How can I calculate the percent composition of CFBrO?

Apr 3, 2015

You can calculate percent composition by using the molar masses of the individual atoms that make up the compound and the molar mass of the compound.

In your case, the molar masses of the atoms that make up the compound $\text{CFBrO}$ are

$\text{C": "12.011 g/mol}$
$\text{F": "18.9984 g/mol}$
$\text{Br": "79.904 g/mol}$
$\text{O": "15.9994 g/mol}$

The molar mass of the compound will be the sum of the molar masses of the atoms it's comprised of

$\text{CFBrO": 12.011 + 18.9984 + 79.904 + 15.9994 = "126.9074 g/mol}$

Now, to get the percent composition of the compound, divide each element's molar mass by the molar mass of the compound and multiply the result by 100 - this will give you each element's contribution to the mass of the compound

$\text{C": "12.011"/"126.9074" * 100 = "9.47%}$

$\text{F": "18.9984"/"126.9074" * 100 = "14.97%}$

$\text{Br": 79.904/126.9074 * 100= "62.96%}$

$\text{O": 15.994/126.9074 * 100 = "12.60%}$

This means that, for every 100 g of $\text{CFBrO}$, you get

$\text{9.47 g}$ $\to$ carbon;
$\text{14.97 g}$ $\to$ fluorine;
$\text{62.96 g}$ $\to$ bromine;
$\text{12.6 g}$ $\to$ oxygen;