"Published in Suisun City, California, USA"
A mixture of CuSO4·5H2O and FeSO4·7H2O weighing 100 g is heated in an oven at 373K (100°C) for the removal of water by hydration (by evaporation). The weight of the mixture after drying (after the removal of water by hydration) is found to be 60 g. Calculate the ratio of CuSO4·5H2O to
FeSO4·7H2O in the original mixture. (Atomic Weights: Cu = 63, S = 32, O = 16, Fe = 56, and H = 1)
Solution:
The given word problem is about the decomposition of the mixture of CuSO4·5H2O and FeSO4·7H2O by heating in an oven to remove water molecules which involves the principles of Stoichiometry. The total amount of the original mixture and the total amount of the mixture after hydration are given in the problem. The total amount of a substance in the reactants or incoming ingredients must be equal to the total amount of a substance in the final products. In short, the Law of Conservation of Mass must be followed all the time. Let's analyze the given word problem as follows
Basis: 100 g of the original mixture
60 g of the mixture after hydration
Let x = be the amount of CuSO4·5H2O
y = be the amount of FeSO4·7H2O
The total amount of a mixture before drying is
Molecular Weight Data:
CuSO4= 159
CuSO4·5H2O = 249
FeSO4= 152
FeSO4·7H2O = 278
Consider the decomposition of CuSO4·5H2O as follows:
The weight of CuSO4 will be equal to
Consider the decomposition of FeSO4·7H2O as follows:
The weight of FeSO4 will be equal to
The total amount of mixture after drying is
Substitute the value of x to the first equation, we have
Substitute the value of y to the first equation, we have
Therefore, the ratio of CuSO4·5H2O to FeSO4·7H2O in the original mixture is
