Question

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

Answer 1

`{("% 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 `"100 g"` of compound.

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

`color(blue)(|bar(ul(color(white)(a/a)"% element i" = "mass of element i"/"total mass of the compound" xx 100color(white)(a/a)|)))`

In your case, the compound is said to contain two elements, bromine, `"Br"`, and magnesium, `"Mg"`.

The total mass of the sample will be

`m_"sample" = m_(Br) + m_(Mg)`

`m_"sample" = "32.0 g" + "4.9 g" = "36.9 g"`

So, you know that `"36.9 g"` of this compound contain `"32.0 g"` of bromine, which means that `"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 `"4.9 g"` of magnesium, which means that `"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 `"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.