Chapter 4, Problem 141CP

Interpretation Introduction

Interpretation: The mass percentages in the sample have to be calculated.

Concept introduction: The mass percentage of a compound can be calculated with the calculated mass of the compound to the total mass of the compound.

$\text{Mass}\text{percent(in}\text{\%)=}\frac{\text{Mass}\text{of}\text{compound(in}\text{g)}}{\text{Total}\text{mass(in}\text{g)}}$

Answer to Problem 141CP

Answer

The mass of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ was found to be $\text{0}\text{.123}\text{g}$ and its mass percent was $\text{60}\text{.0\%}$.  The mass percent of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ in the sample are $\text{39\%}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}\text{61\%}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$

Explanation of Solution

Explanation

Record the given data

Mass of the sample mixture = $\text{0}\text{.205}\text{g}$

Mass of the precipitate         = $\text{0}\text{.298}\text{g}$

The mass of the sample mixture and mass of the precipitate are recorded as shown above.

To calculate the mass of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$

Molar mass of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$= $\text{96}\text{.07}\text{g}$

Molar mass of ${\text{BaSO}}_{\text{4}}$ = $\text{233}\text{.4}\text{g}$

Mass of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$= $\text{0}\text{.298}\text{g}{\text{BaSO}}_{\text{4}}\text{×}\frac{\text{96}\text{.07}\text{g}{\text{SO}}_{\text{4}}{}^{\text{2-}}}{\text{233}\text{.4}\text{g}{\text{BaSO}}_{\text{4}}}$

                       = $\text{0}\text{.123}\text{g}{\text{SO}}_{\text{4}}{}^{\text{2-}}$

The mass of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ is calculated by plugging in the values of mass of precipitate to the molar mass of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ and ${\text{BaSO}}_{\text{4}}$. The mass of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ was found to be $\text{0}\text{.123}\text{g}$.

To calculate the mass percent of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$

Mass of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$= $\text{0}\text{.123}\text{g}$

Mass percent of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$= $\frac{\text{0}\text{.123}\text{g}}{\text{0}\text{.205}\text{g}}\text{×100}$

                                    = $\text{60}\text{.0\%}$

The mass percent of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ is calculated by plugging in the values of mass of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ to the total mass of the sample. The mass percent of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ is found to be $\text{60}\text{.0\%}$.

To calculate the mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$

Molar mass of ${\text{K}}^{\text{+}}$= $\text{39}\text{.10}\text{g}$

Molar mass of ${\text{Na}}^{\text{+}}$= $\text{22}\text{.99}\text{g}$

Assume $\text{100}\text{.0}\text{g}$ of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ mixture there are

$\text{60}\text{.0}\text{g}\text{of}{\text{SO}}_{\text{4}}{}^{\text{2-}}\text{=}\frac{\text{1}\text{mol}}{\text{96}\text{.07}\text{g}}\text{=0}\text{.625}\text{mol}{\text{SO}}_{\text{4}}{}^{\text{2-}}$

Hence,

$\text{2×0}\text{.625=1}\text{.25}\text{mol}\text{of}\text{+1}\text{cations}$ must be present to balance $\text{2-}\text{charge}\text{of}{\text{SO}}_{\text{4}}{}^{\text{2-}}$

Let x= mass of ${\text{K}}^{\text{+}}$

      y= mass of ${\text{Na}}^{\text{+}}$

$\text{x+y=1}\text{.25}\text{g }\mathrm{......}\text{(1)}$

In $\text{100}\text{.0}\text{g}$ of mixture,

Total mass of ${\text{K}}^{\text{+}}$ and ${\text{Na}}^{\text{+}}$= 40g. Then the equation can be given as,

$\text{x}\text{mol}{\text{K}}^{\text{+}}\text{×}\frac{\text{39}\text{.10}\text{g}}{\text{mol}}\text{+y}\text{mol}{\text{Na}}^{\text{+}}\text{×}\frac{\text{22}\text{.99}\text{g}}{\text{mol}}\text{=40}\text{.0}\text{g}\mathrm{......}\text{(2)}$

