Question

0.42 g of an organic compound containing C, H and O gave on combustion 0.942 g of $CO_2$ and 0.231 g of $H_2O$. The empirical formula weight of the compound is (At.wt: $C=12$ u, $H=1$ u, $O=16$ u)

• A. 89 u
• B. 98 u
• C. 79 u
• D. 101 u

Hint:

Always remember: Mass of C in $CO_2$ is $(12/44) \times \text{mass}(CO_2)$; mass of H in $H_2O$ is $(2/18) \times \text{mass}(H_2O)$.

Answer 1

The Correct Option is A

Step 1: Concept
Combustion analysis to determine empirical formula.

Step 2: Meaning
Calculate moles of C and H from $CO_2$ and $H_2O$, then find oxygen mass.

Step 3: Analysis
Moles of $C = \text{mass}(CO_2) / 44 = 0.942 / 44 \approx 0.0214$ mol. Mass of $C = 0.0214 \times 12 = 0.2568$ g. Moles of $H = 2 \times \text{mass}(H_2O) / 18 = 2 \times 0.231 / 18 \approx 0.0257$ mol. Mass of $H = 0.0257 \times 1 = 0.0257$ g. Mass of $O = 0.42 - (0.2568 + 0.0257) = 0.1375$ g. Moles of $O = 0.1375 / 16 \approx 0.0086$ mol. Ratio $C:H:O = 0.0214 : 0.0257 : 0.0086 \approx 2.5 : 3 : 1 \approx 5 : 6 : 2$. Empirical formula $= C_5H_6O_2$. Weight $= 5(12) + 6(1) + 2(16) = 60 + 6 + 32 = 98$ u. Re-evaluating ratio/data: Given options, 98 u is likely the intended calculation.

Step 4: Conclusion
The empirical formula weight is 98 u (Option B).

Final Answer: (B)