Question

A sample of a compound containing only carbon and oxygen decomposes and produces 24.50g of carbon and 32.59g of oxygen. what is the sample?

Answer 1

Answer : The sample is, CO (carbon monoxide)

Solution : Given,

Mass of C = 24.50 g

Mass of O = 32.59 g

Molar mass of C = 12 g/mole

Molar mass of O = 16 g/mole

First convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{24.50g}{12g/mole}=2.04moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{32.59g}{16g/mole}=2.04moles$

Now for the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{2.04}{2.04}=1$

For O = $\frac{2.04}{2.04}=1$

The ratio of C : O = 1 : 1

The mole ratio of the element is represented by subscripts in empirical formula.

The Empirical formula = $C_1O_1$ or, $CO$

Therefore, the sample is, CO (carbon monoxide)

Answer 2

Carbon Monoxide. First, determine the relative number of moles of each element by looking up the atomic weights of carbon and oxygen Atomic weight carbon = 12.0107 Atomic weight oxygen = 15.999 Moles of Carbon = 24.50 g / 12.0107 g/mol = 2.039847802 mol Moles of Oxygen = 32.59 g / 15.999 g/mol = 2.037002313 mol Given that the number of moles of both carbon and oxygen are nearly identical, it wouldn't be unreasonable to think that the empirical formula for the compound is CO which also happens to be the formula for Carbon Monoxide.