Question

An acid contains C, H and O atoms. On combustion analysis, 0.454 g of the acid gives 0.418 g of H$_2$O and 1.023 g of CO$_2$. What is the empirical formula of the acid?

• A. C$_4$H$_5$O$_2$
• B. C$_3$H$_6$O
• C. CH$_2$O
• D. C$_5$H$_8$O

Hint:

In combustion analysis, determine C from CO$_2$, H from H$_2$O, and O by subtraction. Acids often follow COOH-type ratios, so adjust accordingly.

Answer 1

The Correct Option is B

Step 1: Calculate moles of carbon.
From 1.023 g of CO$_2$:
Moles of CO$_2$ = $\dfrac{1.023}{44.01} = 0.02325$ mol.
Each CO$_2$ molecule contains one carbon atom, so moles of C = 0.02325 mol.
Mass of C = $0.02325 \times 12.01 = 0.2793$ g.
Step 2: Calculate moles of hydrogen.
From 0.418 g of H$_2$O:
Moles of H$_2$O = $\dfrac{0.418}{18.02} = 0.02320$ mol.
Each H$_2$O molecule contains 2 H atoms, so moles of H = $2 \times 0.02320 = 0.04640$ mol.
Mass of H = $0.04640 \times 1.008 = 0.04678$ g.
Step 3: Calculate moles of oxygen.
Mass of O = $0.454 - (0.2793 + 0.04678) = 0.1279$ g.
Moles of O = $\dfrac{0.1279}{16.00} = 0.00799$ mol.
Step 4: Simplify ratios.
\[ \text{C: } 0.02325/0.00799 = 2.91, \quad \text{H: } 0.04640/0.00799 = 5.80, \quad \text{O: } 0.00799/0.00799 = 1 \] Ratio ≈ C$_3$H$_6$O.
Step 5: Adjust for acidic nature.
Since it is an acid, multiply by $\dfrac{4}{3}$ → C$_4$H$_5$O$_2$.
Step 6: Conclusion.
Empirical formula = C$_4$H$_5$O$_2$.