Question

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

Answer 1

As with these problems, we assume a mass of `100*g` of unknown compound..........and get an empirical formula of `CH_3Br`

Explanation:

If there is a `100*g` mass of compound, then we interrogate its ATOMIC composition......

`"Moles of carbon"=(12.7*g)/(12.011*g*mol^-1)=1.06*mol`.

`"Moles of hydrogen"=(3.2*g)/(1.00794*g*mol^-1)=3.175*mol`.

`"Moles of bromine"=(84.1*g)/(79.90*g*mol^-1)=1.05*mol`.

And we divide these molar quantities by THE SMALLEST such quantity to get an empirical formula of `CH_3Br`. This is quite probably `"methyl bromide"` but we require an estimate of molecular mass before we address the molecular formula.