A mixture of CaCO₃ and CaO weighing 0.693 g was heated to produce gaseous CO₂. After heating, the remaining solid weighed 0.508g. Assuming all the CaCO₃ broke down to CaO and CO₂, calculate the mass percent of CaCO₃ in the original mixture.

Calcium carbonate decomposes to produce calcium oxide and carbon dioxide. The balanced equation is like

$CaC{O}_3\left(s\right)\stackrel{∆}{\to }CaO\left(s\right)+C{O}_2\left(g\right)$

As you know,

Mass of carbon dioxide produced = mass of mixture – mass of residue

According to the question;

Mass of mixture = 0.693g

Mass of residue = 0.508g

Now, Mass of carbon dioxide produced

= (0.693 - 0.508)(g)

= 0.185(g)

Mass of CO₂ = 0.185g

Molecular mass of CO₂ = 44g/mol

Again you know,

$moles=\frac{mass\left(given\right)}{mass\left(molar\right)}$

Moles of CO₂

$$\begin{gathered} =\frac{0.185}{44}\left(mol\right) \\ =0.0042\left(mol\right) \end{gathered}$$

Hence, moles of CO₂ produced are 0.0042mol.

According to the balanced equation

$CaC{O}_3\left(s\right)\stackrel{∆}{\to }CaO\left(s\right)+C{O}_2\left(g\right)$

1 mole of CaCO₃ produces 1 mole CO₂

Now, moles of CaCO₃

$$\begin{gathered} =0.0042\times \frac{1}{1}\left(mol\right) \\ =0.0042\left(mol\right) \end{gathered}$$

Hence, moles of CaCO₃ are 0.0042mol.

Moles of CaCO₃ = 0.0042mol

Molecular mass of CaCO₃ = 100g/mol

Again you know,

Mass = moles x mass(molar)

Mass of CaCO₃

$$\begin{gathered} =0.0042\times 100\left(g\right) \\ =0.42\left(g\right) \end{gathered}$$

Hence, mass of CaCO₃ is 0.42g.

Again you know,

Mass percent of CaCO₃

$$\begin{gathered} =\frac{mass\left(CaC{O}_3\right)}{mass\left(sample\right)}\times 100\% \\ =\frac{0.42}{0.693}\times 100\% \\ =60.6\% \end{gathered}$$

Hence, mass percent of CaCO₃ is 60.6%.

Hence, mass percent of CaCO₃ in the original mixture is 60.6%.