# How many grams are in 1.98 * 10^21 atoms of Fe?

Mar 5, 2016

The molar mass of Fe is $55.845 g m o {l}^{-} 1$
So it is the mass of 1 mole atom of Fe i.e. $6.023 \cdot {10}^{23}$ atoms of Fe
Hence the mass of$1.98 \cdot {10}^{21}$ atoms of Fe will be = $55.845 \cdot \frac{1.98 \cdot {10}^{21}}{6.023 \cdot {10}^{23}} = 0.55845 \cdot \frac{1.98}{6.023} g \approx 0.1836 g$

Mar 5, 2016

$1.98 \times {10}^{21} \text{atoms Fe}$ has a mass of $\text{0.184 g Fe}$.

#### Explanation:

$\text{1 mol atoms"=6.022xx10^23"atoms}$

$\text{Molar mass Fe"="55.845 g/mol}$

You need to make the following conversions:

$\text{atoms Fe}$$\rightarrow$$\text{mol Fe}$$\rightarrow$$\text{mass Fe}$

Divide the given atoms of iron by $6.022 \times {10}^{23} \text{atoms Fe}$, then multiply by its molar mass.

$1.98 \times {10}^{21} \cancel{\text{atoms Fe"xx(1cancel"mol Fe")/(6.022xx10^23cancel"atoms Fe")xx(55.845"g Fe")/(1cancel"mol Fe")="0.184 g Fe}}$ rounded to three significant figures