# What is the percentage composition of a compound containing 32.0 g of bromine and 4.9 g of magnesium?

Mar 31, 2016

{("% Br" = 87%), ("% Mg" = 13%) :}

#### Explanation:

In order to find a compound's percent composition, all you basically have to do is determine how many grams of its constituent elements you get per $\text{100 g}$ of compound.

So, for a given compound, you can get the percent composition of its constituent element $i$ by using

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{% element i" = "mass of element i"/"total mass of the compound} \times 100 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

In your case, the compound is said to contain two elements, bromine, $\text{Br}$, and magnesium, $\text{Mg}$.

The total mass of the sample will be

${m}_{\text{sample}} = {m}_{B r} + {m}_{M g}$

${m}_{\text{sample" = "32.0 g" + "4.9 g" = "36.9 g}}$

So, you know that $\text{36.9 g}$ of this compound contain $\text{32.0 g}$ of bromine, which means that $\text{100 g}$E of this compound must contain

100color(red)(cancel(color(black)("g compound"))) * "32.0 g Br"/(36.9color(red)(cancel(color(black)("g compound")))) = "86.72 g Br"

Likewise, this sample also contains $\text{4.9 g}$ of magnesium, which means that $\text{100 g}$ of compound must contain

100color(red)(cancel(color(black)("g compound"))) * "4.9 g Mg"/(36.9color(red)(cancel(color(black)("g compound")))) = "13.28 g Mg"

Since percent composition tells you how many grams of each element you get in $\text{100 g}$ of compound, you can say that

"% Br" = color(green)(|bar(ul(color(white)(a/a)"87%"color(white)(a/a)|)))

"% Mg" = color(green)(|bar(ul(color(white)(a/a)"13%"color(white)(a/a)|)))

The answers are rounded to sig figs, the number of sig figs you have for the mass of magnesium.