Question

Limestone consists mainly of the mineral calcite, ${\mathbf{CaCO}}_{\mathbf{3}}$.The carbonate content of 0.5413 g

powdered limestone was measured by suspending the powder in water, adding 10.00 mL of $\mathbf{1}\mathbf{.}\mathbf{396}\mathbf{MHCl}$and heating to dissolve the solid and expel role="math" localid="1654937947195" ${\mathbf{CO}}_{\mathbf{2}}{\mathbf{CaCO}}_{\mathbf{3}}\mathbf{\left(}\mathbf{s}\mathbf{\right)}\mathbf{}\mathbf{+}\mathbf{2}{\mathbf{H}}^{\mathbf{+}}\mathbf{\to }{\mathbf{Ca}}^{\mathbf{2}\mathbf{+}}\mathbf{+}{\mathbf{CO}}_{\mathbf{2}}\mathbf{\uparrow }\mathbf{+}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}$Calcium carbonate FM 100.087

The excess acid required $\mathbf{39}\mathbf{.}\mathbf{96}\mathbf{}\mathbf{mL}$of $\mathbf{0}\mathbf{.}\mathbf{1004}\mathbf{MNaOH}$for complete titration to a phenolphthalein end point. Find the weight percent of calcite in the limestone.

Short answer

The powdered limestone contains $92.3\%$calcite.

Step-by-step solution

Define the formula to find the weight percentage.

$\mathrm{The}\mathrm{weight}\mathrm{percent}\%\mathrm{w}=\frac{\mathrm{mass}\mathrm{of}}{\mathrm{mass}\mathrm{of}\mathrm{solution}}-\times 100$

Step 2:Calculate the number of moles of that reacted with CaCO3

$$\begin{gathered} \mathrm{Given}\mathrm{information} \\ \mathrm{mass}\mathrm{of}\mathrm{powdered}\mathrm{limestone}=0.5413\mathrm{g} \\ \mathrm{volume}\mathrm{of}1.396\mathrm{MJHCI}=10.00\mathrm{mL}=0.01\mathrm{L} \\ \mathrm{volume}\mathrm{of}0.1004\mathrm{MNaOHadded}\mathrm{to}\mathrm{neutralize}\mathrm{the}\mathrm{solution} \\ =39.96\mathrm{mL}=0.0396\mathrm{L} \\ \mathrm{FM}\mathrm{of}{\mathrm{CaCO}}_3=100.087\mathrm{g}/\mathrm{mol} \\ \mathrm{Now}\mathrm{calculate}\mathrm{the}\mathrm{mole}\mathrm{of}\mathrm{HCL}. \\ \mathrm{mole}\mathrm{of}\mathrm{reacted}\mathrm{with}{\mathrm{CaCO}}_3=\mathrm{original}\mathrm{mole}\mathrm{of}\mathrm{HCI}-\mathrm{mole}\mathrm{of}\mathrm{excess}\mathrm{HCI} \\ =(0.01\mathrm{L}\left(1.396\frac{\mathrm{molHCI}}{\mathrm{L}}\right)-(0.0396\mathrm{L}) \\ \left(0.1004\frac{\mathrm{molHCI}}{\mathrm{L}}\right) \\ =9.98416\times {10}^{-3}\mathrm{molHCI} \end{gathered}$$

Calculate the mass of CaCO3.

Using mole concept

$$\begin{gathered} \mathrm{mass}\mathrm{of}{\mathrm{CaCO}}_3=9.98416\times {10}^{-3}\mathrm{molHCI}\left(\frac{1{\mathrm{molCaCO}}_3}{2\mathrm{mollICI}}\right)\left(\frac{100.087{\mathrm{gCaCO}}_3}{1\mathrm{molCaCO}}\right) \\ =0.4996{\mathrm{gCaCO}}_3 \\ \mathrm{The}\mathrm{weight}\mathrm{percent}\mathrm{of}\mathrm{calcite}\mathrm{in}\mathrm{limestone}. \\ \%\mathrm{w}/\mathrm{w}=\frac{\mathrm{mass}\mathrm{of}}{\mathrm{mass}\mathrm{of}\mathrm{solute}}-\times 100 \\ =\frac{0.4996{\mathrm{gCaCO}}_3}{0.5413\mathrm{g}{\mathrm{CaCO}}_3}\times 100 \\ =92.3\% \\ \mathrm{Thus},\mathrm{the}\mathrm{powdered}\mathrm{limestone}\mathrm{contains}92.3\%\mathrm{calcite}. \end{gathered}$$