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

The empirical formula of the hydrocarbon is C₂H₅

Explanation:

The formula for the hydrocarbon is

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

The equation for the chemical reaction is given as follows;

+ (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, = 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, = C₂H₅.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.