# What is the mass, in grams, of a single iridium atom (m_Ir = "192.22 u")?

Feb 16, 2016

$3.1920 \cdot {10}^{- 22} \text{g}$

#### Explanation:

The idea here is that you need to use the definition of the unified atomic mass unit, $\text{u}$, to determine the mass of an iridium atom in grams.

The unified atomic mass unit is defined as $\frac{1}{12}$ of the mass of a neutral unbound atom of carbon-12 and is equivalent to ${\text{1 g mol}}^{- 1}$.

I will not go into the derivation here, but try to remember that the conversion factor that takes you from unified atomic mass units to grams per mole looks like this

${\text{1 u" = "1 g mol}}^{- 1}$

In your case, the mass of a single iridium atom is said to be equal to $\text{192.22 u}$. This means that the molar mass of iridium will be equal to

192.22 color(red)(cancel(color(black)("u"))) * "1 g mol"^(-1)/(1color(red)(cancel(color(black)("u")))) = "192.22 g mol"^(-1)

Finally, use Avogadro's number to help you convert between the mass of one mole of iridium atoms and the mass of a single iridium atom.

"192.22 g"color(red)(cancel(color(black)("mol"^(-1)))) * overbrace((1color(red)(cancel(color(black)("mol"))))/(6.022 * 10^(23)"atoms"))^(color(brown)("Avogadro's number")) = color(green)(3.1920 * 10^(-22)"g")

The answer is rounded to five sig figs.