If a sample of iron oxide has a mass of 1.596g and was found to contain 1.116g of iron and .48g or oxyygen, how would you find the percentage composition of this compound?

1 Answer
Mar 11, 2016

Answer:

Here's what I got.

Explanation:

A compound's percent composition tells you how many grams of each constituent element you get per #"100 g"# of said compound.

This means that you can find the percent composition of a constituent element #"X"# by dividing the mass of #"X"# by the total mass of the compound, and multiplying the result by #100#.

#color(blue)(|bar(ul("% composition X" = "mass of X"/"mass of compound" xx 100|)))#

In your case, you know that you're dealing with a compound that contains iron, #"Fe"#, and oxygen, #"O"#. More specifically, a #"1.596-g"# sample is said to contain

  • #"1.116 g"# of iron
  • #"0.48 g"# of oxygen

To find the percent composition of iron in this sample, divide the mass of iron in the sample by the total mass of the sample, then multiply the result by #100#

#"% iron" = (1.116 color(red)(cancel(color(black)("g"))))/(1.596color(red)(cancel(color(black)("g")))) xx 100 = "69.925% O"#

At this point, you have two ways of finding the percent composition of oxygen. You can either use the same method you used for iron, or you can subtract the percent composition of iron from #100%#.

Since the compound contains only iron and oxygen, it follows that their respective percent compositions must add up to give #100%#.

In this case, you would have

#"% oxygen" = (0.48color(red)(cancel(color(black)("g"))))/(1.596color(red)(cancel(color(black)("g")))) xx 100 = "30.075% O"#

This is equivalent to

#"% oxygen" = 100% - "% iron"#

#"% oxygen" = 100% - 69.925% = "30.075% O"#

I'll leave the answer rounded to three sig figs

#"% Fe: " color(green)(|bar(ul(color(white)(a/a)"69.9%"color(white)(a/a)|)))#

#"% O: " color(green)(|bar(ul(color(white)(a/a)"30.1%"color(white)(a/a)|)))#

This means that every #"100 g"# of this unknown compound will contain #"69.9 g"# of iron and #"30.1 g"# of oxygen.