What is the percentage recovery if 0.98g of copper is left when reacting 0.5g of copper sulphate with excess zinc?
To calculate the percentage recovery, you need to know the initial amount of copper and the final amount of copper obtained after the reaction.
First, determine the initial amount of copper. In this case, the initial amount of copper is given as 0.5g of copper sulfate. However, copper sulfate contains both copper and sulfur, and we need to find out how much of the 0.5g is copper. The molar mass of copper sulfate (CuSO4) is 159.61 g/mol, and the molar mass of copper (Cu) is 63.55 g/mol.
To find the amount of copper in 0.5g of copper sulfate, you can use the following calculation:
(1 mol Cu / 1 mol CuSO4) × (63.55 g Cu / 159.61 g CuSO4) × (0.5 g CuSO4) = 0.198 g Cu
So the initial amount of copper is 0.198g.
Next, subtract the final amount of copper from the initial amount to find the amount of copper that reacted:
0.198 g - 0.98 g = -0.782 g
The negative value indicates that there was excess copper left after the reaction.
Finally, to calculate the percentage recovery, use the formula:
Percentage Recovery = (Final amount / Initial amount) × 100%
In this case, the final amount of copper is 0.98g, and the initial amount is 0.198g. Plug these values into the formula:
Percentage Recovery = (0.98 g / 0.198 g) × 100% = 494.95%
Therefore, the percentage recovery of copper is 494.95%.