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

1 Answer
Mar 31, 2016

Answer:

#{("% 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.