   Chapter 3, Problem 173CP

Chapter
Section
Textbook Problem

A compound contains only carbon, hydrogen, nitrogen, and oxygen. Combustion of 0.157 g of the compound produced 0.213 g CO2 and 0.0310 g H2O. In another experiment, it is found that 0.103 g of the compound produces 0.0230 g NH3. What is the empirical formula of the compound? Hint: Combustion involves reacting with excess O2. Assume that all the carbon ends up in CO2 and all the hydrogen ends up in H2O. Also assume that all the nitrogen ends up in the NH3 in the second experiment.

Interpretation Introduction

Interpretation: The given compound contains carbon, hydrogen, oxygen and nitrogen. The empirical formula is to be calculated for the given compound.

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 given compound.

Explanation

Given

Combustion of 0.157 g  given compound produces 0.213 g  of carbon dioxide (CO2)  and 0.0310 g  of water (H2O). The 0.103 g of given compound produces 0.0230 g of NH3.

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

The atomic mass of nitrogen (N)  is 14.006 g/mol

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

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

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

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

The molar mass of NH3  is 17.03 g/mol.

Formula

The mass of atom for the given compound is calculated using the formula,

Mass of atom = Mass of compound× Atomic mass of atomMolar mass of compound (1)

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

Mass of C = Mass of CO2× Atomic mass of CMolar mass of CO2

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

Mass of H  = Mass of H2O× Atomic mass of HMolar mass of H2O =0.0310 g× 2(1.008) g/mol18.015 g/mol =3.47×103 g

Substitute the values of mass, molar mass of NH3 and atomic mass of nitrogen in equation (1).

Mass of N  = Mass of NH3× Atomic mass of HMolar mass of NH3 =0.0230 g× 14.006 g/mol17.03 g/mol =0.0189 g

Since, 0.103 gof given compound contains 0.0186 g of nitrogen. Therefore, 0.157 g given compound contains,

157 g×0.0189 g0.103 g = 0.0288 g of N atom

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

0.157 g - (0.0581 g + 3.47×103 g + 0.0288 g) = 0.0667 g

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

Moles of atom = Given mass of atomAtomic mass (2)

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

Moles of C  Given mass of CAtomic mass =(0

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