in order to produce hydrogen gas, producing 10dm3 of H2 at standard pressure and 298k. What were the atom economy and percentage yield of the reaction in terms of volume?
Working Out
2Na + 2HCl → 2NaCl + H2
Atom Economy (Desired Product: H2) Mass of Desired (H2) = 2.0
Total Mass = 2(58.5) + 2.0 = 119.0
Atom Economy = (2.0 / 119.0) × 100 = 1.68%
Percentage Yield (Volume) n(Na) = 50.0 / 23.0 = 2.174 mol
n(H2) theoretical = 2.174 / 2 = 1.087 mol
Theoretical Volume (PV=nRT) = (1.087 × 8.31 × 298) / 100000 = 0.0269 m3 = 26.9 dm3
Yield = (10 / 26.9) × 100 = 37.2%
2. 1 litre of propene gas was reacted against excess hydrogen chloride at 25oC and 101kPa. It produced 2.54g of chloropropane. What was the percentage yield?
Working Out
Step 1: Moles of Propene n = (PV) / (RT) = (101000 × 0.001) / (8.31 × 298) = 0.0408 mol
Step 2: Theoretical Mass Mr(Chloropropane) = 78.5
Theoretical Mass = 0.0408 × 78.5 = 3.20 g
Step 3: Percentage Yield Yield = (2.54 / 3.20) × 100
Answer = 79.4%
3. During the production of sulfuric acid, a series of chemical reactions take place:
Sulfur + oxygen → sulfur dioxide
Sulfur dioxide + oxygen → sulfur trioxide
Sulfur trioxide + sulfuric acid → oleum
Oleum + water → sulfuric acid
a) What is the percentage yield if you produce 55kg of sulfuric acid from 20kg of sulfur?
Step 1: Moles of Sulfur n(S) = 20,000 / 32.1 = 623.05 mol
Step 2: Theoretical Yield H2SO4 Ratio S:H2SO4 is 1:1 across the sequence.
Mass = 623.05 × 98.1 = 61,121 g = 61.1 kg
Step 3: Percentage Yield Yield = (55 / 61.1) × 100
Answer = 90.0%
b) What is the atom economy?
The process involves sequential addition reactions where all reactants ultimately become part of the product (Sulfuric Acid). No waste products are formed in the stoichiometry.
Answer = 100%

Hard Questions

1. 40.0g of ethene was reacted with excess steam. The organic product was reacted with excess dichromate under reflux. The volume of the product of this second reaction was made up to 200cm3. 15.0cm3 of the product was titrated against 0.5moldm-3 sodium hydroxide where it was found to require 20.0cm3 of sodium hydroxide to neutralise.
Assuming that the titration was completely accurate, what was the percentage yield of ethanoic acid?
Working Out
Step 1: Moles of Acid in 15cm3 sample Reaction: CH3COOH + NaOH → CH3COONa + H2O
n(NaOH) = 0.5 × 0.020 = 0.01 mol
n(Ethanoic Acid) = 0.01 mol
Step 2: Total Actual Moles (in 200cm3) Scale Factor = 200 / 15 = 13.33
n(Total) = 0.01 × 13.33 = 0.1333 mol
Step 3: Theoretical Moles n(Ethene) = 40.0 / 28.0 = 1.428 mol
Sequence: Ethene → Ethanol → Ethanoic Acid (1:1 ratio)
Theoretical n(Ethanoic Acid) = 1.428 mol
Step 4: Percentage Yield Yield = (0.1333 / 1.428) × 100
Answer = 9.33%
2. Rutile, a common titanium (iv) oxide ore, is extracted in a two-step process. The ore is first reacted with chlorine and carbon. In this process the chlorine is reduced whereas the carbon is oxidised to form carbon monoxide. The titanium does not undergo either reduction or oxidation. In the second step, the titanium containing compound is reacted with magnesium in a redox reaction.
a) If you start with 50 tonnes of rutile and produce 90 tonnes of titanium compound in the first step, what is the percentage yield of the first step?
TiO2 + 2Cl2 + 2C → TiCl4 + 2CO
Step 1: Theoretical Yield n(TiO2) = 50,000,000 / 79.9 = 625,782 mol
The titanium compound produced is TiCl4 (Mr = 189.9)
Theoretical Mass = 625,782 × 189.9 = 118.84 tonnes
Step 2: Percentage Yield Yield = (90 / 118.84) × 100
Answer = 75.7%
b) The second step typically has a percentage yield of 66%. How much titanium would you produce from 50 tonnes of rutile?
Calculation Theoretical mass of pure Ti from 50t Rutile:
n(Ti) = 625,782 mol
Mass(Ti) = 625,782 × 47.9 = 29.97 tonnes

Overall Yield = Yield 1 × Yield 2 = 0.757 × 0.66 = 0.4996 (49.96%)
Actual Mass = 29.97 × 0.4996 = 14.97 tonnes
Answer = 15.0 tonnes
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