Question

Burning a compound of calcium, carbon, and nitrogen in oxygen in a combustion train generates calcium oxide$\left(\mathbf{CaO}\right)$ , carbon dioxide$\left({\mathbf{CO}}_{\mathbf{2}}\right)$ , nitrogen dioxide$\left({\mathbf{NO}}_{\mathbf{2}}\right)$ , and no other substances. A small sample gives 2.389 g CaO, 1.876$\left({\mathbf{CO}}_{\mathbf{2}}\right)$ g , and 3.921 g $\left({\mathbf{NO}}_{\mathbf{2}}\right)$. Determine the empirical formula of the compound.

Short answer

The empirical formula of the compound is ${\text{CCaN}}_{\text{2}}$.

Step-by-step solution

Empirical formula

Empirical formulas are formulas that exhibit the ratio of elements present in a particular compound. It does not show the actual count of atoms in a molecule. The ratios are indicated as subscripts next to a particular element’s symbol.

Determining the empirical formula of the compound

The given mass of calcium oxide is 2.389 g.

As the molar mass of CaO is 56.0774 g/mol, therefore the number of moles of calciumoxide is:

$$\begin{gathered} {\text{n}}_{\text{CaO}}\text{=}\frac{\text{2.389g}}{{\text{56.0774gmol}}^{\text{- 1}}} \\ \text{= 0.043mol} \end{gathered}$$

One mole of CaO contains one mole of calcium. Hence, 0.043 mol of CaO contains 0.043 mol of calcium.

role="math" localid="1663731706415" $$\begin{gathered} \text{CaO}\to \text{Ca + O} \\ \text{111} \end{gathered}$$

The given mass of carbon dioxide is 1.876 g.

As the molar mass of${\text{CO}}_2$is 44.01 g/mol, therefore thenumber of moles of carbondioxideis:

$$\begin{gathered} {\text{n}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{1.876g}}{{\text{44.01gmol}}^{\text{- 1}}} \\ \text{= 0.043mol} \end{gathered}$$

1 mole of${\text{CO}}_2$contains one mole of carbon. Hence, 0.043 mol of carbon dioxide will contain 0.043 mol of carbon.

$$\begin{gathered} {\text{CO}}_2\to \text{C + 2O} \\ \text{112} \end{gathered}$$

The given mass of${\text{NO}}_2$ is 3.921 g, and the molar mass is 46.0055 g/mol, therefore the number of moles of nitrogen dioxideis:

$$\begin{gathered} {\text{n}}_{{\text{NO}}_{\text{2}}}\text{=}\frac{\text{3.921g}}{{\text{46.0055gmol}}^{\text{- 1}}} \\ \text{= 0.085mol} \end{gathered}$$

One mole of ${\text{NO}}_2$contains one mole of nitrogen. Hence, 0.085 mol of nitrogen dioxide contains 0.085 mol of nitrogen.

The ratio of moles of C:Ca:N is 0.043:0.043:0.085.

This ratio can be expressed in the whole number as 1:1:2.

Hence, the empirical formula of the compound is${\text{CCaN}}_{\text{2}}$ .