Quantitative analysis of an organic compound (X) shows following % composition.
C : 14.5 %
Cl : 64.46%
H: 1.8 %
(Empirical formula mass of the compound (X) is _________ × 10⁻¹
(Given molar mass in g mol⁻¹ of C: 12, H: 1, O: 16, Cl: 35.5)

Solution

Chemistry ma’am · 3 steps

The percentages add up to only 80.76, so the missing 19.24% is oxygen. Do not leave it out.

1. Find the missing element. C, Cl and H add up to 14.5 + 64.46 + 1.8 = 80.76%. The oxygen mass is given for a reason, so the remaining 100 - 80.76 = 19.24% is oxygen.

   Element	Mass in 100 g	Moles	Divide by 1.2025	Times 2
   C	14.5 g	14.5 / 12 = 1.208	1.00	2
   H	1.8 g	1.8 / 1 = 1.8	1.50	3
   Cl	64.46 g	64.46 / 35.5 = 1.816	1.51	3
   O	19.24 g	19.24 / 16 = 1.2025	1	2

2. Whole-number ratio. Dividing by the smallest value gives 1 : 1.5 : 1.5 : 1. Multiply by 2 to clear the halves, so the empirical formula is C₂H₃Cl₃O₂.

   2(12) + 3(1) + 3(35.5) + 2(16) = 24 + 3 + 106.5 + 32 = 165.5 g mol⁻¹

3. So. The empirical formula mass is 165.5 = 1655 × 10⁻¹, so the answer is 1655.

The common slip: 1335. Ignoring the 19.24% oxygen gives C₂H₃Cl₃ with mass 133.5, which is reported as 1335.

The idea to keep. Always add the given percentages first. If they fall short of 100, the gap is usually oxygen. Then divide by the smallest mole value, and multiply to clear any 0.5.

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