# Question #ba64c

Sep 6, 2017

Let's underthink the problem....suppose you had $\text{3 dozen}$ individual water molecules, i.e. $36$ molecules......

#### Explanation:

Do you agree that in this scenario we gots 36 individual oxygen atoms, and 72 hydrogen atoms? Yea or nay?

The question here specified $3.4 \cdot m o l$ water molecules, which has a mass of $3.4 \cdot m o l \times 18.01 \cdot g \cdot m o {l}^{-} 1 = 61.2 \cdot g$ of water......

But a mole is simply a numerical quantity....i.e. ${N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$.

And so back to your problem, we gots....

$3.4 \times {N}_{A}$ $\text{oxygen atoms}$ $=$ $2.05 \times {10}^{24}$ $\text{oxygen atoms}$.

And......

$2 \times 3.4 \times {N}_{A}$ $\text{hydrogen atoms}$ $=$ $4.09 \times {10}^{24}$ $\text{hydrogen atoms}$.

And the moral, the mole is simply a number, admittedly a very large number.......with the property that a mole of $\text{hydrogen atoms}$ has a mass of $1 \cdot g$.