   Chapter 3, Problem 96E

Chapter
Section
Textbook Problem

A compound contains only carbon, hydrogen, and oxygen. Combustion of 10.68 mg of the compound yields 16.01 mg CO2 and 4.37 mg H2O. The molar mass of the compound is 176.1 g/mol. What are the empirical and molecular formulas of the compound?

Interpretation Introduction

Interpretation: The molar mass of the given compound that contains carbon, hydrogen and oxygen. The empirical and molecular formula is to be calculated of given 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 given compound.

Explanation

To determine: The empirical formula of given compound.

Given

Combustion of 10.68mg given compound produces 16.01mg of carbon dioxide (CO2) and 4.37mg of water (H2O) .

Total mass of given compound is 10.68mg .

The atomic mass of carbon (C) is 12.01g/mol .

The atomic mass of oxygen (O) is 15.999g/mol .

The atomic mass of hydrogen (H) is 1.008g/mol .

The molar mass of carbon dioxide (CO2) is 44.008g/mol .

The molar mass of water (H2O) is 18.015g/mol .

Formula

The mass of atom in 10.68×103g of given compound is calculated using the formula,

Massofatom=Massofcompound×AtomicmassofatomMolarmassofcompound (1)

Substitute the values of mass, molar mass of carbon dioxide and atomic mass of carbon in above equation.

MassofC=MassofCO2×AtomicmassofCMolarmassofCO2=16.01mg×12.01g/mol44.008g/mol=4.37mg

Substitute the values of mass, molar mass of water and atomic mass of hydrogen in equation (1).

MassofH=MassofH2O×AtomicmassofHMolarmassofH2O=4.37mg×2(1.008)g/mol18.015g/mol=0.489mg

Since, the given compound is composed of carbon, hydrogen and oxygen. Therefore, mass of oxygen is,

10.68mg(4.37mg+0.489mg)=5.821mg

The number of moles in each element is calculated by using the formula,

Molesofatom=GivenmassofatomAtomicmass (2)

Substitute the values of mass and atomic mass of carbon in above equation.

MolesofC=GivenmassofCAtomicmass=(4.37mg12

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