# 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

#### Answer:

Here's what I got.

#### Explanation:

A compound's **percent composition** tells you how many grams of each constituent element you get **per**

This means that you can find the percent composition of a constituent element **total mass** of the compound, and multiplying the result by

#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*, *oxygen*,

#"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

#"% 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

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

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**