Question #c4c4e

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Answer 1 · 2016-10-03T08:54:31

The empirical formula given should be $C_{2} H_{4} O$ $C_2H_4O$ .

Then let the molecular formula be $(C_{2} H_{4} O) n$ $(C_2H_4O)n$ .

Considering atomic masses as

$C \to 12 g/mol$ $C->12"g/mol"$

$O \to 16 g/mol$ $O->16"g/mol"$

$H \to 1 g/mol$ $H->1"g/mol"$

We can get calculated molar mass of the compound is $(2 \cdot 12 + 4 \cdot 1 + 16 \cdot 1) n = 44 n g/mol$ $(2*12+4*1+16*1)n=44n" g/mol"$

Equating it with given molar mass we get

$44 n = 88.12$ $44n=88.12$

$\Rightarrow n = \frac{88.12}{44} \approx 2$ $=>n=88.12/44~~2$

Hence the MF of the compound is

$C_{4} H_{8} O_{2}$ $C_4H_8O_2$