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Questions on reacting masses, percentage yield and atom economy

Reacting mass calculations from a balanced equation, the limiting reagent and reagents in excess, theoretical and percentage yield, where yield is lost in a preparation and over several steps, atom economy and why it is not the same as yield, why addition reactions have an atom economy of 100%, and the economic and environmental case for high atom economy, with the two industrial routes to ibuprofen, and why chemists aim for processes that need no solvent.

21 original questions · 65 marks · the reacting masses, percentage yield and atom economy notes · Amount of substance · Flashcards for this unit

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• 6 marks · Compare · RMP-Q14

  Compare the hydration of ethene with the fermentation of glucose as industrial methods of making ethanol. Consider atom economy, yield, raw materials, rate, energy and the purity of the product.

  Mark scheme

  Indicative content

  1. hydration has an atom economy of 100% because ethanol is the only product, whereas fermentation has 51.1% because carbon dioxide is formed as well
  2. only about 5% of the ethene reacts on each pass, but unreacted ethene is recycled so the overall conversion is about 95%; fermentation stops when the ethanol reaches about 15%
  3. hydration uses ethene from crude oil, a finite resource, whereas fermentation uses sugars from crops that can be grown again each year
  4. hydration is fast and continuous over a phosphoric acid catalyst, whereas fermentation is a slow batch process that takes days at about 35 °C
  5. hydration needs high temperature and pressure, which costs energy, whereas fermentation runs warm at atmospheric pressure but needs energy for distillation
  6. hydration gives nearly pure ethanol, whereas fermentation gives a dilute aqueous solution that must be fractionally distilled before use

  Levels

  • Level 3 (5–6 marks) all six aspects named in the question are compared, each with both processes in the same point and a correct figure where one applies, such as the 100% and 51.1% atom economies and the 95% conversion with recycling; no point describes one process without the other
  • Level 2 (3–4 marks) four or five of the aspects are compared with both processes in each point, or all six appear and some describe one process only or carry a wrong figure
  • Level 1 (1–2 marks) two or three statements, mostly about one process; fermentation is usually called the 'greener' method on every count, with no figures

• 4 marks · Calculate · RMP-Q2

  Calculate the mass of waste for each kilogram of a drug intermediate made by route P and by route Q, from the data given in the table, and state which route makes less waste. Assume reactant that does not become product is not recovered.

  Measure	Route P	Route Q
  Atom economy	60%	45%
  Percentage yield	80%	90%
  Solvent used per kg of product	10 kg, 95% recovered	None

  Mark scheme

  1. M1 route P: reactants per kg of product = 1.00 ÷ 0.60 ÷ 0.80 = 2.08 kg, so 1.08 kg of reactant waste
  2. M1 route Q: reactants per kg of product = 1.00 ÷ 0.45 ÷ 0.90 = 2.47 kg, so 1.47 kg of reactant waste
  3. M1 solvent lost by route P = 5% of 10 kg = 0.50 kg, added to its reactant waste
  4. A1 route P 1.6 kg and route Q 1.5 kg of waste per kg, to two significant figures as the data are given (accept 1.58 and 1.47 kg), so route Q makes less waste despite its lower atom economy; allow error carried forward

• 4 marks · Calculate · RMP-Q4

  Calculate the volume of hydrogen at room temperature and pressure formed when 0.486 g of magnesium is added to 20.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). The molar gas volume is 24.0 dm³ mol⁻¹; Mg is 24.3.

  Mark scheme

  1. M1 n(Mg) = 0.486 ÷ 24.3 = 0.0200 mol and n(HCl) = 1.00 × 0.0200 = 0.0200 mol
  2. M1 0.0200 mol of Mg would need 0.0400 mol of HCl, so the hydrochloric acid is the limiting reagent
  3. M1 n(H₂) = 0.0200 ÷ 2 = 0.0100 mol
  4. A1 volume of hydrogen = 0.0100 × 24.0 = 0.240 dm³, to three significant figures

• 4 marks · Evaluate · RMP-Q15

  Evaluate the claim that a process with a percentage yield of 95% must be more sustainable than one with a percentage yield of 70%.

  Mark scheme

  1. B1 percentage yield measures only how much of the theoretical product is collected
  2. B1 a process with a high yield can still have a low atom economy, turning much of its reactant mass into by-products
  3. B1 sustainability also depends on energy use, whether the raw materials are renewable, and whether by-products are used or sold
  4. B1 so the claim is not justified on yield alone; the two processes need comparing on atom economy and these other factors as well

• 4 marks · Calculate · RMP-Q16

  Calculate the percentage yield of ethyl propanoate from a preparation whose masses are given in the table: C₂H₅OH(l) + C₂H₅COOH(l) ⇌ C₂H₅COOC₂H₅(l) + H₂O(l).

