Chemistry
Chemistry
3rd Edition
ISBN: 9780073402734
Author: Julia Burdge
Publisher: MCGRAW-HILL HIGHER EDUCATION
Question
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Chapter 3, Problem 154AP
Interpretation Introduction

Interpretation:

The empirical formula of the compound is to be calculated.

Concept introduction:

The number of moles is defined as the ratio of mass to the molar mass:

n=mM

Here, n is the number of moles, m is the mass, and M is the molar mass.

Molar mass is calculated by adding the masses of each element multiplied by their number of atoms present (given in subscript). Its S.I. unit is g/mol.

The mass of compound can be calculated as m=n×M.

Here, n is the number of moles, m is the mass, and M is the molar mass.

the Mass of carbon can be calculated as mC=nCO2×MC×nC.

The mass of hydrogen can be calculated as mH=nH2O×MH×nH.

Expert Solution & Answer
Check Mark

Answer to Problem 154AP

Solution: The empirical formula of the first compound is PbC8H20 and for thesecond compound is C5H12O

Explanation of Solution

Given information:

For gasoline

mCO2=55.9 gmH2O=28.61 g

For methyl tert-butyl ether

mCO2=30.2 gmH2O=14.8 g

The molar mass of H2O is 18 g/mol

Calculate the number of moles of water as

nH2O=mH2OMH2O

Substitute 28.61 g for mH2O and 18 g/mol for MH2O in the equation above

nH2O=28.61 g18 g/mol=1.589 mol

The molar mass of CO2 is 44 g/mol.

Calculate the number of moles of carbon dioxide as

nCO2=mCO2MCO2

Substitute 55.9 g for mCO2 and 44 g/mol for MCO2 in the equation above

nCO2=55.9 g44 g/mol=1.27 mol

The molar mass of C is 12 g/mol.

For carbon in CO2, the number of moles is one.

Calculate the mass of carbon as

mC=nCO2×MC×nC

Substitute 1.27 mol for nCO2, 12 g/mol for MC, and 1 for nC in the equation above

mC=1.27 mol×12 g/mol×1=15.25 g

The molar mass of H is 1 g/mol.

For hydrogen in H2O, the number of moles is two.

Calculate the mass of hydrogen as

mH=nH2O×MH×nH

Substitute 1.589 mol for nH2O, 1 g/mol for MH, and 2 for nH in the equation above

mH=1.589 mol×1 g/mol×2=3.178 g

The mass of the given sample is 51.36 g.

Calculate the mass of lead as

51.36 g=mass C+mass H+massPb

Substitute 3.178 g for mass H and 15.25 g for mass C in the equation above

51.36 g = 15.25 g+3.201 g+mass Pb=18.451 g+mass Pb

Rearrange the above equation for mass Pb as

Mass Pb=51.36 g18.451 gMass Pb = 32.91 g

Now, the number of moles of each element is to be calculated as

The molar mass of C is 12 g/mol.

Calculate the number of moles of carbon as

nC=mCMC

Substitute 15.25 g for mC and 12 g/mol for MC in the equation above

nC=15.25 g12 g/mol=1.27 mol

The molar mass of hydrogen is 1 g/mol.

Calculate the mass of hydrogen as

nH=mHMH

Substitute 3.178 g for mH and 1 g/mol for MH in the equation above

nH=3.178 g1 g/mol=3.17 mol

The molar mass of lead is 207.2 g/mol.

Calculate the mass of lead as

nPb=mPbMPb

Substitute 32.91 g for mPb and 207.2 g/mol for MPb in the equation above

nPb=32.91 g207.2 g/mol=0.1588 mol

So, the molecular formula of the compound can be written as

Pb0.1588C1.270H3.176

As the empirical formula are always has whole number multiples, on simplifying the number of moles of each element, it is calculated as

For H

nH=3.1760.158820

For C

nC=1.2700.15888

For Pb

nPb=0.15880.1588=1

So, the empirical formula of the compound is PbC8H20.

Now, for tert-butyl ether:

The molar mass of H2O is 18 g/mol

Calculate the number of moles of water as

nH2O=mH2OMH2O

Substitute 14.8 g for mH2O and 18 g/mol for MH2O in the equation above

nH2O=14.8 g18 g/mol=0.822 mol

The molar mass of CO2 is 44 g/mol.

Calculate the number of moles of carbon dioxide as

nCO2=mCO2MCO2

Substitute 30.2 g for mCO2 and 44 g/mol for MCO2 in the equation above

nCO2=30.2 g44 g/mol=0.686 mol

The molar mass of C is 12 g/mol.

For carbon in CO2, the number of moles is one.

Calculate the mass of carbon as

mC=nCO2×MC×nC

Substitute 0.686 mol for nCO2, 12 g/mol for MC, and 1 for nC in the equation above

mC=0.686 mol×12 g/mol×1=8.24 g

The molar mass of H is 1 g/mol.

For hydrogen in H2O, the number of moles is two.

Calculate the mass of hydrogen as

mH=nH2O×MH×nH

Substitute 0.822 mol for nH2O, 1 g/mol for MH, and 2 for nH in the equation above

mH=0.822 mol×1 g/mol×2=1.6 g

The mass of the given sample is 12.1 g.

Calculate the mass of oxygen as

mass C+mass H+mass O=12.1 g

Substitute 1.6 g for mass H and 8.24 g for mass H in the equation above

12.1 g=  8.24 g+1.6 g+mass O=9.9 g+mass O

Rearrange the above equation for mass O as

mass O=12.1 g9.9 gmass O = 2.2 g

Now, the number of moles of each element is to be calculated as

The molar mass of carbon is 12 g/mol.

Calculate the mass of carbon as

nC=mCMC

Substitute 8.24 g for mC and 12 g/mol for MC in the above equation

nC=8.24 g12 g/mol=0.686 mol

The molar mass of hydrogen is 1 g/mol.

Calculate the mass of hydrogen as

nH=mHMH

Substitute 1.6 g for mH and 1 g/mol for MH in the equation above

nH=1.6 g1 g/mol=1.6 mol

The molar mass of oxygen is 16 g/mol.

Calculate the mass of oxygen as

nO=mOMO

Substitute 2.2 g for mO and 16.00 g/mol for MO in the equation above

nO=2.2 g16 g/mol=0.14 mol

So, the molecular formula of the compound is C0.686H1.65O0.14.

As the empirical formula are always whole number multiples, on simplifying the number of moles of each element, it is calculated as

For H

nH=1.650.1412

For C

nC=0.6860.145

For O

nPb=0.140.14=1

So, the empirical formula of the compound is C5H12O.

Conclusion

The empirical formula of the first compound is PbC8H20 and for the second compound is C5H12O.

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Chapter 3 Solutions

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