• # question_answer A mixture of ${{\operatorname{CuSO}}_{4}}.5{{H}_{2}}\operatorname{O}$ and ${{\operatorname{MgSO}}_{4}}.7{{H}_{2}}O$ is heated until all the water is lost. If 5.020 g of the mixture gives 2.988 g of the anhydrous salts, what is the percent by mass of${{\operatorname{CuSO}}_{4}}.5{{H}_{2}}\operatorname{O}$ in the mixture? A) 65.86 B) 70.86 C) 75.45 D) 79.25

 The mass of water lost upon heating the mixture is $\left( 5.020 g-2.988 g \right) = 2.032 g$water. If$\operatorname{x}=mass\,of\,CuS{{O}_{4}}.5{{H}_{2}}O$, then the mass of ${{\operatorname{MgSO}}_{4}}7{{H}_{2}}O\,is\,(5.020-x)g.$We can calculate the amount of water lost by each salt based on the mass % of water in each hydrate. We can write:
 $(mass\,\,CuS{{O}_{4}}.5{{H}_{2}}O)$(%${{H}_{2}}O$)+$(mass\,MgS{{O}_{4}}.7{{H}_{2}}O)$(%${{H}_{2}}O$) = total mass ${{H}_{2}}O=2.032g\,{{H}_{2}}O$ Calculate the $%{{H}_{2}}O$in each hydrate. %${{H}_{2}}O=\left( CuS{{O}_{4}}.5{{H}_{2}}O \right)$ $=\frac{\left( 5 \right)\left( 18.02g \right)}{249.7g}\times 100%$% = 36.08%${{H}_{2}}O$ %${{H}_{2}}O=\left( MgS{{O}_{4}}.7{{H}_{2}}O \right)$ $=\frac{\left( 7 \right)\left( 18.02g \right)}{246.5g}\times 100%$% = 51.17%${{H}_{2}}O$
 Substituting into the equation above $(x)(0.3608)+(5.020- x)(0.5117)= 2.032 g$ $0.1509\text{ }x=0.5367$ $x=3.557g=mass of\,CuS{{O}_{4}}. 5{{H}_{2}}O$ Finally, the percent by mass of$CuS{{O}_{4}}$. $5{{H}_{2}}O$In the mixture is:$\frac{3.557g}{5.020g}\times 100%$% $=70.86%$%