Question

A hydrate of CuSO4 has a mass of 12.98 g before heating. After heating, the mass of the anhydrous compound is found to be 9.70 g.
Explain how you would determine the formula of the hydrate and then write out the full name of the hydrate.

Answer 1

Explanation of the strategy: 1) calculate the mass of water and convert to number of moles, 2) convert the mass of anhydrous CuSO4 to moles, and, 3)calculate the mole ratio of water to CuSO4 anhydrous

1) Calculate the mass of water:

mass of water = mass of the hydrate CuSO4 - mass of the anhydrous compound

mass of water = 12.98 g - 9.70 g = 3.28g

2) Calculate the number of moles of water

number of moles = mass in grams / molar mass

molar mass of water = 18.01 g/mol

number of moles of water = 3.28 g / 18.01 g/mol = 0.182 mol

3) Calculate the number of moles of CuSO4 anhydrous

number of moles = mass in grams / molar mass

molar mass of Cu SO4 = 159.6 g/mol

number of moles of CuSO4 = 9.70g / 159.6 g/mol = 0.0608 moles

4) Calculate the ratio moles of water / moles of CuSO4

ratio = moles of water / moles of CuSO4 = 0.182 / 0.0608 = 2.99 ≈ 3

Therefore the molecular formula is CuSO4 . 3H2O

Name: copper(II) sulfate trihydrate.

Answer 2

The solution follows as;

Given:

CuSO4 – 9.70 (weight)

Water / H2O – 12.98 – 9.70 = 3.28 (weight)

Molecular weight of CuSO4 – 160

Molecular weight of H2O – 18

 

To solve;

= weight of CuSO4 x MW of CuSO4

= 970 / 160

= 328/x*18

= 6x = 18

= x – 3

= CuSO4*3H2O