Chapter 10, Problem 110AP

Interpretation Introduction

Interpretation:

The value of x in the molecular formula is to be calculated.

Concept introduction:

The ideal gas equation elaborates the physical properties of gases by relating the pressure, volume, temperature, and number of moles with each other with the help of four gas laws. This can be shown by:

$PV=nRT$

Here, $n$ is the number of moles, $R$ is the gas constant, $T$ is the temperature, $V$ is thevolume, and $P$ is the pressure.

The formula for conversion of temperature from degree Celsius to Kelvin is represented as

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15$

Answer to Problem 110AP

Solution: ${\text{MgSO}}_4.7{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

Given information:

Mass $m=54.2g$

Pressure $P=24.8\text{ atm}$

Temperature $T=120°\text{C}$

Volume $V=2\text{ L}$

The temperature is $120°\text{C}$.

The conversion of temperature from degree Celsius to Kelvin can be done by using the formula given below:

$\begin{array}{c}T\left(\text{K}\right)=T\left(°\text{C}\right)+273.15\\ =\left(120+273.15\right)\\ =393.15\text{ K}\end{array}$

The number of moles for $1\text{ L}$ of volume, that is, molar volume is

The equation for an ideal gas is as follows:

$PV=n_{{\text{H}}_2\text{O}}RT$

Rearrange the above equation for number of moles as follows:

$n_{{\text{H}}_2\text{O}}=\frac{PV}{RT}$

Substitute $0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}$ for $R$, $393.15\text{ K}$ for $T$, $24.8\text{ atm}$ for $P$, and $2\text{ L}$ for $V$ in the above equation

$\begin{array}{c}{n}_{{\text{H}}_2\text{O}}=\frac{\left(24.8\cancel{\text{atm}}\right)\left(\text{2 L}\right)}{\left(0.08206\cancel{\text{L}}\text{.}\cancel{\text{atm}}/\cancel{\text{K}}\text{.mol}\right)\left(393.15\cancel{\text{K}}\right)}\\ =\left(\frac{49.6}{32.26}\right)\text{ mol}\\ =1.537\text{ mol}\end{array}$ $\begin{array}{c}{n}_{{\text{H}}_2\text{O}}=\frac{\left(24.8\text{ atm}\right)\left(\text{2 L}\right)}{\left(0.08206\text{ L}\text{.atm}/\text{K}\text{.mol}\right)\left(393.15\text{ K}\right)}\\ =\left(\frac{49.6}{32.26}\right)\text{ mol}\\ =1.537\text{ mol}\end{array}$

Calculate the mass of water as follows:

$m_{{\text{H}}_{\text{2}}\text{O}}=n_{{\text{H}}_{\text{2}}\text{O}}\times M_{{\text{H}}_{\text{2}}\text{O}}$

Substitute $1.537\text{ mol}$ for $n_{{\text{H}}_{\text{2}}\text{O}}$ and $18\text{ g}/\text{mol}$ for $M_{{\text{H}}_{\text{2}}\text{O}}$ in the above equation

$\begin{array}{c}{m}_{{\text{H}}_{\text{2}}\text{O}}=1.537\cancel{\text{mol}}\times 18\text{ g}/\cancel{\text{mol}}\\ =27.6\text{ g}\end{array}$ $\begin{array}{c}{m}_{{\text{H}}_{\text{2}}\text{O}}=1.537\text{ mol}\times 18\text{ g}/\text{mol}\\ =27.6\text{ g}\end{array}$

The mass of thesample is thesum of mass of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{MgSO}}_4$, which is shown as follows:

$m_{\text{sample}}=m_{{\text{H}}_{\text{2}}\text{O}}+m_{{\text{MgSO}}_{\text{4}}}$

Rearrange the above equation for $m_{{\text{MgSO}}_{\text{4}}}$ as follows:

$m_{{\text{MgSO}}_{\text{4}}}=m_{\text{sample}}-m_{{\text{H}}_{\text{2}}\text{O}}$

Substitute $54.2\text{ g}$ for $m_{\text{sample}}$ and $27.6\text{ g}$ for $m_{{\text{H}}_{\text{2}}\text{O}}$ in the above equation

$\begin{array}{c}{m}_{{\text{MgSO}}_{\text{4}}}=54.2\text{ g}-27.6\text{ g}\\ =26.4\text{ g}\end{array}$

Calculate the moles of ${\text{MgSO}}_4$ as follows:

$n_{{\text{MgSO}}_4}=\frac{m_{{\text{MgSO}}_4}}{M_{{\text{MgSO}}_4}}$

Substitute $26.4\text{ g}$ for $m_{{\text{MgSO}}_4}$ and $120.4\text{ g}/\text{mol}$ for $M_{{\text{MgSO}}_4}$ in the above equation

$\begin{array}{c}{n}_{{\text{MgSO}}_4}=\frac{26.4\cancel{\text{g}}}{120.4\cancel{\text{g}}/\text{mol}}\\ =0.22\text{ mol}\end{array}$ $\begin{array}{c}{n}_{{\text{MgSO}}_4}=\frac{26.4\text{ g}}{120.4\text{ g}/\text{mol}}\\ =0.22\text{ mol}\end{array}$

Now, the mole ratio of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{MgSO}}_4$ is

$\begin{array}{c}\frac{n_{{\text{H}}_2\text{O}}}{n_{{\text{MgSO}}_4}}=\frac{1.54\cancel{\text{mol}}}{0.22\cancel{\text{mol}}}\\ =7\end{array}$ $\begin{array}{c}\frac{n_{{\text{H}}_2\text{O}}}{n_{{\text{MgSO}}_4}}=\frac{1.54\text{ mol}}{0.22\text{ mol}}\\ =7\end{array}$

As the mole ratio of ${\text{H}}_{\text{2}}\text{O}$ and ${\text{MgSO}}_4$ is $7:1$, the value of x in the compound ${\text{MgSO}}_4.x{\text{H}}_{\text{2}}\text{O}$ will be $7$.

So, the molecular formula of the compound is ${\text{MgSO}}_4{\text{.7H}}_{\text{2}}\text{O}$.

Conclusion

The value of x in the compound ${\text{MgSO}}_4.x{\text{H}}_{\text{2}}\text{O}$ is $7$.