The empirical formula of a compound is CH. The experimental molar mass is 78.12 g/mol. How do you determine the molecular formula of the compound?

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Answer 1 · 2017-01-19T03:07:47

The $molecular formula$ $"molecular formula"$ is always a whole number multiple of the $empirical formula$ $"empirical formula"$ .

And thus: $molecular formula = n \times (empirical formula)$ $"molecular formula"=nxx"(empirical formula)"$ .

We take the given empirical formula, work out its mass, and then solve for $n$ $n$ .

So $78.23 \cdot g \cdot m o l^{- 1} = n \times (12.011 + 1.00794) \cdot g \cdot m o l^{- 1}$ $78.23*g*mol^-1=nxx(12.011+1.00794)*g*mol^-1$

Where of course, $12.011$ $12.011$ and $1.00794$ $1.00794$ are the atomic masses of carbon and hydrogen in $g \cdot m o l^{- 1}$ $g*mol^-1$ ,

$n = \frac{78.23 \cdot g \cdot m o l^{- 1}}{(12.011 + 1.00794) \cdot g \cdot m o l^{- 1}} = 6$ $n=(78.23*g*mol^-1)/((12.011+1.00794)*g*mol^-1)=6$

And thus the molecular formula $\equiv$ $-=$ ${(C H)}_{6}$ $(CH)_6$ or $C_{6} H_{6}$ $C_6H_6$ . What is the likely name of this molecule?