Explanation To determine : The empirical formula of cumene. Given Combustion of 47 .6 mg cumene produces some amount of carbon dioxide ( CO 2 ) and 42 .8 mg of water ( H 2 O ) . The atomic mass of carbon ( C ) is 12 .01 g/mol . The atomic mass of hydrogen ( H ) is 1 .008 g/mol . The molar mass of carbon dioxide ( CO 2 ) is 44 .008 g/mol . The molar mass of water ( H 2 O ) is 18 .015 g/mol . Formula The number of moles is calculated by using the formula, Number of Moles = Given mass Molar mass Substitute the values of mass and molar mass of water in above equation. Moles of H 2 O = Given mass of H 2 O Molar mass = ( 42 .8 mg 18 .015 g/mol ) = 2 .375 m-mol Since, one mole of H 2 O is equal to two moles of hydrogen, Therefore, 2 .375 m-mol of H 2 O is equal to, 2 × 2 .375 m-mol = 4 .75 m-mol of hydrogen ( H ) . The mass of atom is calculated by using the formula, Number of Moles = Mass of atom Molar mass Mass of atom = Number of Moles × Molar mass Substitute the values of number of moles and molar mass of hydrogen in above equation. Mass of atom = Number of Moles × Molar mass = 4 .75 m-mol × 1 .008 g/mol = 4 .788 mg The given compound is composed of carbon and hydrogen. A 47 .6 mg of the cumene contains 4 .788 mg of H . The mass of carbon is, 47 .6 mg − 4 .788 mg = 42 .812 mg The number of moles of atom calculated by using the formula, Moles of atom = Mass of atom Atomic mass Substitute the values of mass and molar mass of carbon in above equation Moles of C = Mass of C Atomic mass = 42

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Chapter 5, Problem 93E

Interpretation Introduction

Interpretation: The range of molar mass of cumene that contains only carbon and hydrogen is given. The empirical and molecular formula is to be calculated for cumene compound from the given data.

Concept introduction: The empirical formula is a formula which gives elemental composition of a compound. It is the smallest whole number ratio of atoms of each element.

To determine: The empirical and molecular formula of cumene.

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Chapter 5, Problem 92E

Chapter 5, Problem 94E

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