Question

Carbon and oxygen form no fewer than five different binary compounds. The mass percentages of carbon in the five compounds are as follows: A, 27.29; B, 42.88; C, 50.02; D, 52.97; and E, 65.24. Determine the empirical formulas of the five compounds.

Short answer

The empirical formula of the compounds can be given as:

Compound A-${\text{CO}}_{\text{2}}$ , Compound B-CO, Compound C-${\text{C}}_{\text{4}}{\text{O}}_{\text{3}}$ , Compound D-${\text{C}}_{\text{3}}{\text{O}}_{\text{2}}$ , Compound E-${\text{C}}_{\text{5}}{\text{O}}_{\text{2}}$ .

Step-by-step solution

Empirical formula

Empirical formulas are the simplest formulas of a particular compound. It is the formula of a substance indicated with the smallest integer subscript.

Determining the empirical formula of these compounds

The mass of carbon is 27.29 g.

The number of moles of carbon is:

$$\begin{gathered} \text{Number of moles =}\frac{\text{27.29 g}}{{\text{12 gmol}}^{\text{- 1}}} \\ \text{= 2.27mol} \end{gathered}$$

As 27.29% of the total percentage is carbon and the compound is made up of only carbon, the remaining 72.71 g should be oxygen.

The mass of oxygen is 72.71 g.

The number of moles of oxygen is:

$$\begin{gathered} \text{Number of moles =}\frac{\text{72.71g}}{{\text{16 g mol}}^{\text{- 1}}} \\ \text{= 4.54mol} \end{gathered}$$

The ratio of the moles of C:O is 2.27:4.54. This ratio can be rounded off as 1:2.

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

The calculation for compound B can be given as:

The percentage mass of carbon in the compound B is 42.88%. Hence the sample of 100 g contains 42.88 g of carbon.

The calculation can be given as:

The mass of carbon is 42.88 g.

The number of moles of carbon is:

$$\begin{gathered} \text{Moles of carbon =}\frac{\text{42.88 g}}{{\text{12 g mol}}^{\text{- 1}}} \\ \text{= 3.57mol} \end{gathered}$$

The ratio of moles of carbon:oxygen is 3.57:3.57. When simplified to integer numbers the ratio becomes 1:1.

The empirical formula of compound B is CO.

The calculation for compound C can be given as:

The percentage mass of carbon in the compound B is 50.02%. Hence the sample of 100 g contains 50.02 g of carbon.

The calculation can be given as:

The mass of carbon is 50.02 g.

The number of moles of carbon is:

$$\begin{gathered} \text{Moles of carbon =}\frac{\text{50.02 g}}{{\text{12 g mol}}^{\text{- 1}}} \\ \text{= 4.17mol} \end{gathered}$$

As 50.02% of the total percentage is carbon and the compound is made up of only carbon and oxygen, the remaining 49.98 g should be oxygen.

The mass of oxygen is 49.98 g.

The number of moles of oxygen is:

$$\begin{gathered} \text{Number of moles =}\frac{\text{49.98 g}}{{\text{16 g mol}}^{\text{- 1}}} \\ \text{= 3.12mol} \end{gathered}$$

The ratio of moles of carbon:oxygen is 4.17:3.12. When simplified the ratio becomes 1.34:1. This becomes equal to 4.02:3. The ratio can be rounded as C:O as 4:3.

Hence, the empirical formula of the compound C is${\text{C}}_{\text{4}}{\text{O}}_{\text{3}}$ .

The calculation for compound D can be given as:

The percentage mass of carbon in the compound B is 52.97%. Hence the sample of 100 g contains 52.97 g of carbon.

The calculation can be given as:

The mass of carbon is 52.97 g.

The number of moles of carbon is:

$$\begin{gathered} \text{Moles of carbon =}\frac{\text{52.97 g}}{{\text{12 g mol}}^{\text{- 1}}} \\ \text{= 4.41mol} \end{gathered}$$

As 52.97% of the total percentage is carbon and the compound is made up of only carbon and oxygen, the remaining 47.03 g should be oxygen.

The mass of oxygen is 47.03 g.

The number of moles of oxygen is:

$$\begin{gathered} \text{Number of moles =}\frac{\text{47.03 g}}{{\text{16 g mol}}^{\text{- 1}}} \\ \text{= 2.94mol} \\ \backslash \mathrm{endgathered} \end{gathered}$$

The ratio of moles of carbon:oxygen is 4.41:2.94. This becomes equal to 1.5:1. The ratio can be rounded as C:O as 3:2.

Hence, the empirical formula of the compound D is ${\text{C}}_3{\text{O}}_2$.

The calculation for compound E can be given as:

The percentage mass of carbon in the compound E is 65.24%. Hence the sample of 100 g contains 65.24 g of carbon.

The calculation can be given as:

The mass of carbon is 65.24 g.

The number of moles of carbon is:

$$\begin{gathered} \text{Moles of carbon =}\frac{\text{65.24 g}}{{\text{12 g mol}}^{\text{- 1}}} \\ \text{= 5.44mol} \end{gathered}$$

As 65.24% of the total percentage is carbon and the compound is made up of only carbon and oxygen, the remaining 34.76 g should be oxygen.

The mass of oxygen is 34.76 g.

The number of moles of oxygen is:

$$\begin{gathered} \text{Number of moles =}\frac{\text{34.76 g}}{{\text{16 g mol}}^{\text{- 1}}} \\ \text{= 2.17mol} \end{gathered}$$

The ratio of moles of carbon: oxygen is 5.44:2.17. This becomes equal to 2.5:1. The ratio can be rounded as C:O as 5:2.

Hence, the empirical formula of the compound E is ${\text{C}}_5{\text{O}}_2$.

The empirical formula of the compounds can be given as:

Compound A-${\text{CO}}_{\text{2}}$ , Compound B-CO, Compound C- ${\text{C}}_{\text{4}}{\text{O}}_{\text{3}}$, Compound D-${\text{C}}_{\text{3}}{\text{O}}_{\text{2}}$ , Compound E-${\text{C}}_5{\text{O}}_2$ .