Chapter 10, Problem 183A

Determine the percent composition of ${\text{MgCO}}_{\text{3}}{\text{•5H}}_{\text{2}}\text{O}$ and draw a pie graph to represent the hydrate.

Interpretation Introduction

Interpretation:

Percentage composition of MgCO₃.5H₂O need to be determined and the hydrate must be represented with a pie graph.

Concept introduction:

Percentage of an element in a compound can be calculated as follows:

Percentage of an element in a compound $\text{=}\frac{\text{Number of atom}\times \text{Atomic mass of the element}}{\text{Molar mass of the compound}}\times 100\%$

Answer to Problem 183A

Percentage composition of MgCO₃.5H₂O is 13.79 % Mg, 73.56 % O, 6.90 % C and 5.75 % H.

Thus, pie chart is represented as follows:

Explanation of Solution

Molar mass of MgCO₃.5H₂O is 174 g/mol.

Atomic mass of Mg, O, C and H are 24 amu, 16 amu, 12 amu and 1 amu respectively.

The mass percentage of Mg, O and C can be calculated as follows:

% of Mg $=\frac{1\times 24\times 100\%}{174}=\frac{2400}{174}\%=13.79\%$

% of O $=\frac{8\times 16\times 100\%}{174}=\frac{12800}{174}\%=73.56\%$

% of C $=\frac{1\times 12\times 100\%}{174}=\frac{1200}{174}\%=6.90\%$

% of H $=\frac{10\times 1\times 100\%}{174}=\frac{1000}{174}\%=5.75\%$

Thus, % of all the elements are shown in the pie chart which shows that O is maximum and H is minimum.

Conclusion

Thus, 13.79 % Mg, 73.56 % O, 6.90 % C and 5.75 % H is present in MgCO₃.5H₂O.