# A compound is found to be 12.7% carbon, 3.2% hydrogen, and 84.1% bromine. What is its empirical formula?

Apr 29, 2017

As with these problems, we assume a mass of $100 \cdot g$ of unknown compound..........and get an empirical formula of $C {H}_{3} B r$

#### Explanation:

If there is a $100 \cdot g$ mass of compound, then we interrogate its ATOMIC composition......

$\text{Moles of carbon} = \frac{12.7 \cdot g}{12.011 \cdot g \cdot m o {l}^{-} 1} = 1.06 \cdot m o l$.

$\text{Moles of hydrogen} = \frac{3.2 \cdot g}{1.00794 \cdot g \cdot m o {l}^{-} 1} = 3.175 \cdot m o l$.

$\text{Moles of bromine} = \frac{84.1 \cdot g}{79.90 \cdot g \cdot m o {l}^{-} 1} = 1.05 \cdot m o l$.

And we divide these molar quantities by THE SMALLEST such quantity to get an empirical formula of $C {H}_{3} B r$. This is quite probably $\text{methyl bromide}$ but we require an estimate of molecular mass before we address the molecular formula.