# How many atoms are present in a 112 g sample of CaBr_2?

Jan 3, 2018

$\approx 0.56$ moles or $\approx 3.37 \cdot {10}^{23}$ atoms

#### Explanation:

Molar mass of $C a B {r}_{2}$ is $199.886$ amu

Moles = $\text{Mass"/"Molar Mass}$

In this case, mass = 112g, molar mass = 199.886

$\therefore$ Moles of $C a B {r}_{2}$ = $\frac{112}{199.886} \approx 0.56$

$1$ mole = $6.02 \cdot {10}^{23}$ atoms (Avogadro's number)

$\therefore 0.56$ moles = $3.37 \cdot {10}^{23}$ atoms