Question

A mixture of $CuS{O_4}.5{H_2}O$ and $MgS{O_4}.7{H_2}O$ was heated until all the water was driven off. If $5.0g$ of mixture gave 3 g of anhydrous salt, what was the percentage by mass of $CuS{O_4}.5{H_2}O$ in the original mixture?
A.$74.4$
B.70
C.80
D.90

Answer 1

The knowledge of mass percent in the chapter of Mole Concept and Stoichiometry is important to solve this question. We shall put the appropriate values in the given equation to calculate the mass percentage.

The formula used:
Mass percent = $\dfrac{{{\text{Component's mass}}}}{{{\text{total mass}}}}{ \times 100}$

Complete step by step answer:
Let the mass percent of $CuS{O_4}.5{H_2}O$ in the mixture be $x\%$,
Then that of $MgS{O_4}.7{H_2}O$ will be $\left( {100 - x} \right)\%$.
Thus, 5 g of the mixture will have $5x\%$ $CuS{O_4}.5{H_2}O$ and $5\left( {100 - x} \right)\%$ of $MgS{O_4}.7{H_2}O$.
\Rightarrow $$$0.05x$$g ofCuS{O_4}.5{H_2}O$$ \Rightarrow \left( {5 - 0.05x} \right)g of $$MgS{O_4}.7{H_2}O$$ Molecular mass ofCuS{O_4}.5{H_2}O$=$\left[ {63.54 + 32 + \left( {16 \times 4} \right)} \right] + 5\left[ {16 + \left( {2 \times 1} \right)} \right]$=$249.7$${\text{gmo}}{{\text{l}}^{{\text{ - 1}}}} Molecular mass of $$MgS{O_4}.7{H_2}O$$ =\left[ {24 + 32 + \left( {16 \times 4} \right)} \right] + 7\left[ {16 + \left( {2 \times 1} \right)} \right]$=$246.5$${\text{gmo}}{{\text{l}}^{{\text{ - 1}}}}$No. of moles of$CuS{O_4}.5{H_2}O$=$\dfrac{{0.05x}}{{249.7}} = 2.00 \times {10^{ - 4}}xmoles. No. of moles of $$MgS{O_4}.7{H_2}O$$ =\dfrac{{5 - 0.05x}}{{246.5}}$moles. Therefore total no. of moles of water that was released after evaporation =$5 \times 2.00 \times {10^{ - 4}}x$+$7 \times \left( {\dfrac{{5 - 0.05x}}{{246.5}}} \right)$$ \Rightarrow {10^{ - 3}}x + \dfrac{{5.0 - 0.05x}}{{35.22}}$moles of water. The mass of water that was released was =$5 - 3 = 2{\text{ g}}$Converting it to moles we get,${\text{n}} = \dfrac{2}{{18}} = 0.111{\text{ moles}}$Therefore,${10^{ - 3}}x + \dfrac{{5.0 - 0.05x}}{{35.22}} = 0.111$$ \Rightarrow 3.52 \times {10^{ - 2}}x + 5 - 0.05x = 3.9133$Solving this, we get:$ \Rightarrow 0.01478x = 1.0867$Therefore,$x = 74.4% $So the mass of$CuS{O_4}.5{H_2}O$in the original mixture was$74.4% $.

Hence, the correct option is option A.

Notes:
The amount of $MgS{O_4}.7{H_2}O$ in the mixture was, $\left( {100 - 74.4} \right)$ = $25.6\%$
Mass percent has no unit because it is a ratio of two similar quantities.
The steps to calculate mass percentage are as (a) Calculate the molecular masses of the total compounds and present in the mixture. This will be the denominator in the equation.
The mass of the compound in question will be the numerator of the mass percent.
When you have calculated both the numerator and denominator, put their values to get the mass percent of the compound.