# Question fb39d

Sep 22, 2015

41.7%

#### Explanation:

To get the percent composition of water in the copper(II) pentahydrate sample, you need to divide the mass of water that was driven off by heating by the total mass of the sample, and multiply this ratio by 100.

In your case, you know that the mass of the pentahydrate consists of the mass of the anhydrous salt and the mass of the water of hydration.

This means that you can find the mass of the water of hydration by

${m}_{\text{pentahydrate" = m_"anhydrous" + m_"water}}$

${m}_{\text{water" = m_"pentahydrate" - m_"anhydrous}}$

${m}_{\text{water" = 1.30 * 10^0"g" - 7.5800 * 10^(-1)"g" = "0.542 g}}$

The percent composition of water in th pentahydrate will thus be

(0.542color(red)(cancel(color(black)("g"))))/(1.30color(red)(cancel(color(black)("g")))) xx 100 = color(green)(41.7%)

This value is a little high, because the actual percent composition of water in copper(II) sulfate pentahydrate, $\text{CuSO"""_4 * 5"H"_2"O}$, is

(5 * 18.02color(red)(cancel(color(black)("g/mol"))))/(249.69color(red)(cancel(color(black)("g/mol")))) xx 100 = 36.1%#

Here is a video which shows the steps for this calculation.

Video from: Noel Pauller