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

1 Answer
Apr 29, 2017

Answer:

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.