An hydrocarbon is $92.3 %$ $92.3%$ with respect to carbon. What is its empirical formula? And if the mass of the molecule is $26 \cdot g \cdot m o l^{- 1}$ $26gmol^-1$ , what is its molecular formula?

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Answer 1 · 2016-04-28T14:56:44

It is likely acetylene.

We assume 100 g of compound, and divide thru by the atomic masses:

$\frac{92.3 \cdot g}{12.011 \cdot g \cdot m o l^{- 1}}$ $(92.3*g)/(12.011*g*mol^-1)$ $=$ $=$ $7.65 \cdot m o l \cdot C$ $7.65*mol*C$

$\frac{7.69 .3 \cdot g}{1.00794 \cdot g \cdot m o l^{- 1}}$ $(7.69.3*g)/(1.00794*g*mol^-1)$ $=$ $=$ $7.65 \cdot m o l \cdot H$ $7.65*mol*H$

So, clearly, the empirical formula is $C H$ $CH$ .

But we have been given a molecular mass, and it is a fact that the molecular formula is always a multiple of the empirical formula:

$Molecular formula$ $"Molecular formula"$ $=$ $=$ $n \times (empirical formula)$ $nxx("empirical formula")$

$26 \cdot g \cdot m o l^{- 1}$ $26*g*mol^-1$ $=$ $=$ $n \times (12.011 + 1, 00794) \cdot g \cdot m o l^{- 1}$ $nxx(12.011+1,00794)*g*mol^-1$

Clearly, $n$ $n$ $=$ $=$ $2$ $2$ , and molecular formula $=$ $=$ $C_{2} H_{2}$ $C_2H_2$ , likely acetylene, $H C \equiv C H$ $HC-=CH$ .

An hydrocarbon is 92.3%92.3%92.3% with respect to carbon. What is its empirical formula? And if the mass of the molecule is 26*g*mol^-126⋅g⋅mol−126*g*mol^-1, what is its molecular formula?