percentage Yield and Atom economy

Define percentage yield.

The percentage yield is the actual yield of a reaction compared with the theoretical yield, expressed as a percentage. It is a measure of the efficiency of a reaction in converting reactants into products.

State the equation for percentage yield.

Percentage yield = (actual yield / theoretical yield) × 100.

Define theoretical yield.

The theoretical yield is the maximum possible mass of product that could be formed from a given amount of reactants, assuming that the reaction goes to completion with 100% efficiency.

Define actual yield.

The actual yield is the mass of product that is actually obtained from a reaction in practice. It is always less than the theoretical yield due to losses, incomplete reactions or side reactions.

State the equation for atom economy.

Atom economy = (Mr of desired product / sum of Mr of all products) × 100. OR Atom economy = (Mr of desired product / sum of Mr of all reactants) × 100 (for reactions with 100% atom economy).

Define atom economy.

Atom economy is a measure of how much of the total mass of reactants is converted into the desired product. It is calculated as the molecular mass of the desired product divided by the molecular mass of all products, multiplied by 100.

Explain why a high atom economy is beneficial.

A high atom economy means there is less waste produced. This increases the sustainability of the chemical process as fewer resources are wasted and less waste needs to be disposed of. It also means more of the starting materials are converted into useful products.

State the atom economy for addition reactions.

Addition reactions have 100% atom economy because there is only one product (the desired product) and no by-products. All atoms in the reactants are converted into useful products.

State the atom economy for substitution and elimination reactions.

Substitution and elimination reactions have atom economies less than 100% because other products (by-products) are formed in addition to the desired product.

Explain why addition reactions have 100% atom economy.

In an addition reaction, all atoms from the reactants are incorporated into the single product. There are no by-products, so no atoms are wasted. The mass of the product equals the sum of the masses of all reactants.

Explain how atom economy can be improved.

Atom economy can be improved by finding a use for the by-product (waste product), using alternative reactants or reactions that produce fewer by-products, or using a catalyst which allows the reaction to proceed by an alternative route with higher atom economy.

Explain how percentage yield and atom economy are different.

Percentage yield measures the efficiency of converting reactants into products (how much product is actually obtained compared to the theoretical maximum). Atom economy measures how much of the mass of reactants ends up in the desired product (how much waste is produced). A reaction can have a high percentage yield but a low atom economy.

Calculate the atom economy for CaCO₃ → CaO + CO₂.

Desired product: CaO (Mr = 56.1). All products: CaO + CO₂ (Mr = 56.1 + 44.0 = 100.1). Atom economy = 56.1 / 100.1 × 100 = 56.0%.

Calculate the atom economy for 2Fe₂O₃ + 3C → 4Fe + 3CO₂.

Desired product: Fe (Mr = 55.8, ×4 = 223.2). All products: Fe + CO₂ (Mr = 223.2 + 132.0 = 355.2). Atom economy = 223.2 / 355.2 × 100 = 62.8%.

Calculate the atom economy for C₂H₄ + Cl₂ → C₂H₄Cl₂.

Desired product: C₂H₄Cl₂ (Mr = 99.0). All products: C₂H₄Cl₂ (Mr = 99.0). Atom economy = 99.0 / 99.0 × 100 = 100%.

Calculate the atom economy for C₆H₆ + HNO₃ → C₆H₅NO₂ + H₂O.

Desired product: C₆H₅NO₂ (Mr = 123.0). All products: C₆H₅NO₂ + H₂O (Mr = 123.0 + 18.0 = 141.0). Atom economy = 123.0 / 141.0 × 100 = 87.2%.

Calculate the atom economy for CH₃COCl + C₂H₅NH₂ → CH₃CONHC₂H₅ + HCl.

Desired product: CH₃CONHC₂H₅ (Mr = 87.0). All products: CH₃CONHC₂H₅ + HCl (Mr = 87.0 + 36.5 = 123.5). Atom economy = 87.0 / 123.5 × 100 = 70.4%.

Calculate the atom economy for C₂H₅Cl + NaOH → C₂H₅OH + NaCl.

Desired product: C₂H₅OH (Mr = 46.0). All products: C₂H₅OH + NaCl (Mr = 46.0 + 58.5 = 104.5). Atom economy = 46.0 / 104.5 × 100 = 44.0%.

Calculate the atom economy for C₂H₅Cl + NaOH → C₂H₄ + H₂O + NaCl.

Desired product: C₂H₄ (Mr = 28.0). All products: C₂H₄ + H₂O + NaCl (Mr = 28.0 + 18.0 + 58.5 = 104.5). Atom economy = 28.0 / 104.5 × 100 = 26.8%.

Calculate the atom economy for the fermentation of glucose: C₆H₁₂O₆ → 2C₂H₅OH + 2CO₂.

Desired product: C₂H₅OH (Mr = 46.0, ×2 = 92.0). All products: ethanol + CO₂ (Mr = 92.0 + 88.0 = 180.0). Atom economy = 92.0 / 180.0 × 100 = 51.1%.

