Question

A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Balance the chemical equation

Answer 1

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Step-by-step explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂:

`1.039gCO_(2)*(1molCO_(2))/(44gCO_(2)) *(1molC)/(1molCO_(2)) *(12gC)/(1molC) =0.2834gC`

All of the hydrogens atoms in the compound ended up becoming H₂O, so the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O:

`0.6369g*(1molH_(2)O)/(18gH_(2)O) *(1molH)/(1molH_(2)O) *(1gH)/(1molH) =0.0354gH`

Because the compound is composed only by C, H and O, the mass of O in the compound can be calculated by substraction:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

• mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C

• mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H

• mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

• The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O