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

May 1, 2017

This may be simplified into a table to help answer this question

#### Explanation:

First, we assumed the percentage to be in grams (like mass)
then we take the percentage and divide by the molar masses of each element to get the no. of mol of each element
Afterwards, divide all of the results by the smallest mol value in the table, which gives us the simplest molar ratio

May 1, 2017

$\text{Empirical formula} = C {H}_{3} B r$.

#### Explanation:

As is usual in these problems, we ASSUME a mass of $100 \cdot g$, and then work out the atomic composition:

$\text{Moles of carbon} = \frac{12.7 \cdot g}{12.01 \cdot g \cdot m o {l}^{-} 1} = 1.05 \cdot m o l$.

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

$\text{Moles of bromine} = \frac{84.1 \cdot g}{79.9 \cdot g \cdot m o {l}^{-} 1} = 1.05 \cdot m o l$.

And then we divide thru by the smallest molar quantity, to get an empirical formula, the simplest whole number ratio defining constitutent atoms in a species as, $C {H}_{3} B r$.

Just to add, that we calculated the empirical formula. The molecule COULD be methyl bromide, in which case the question is highly unreasonable. You do not perform microanalysis, i.e. %C,%H,%N composition on LIQUIDS; you can do so on solids.