Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Textbook Question
Chapter 8, Problem 139AE

An organic compound contains C, H, N, and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2766 g CO2 and 0.0991 g H2O. A sample of 0.4831 g of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129). At STP, 27.6 mL of dry N2 was obtained. In a third experiment, the density of the compound as a gas was found to be 4.02 g/L at 127°C and 256 torr. What are the empirical and molecular formulas of the compound?

Expert Solution & Answer
Check Mark
Interpretation Introduction

Interpretation: From the given data, molecular formula and molecular formula of the compound should be determined.

Concept introduction:

  • Empirical formula can be determined from the mass per cent as given below
    1. 1. Convert the mass per cent to gram
    2. 2. Determine the number of moles of each elements
    3. 3. Divide the mole value obtained by smallest mole value
    4. 4. Write empirical formula by mentioning the numbers after writing the symbols of respective elements
  • Molecular formula can be write by the following steps
    1. 1. Determine the empirical formula mass
    2. 2. Divide the molar mass by empirical formula mass

      MolarmassEmpiricalformula=n

    3. 3. Multiply the empirical formula by n
  • Mass%ofanelement=MassofthatelementinthecompoundMolarmassofthecompound×100
  • According to ideal gas equation,

    Pressure×Volume=Mass×R×TemperatureMolecularmass

  • Ideal gas equation in terms of Density,

Density=Pressure×Molecular massR×Temperature

Answer to Problem 139AE

Answer

Molecularformulaofthecompound=C24H42N2O2

Explanation of Solution

Explanation

  • To determine: The mass percent from the given data

73.78%Carbon10.9%Hydrogen7.14%Nitrogen8.2% Oxygen

Mass % of carbon

Since,completecombustionof0.1023gofthecompoundproduced0.2766mgCO2and0.0991gH2ONumber ofmolesofcarbon dioxide =GivenmassMolecularmassSince,0.1023gof thecompoundis combusted to produce0.2766gCO2Givenmass=0.2766gMolecularmassofCO2=44.01gNumber ofmolesofcarbon dioxide = 0.2766gCO244.01gCO2=0.00628molEverymoleofCO2willcontains1moleofcarbonand2molesofoxygen.Since,everymoleofCO2containsonemoleofcarbonNumberofmolesofcarbon=0.00628molMassofcarboninthecompound=Number ofmolesofcarbon×atomic massofcarbonMassofcarboninthecompound=0.00628mol×12.01gC=0.0757mgC =7.548 ×10-2gCMass%ofanelement=MassofthatelementinthecompoundMolarmassofthecompound×100Since,0.1023 gof thecompoundis combusted to produce0.2766gCO2Molarmassofthecompound=0.1023gMass%ofcarbon=7.548 ×10-2gC0.1023g×100=73.78%C

Mass % of Hydrogen

Since,completecombustionof0.1023mgofthecompoundproduced0.2766gCO2and0.0991gH2ONumber ofmolesofwater =GivenmassMolecularmassSince,0.1023gof thecompoundis combusted to produce0.0991gH2OGivenmass=0.0991gMolecularmassofH2O= 18.02 mgNumber ofmolesofwater = 0.0991g18.02g=0.00550molEverymoleofH2Ocontains2molesofHydrogenand1moleofoxygen.NumberofmolesofHydrogen= 2×0.005500.011molMassofHydrogeninthecompound=Number ofmolesofHydrogen×AtomicmassofHydrogenMassofHydrogeninthecompound=0.011mol×1.008mgH=0.011gHMass%ofanelement=MassofthatelementinthecompoundMolarmassofthecompound×100Since,0.1023gof thecompoundis combusted to produce0.0991gH2OMolarmassofthecompound=0.1023gMass%ofHydrogen=0.011gH0.1023g×100=10.9%H

Mass % of Nitrogen

The pressure of N2 is1.0 atm

The volume of N2 gas is 27.6mL=27.6×10-3L

The temperature of N2 gas is 273K.

The equation for finding number of moles of a substance,

According to ideal gas equation,

Numberofmoles=Pressure×VolumeR×Temperature

Numberofmoles=1.0atm×27.6×10-3L0.08206Latm/Kmol×273K =1.23×10-3mol

Numberofmolesof Nitroogen=1.23×10-3molMassof Nitrogenin thecompound=Number ofmolesofNitrogen×MolecularmassofnitrogenMassofNitrogeninthecompound=1.23×10-3mol×28.02g=3.45×10-2gMass%ofanelement=MassofthatelementinthecompoundMolarmassofthecompound×100Molarmassofthecompound=0.4831gMass%ofNitrogen=3.45×10-2g0.4831g×100=7.14%NMass%ofoxgen=100.0-(73.78+10.9+7.14)=8.2%O

  • To convert: The mass percent to gram

73.78%Carbon= 73.78g10.9%Hydrogen =10.9 g7.14%Nitrogen = 7.14g8.2% Oxygen=8.2 g

First step to determine the empirical formula is to convert the mass percent to gram.

