Question

Geoffrey used 2.498g of a mixture of CuSO4 x 5H20 and salicylic acid in this experiment. He recovered 1.324g of salicylic acid and 1.247 g of CuSO4 x 5H2O. What is the % salicylic acid in the original mixture? What is the % CuSO4 x 5H2O in the original mixture?

Answer 1

Geoffrey used 2.498 g of a mixture of CuSO₄×5H₂O and salicylic acid in this experiment. He recovered 1.324 g of salicylic acid and 1.247 g of CuSO₄×5H₂O. What is the % of salicylic acid in the original mixture? What is the % of CuSO₄×5H₂O in the original mixture?

Solution:

According to the law of conservation of mass, mass of the initial mixture should be bigger (in case of losses during the separation of individual substances) or equal to the total mass of recovered substances. In our case, this condition is not satisfied: 2.498 g < 1.324 g + 1.247 g.

We can make a conclusion that there is a mistake in the condition of this task. We will assume that mass of mixture of CuSO₄×5H₂O and salicylic acid before the separation was 3.498 g.

The percentage of salicylic acid can be calculated in such way:

3.498g of original mixture is equal to 100 %

1.324g of salicylic acid is equal to X %

Then, X = 1.324*100/3.498 = 37.85 %.

The percentage of CuSO₄×5H₂O will be:

3.498g of original mixture is equal to 100 %

1.247 g of CuSO₄ × 5H₂O equal to Y %

Then, Y = 1.247*100/3.498 = 35.65%

Also, we can make a conclusion that in the initial mixture was 100 - 37.85 - 35.65 = 26.5% of impurities.