How many grams are there in 1.18 * 10^26 atoms of iridium?

Feb 1, 2016

$1.18 \times {10}^{26}$ atoms of iridium have a mass of 37700 g.

Explanation:

We must first determine the molar mass of iridium $\left(\text{Ir}\right)$, which is its relative atomic mass on the periodic table in g/mol.

The molar mass of $\text{Ir}$$=$$\text{192.217 g/mol}$.

$\text{1 mol Ir} = 6.022 \times {10}^{23}$ atoms $\text{Ir}$

To determine the mass of $1.18 \times {10}^{26}$ atoms of $\text{Ir}$, divide the given number of atoms by $6.022 \times {10}^{23}$ atoms/mol to get moles of $\text{Ir}$, then multiply times the molar mass of $\text{Ir}$ to get mass of $\text{Ir}$ in grams.

$1.18 \times {10}^{26} \cancel{\text{atoms Ir"xx(1cancel"mol Ir")/(6.022xx10^23cancel"atoms Ir")xx(192.217"g Ir")/(1cancel"mol Ir")=37700 "g Ir}}$ rounded to three significant figures.