- Created by: :) PurpleJaguar (: - Team GR
- Created on: 30-04-14 15:41
Q) A sample of hydrated copper sulphate (CuSO4.xH2O) was weighed heated, cooled and then reweighed again. 1.738g of the hydrated copper sulphate gave a residue of 1.112g of copper sulphate (CuSO4). Calculate the value of x.
First step: n(CuSO4) = Mass / Mr = 1.112/ 159.6 = 0.00679mol
Second step: n(H2O) = Mass / Mr = The mass of water will be the mass of copper sulphate minus the mass of residue left, 1.738 - 1.112 = 0.626g
So n(H2O) = Mass /Mr = 0.626 / 18 = 0.0348mol
Third step: 0.0348 / 0.00697 = 5 So x = 5
Each step explained:
Step 1: Calculate the number of moles of copper sulphate present in the experiment.
Step 2: Calculate the number of moles of water present. You first have to work out the amount in grams (even though liquids are measured in ml) of water because you are not provided that information in the question.
Step 3: Work out x by dividing the amount of moles in water with the compound.