  Substance	Mass / g	Mr
  Ethanol, used	2.30	46.0
  Propanoic acid, used	6.00	74.0
  Ethyl propanoate, collected	3.57	102.0

  Mark scheme

  1. M1 n(ethanol) = 2.30 ÷ 46.0 = 0.0500 mol and n(propanoic acid) = 6.00 ÷ 74.0 = 0.0811 mol
  2. M1 the ratio is 1 : 1, so ethanol is the limiting reagent
  3. M1 theoretical mass of ester = 0.0500 × 102.0 = 5.10 g
  4. A1 percentage yield = 3.57 ÷ 5.10 × 100 = 70.0%, to three significant figures

• 4 marks · Calculate · RMP-Q18

  Calculate the percentage yield when 4.60 g of sodium reacts with 5.68 g of chlorine and 8.42 g of sodium chloride is collected: 2Na(s) + Cl₂(g) → 2NaCl(s). Use Na 23.0, Cl 35.5.

  Mark scheme

  1. M1 n(Na) = 4.60 ÷ 23.0 = 0.200 mol and n(Cl₂) = 5.68 ÷ 71.0 = 0.0800 mol
  2. M1 dividing by the coefficients gives Na 0.100 and Cl₂ 0.0800, so chlorine is the limiting reagent
  3. M1 n(NaCl) = 2 × 0.0800 = 0.160 mol, so the theoretical mass = 0.160 × 58.5 = 9.36 g
  4. A1 percentage yield = 8.42 ÷ 9.36 × 100 = 90.0%, to three significant figures

• 3 marks · Calculate · RMP-Q3

  Calculate the maximum mass of iron that can be made from 500 g of iron(III) oxide: Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g). Use Fe 55.8, O 16.0.

  Mark scheme

  1. M1 n(Fe₂O₃) = 500 ÷ 159.6 = 3.133 mol
  2. M1 n(Fe) = 2 × 3.133 = 6.266 mol, and mass = 6.266 × 55.8
  3. A1 mass of iron = 3.50 × 10² g, to three significant figures

• 3 marks · Calculate · RMP-Q5

  Calculate the percentage yield of calcium oxide from the thermal decomposition of calcium carbonate, CaCO₃(s) → CaO(s) + CO₂(g), using the data in the table.

  Substance	Mass / g	Mr
  Calcium carbonate, heated	10.0	100.1
  Calcium oxide, collected	5.04	56.1

  Mark scheme

  1. M1 n(CaCO₃) = 10.0 ÷ 100.1 = 0.0999 mol, so n(CaO) = 0.0999 mol
  2. M1 theoretical mass of CaO = 0.0999 × 56.1 = 5.604 g
  3. A1 percentage yield = 5.04 ÷ 5.604 × 100 = 89.9%, to three significant figures

• 3 marks · Calculate · RMP-Q6

  Calculate the atom economy for iron in the reaction Fe₂O₃(s) + 3CO(g) → 2Fe(l) + 3CO₂(g). Use Fe 55.8, C 12.0, O 16.0.

  Mark scheme

  1. M1 mass of desired product = 2 × 55.8 = 111.6
  2. M1 total mass of reactants = 159.6 + 3 × 28.0 = 243.6
  3. A1 atom economy = 111.6 ÷ 243.6 × 100 = 45.8%, to three significant figures

• 3 marks · Calculate · RMP-Q7

  Calculate the overall percentage yield of the three-step synthesis in the table, and the mass of final product (Mr 150.0) made from 0.100 mol of starting material, one mole of product forming from each mole of starting material.

  Step	1	2	3
  Yield / %	85	70	90

  Mark scheme

  1. M1 overall yield = 0.85 × 0.70 × 0.90 = 0.5355
  2. A1 overall percentage yield = 54%, to two significant figures as the step yields are given; accept 53.6%
  3. A1 mass = 0.100 × 150.0 × 0.5355 = 8.0 g, to two significant figures; accept 8.03 g

• 3 marks · Explain · RMP-Q8

  Explain why the percentage yield of a preparation carried out in a school laboratory is less than 100%.

