Question #27fd7

The idea here is that you need to use the molar mass of ferrous oxide, #"FeO"#, to find the percent composition of iron in this compound.

Once you know how many grams of iron you get per #"100 g"# of ferrous oxide, you can calculate how many grams you get in #"5 g"#.

So, ferrous oxide has a molar mass of #"71.844 g mol"^(-1)#, which means that one mole of ferrous oxide has a mass of #"71.844 g"#.

You also know that one mole of ferrous oxide contains

• one mole of iron, #1 xx "Fe"#
• one mole of oxygen, #1 xx "O"#

Elemental iron has a molar mass of #"55.845 g mol"^(-1)#, which of course means that one mole of iron has a mass of #"55.845 g"#.

This means that every #"71.844 g"# of ferrous oxide will contain #"55.845 g"# of iron. As a result, the percent composition of iron will be

#"% Fe" = (55.845 color(red)(cancel(color(black)("g"))))/(71.844color(red)(cancel(color(black)("g")))) xx 100 = "77.73% Fe"#

So, if every #"100 g"# of ferrous oxide contain #"77.73 g"# of iron, it follows that #"5 g"# of ferrous oxide will contain

#5 color(red)(cancel(color(black)("g FeO"))) * "77.73 g Fe"/(100color(red)(cancel(color(black)("g FeO")))) = "3.8865 g Fe"#

Rounded to one sig fig, the number of sig figs you have for the mass of ferrous oxide, the answer will be

#"mass of Fe" = color(green)(|bar(ul(color(white)(a/a)"4 g"color(white)(a/a)|)))#