   # An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of −5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte. ### 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 10, Problem 103AE
Textbook Problem
84 views

## An unknown compound contains only carbon, hydrogen, and oxygen. Combustion analysis of the compound gives mass percents of 31.57% C and 5.30% H. The molar mass is determined by measuring the freezing-point depression of an aqueous solution. A freezing point of −5.20°C is recorded for a solution made by dissolving 10.56 g of the compound in 25.0 g water. Determine the empirical formula, molar mass, and molecular formula of the compound. Assume that the compound is a nonelectrolyte.

Interpretation Introduction

Interpretation: The molar mass, empirical formula and molecular formula of the unknown compound has to be determined.

Concept Introduction:

The ratio of the elements present in a compound and not the arrangement of the atoms are called as Empirical formula

Molar mass of a compound can be given by mass of a substance to the amount of the substance. It can be given by expression,

Molarmass=MassofsubstanceAmountofsubstance

Molecular formula is the representation of sum of number of atoms and molecules, not their arrangement in structure.

### Explanation of Solution

Record the given info

Mass percent of Carbon     = 31.57%

Mass percent of Hydrogen = 5.30%

Mass of the compound      = 10.56g

Mass of Water                  = 25.0g

Freezing point depression = -5.20°C

To calculate the moles of Carbon, Hydrogen and Oxygen.

Molecular Weight of Carbon = 12.01g

Molecular Weight of Hydrogen = 1.008g

Molecular weight of Oxygen = 16.00g

Moles of Carbon = 31.57g1molC12.01g=2.629moles

Moles of Hydrogen = 5.301molH1.008g=5.26moles

Moles of Oxygen = 63.131molO16.00g=3.946moles

Moles of Carbon = 2.629moles

Moles of Hydrogen = 5.26moles

Moles of Oxygen = 3.946moles

To write the empirical formula of the Unknown compound

Moles of Carbon = 2.629moles

Moles of Hydrogen = 5.26moles

Moles of Oxygen = 3.946moles

In order, to get empirical formula of the Unknown compound, the moles of individual elements is divided by the smallest number of moles

Empirical formula of the compound = C = 2.6292.629=1.000

H = 5.262.629=2.00

O = 3.9462.629=1.501

The empirical formula of the compound = C2H4O3

Record the given info

Mass percent of Carbon     = 31.57%

Mass percent of Hydrogen = 5.30%

Mass of the compound      = 10.56g

Mass of Water                  = 25

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