# How do I work out how many molecules there are in a 1*g mass of water?

Sep 10, 2017

$\text{Do it dimensionally......}$

#### Explanation:

What do I mean by $\text{dimensionally}$?

Suppose we have a $1.00 \cdot g$ mass of water. Now we know that $\text{Avogadro's number of water molecules}$ has a mass of $18.01 \cdot g$, and we would typically write this as......

$\text{Molar mass of water} = 18.01 \cdot g \cdot m o {l}^{-} 1$

And if I wanted to work out the MOLAR quantity, I would perform the division by leaving the UNITS in......i.e.

$\text{Moles of water} = \frac{1.00 \cdot g}{18.01 \cdot g \cdot m o {l}^{-} 1}$, and of course we can do some cancellation of units.....

$\text{Moles of water} = \frac{1.00 \cdot \cancel{g}}{18.01 \cdot \cancel{g} \cdot m o {l}^{-} 1}$

$= 0.0555 \cdot \frac{1}{m o {l}^{-} 1} = 0.0555 \cdot \frac{1}{\frac{1}{m o l}}$ because ${x}^{-} 1 \equiv \frac{1}{x}$, and

$= 0.0555 \cdot \frac{1}{m o {l}^{-} 1} = 0.0555 \cdot \frac{1}{\frac{1}{m o l}} = 0.0555 \cdot m o l$.

And as for volumes, we need a $\text{density}$, $\rho$, the which for chemists is typically quoted as $g \cdot m {L}^{-} 1 \equiv g \cdot c {m}^{-} 3$.

By definition, $\rho = \text{Mass"/"Volume}$ i.e. mass per unit volume.

Here ${\rho}_{{H}_{2} O} = \frac{\text{Molar quantity"xx"Molar mass}}{1 \cdot m L}$

$= \frac{0.0555 \cdot \cancel{m o l} \times 18.01 \cdot g \cdot \cancel{m o {l}^{-} 1}}{1 \cdot m L}$

$1 \cdot g \cdot m L$; dimensionally consistent as required. Of course, the $\text{density}$ needs to be measured......