# How many "hydrogen ATOMS" are present in a 23.7*g mass of "methane"?

Oct 3, 2016

There are approx. ${N}_{A} \times 4 \times 2$ hydrogen atoms in such a mass of methane.

#### Explanation:

In $16.04 \cdot g$ methane, there are ${N}_{A}$ methane molecules, where ${N}_{A} = \text{Avogadro's number} = 6.022 \times {10}^{23}$.

And so $\text{moles of methane}$ $=$ $\frac{23.7 \cdot g}{12.04 \cdot g \cdot m o {l}^{-} 1}$ $=$ $1.97 \cdot m o l$

And thus $\text{atoms of hydrogen}$ $=$ $\frac{23.7 \cdot g}{12.04 \cdot g \cdot m o {l}^{-} 1} \times \text{4 H atoms"*mol^-1xxN_A" H atoms}$

$\cong$ $4.80 \times {10}^{24} \text{ hydrogen atoms}$