What is the percent composition of water in "Na"_2"SO"_4 * 10"H"_2"O"?

Jan 3, 2017

Sodium sulfate decahydrate is 55.9 %"H"_2"O".

Explanation:

Your goal here is to figure out how many grams of water you have in $\text{100 g}$ of sodium sulfate decahydrate, $\text{Na"_2"SO"_4 * 10"H"_2"O}$.

To do that, start with what you know. More specifically, start with the fact that $1$ mole of sodium sulfate decahydrate contains

• one mole of sodium sulfate, $1 \times {\text{Na"_2"SO}}_{4}$
• ten moles of water, $10 \times \text{H"_2"O}$

You also know that sodium sulfate has a molar mass of ${\text{142.04 g mol}}^{- 1}$ and that water has a molar mass of ${\text{18.015 g mol}}^{- 1}$. This means that $1$ mole of sodium sulfate decahydrate contains

• $1 \times \text{142.04 g } \to$ sodium sulfate
• $10 \times \text{18.015 g } \to$ water

and has a total mass of

$\text{142.04 g"color(white)(.) + 10 xx "18.105 g" = "322.19 g}$

This means that $\text{100 g}$ of sodium sulfate decahydrate will contain

100 color(red)(cancel(color(black)("g Na"_ 2"SO"_ 4))) * (10 xx "18.015 g H" _ 2"O")/(322.19 color(red)(cancel(color(black)("g Na"_ 2"SO"_ 4)))) = "55.9 g H"_2"O"

Therefore, you can say that the percent composition of water in this hydrate is equal to

color(darkgreen)(ul(color(black)("% H"_2"O" = 55.9%)))

Keep in mind that the % sign means out of $\text{100 g}$ of hydrate.