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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

Benzoquinone, a chemical used in the dye industry and in photography, is an organic compound containing only C, H, and O. What is the empirical formula of the compound if 0.105 g of the compound gives 0.257 g of CO2 and 0.0350 g of H2O when burned completely in oxygen?

Interpretation Introduction

Interpretation:

The empirical formula of benzoquinone has to be determined.

Concept introduction:

  • Empirical formula of a compound represents the smallest whole number relative ratio of elements in that compound.
  • Equation for finding Molecular formula from the empirical formula,

  MolarmassEmpiricalformula mass × Empirical formula

  • The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
Explanation

Given that 0.105g of benzoquinone burns in oxygen then give 0.257g of CO2 and 0.0350g of H2O

From the given masses the amount of CO2 and H2O can be calculated.

Thus, amount of CO2 and H2O are,

  0.257gCO2×1molCO244.010gCO2=0.00583molCO20.0350gH2O×1molH2O18.015gH2O=0.00194molH2O

For every mole of CO2 isolated, 1 mol of Carbon must have been present in the benzoquinone.

Hence,

  0.00583molCO2×1 mol C in menthol1 mol CO2=0.00583molC

For every mole of H2O isolated, 2 mol of Hydrogen must have been present in the unknown compound.

Therefore,

  0.00194molH2O×2 mol H in menthol1 mol H2O=0.00388molH

The mass of the substance can be determined by multiplying the amount of substance with its molar mass.

The mass of carbon and Hydrogen are 0.0699gand0.0039g respectively

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