Question

Dissolving \(3.00 \mathrm{~g}\) of an impure sample of calcium carbonate in hydrochloric acid produced \(0.656 \mathrm{~L}\) of carbon dioxide (measured at \(20.0^{\circ} \mathrm{C}\) and \(792 \mathrm{mmHg}\) ). Calculate the percent by mass of calcium carbonate in the sample. State any assumptions.

Short answer

The percent by mass of \(CaCO_3\) in the impure sample is the calculated mass percent from Step 3.

Step-by-step solution

Calculate the amount in moles of \(CO_2\) produced

The ideal gas law is \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is temperature. Convert \(T\) to the Kelvin scale by adding 273.15 to the Celsius temperature. Convert \(P\) to atmospheres by dividing the given mmHg by 760. Use \(R\) as \(0.0821 \mathrm{L \cdot atm/(mol \cdot K)}\). Solve for \(n\), the number of moles of \(CO_2\).

Determine the mass of \(CaCO_3\) that would produce this amount of \(CO_2\)

In the reaction \(CaCO_3 + 2HCl \rightarrow CaCl_2 + H_2O + CO_2\), one mole of \(CaCO_3\) produces one mole of \(CO_2\). Therefore, the moles of \(CaCO_3\) equal the moles of \(CO_2\). Use the molar mass of \(CaCO_3\) (100.0869 g/mol) to find the mass that would produce the calculated moles of \(CO_2\).

Calculate the percent by mass of \(CaCO_3\) in the impure sample.

The mass percent is defined as \(\frac{mass \ of \ the \ component}{total \ mass} \times 100\%\). The mass of the component is the mass of \(CaCO_3\) calculated in Step 2, and the total mass is the mass of the impure sample given in the problem (3.00 g). Calculate the mass percent of \(CaCO_3\).