What would be considered the molecular formula for the compound that has an empirical formula of $C H_{4}$ $CH_4$ but a molar mass of 80 g/mol?

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Answer 1 · 2017-01-22T12:41:58

Empirical formula means the ratio $C \div H = 1 \div 4$ $CdivH=1div4$

If we translate this to atomic mass units, the ratio goes to:
$1 \times 12 \div 4 \times 1 = 12 \div 4$ $1xx12div4xx1=12div4$

The sum of the ratios is $16$ $16$ and this goes into $80$ $80$ five times.

So the molecular formula would be $C_{1 \times 5} H_{4 \times 5} = C_{5} H_{20}$ $C_(1xx5)H_(4xx5)=C_5H_20$

Check : $5 \times 12 + 20 \times 1 = 80$ $5xx12+20xx1=80$

Note : This is a non-existing compound, as $5 C$ $5C$ atoms can at most hold $12 H$ $12H$ atoms (matter of valencies).

What would be considered the molecular formula for the compound that has an empirical formula of CH4CH_4 but a molar mass of 80 g/mol?