Calculate the atom economy for C(s) + 2H₂O(g) → 2H₂(g) + CO₂(g).

Desired product: H₂ (Mr = 2.0, ×2 = 4.0). All products: H₂ + CO₂ (Mr = 4.0 + 44.0 = 48.0). Atom economy = 4.0 / 48.0 × 100 = 8.3%.

Calculate the atom economy for CH₃CH₂CH(OH)CH₃ → CH₃CH=CHCH₃ + H₂O.

Desired product: but-2-ene (Mr = 56.0). All products: but-2-ene + H₂O (Mr = 56.0 + 18.0 = 74.0). Atom economy = 56.0 / 74.0 × 100 = 75.7%.

Calculate the atom economy for CH₃CHICH₃ + NaOH → CH₃CH(OH)CH₃ + NaI.

Desired product: propan-2-ol (Mr = 60.0). All products: propan-2-ol + NaI (Mr = 60.0 + 149.9 = 209.9). Atom economy = 60.0 / 209.9 × 100 = 28.6%.

Calculate the atom economy for CH₃CHICH₃ + KOH → CH₃CH(OH)CH₃ + KI.

Desired product: propan-2-ol (Mr = 60.0). All products: propan-2-ol + KI (Mr = 60.0 + 166.0 = 226.0). Atom economy = 60.0 / 226.0 × 100 = 26.5%.

Calculate the atom economy for 2NH₄Cl + Ca(OH)₂ → CaCl₂ + 2NH₃ + 2H₂O.

Desired product: NH₃ (Mr = 17.0, ×2 = 34.0). All products: CaCl₂ + NH₃ + H₂O (Mr = 111.0 + 34.0 + 36.0 = 181.0). Atom economy = 34.0 / 181.0 × 100 = 18.8%.

Explain why using KOH instead of NaOH in a reaction reduces the atom economy.

KOH has a higher Mr than NaOH. The by-product (KI) has a higher Mr than NaI. The desired product has the same Mr. The total mass of products increases, so the atom economy decreases.

Explain how the atom economy of the fermentation of glucose could be increased.

Find a use for the CO₂ by-product (e.g. in carbonated drinks, fire extinguishers, or as a chemical feedstock). This means the CO₂ is no longer waste, so the atom economy increases.

Explain how the atom economy of the Haber process (N₂ + 3H₂ → 2NH₃) is 100%.

The reaction is an addition reaction (forming a single product). There are no by-products. All atoms from the reactants are incorporated into the ammonia product.

Calculate the percentage yield for the production of CaO from CaCO₃.

Moles CaCO₃ = 50000 / 100.1 = 499.5 mol. Theoretical moles CaO = 499.5 mol. Theoretical mass CaO = 499.5 × 56.1 = 28022 g = 28.02 kg. Percentage yield = (21.0 / 28.02) × 100 = 74.9%.

Calculate the mass of compound A from phenylamine with 61.0% yield.

Moles phenylamine = 3.00 / 93.0 = 0.0323 mol. Theoretical moles compound A = 0.0323 mol. Theoretical mass compound A = 0.0323 × 135.0 = 4.35 g. Actual mass = 4.35 × 61.0 / 100 = 2.66 g.

Calculate the mass of 2-bromobutane needed to produce 3.552 g of butan-2-ol with 80.0% yield.

Theoretical mass butan-2-ol = 3.552 / 0.800 = 4.440 g. Moles butan-2-ol = 4.440 / 74.0 = 0.0600 mol. Moles 2-bromobutane = 0.0600 mol (1:1 ratio). Mass 2-bromobutane = 0.0600 × 136.9 = 8.21 g.

Calculate the percentage yield of urea from ammonia.

Moles NH₃ = 1.00 tonnes = 1.00 × 10⁶ g / 17.0 = 58823.5 mol. From equation: 2 mol NH₃ → 1 mol urea. Theoretical moles urea = 58823.5 / 2 = 29411.8 mol. Theoretical mass urea = 29411.8 × 60.0 = 1764706 g = 1.7647 tonnes. Percentage yield = (1.35 / 1.7647) × 100 = 76.5%.

Calculate the atom economy for the production of urea: 2NH₃ + CO₂ → NH₂CONH₂ + H₂O.

Desired product: urea (Mr = 60.0). All products: urea + H₂O (Mr = 60.0 + 18.0 = 78.0). Atom economy = 60.0 / 78.0 × 100 = 76.9%.

Calculate the atom economy for C₄H₉OH + KBr + H₂SO₄ → C₄H₉Br + KHSO₄ + H₂O.

Desired product: C₄H₉Br (Mr = 136.9). All products: C₄H₉Br + KHSO₄ + H₂O (Mr = 136.9 + 136.2 + 18.0 = 291.1). Atom economy = 136.9 / 291.1 × 100 = 47.0%.

