   # A compound contains only C, H, and N. Combustion of 35.0 mg of the compound produces 33.5 mg CO 2 and 41.1 mg H 2 O. What is the empirical formula of the compound? ### 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 92E
Textbook Problem
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## A compound contains only C, H, and N. Combustion of 35.0 mg of the compound produces 33.5 mg CO2 and 41.1 mg H2O. What is the empirical formula of the compound?

Interpretation Introduction

Interpretation: The mass of a compound and the products CO2 and H2O formed in the combustion reaction of that compound is given. By using these values, the empirical formula is to be calculated.

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 the given compound.

### Explanation of Solution

Given

Combustion of 35.0mg of given compound produces 33.5mg of carbon dioxide (CO2) and 41.1mg of water (H2O) .

The conversion of milligram (mg) into gram (g) is done as,

1mg=103g

Hence, the conversion of 35.0mg into gram is,

35.0mg=(35×103)g=35×103g

The conversion of 33.5mg into gram is,

33.5mg=(33.5×103)g=3.35×102g

The conversion of 41.1mg into gram is,

41.1mg=(41.1×103)g=4.11×102g

Total mass of given compound is 35×103g .

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

The atomic mass of nitrogen (N) is 14.006g/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 35×103g 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=3.35×102g×12.01g/mol44.008g/mol=9.14×103g

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

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