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Answer 1 · 2017-11-03T17:23:19

The molecular formula is $C_{12} H_{10} O_{2}$ $C_12H_10O_2$ .

In this question, we have given a compound with molecular formula $C_{6 x} H_{5 x} O_{x}$ $C_(6x)H_(5x)O_x$ . It can also be written as:

$x C_{6} H_{5} O$ $xC_(6)H_(5)O$

where x is an empirical unit and $C_{6} H_{5} O$ $C_(6)H_(5)O$ is an empirical formula.

Here, we can use an empirical unit formula to find the value of x.

$x = \frac{molecular mass}{empirical mass}$ $x=frac("molecular mass")("empirical mass")$

The empirical mass of $C_{6} H_{5} O$ $C_(6)H_(5)O$ is given by:

$= 6 x 12.01 g/mol + 5 x 1.008 g/mol + 1 x 16 g/mol = 93.1 g/mol$ $=6x12.01 "g/mol"+ 5x1.008 "g/mol"+ 1x16"g/mol" = 93.1 "g/mol"$

Now,

$x = \frac{186 g/mol}{93.1 g/mol} = 1.997 \approx 2$ $x=frac("186 g/mol")("93.1 g/mol") = 1.997 \approx 2$

Therefore the molecular formula is: $C_{12} H_{10} O_{2}$ $C_(12)H_(10)O_2$ .