Here the given gasses are carbon, hydrogen

Percentage composition of carbon, hydrogen, nitrogen and oxygen out of 100 g of compound

Are73.78%, 10.9%, 7.14%and 8.2%   respectively.

  • To determine: the number of moles of each element

Numberofmolesofcarbon=6.143molNumberofmolesofhydrogen=10.8molNumberofmolesofnitrogen=0.510molNumberofmolesofoxygen=0.51mol

 Numberofmoles from its given mass is,Numberofmoles=GivenmassingramMolecularmassMassesofcarbon,nitrgen,hydrgrogen and oxygenaregivenabove.MolecularmassesofCarbon=73.78gNitogen=7.14gHydrogen=10.9gOxygen =8.2 gNumberofmolesofcarbon=73.78g12.01g=6.143molNumberofmolesofhydrogen=10.9g1.008g=10.8molNumberofmolesofnitrogen=7.14g14.01g=0.510molNumberofmolesofoxygen=8.2g16.00=0.51mol                      

  • To divide:  the mole value obtained by smallest mole value

Inthecaseofcarbon,Ratioofmolevaluetosmallestmolevalue=12.00Inthecaseofhydrogen,Ratioofmolevaluetosmallestmolevalue=21.00Inthecaseofnitrogen,Ratioofmolevaluetosmallestmolevalue=1.00Inthecaseofoxygen,Ratioofmolevaluetosmallestmolevalue=1.00

From the mole values, smallest mole value is 0.51

Numberofmolesofcarbon=6.143molNumberofmolesofhydrogen=10.8molNumberofmolesofnitrogen=0.510molNumberofmolesofoxygen=0.51molInthecaseofcarbon,Ratioofmolevaluetosmallestmolevalue=6.1430.51=12.00Inthecaseofhydrogen,Ratioofmolevaluetosmallestmolevalue10.80.51=21.00Inthecaseofnitrogen,Ratioofmolevaluetosmallestmolevalue=0.510.51=1.00Inthecaseofoxygen,Ratioofmolevaluetosmallestmolevalue=0.510.51=1.00

  • To write: the empirical formula by mentioning the numbers after writing the symbols of respective elements

The empirical formula is C12H21NO

Empirical formula can be determined by dividing the mole value of each gas to smallest mole value gas.

RatioofmolevaluetosmallestmolevalueofeachgasesisC:H:N:O=12:21:1:1

So, the empirical formula is C12H21NO

  • To find: molecular formula of the given compound

Determine the molar mass of the given compound.

Molar mass of the given compound is =392g/mol

Ideal gasequation in terms of Density,                                                  Density=Pressure×Molecular massR×TemperatureMolecular mass=Density×R×TemperaturePressureHere,Density=4.02g/LR=0.08206LatmKmolAtSTPT=127°C= 400K Since,°C+273=K,127°C+273=400KP=256torr=0.33atmSince,1atm=760torr256torr×1atm760torr=0.33atmByaddingthegivenvaluestotheequation,Molecular mass=4.02g/L×0.08206LatmKmol×400K0.33atm=392g/mol

  • To determine: the empirical formula mass

Empirical formula mass =195g/mol

EmpiricalformulaisC12H21NO.TheempiricalformulamassofC12H21NO  12×12+ 1×21+1×14+1×16 = 195g/mol

  • To divide: the molar mass by empirical formula

nMolarmassEmpiricalformula=2

n=MolarmassEmpiricalformulaMolar mass of the given compound is=392g/molEmpirical formula mass=195g/moln=392g/mol195g/mol=2

  • To multiply: the empirical formula by n

By multiplying the empirical formula by n,

molecularformulaofthecompound=C24H42N2O2

By multiplying the empirical formula by n molecular formula of the compound should obtain.

n=2andempiricalofthecompound=C12H21NO(C12H21NO)×2=C24H42N2O2So,themolecularformulaofthecompoundisC24H42N2O2

Conclusion

Conclusion

Empirical formula of the compound is determined from the mass per cent as given below

1. Converted the mass per cent to gram

2. Determined the number of moles of each element

3. Divided the mole value obtained by smallest mole value

4. Wrote empirical formula by mentioning the numbers after writing the symbols of respective elements

Molecular formula of the given compound is determined by the following steps

1. Determined the empirical formula mass

2. Determined the molar mass from rate effusion

3. Divided the molar mass by empirical formula mass

MolarmassEmpiricalformula=n

  1. 4. Multiplied the empirical formula by n

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Chapter 8 Solutions

Chemistry: An Atoms First Approach

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