Solving equation (1) and (2),

$\text{x+y=1}\text{.25}\text{g}$ and $\text{x}\text{mol}{\text{K}}^{\text{+}}\text{×}\frac{\text{39}\text{.10}\text{g}}{\text{mol}}\text{+y}\text{mol}{\text{Na}}^{\text{+}}\text{×}\frac{\text{22}\text{.99}\text{g}}{\text{mol}}\text{=40}\text{.0}\text{g}$

$\begin{array}{l}\text{x=1}\text{.25-y}\\ \text{substitute}\text{the}\text{value}\text{of}\text{x}\text{in}\text{(2)}\\ \\ \text{39}\text{.10(1}\text{.25)y+(22}\text{.99)y=40}\text{.0}\\ \\ \text{48}\text{.9-(39}\text{.10)y+(22}\text{.99)y=40}\text{.0}\\ \\ \text{-(16}\text{.11)y=-8}\text{.9}\\ \\ \text{y=0}\text{.55}\text{mol}{\text{Na}}^{\text{+}}\\ \\ \text{Therefore,}\\ \\ \text{x=1}\text{.25-0}\text{.55}\\ \text{=0}\text{.70}\text{mol}{\text{K}}^{\text{+}}\end{array}$

Therefore,

$\text{0}\text{.70}\text{mol}{\text{K}}^{\text{+}}\text{×}\frac{\text{1}\text{mol}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{2}\text{mol}}\text{=0}\text{.35}\text{mol}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$

$\text{0}\text{.35}\text{mol}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{×}\frac{\text{174}\text{.27}\text{g}}{\text{mol}}\text{=61}\text{g}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$

$\begin{array}{l}\text{Mass}{\text{of Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=100-61}\\ \text{=39}\text{g}\end{array}$

$\begin{array}{l}\text{Mass}\text{of}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=61}\text{g}\\ \text{Mass}\text{of}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=39}\text{g}\end{array}$

The mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated by plugging in the values of their respective masses to the total assumed mass. The moles of individual ions and moles of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ are also calculated to get their respective masses. The masses of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ are $\text{61}\text{g}\text{and}\text{39}\text{g}$ respectively.

To calculate the mass percent of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$

$\begin{array}{l}\text{Mass}\text{of}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=61}\text{g}\\ \text{Mass}\text{of}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=39}\text{g}\end{array}$

Assume total mass = $\text{100}\text{g}$

$\begin{array}{l}\text{Mass}\text{percent}\text{of}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\frac{\text{61}}{\text{100}}\text{×100}\\ \text{ =61\%}\end{array}$

$\begin{array}{l}\text{Mass}\text{percent}\text{of}{\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\frac{\text{39}}{\text{100}}\text{×100}\\ \text{=39\%}\end{array}$

The mass percent of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated by plugging in the values of calculated mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ to the total mass of the mixture. The mass percent of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ is found to be $\text{61\%}\text{and}\text{39\%}$ .

Conclusion

Conclusion

The mass and mass percent of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ were calculated using the molar masses of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$, ${\text{BaSO}}_{\text{4}}$ and mass of precipitate with the total mass of the sample. The mass percent of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ was found to be $\text{60}\text{.0\%}$ and mass of ${\text{SO}}_{\text{4}}{}^{\text{2-}}$ was found to be $\text{0}\text{.123}\text{g}$.

The mass percent of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ was calculated by using the calculated mass of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ to the total mass of the mixture. The mass percent of ${\text{Na}}_{\text{2}}{\text{SO}}_{\text{4}}\text{and}{\text{K}}_{\text{2}}{\text{SO}}_{\text{4}}$ was found to be $\text{61\%}\text{and}\text{39\%}$ .