Suggest a reactant that could improve the atom economy of making 1-bromobutane.

Use HBr instead of KBr and H₂SO₄. The reaction would be C₄H₉OH + HBr → C₄H₉Br + H₂O. Atom economy = 136.9 / (136.9 + 18.0) × 100 = 88.4%. This has a higher atom economy than using KBr and H₂SO₄.

Calculate the percentage yield of 1-bromobutane from butan-1-ol.

Moles butan-1-ol = 5.92 / 74.0 = 0.0800 mol. Theoretical moles 1-bromobutane = 0.0800 mol. Theoretical mass 1-bromobutane = 0.0800 × 136.9 = 10.95 g. Percentage yield = (9.72 / 10.95) × 100 = 88.8%.

Explain why a reaction with 100% atom economy does not necessarily have 100% percentage yield.

Atom economy measures the proportion of atoms that end up in the desired product (theoretical maximum). Percentage yield measures how much product is actually obtained compared to the theoretical maximum. Even with 100% atom economy, the reaction may not go to completion, or product may be lost during purification, so the percentage yield may be less than 100%.

Explain why a reaction with a high percentage yield may still have a low atom economy.

Percentage yield measures how efficiently reactants are converted to products (including by-products). Atom economy measures how much of the reactants' mass ends up in the desired product. A reaction can convert 100% of the reactants into products (high percentage yield) but the desired product may be only a small fraction of the total mass of products (low atom economy).

Explain why sustainable development involves maximising atom economy.

Sustainable development involves maximising our use of resources available and reducing waste products when possible. High atom economy means less waste is produced, fewer resources are consumed, and less energy is required for waste disposal.

Explain why addition reactions are preferred for sustainability.

Addition reactions have 100% atom economy. All atoms from the reactants are incorporated into the desired product. There are no by-products, so no waste is produced. This is more sustainable than substitution or elimination reactions.

Explain why catalysts are beneficial for sustainability.

Catalysts allow reactions to proceed via alternative routes with lower activation energy, often at lower temperatures. This reduces energy demand and CO₂ emissions. Catalysts also often allow reactions to be more selective, reducing the formation of unwanted by-products and increasing atom economy.

Explain why finding a use for a by-product improves atom economy.

If a by-product is used rather than discarded as waste, it is no longer considered waste. The atom economy calculation considers all products; if all products are useful, the atom economy increases.

Explain why the atom economy for the formation of ethanol by fermentation is low.

Glucose fermentation produces ethanol and CO₂. The desired product (ethanol) is only 51% of the total mass of products. The other 49% is CO₂ waste. This is a low atom economy.

Explain why the atom economy for the production of hydrogen from carbon and steam is low.

The reaction is C + 2H₂O → 2H₂ + CO₂. The desired product (H₂) is only 8.3% of the total mass of products. The other 91.7% is CO₂ waste. This is a very low atom economy.

Explain why the atom economy for the Haber process is 100%.

N₂ + 3H₂ → 2NH₃. This is an addition reaction with only one product (NH₃). All atoms from the reactants are incorporated into the product, so the atom economy is 100%.

Explain why the atom economy for the production of 1-bromobutane using HBr is higher than using KBr and H₂SO₄.

With HBr: C₄H₉OH + HBr → C₄H₉Br + H₂O. Only one by-product (H₂O). With KBr and H₂SO₄: C₄H₉OH + KBr + H₂SO₄ → C₄H₉Br + KHSO₄ + H₂O. Two by-products (KHSO₄ and H₂O). Using HBr produces less waste, so the atom economy is higher.

Explain why percentage yield can never exceed 100%.

The theoretical yield is the maximum possible amount of product that could be formed based on the stoichiometry of the reaction. The actual yield cannot exceed this maximum because it is limited by the amount of reactants present. If the actual yield exceeded the theoretical yield, it would violate the law of conservation of mass.

Explain why the actual yield is always less than the theoretical yield.

In practice, reactions rarely go to completion. Product may be lost during purification or transfer. Side reactions may consume some of the reactants. Some product may remain in solution or on apparatus. These factors all reduce the actual yield.

Calculate the mass of CaO produced from 50 kg CaCO₃ with 74.9% yield.

Moles CaCO₃ = 50000 / 100.1 = 499.5 mol. Theoretical moles CaO = 499.5 mol. Theoretical mass CaO = 499.5 × 56.1 = 28022 g = 28.02 kg. Actual mass = 28.02 × 0.749 = 20.98 kg (≈ 21 kg as given in the question).

Calculate the theoretical yield of urea from 1.00 tonne of ammonia.

Moles NH₃ = 1.00 × 10⁶ / 17.0 = 58823.5 mol. From equation: 2 mol NH₃ → 1 mol urea. Theoretical moles urea = 29411.8 mol. Theoretical mass urea = 29411.8 × 60.0 = 1764706 g = 1.7647 tonnes.