A gaseous hydrocarbon gives upon combustion 0.72 gram of water in 3.08 gram of CO2 the empirical formula of the hydrocarbon is
Correct answer: C. C7H8
- A. C3H4
- B. C6H5
- C. C7H8
- D. C2H4
Explanation
Mr of H2O= 18g/molMr of CO2= 44g/molMole of H2O= 0.72/18 = 0.04molMole of CO2= 3.08/44 = 0.07molFor a balanced chemical equation for combustion of any hydrocarbon the number of moles of CO2 is equal to number of carbon atoms in the hydrocarbon and the half number of moles of H2O is equal to hydrogen atoms in the hydrocarbon.
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Mole calculations connect mass, number of particles and Avogadro's number, while balanced equations provide the mole ratios used in stoichiometry. Questions cover limiting and excess reactants, theoretical yield and percentage yield, including identifying which reactant is consumed first and comparing the actual product with the maximum possible product.
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