Chapter 5, Problem 5.153QP

Interpretation Introduction

Interpretation:

The percent composition by mass of ${\text{MgCO}}_{\text{3}}$ and ${\text{CaCO}}_{\text{3}}$ mixture has to be calculated.

Concept Introduction:

Ideal gas is the most usually used form of the ideal gas equation, which describes the relationship among the four variables $\text{P, V, n, and T}$.  An ideal gas is a hypothetical sample of gas whose pressure-volume-temperature behavior is predicted accurately by the ideal gas equation.

  $\text{PV = nRT}$

Answer to Problem 5.153QP

The percent composition by mass of ${\text{CaCO}}_{\text{3}}$ is $6.15\%$

The percent composition by mass of ${\text{MgCO}}_{\text{3}}$ is $93.9\%$

Explanation of Solution

Given data,

${\text{MgCO}}_{\text{3}}$ and ${\text{CaCO}}_{\text{3}}$ mixture of mass is $6.26g$

$\begin{array}{l}\text{P}\text{=}\text{1}\text{.12}\text{atm}\\ \text{V=1}\text{.73}\text{L}\\ \text{T}\text{=}\text{48+273=}\text{321K}\end{array}$

Calculate the moles of ${\text{CO}}_{\text{2}}$ produced using the ideal gas equation.  We carry an extra significant figure throughout this calculation to limit rounding errors.

  $\begin{array}{l}{\text{ n}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{PV}}{\text{RT}}\\ \\ \text{=}\frac{\text{(1}\text{.24}\text{atm)(1}\text{.73}\text{L)}}{\left(\text{0}\text{.0821}\frac{\text{L}\text{.}\text{atm}}{\text{K}\text{.}\text{mol}}\right)\text{(48+}\text{273)}\text{K}}\text{=}\text{0}\text{.07352}\text{mol}{\text{CO}}_{\text{2}}\end{array}$

The moles of ${\text{CO}}_{\text{2}}$ produced are calculated by plugging in the given volume, pressure, gas constant and temperature.  The moles of ${\text{CO}}_{\text{2}}$ produced was found to

  $\text{0}\text{.07352}\text{mol}{\text{CO}}_{\text{2}}$

Since the mole ratio between ${\text{CO}}_{\text{2}}$ and both reactants $\left({\text{CaCO}}_{\text{3}}{\text{ and MgCO}}_{\text{3}}\right)$ is 1:1, $\text{0}\text{.07352}\text{mol}$ of the mixture must have reacted. We can write

  $\text{mol}{\text{CaCO}}_{\text{3}}\text{+}\text{mol}{\text{MgCO}}_{\text{3}}\text{=}\text{0}\text{.07352}\text{mol}$

${\text{Let x be the mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ in the mixture, then (7}{\text{.63 - x) is the mass of MgCO}}_{\text{3 }}\text{in the mixture}\text{.}$

  $\begin{array}{l}\left(\text{x}\text{g}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}{\text{106}\text{.0}\text{g}{\text{Na}}_{\text{2}}{\text{CO}}_{\text{3}}}\right)\text{+}\left[\text{(7}\text{.63-x)g}{\text{MgCO}}_{\text{3}}\text{×}\frac{\text{1}{\text{molMgCO}}_{\text{3}}}{\text{84}\text{.32}\text{g}{\text{MgCO}}_{\text{3}}}\right]\\ \text{=}\text{0}\text{.08436}\text{mol}\end{array}$

  $\text{x = 2}{\text{.527 g = mass of Na}}_{\text{2}}{\text{CO}}_{\text{3}}\text{ in the mixture}$

$\text{Let}\text{x}\text{g}\text{=}{\text{mass of CaCO}}_{\text{3}}\text{in the mixture, then (6}\text{.26-x)g}\text{=}{\text{mass of MgCO}}_{\text{3}}$ in the mixture.

We can write

  $\begin{array}{l}\left[\text{x}\text{g}\text{}{\text{CaCO}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}{\text{CaCO}}_{\text{3}}}{\text{100}\text{.1}\text{g}{\text{CaCO}}_{\text{3}}}\right]\text{+}\left[\left(\text{6}\text{.26-x}\right)\text{g}\right]{\text{MgCO}}_{\text{3}}\text{×}\frac{\text{1}\text{mol}{\text{MgCO}}_{\text{3}}}{\text{84}\text{.32}\text{g}{\text{MgCO}}_{\text{3}}}\text{=}\text{0}\text{.07352}\text{mol}\\ \\ \text{0}\text{.009990x}\text{-}\text{0}\text{.01186x}\text{+}\text{0}\text{.07424}\text{=}\text{0}\text{.07352}\\ \text{x}\text{=}\text{0}\text{.385}\text{g}\\ \\ \text{=}\text{mass}\text{of}{\text{CaCO}}_{\text{3}}\text{in}\text{the}\text{mixture}\\ \text{mass}\text{of}{\text{MgCO}}_{\text{3}}\text{in}\text{the}\text{mixture}\text{=}\text{6}\text{.26}\text{-x}\text{=}\text{5}\text{.88}\text{g}\end{array}$

The percent composition by mass in the mixture are

  $\begin{array}{l}{\text{CaCO}}_{\text{3}}\text{:}\frac{\text{0}\text{.385 g}}{\text{6}\text{.26 g}}\text{×}\text{100\%}\text{=}\text{6}\text{.15\%}\\ {\text{MgCO}}_{\text{3}}\text{:}\frac{\text{5}\text{.88 g}}{\text{6}\text{.26 g}}\text{×}\text{100\%}\text{=}\text{93}\text{.9\%}\end{array}$

Conclusion

The percent composition by mass of ${\text{CaCO}}_{\text{3}}$ was calculated as $6.15\%$

The percent composition by mass of ${\text{MgCO}}_{\text{3}}$ was calculated as $93.9\%$