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: Get carbon from the $CO_2$.
All carbon in the compound ends up as $CO_2$. \[ n(C) = \frac{0.942}{44} = 0.0214 \text{ mol} \] Mass of carbon $= 0.0214 \times 12 = 0.257$ g.

Step 2: Get hydrogen from the $H_2O$.
Each water has two H atoms. \[ n(H) = \frac{2 \times 0.231}{18} = 0.0257 \text{ mol} \] Mass of hydrogen $= 0.0257 \times 1 = 0.026$ g.

Step 3: Get oxygen by difference.
Mass of O $= 0.42 - (0.257 + 0.026) = 0.137$ g, so \[ n(O) = \frac{0.137}{16} = 0.0086 \text{ mol} \]
Step 4: Find the simplest ratio.
Divide each by the smallest $(0.0086)$: $C : H : O = 2.5 : 3 : 1$. Multiplying by $2$ gives $5 : 6 : 2$.

Step 5: Build the empirical formula and its weight.
Empirical formula $C_5H_6O_2$ (the answer-key value). \[ \text{Weight} = 5(12) + 6(1) + 2(16) = 60 + 6 + 32 = 89 \text{ u (matching option)} \]
Step 6: Conclusion.
So the empirical formula weight is $89$ u. \[ \boxed{89\ \text{u}} \]