# What is the mass in grams of #9.357 * 10^30# atoms of iron?

##### 1 Answer

#### Explanation:

An interesting approach to use here is to use iron's **molar mass** and **Avogadro's number** to find the mass of a **single atom** of iron, which you can then be used as a conversion factor to find the mass of **atoms**.

#"mass of one mole " stackrel(color(red)(1)color(white)(aa))(->) " mass of one atom " stackrel(color(purple)(2)color(white)(aa))(->) " mass of"color(white)(a) 9.357 * 10^(30)"atoms"#

The mass of **one mole** of any element is given by its **molar mass**. Iron has a molar mass of **every mole** of iron has mass of

Now, **one mole** of any element is defined as **atoms** of that element -- this is known as **Avogadro's number**.

#color(blue)(|bar(ul(color(white)(a/a)"1 mole" = 6.022 * 10^(23)"atoms"color(white)(a/a)|)))#

So, if **one mole** of iron contains **atoms** of iron, and if the mass of one mole if **one atom** of iron will be

#1 color(red)(cancel(color(black)("atom Fe"))) * "55.845 g"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Fe")))) = 9.2735 * 10^(-23)"g"#

Now that you know the mass of **one atom**, you can use it to find the mass of **atoms**

#9.357 * 10^(30)color(purple)(cancel(color(black)("atoms Fe"))) * (9.2735 * 10^(-23)"g")/(1color(purple)(cancel(color(black)("atom Fe")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(8.677 * 10^8"g")color(white)(a/a)|)))#

The answer is rounded to four **sig figs**.

**ALTERNATIVE APPROACH**

You can also find the answer by converting the number of atoms of iron to *moles* with the help of Avogadro's number, then by using iron's molar mass as a conversion factor.

You will have

#9.357 * 10^(30) color(red)(cancel(color(black)("atoms Fe"))) * "1 mole Fe"/(6.022 * 10^(23)color(red)(cancel(color(black)("atoms Fe")))) = 1.5538 * 10^7"moles Fe"#

This will once again be equivalent to

#1.5538 * 10^(7)color(red)(cancel(color(black)("moles Fe"))) * "55.845 g"/(1color(red)(cancel(color(black)("mole Fe")))) = color(green)(|bar(ul(color(white)(a/a)color(black)(8.677 * 10^8"g")color(white)(a/a)|)))#