The elemental composition of a compound is 54.2 % C, 9.2 % H and 36.6 % O. If the molar mass of the compound is 132 g mol⁻¹, the molecular formula of the compound is : [Given : The relative atomic mass of C: H: O=12: 1: 16 ]

Solution

Chemistry ma’am · 3 steps

Turn the percentages into a mole ratio first, then let the molar mass tell you how many times to scale it up.

1. Take 100 g. In 100 g of the compound the percentages become grams. Divide each mass by its atomic mass to get moles, then divide by the smallest number of moles.

   Element	Mass in 100 g	Moles	Divide by 2.29
   C	54.2 g	54.2/12 = 4.52	1.97 ≈ 2
   H	9.2 g	9.2/1 = 9.2	4.02 ≈ 4
   O	36.6 g	36.6/16 = 2.29	1

2. Empirical formula. The ratio C : H : O is 2 : 4 : 1, so the empirical formula is C₂H₄O. Its mass is 24 + 4 + 16 = 44 g mol⁻¹.

   n = (molar mass)/(empirical formula mass) = 132/44 = 3

3. So. The molecular formula is (C₂H₄O)₃ = C₆H₁₂O₃. Check: 72 + 12 + 48 = 132 g mol⁻¹, so the answer is (d).

The idea to keep. Percent to grams, grams to moles, divide by the smallest: that gives the empirical formula. Then molar mass divided by empirical mass gives the multiplier n.

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