Question

7. A certain hydrocarbon, CxHy, is burned (reacts with O2 gas) and produces 1.955 g of CO2 for every

1.000 g of H20. What is the empirical formula of this hydrocarbon?
CxHy + (X + y/4) O2 -> xCO2 + y/2 H2O

Answer 1

The empirical formula of the hydrocarbon is C₂H₅

Step-by-step explanation:

The formula for the hydrocarbon is
`C_xH_y`

The mass of CO₂ produced per 1,000 g of H₂O = 1,955 g

The equation for the chemical reaction is given as follows;

`C_xH_y` + (x + y/4) O₂ → XCO₂ + y/2H₂O

From the given chemical equation, counting the number of atoms on both sides of the equation, we have;

The molar mass of CO₂ = 44.01 g/mol

The molar mass of H₂O = 18.01528 g/mol

The number of moles of H₂O in 1,000 g of H₂O = 1,000 g/(18.01528 g/mol) = 55.5084351 moles

The number of moles of CO₂ in 1,955 g of H₂O = 1,955 g/(44.01 g/mol) = 44.4217223 moles

Therefore, given that X moles of CO₂ is produced alongside Y/2 moles of H₂O. we have;

X = 44.4217223, Y/2 = 55.5084351

∴ Y = 2 × 55.5084351 = 111.0168702

The ratio of X to Y = X/Y = 44.4217223/111.0168702 = 0.40013488238

∴ The ratio of X to Y = X/Y ≈ 0.4 = 4/10

X/Y ≈ 4/10

The empirical formula is the representation of molecular formula in the smallest whole number ratio of the elements of the molecules

Therefore, when X = 4, Y = 10, from which we have the smallest ratio as;

When X = 2, Y = 5

The empirical formula of the hydrocarbon is therefore,
`C_xH_y` = C₂H₅

The given chemical equation becomes;

C₂H₅ + (2 + 5/4) O₂ → 2CO₂ + 5/2H₂O

C₂H₅ + 3.25 O₂ → 2CO₂ + 2.5 H₂O

We then have;

4C₂H₅ + 13 O₂ → 8CO₂ + 10 H₂O

The empirical formula of the hydrocarbon,
`C_xH_y` = C₂H₅.