# What percent of deuterated water is contained in a sample that has an average relative molecular mass of "19 g/mol" and consists of some "H"_2""^(16)"O" and some "D"_2""^(16)"O"?

Jun 17, 2016

50%

#### Explanation:

The idea here is that you need to use the fact that deuterium, $\text{D}$, which is one of the three naturally occurring isotopes of hydrogen, $\text{H}$, contains two nucleons in its nucleus.

$\textcolor{w h i t e}{}$

As you know, the relative atomic mass of an atom, ${A}_{r}$, is calculated by dividing the atomic mass of the atom, ${m}_{a}$, by $\frac{1}{12} \text{th}$ the mass of an atom of $\text{^12"C}$, which is equivalent to the mass of one nucleon, i.e. one proton or one neutron.

The relative atomic mass of hydrogen is approximately equal to $1$ because its nucleus contains one proton. Since the nucleus of a deuterium atom contains a proton and a neutron, its relative atomic mass will be equal to $2$.

Now, water, $\text{H"_2"O}$, contains two hydrogen atoms and one oxygen atom. Heavy water, $\text{D"_2"O}$, contains two deuterium atoms and one oxygen atom.

This means that the difference between the relative molecular mass of water and the relative molecular mass of heavy water will be equal to twice the difference between ${A}_{r}$ of hydrogen and ${A}_{r}$ of deuterium.

The relative atomic mass of oxygen is $16$, which means that the relative molecular masses of water and of heavy water are

${A}_{\text{r H"_2"O}} = 2 \times 1 + 16 = 18$

${A}_{\text{r D"_2"O}} = 2 \times 2 + 16 = 20$

Your sample has an average relative molecular mass of $19$, which can only mean that it contains equal amounts of water and of heavy water.

Mathematically, you can show this by using a system of two equations that have $x$ as the decimal abundance of $\text{H"_2"O}$ and $y$ as the decimal abundance of $\text{D"_2"O}$

$\left\{\begin{matrix}19 = x \cdot 18 + y \cdot 20 \\ \textcolor{w h i t e}{9} 1 = x + y\end{matrix}\right.$

This will get you

$19 = 18 x + 20 - 20 x$

$2 x = 1 \implies x = \frac{1}{2}$

This means that

$y = 1 - \frac{1}{2} = \frac{1}{2}$

and thus your sample contains 50% "H"_2"O" and 50% "D"_2"O".