# How many carbon atoms in a 5.85*g mass of CBr_4?

Aug 26, 2017

Approx. $1 \times {10}^{22} \cdot \text{carbon atoms}$.

#### Explanation:

We got $5.85 \cdot g$ of $C B {r}_{4}$.

This represents a molar quantity of .................

$\frac{5.85 \cdot g}{331.63 \cdot g \cdot m o {l}^{-} 1} = 0.0176 \cdot m o l$

And of course in such a molar quantity, there are $4 \times 0.0176 \cdot m o l$ of bromine, and $1 \times 0.0176 \cdot m o l$ of carbon. Are you with me?

And so we take the product of the molar quantity and Avogadro's number, to get.......

$0.0176 \cdot m o l \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1 = 1.06 \times {10}^{22}$ individual carbon atoms.....