MCQ
An organic compound has $\%$ of $C$ and $\%$ of $H$ in the ratio $6 : 1$ and $\%$ of $C$ and $\%$ of $O$ in the ratio $3 : 4$. The compound is
- ✓$HCHO$
- B$C{H_3}OH$
- C$C{H_3}C{H_2}OH$
- D$(COOH)_2$
The number of moles of $C =\frac{6 / g }{12 g / mol }=0.5\, mol$.
The number of moles of $H =\frac{1 / g }{1 g / mol }=1 \,mol$.
The number of moles of $O =\frac{8 / g }{16 g / mol }=0.5 \,mol$.
The mole ratio $C : H : O =0.5: 1: 0.5=1: 2: 1$
Hence, the compound is $CH _2 O$ or $HCHO$
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