# A hydrogen atom is 84% carbon, by mass. Its relative molecular mass is 100. What is it's empirical formula?

Feb 23, 2017

$\text{A hydrocarbon is 84% carbon by mass...........}$

#### Explanation:

If that is what the question said, then the empirical formula of the hydrocarbon is ${C}_{7} {H}_{16}$. How do we know?

$\left(i\right)$ We assume a mass of $100 \cdot g$ of hydrocarbon.

$\left(i i\right)$ We work out its atomic composition on the basis of the atomic percentages:

$\text{Moles of carbon}$ $\equiv$ $\frac{84.0 \cdot g}{12.01 \cdot g \cdot m o {l}^{-} 1} = 7 \cdot m o l$

$\text{Moles of hydrogen}$ $\equiv$ $\frac{16.0 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 16 \cdot m o l$

How did I know that there were $16 \cdot g$ of hydrogen based solely on the GIVEN $\text{% percentage composition of carbon?}$

On this basis, the empirical formula, the simplest whole number representing constituent atoms in a species is ${C}_{7} {H}_{16}$. Clearly the molecular formula is identical.

It is a fact that the $\text{molecular formula}$ is always a whole number multiple of the $\text{empirical formula}$.

Thus $\text{molecular formula}$ $=$ $n \times \text{empirical formula}$. But we have been given an estimate of molecular mass. So.............

So $100 \cdot \text{amu} \equiv \left({C}_{7} {H}_{16}\right) \times n$

Thus $100 \cdot \text{amu"-=(7xx12.011+16xx1.00794)*"amu} \times n$

We solve for $n$ (how), and find $n \equiv 1$.

$\text{And thus the molecular formula} \equiv {C}_{7} {H}_{16}$.