Step 1

Given:

Mass of sample = 1.000 g

Mass of CO₂ formed = 2.579 g

Mass of H₂O formed = 0.6536 g

Molar mass of CO₂ = 44.011 g/mol

Molar mass of H₂O = 18.02 g/mol

Molar mass of carbon = 12.011 g/mol

Molar mass of hydrogen = 1.016 g/mol

Molar mass of oxygen = 16.00 g/mol

Step 2

1 mole of CO₂ contains 1 mol of carbon. Therefore, the mass of carbon is calculated as:

$Mass of C = 2.579 g {CO}_{2} \times \frac{1 mol {CO}_{2}}{44.011 g {CO}_{2}} \times \frac{1 mol C}{1 mol {CO}_{2}} \times \frac{12.011 g C}{1 mol C} = 0.7038 g$

1 mole of H₂O contains 2 mol of hydrogen. Therefore, the mass of hydrogen is calculated as:

$Mass of H = 0.6536 g H_{2} O \times \frac{1 mol H_{2} O}{18.02 g H_{2} O} \times \frac{2 mol H}{1 mol H_{2} O} \times \frac{1.016 g H}{1 mol H} = 0.0737 g$

Mass of oxygen is calculated as:

$Mass of oxygen = 1.000 g - (0.7038 g + 0.0737 g) = 0.2225 g$

Step 3

The mole of carbon, hydrogen and oxygen present in the sample is calculated as:

$Mol of C = 0.7038 g C \times \frac{1 mol C}{12.011 g C} = 0.0586 mol Mol of H = 0.0737 g H \times \frac{1 mol H}{1.016 g H} = 0.0725 mol Mol of O = 0.2225 g O \times \frac{1 mol O}{16.00 g O} = 0.0139 mol$

Divide the mole from the smallest mole value to find out the mole ratio of each element present in the compound.

$Mol of C : Mol of H : Mol of O \frac{0.0586 mol}{0.0139 mol} : \frac{0.0725 mol}{0.0139 mol} : \frac{0.0139 mol}{0.0139 mol} 4 : 5 : 1$

Therefore, the empirical formula of compound is C₄H₅O.

A 1.000 g sample of an unknown compound containing only carbon, hydrogen, and oxygen was analyzed through combustion analysis and yielded 2.579 g of CO₂ and 0.6536 g of H₂O. What is the compound's empirical formula?

Public Answer