# Question 27fd7

Apr 16, 2016

$\text{4 g}$

#### Explanation:

The idea here is that you need to use the molar mass of ferrous oxide, $\text{FeO}$, to find the percent composition of iron in this compound.

Once you know how many grams of iron you get per $\text{100 g}$ of ferrous oxide, you can calculate how many grams you get in $\text{5 g}$.

So, ferrous oxide has a molar mass of ${\text{71.844 g mol}}^{- 1}$, which means that one mole of ferrous oxide has a mass of $\text{71.844 g}$.

You also know that one mole of ferrous oxide contains

• one mole of iron, $1 \times \text{Fe}$
• one mole of oxygen, $1 \times \text{O}$

Elemental iron has a molar mass of ${\text{55.845 g mol}}^{- 1}$, which of course means that one mole of iron has a mass of $\text{55.845 g}$.

This means that every $\text{71.844 g}$ of ferrous oxide will contain $\text{55.845 g}$ of iron. As a result, the percent composition of iron will be

$\text{% Fe" = (55.845 color(red)(cancel(color(black)("g"))))/(71.844color(red)(cancel(color(black)("g")))) xx 100 = "77.73% Fe}$

So, if every $\text{100 g}$ of ferrous oxide contain $\text{77.73 g}$ of iron, it follows that $\text{5 g}$ 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)|)))#