   # One of the components that make up common table sugar is fructose, a compound that contains only carbon, hydrogen, and oxygen. Complete combustion of 1.50 g of fructose produced 2.20 g of carbon dioxide and 0.900 g of water. What is the empirical formula of fructose? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 5, Problem 91E
Textbook Problem
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## One of the components that make up common table sugar is fructose, a compound that contains only carbon, hydrogen, and oxygen. Complete combustion of 1.50 g of fructose produced 2.20 g of carbon dioxide and 0.900 g of water. What is the empirical formula of fructose?

Interpretation Introduction

Interpretation: The empirical formula is to be calculated for fructose 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 formula of fructose.

### Explanation of Solution

Given

Combustion of 1.50g fructose produces 2.20g of carbon dioxide (CO2) and 0.900g of water (H2O) .

Total mass of fructose is 1.50g .

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 1.50g of fructose 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=2.20g×12.01g/mol44.008g/mol=0.6004g

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

MassofH=MassofH2O×AtomicmassofHMolarmassofH2O=0.900g×2(1.008)g/mol18.015g/mol=0

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