Methylation of 10 g of benzene gave 9.2 g of toluene. Calculate the percentage yield of toluene __________. (Nearest integer)

Solution

Chemistry ma’am · 3 steps

Yield is about moles, not grams: find how much toluene 10 g of benzene could give, then compare.

1. Mole ratio. Methylation adds one CH₃ group: C₆H₆ → C₆H₅CH₃. One mole of benzene gives one mole of toluene. The molar masses are 78 g mol⁻¹ for benzene and 92 g mol⁻¹ for toluene.
2. Theoretical yield. 10 g of benzene is 10/78 = 0.128 mol. At full conversion it gives 0.128 mol of toluene, which is 10/78 × 92 = 11.8 g.

   % yield = actual/theoretical × 100 = 9.2/(10/78 × 92) × 100 = (9.2 × 78)/920 × 100

3. So. (9.2 × 78)/920 = 0.78, so the percentage yield of toluene is 78.

The common slip: 92. Dividing 9.2 g of toluene by 10 g of benzene directly gives 92, but grams of two different compounds cannot be compared until you convert through moles.

The idea to keep. Percentage yield always compares actual product with the theoretical product from the balanced equation. Convert the reactant to moles first, then to grams of product.

Answer from the NTA final answer key. Solution by PrepWiser.