Chapter 2, Problem 119GQ

Interpretation Introduction

Interpretation: The Molar mass, mass percent of each elements and the mass of copper and water in 10.5g of the given compound should be calculated.

Concept Introduction:

• $\text{Mass}=\text{Density}\text{×}\text{Volume}$
• $\text{Mole}\text{of}\text{substance}=\frac{\text{Gram}\text{of}\text{substance}}{\text{Molar}\text{mass}}$
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol)
• Mass percentage is one way to represent the concentration of an element in a compound or a component in a mixture. Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

Answer to Problem 119GQ

Molar mass of given compound $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$ is $245.75g/mol$

$\begin{array}{cc}\text{Elements}\text{in}\text{Cu}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{SO}}_{\text{4}}{\text{.H}}_{\text{2}}\text{O}& Masspercentage\\ Cu& 25.86\%\\ N& 22.8\%\\ H& 5.69\%\\ O& 32.55\%\\ S& 13.02\%\end{array}$

In 10.5 g of compound there are $2.72gCuand0.77g{H}_{2}O$

Explanation of Solution

The molar mass of the given compound is as follows,

Molecular formula of the given compound is $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$

Molar mass of $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$ can be calculated as follows,

$\left(\text{1×}\underset{molarmass\text{of}\text{Cu}}{63.55g/mol}\right)\text{+}\left(\text{4×}\underset{\text{molar}\text{mass}\text{of}N{\text{H}}_3}{17.031}\right)\text{+}\left(\text{1×}\underset{\text{molar}\text{mass}\text{of}S{O}_4}{96.06}\right)\text{+}\left(\text{1×}\underset{\text{molar}\text{mass}\text{of}{H}_2\text{O}}{18.01}\right)\text{=}245.75\text{g/mol}$

The mass percent of each element in $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$ is as follows,

Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.

The molecular formula of the given compound is $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$

And this compound contains $1Copper,4nitrogen,5oxygen,14hydrogenand1sulfur$

Molar mass of Copper is $\text{1}\text{×}\text{ 63}\text{.55}\text{=}\text{63}\text{.55}\text{g/mol}$

Mass percent of copper in $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$ calculated as follows,

$\begin{array}{c}\text{Cu}\text{in}\text{Cu}{\left(N{H}_3\right)}_{4}S{O}_4{\text{.H}}_2\text{O}\text{=}\frac{\text{Molar}\text{mass}\text{of}\text{copper}}{\text{Molar}\text{mass}\text{of}\text{Cu}{\left(N{H}_3\right)}_{4}S{O}_4{\text{.H}}_2\text{O}}\text{×100\%}\\ \text{=}\frac{63.55}{245.75}\text{×100\%}\\ \text{=}25.86\text{\%}\end{array}$

Similarly, Molar mass of Nitrogen in compound is $4\text{×14}\text{=}56\text{g/mol}$

$\begin{array}{c}\text{Mass}\text{percent}\text{of}\text{Nitrogen}\text{in}{\text{C}}_{\text{18}}{\text{H}}_{\text{27}}{\text{NO}}_{\text{3}}\text{=}\frac{\text{Molar}\text{mass}\text{of}\text{Nitrogen}}{\text{Molar}\text{mass}\text{of}\text{Cu}{\left(N{H}_3\right)}_{4}S{O}_4{\text{.H}}_2\text{O}}\text{×100\%}\\ \text{=}\frac{56}{245.75}\text{×100}\%\\ \text{=}22.8\text{\%}\end{array}$

Molar mass of Hydrogen in the given compound is $\text{1}\text{×14}\text{=}\text{14}\text{g/mol}$

$\begin{array}{c}\text{Mass}\text{percent}\text{of}\text{nirogen}\text{in}\text{Cu}{\left(N{H}_3\right)}_{4}S{O}_4{\text{.H}}_2\text{O}\text{=}\frac{\text{Molar}\text{mass}\text{of}hydrogen}{\text{Molar}\text{mass}\text{of}\text{Cu}{\left(N{H}_3\right)}_{4}S{O}_4{\text{.H}}_2\text{O}}\text{×100}\\ \text{=}\frac{\text{14}}{245.75}\text{×100}\\ \text{=}5.69\text{\%}\end{array}$ Molar mass of oxygen in the given compound is $5\text{×16}\text{=}80\text{g/mol}$

$\begin{array}{c}\text{Mass}\text{percent}\text{of}\text{oxygen}\text{in}\text{Cu}{\left(N{H}_3\right)}_{4}S{O}_4{\text{.H}}_2\text{O=}\frac{\text{Molar}\text{mass}\text{of}\text{oxygen}}{\text{Molar}\text{mass}\text{of}\text{Cu}{\left(N{H}_3\right)}_{4}S{O}_4{\text{.H}}_2\text{O}}\text{×100}\\ \text{=}\frac{\text{80}}{245.75}\text{×100}\\ \text{=}32.55\text{\%}\end{array}$ Molar mass of sulfur in the compound is $1\times 32g/mol\text{=}32\text{g/mol}$.

$\begin{array}{c}\text{Mass}\text{percent}\text{of}\text{sulfur}\text{in}\text{Cu}{\left(N{H}_3\right)}_{4}S{O}_4{\text{.H}}_2\text{O=}\frac{\text{Molar}\text{mass}\text{of}\text{sulfur}}{\text{Molar}\text{mass}\text{of}\text{Cu}{\left(N{H}_3\right)}_{4}S{O}_4{\text{.H}}_2\text{O}}\text{×100}\\ \text{=}\frac{32}{245.75}\text{×100}\\ \text{=}13.02\text{\%}\end{array}$

The mass of copper and water in 10.5 g $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$ is as follows,

Mass percent of copper in $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$ is $25.86\%$

Then, the mass of copper in 10.5 of $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$,

        $\frac{25.86}{100}\times 10.5g=2.7g$

Similarly,

Mass percent of water in $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$ is $7.33\%$

Then, the mass of water in 10.5 of $Cu{\left(N{H}_3\right)}_{4}S{O}_4.H_{2}O$,

        $\frac{7.33}{100}\times 10.5g=0.77g$

Conclusion

Molar mass, mass percent of each elements and the mass of copper and water in 10.5g of the given compound were calculated.