  Mark scheme

  1. B1 the reaction may not go to completion, for example because it reaches equilibrium
  2. B1 side reactions turn some of the reactant into other products
  3. B1 product is lost on glassware, during transfer and in purification steps such as filtration, recrystallisation or distillation

• 3 marks · Explain · RMP-Q10

  Explain how the hydration of ethene can have an atom economy of 100% while only about 5% of the ethene reacts on each pass over the catalyst.

  Mark scheme

  1. B1 atom economy is calculated from the equation and depends only on which products form, not on how much reacts
  2. B1 C₂H₄(g) + H₂O(g) → C₂H₅OH(g) has ethanol as its only product, so the atom economy is 100%
  3. B1 the 5% per pass is a measure of conversion or yield, which is limited by the equilibrium and conditions; unreacted ethene is recycled

• 3 marks · Suggest · RMP-Q13

  Suggest three reasons why a student who expects 8.80 g of ethyl ethanoate from a preparation collects only 5.94 g.

  Mark scheme

  1. B1 esterification is reversible, so the mixture reaches equilibrium with some ethanol and ethanoic acid unreacted
  2. B1 some ester is lost in the separating funnel and washings, or left on the drying agent
  3. B1 some ester is left in the flask and still head, or evaporates, during distillation

• 3 marks · Calculate · RMP-Q17

  Calculate the atom economy for chloromethane in the reaction CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g), and the mass of hydrogen chloride formed for each kilogram of chloromethane. Use C 12.0, H 1.0, Cl 35.5.

  Mark scheme

  1. M1 Mr of CH₃Cl = 50.5 and total Mr of the reactants = 16.0 + 71.0 = 87.0
  2. A1 atom economy = 50.5 ÷ 87.0 × 100 = 58.0%, to three significant figures
  3. A1 mass of HCl = 1000 × 36.5 ÷ 50.5 = 723 g per kilogram, to three significant figures; accept 0.723 kg

• 3 marks · Explain · RMP-Q19

  Explain why chemists designing a manufacturing process aim for one that needs no solvent.

  Mark scheme

  1. B1 a solvent must be bought and then separated from the product, so it adds cost and waste
  2. B1 recovering the solvent by distillation uses energy, and solvent that is not recovered must be disposed of
  3. B1 many organic solvents are flammable or toxic and escape as volatile organic compounds, so removing them reduces the hazard

• 3 marks · Justify · RMP-Q21

  Justify the use of an excess of magnesium carbonate, rather than an excess of sulfuric acid, when magnesium sulfate crystals are prepared from the two.

  Mark scheme

  1. B1 magnesium carbonate is insoluble, so when the acid is used up the unreacted excess stays as a solid and is removed by filtration
  2. B1 the acid is the limiting reagent, so none is left in the filtrate to contaminate the crystals; an excess of acid cannot be filtered off and would stay in the solution
  3. B1 the end of the effervescence shows that all the acid has reacted, so the filtrate holds only magnesium sulfate and water and evaporation gives a pure product

• 2 marks · Explain · RMP-Q9

  Explain why every addition reaction has an atom economy of 100%.

  Mark scheme

  1. B1 in an addition reaction the reactants join to form a single product
  2. B1 all the atoms of the reactants end up in the desired product, so its mass equals the total mass of the reactants

• 2 marks · Explain · RMP-Q11

  Explain why a chemist making an ester often uses an excess of the cheaper reactant.

  Mark scheme

  1. B1 an excess of the cheaper reactant makes the expensive reactant the limiting reagent, so as much of it as possible reacts
  2. B1 esterification is an equilibrium, and an excess of one reactant moves the position of equilibrium towards the ester, raising the yield

• 2 marks · Suggest · RMP-Q12

  Suggest two ways in which a company could reduce the waste from a process that has a low atom economy.

  Mark scheme

  1. B1 find a use for the by-product, or sell it, so that it is not disposed of as waste
  2. B1 change to a different route, for example one with fewer steps or an addition reaction, that has a higher atom economy

• 2 marks · Suggest · RMP-Q20

  Suggest why the atom economy of a pharmaceutical synthesis can make it look much less wasteful than it is.

  Mark scheme

  1. B1 atom economy is calculated from the balanced equation, and solvents, washes and recrystallisation liquids are not in the equation
  2. B1 the mass of solvent used can be many times the mass of the product, so most of the waste is not counted

• 1 mark · State · RMP-Q1

  State what is meant by the limiting reagent in a reaction.

  Mark scheme

  1. B1 the reactant that is completely used up first, and so determines the maximum amount of product

Practise reacting masses, percentage yield and atom economy one question at a time