In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is ________%. (Round off to the Nearest Integer).
(Given atomic mass : C : 12.0 u, H : 1.0 u, O : 16.0 u, N : 14.0 u)

Solution

Chemistry ma’am · 3 steps

Compare the nitrobenzene you got with the nitrobenzene you should have got, both worked out through moles.

1. Moles of benzene. Molar mass of C₆H₆ is 72 + 6 = 78 g mol⁻¹. So 3.9 g is 3.9/78 = 0.05 mol.
2. Theoretical yield. Each benzene gives one nitrobenzene, so at most 0.05 mol of C₆H₅NO₂ can form. Its molar mass is 72 + 5 + 14 + 32 = 123 g mol⁻¹, so the expected mass is 0.05 × 123 = 6.15 g.

   % yield = (actual mass)/(theoretical mass) × 100 = 4.92/6.15 × 100 = 80

3. So. The percentage yield of nitrobenzene is 80%.

The common slip: 126. Dividing the product mass by the benzene mass directly gives 4.92/3.9 × 100 ≈ 126, which ignores that the two compounds have different molar masses.

The idea to keep. Percentage yield always compares the product with the product. Use the mole ratio to find how much product you should get, then divide the actual mass by it.

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