Question

On complete combustion 1.0 g of an organic compound (X) gave 1.46 g of CO₂ and 0.567 g of H₂O. The empirical formula mass of compound (X) is:
(Given molar mass in g mol\(^{-1}\): C: 12, H: 1, O: 16)

• A. 30
• B. 45
• C. 60
• D. 15

Hint:

When calculating the empirical formula, first determine the moles of each element from the given combustion data. Then, divide each element’s mass by its atomic mass to obtain the ratio of atoms.

Answer 1

The Correct Option is A

Moles of 'C' = \( n_{\text{CO}_2} = \frac{1.46}{44} = 0.033 \) 
Moles of 'C' = \( W_c = 0.033 \times 12 = 0.396 \) 
Moles of 'H' = \( 2 \times n_{\text{H}_2\text{O}} = 2 \times \frac{0.567}{18} = 0.063 \) 
Mass of 'H' = \( 0.0063 \) Mass of Oxygen \( O \) = \( 1 - (W_c + W_h) = 1 - (0.033 \times 12 + 0.063 \times 1) = 0.541 \) 
Moles of 'O' = \( \frac{0.541}{16} = 0.033 \) 
Empirical formula = \( \text{CH}_2\text{O} \) 
Empirical formula mass = 30