The empirical formula of the compound is CH₃
The empirical formula is determined as follows:
Given the general equation of the combustion of hydrocarbons as follows:
CxHy + O₂ → x CO₂ + y H₂O
mass of CO₂ produced = 3.54 g
molar mass of CO₂ = 44 g/mol
moles of C in the sample = 3.54/44 = 0.08 moles
molar mass of C = 12 g/mol
mass of C in the sample = 0.08 * 12 = 0.96 g
mass of H₂O produced = 1.09 g
molar mass of water = 18 g/mol
moles of H in the sample = 1.09 /18 * 2
moles of H in the sample = 0.12 moles
Ratio of C : H = 0.08 : 0.12
Ratio of C : H = 1 : 3
Therefore, the empirical formula of compound = CH₃
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