The percentage proportion of carbon, hydrogen, nitrogen and oxygen in an organic compound are $62.07 %$ , $10.34 %$ , $14.0 %$ , $13.59 %$ , respectively. Find its empirical formula. If its molecular mass is $144$ gram $m o l^{- 1}$ , find its molecular formula.

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Solution

100 g of compound will contain 62.07 g of C, 10.34 g of H, 14.0 g of N and 13.59 g of O.
The atomic masses of C, H, N and O are 12 g/mol, 1 g/mol, 14 g/mol and 16 g/mol respectively.
The number of moles of C $= \frac{62.07 g}{12 g / m o l} = 5.17 m o l$ .
The number of moles of H $= \frac{10.34 g}{1 g / m o l} = 10.34 m o l$ .
The number of moles of N $= \frac{14 g}{14 g / m o l} = 1 m o l$ .
The number of moles of O $= \frac{13.59 g}{16 g / m o l} = 0.85 m o l$ .
The mole ratio $C : H : N : O = 5.17 : 10.34 : 1 : 0.85 ≃ 5 : 10 : 1 : 1$
Hence, the empirical formula is $C_{5} H_{10} N O$
The empirical formula mass $= 5 (12) + 10 (1) + 14 + 16 = 100 g / m o l$
The ratio of molecular formula mass to empirical formula mass $= \frac{144}{100} = 1.44$
Hence, the molecular formula of organic compound is $(C_{5} H_{10} N O)_{1.44}$ .

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The percentage proportion of carbon, hydrogen, nitrogen and oxygen in an organic compound are 62.07%, 10.34%, 14.0%, 13.59%, respectively. Find its empirical formula. If its molecular mass is 144 gram mol−1, find its molecular formula.

The percentage proportion of carbon, hydrogen, nitrogen and oxygen in an organic compound are $62.07 %$ , $10.34 %$ , $14.0 %$ , $13.59 %$ , respectively. Find its empirical formula. If its molecular mass is $144$ gram $m o l^{- 1}$ , find its molecular formula.