What is the percent composition of water in magnesium sulfate heptahydrate?
The formula of our compound is: MgSO₄.7H₂O
Before we find the percent composition we need the molar mass of our compound. To calculate it, we will use the periodict table and look for the atomic mass of each element.
Mg: 24.31 amu
S: 32.07 amu
O: 16.00 amu
H: 1.01 amu
We can calcalute the molar mass like this:
molar mass of MgSO₄.7H₂O = 24.31 + 32.07 + 4 * 16.00 + 7 * (2* 1.01 + 16.00)
molar mass of MgSO₄.7H₂O = 246.52 g/mol
To find the percent composition of water in our compound we will suppose that we have a certain amount of it. For example we can suppose that we have 1 mol of it. So, if we have one mol of MgSO₄.7H₂O, how many grams do we have?
1 mol of MgSO₄.7H₂O * 246.52 g/mol = 246.52 g of MgSO₄.7H₂O
So the total mass of our sample is 246.52 g.
Then, if we look at the formula of our compound we can see that in 1 molecule of MgSO₄.7H₂O we have 7 molecules of water. So in one mol of our compound we have 7 moles of water. So, let's find the mass of water that we have in our sample:
molar mass of H₂O = 2 * 1.01 + 16.00
molar mass of H₂O = 18.02 g/mol
mass of H₂O = 7 moles of H₂O * 18.02 g/mol = 126.14 g of H₂O
mass of H₂O = 126.14 g
We have 126.14 g of H₂O in 246.52 g of MgSO₄.7H₂O. To find the percentage we have to:
% of H₂O = 126.14 g / 246.52 g * 100 = 51.17 %
% of H₂O = 51.17 %
The percent composition of water in magnesium sulfate heptahydrate is 51.17 %