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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Chapter 8, Problem 137AE
Interpretation Introduction

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

Concept introduction:

  • According to Thomas Graham the rate of effusion of a gas is inversely proportional to the square root of the mass of its particles.

    The relative rate of effusion of two gases at the same temperature and pressure are inverse ratio of the square root of the masses of the gases particles. That is,

Rateofeffusionforgas1Rateofeffusionforgas2=M2M1orRate1Rate2=(M2M1)1/2

M1andM2arethemolarmassesofgas1andgas2

This equation is known as Graham’s law of effusion.

  • 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

Expert Solution & Answer
Check Mark

Answer to Problem 137AE

Answer

Empiricalformulaandmolecularformulaofthecompound=C2H3N

Explanation of Solution

Explanation

  • To find: the empirical formula of the compound

Convert the mass percent to gram

58.51%Carbon=58.51gC7.37%Hydrogen=7.37gH34.12%Nitrogen=34.12gN

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

Here the given gasses are carbon, hydrogen, nitrogen

Percentage composition of carbon and hydrogen are, 58.51% and 7.37% respectively.

Out of 100 g of compound, percent composition of Nitrogen is 34.12%

  • To determine: the number of moles of each element

Numberofmolesofcarbon=4.872molNumberofmolesofhydrogen=7.31molNumberofmolesofnitrogen=2.435mol

 Numberofmoles from its given mass is,Numberofmoles=GivenmassingramMolecularmassMassesofcarbon,nitrgen,andhydrgrogenaregivenabove.MolecularmassesofCarbon=12.01gNitogen=14.01gHydrogen=1.008gNumberofmolesofcarbon=58.51g12.01g=4.872molNumberofmolesofhydrogen=7.37g1.008g=7.31molNumberofmolesofnitrogen=34.12g14.01g=2.435mol                      

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

Inthecaseofcarbon,Ratioofmolevaluetosmallestmolevalue=2.001Inthecaseofhydrogen,Ratioofmolevaluetosmallestmolevalue=3.00Inthecaseofnitrogen,Ratioofmolevaluetosmallestmolevalue=1.00

From the mole values, smallest mole value is 2.435

Numberofmolesofcarbon=4.872molNumberofmolesofhydrogen=7.31molNumberofmolesofnitrogen=2.435molInthecaseofcarbon,Ratioofmolevaluetosmallestmolevalue=4.8722.435=2.001Inthecaseofhydrogen,Ratioofmolevaluetosmallestmolevalue7.312.435=3.00Inthecaseofnitrogen,Ratioofmolevaluetosmallestmolevalue=2.4352.435=1.00

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

The empirical formula is C2H3N

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

RatioofmolevaluetosmallestmolevalueofeachgasesisC:H:N=2:3:1

So, the empirical formula is C2H3N

  • To find: molecular formula of the given compound

Determine the molar mass of the given compound.

Molar mass of the given compound is =41.0g/mol

AccordingtoThomasGraham,Rateofeffusionforgas1Rateofeffusionforgas2=M2M1orRate1Rate2=(M2M1)1/2M1andM2arethemolarmassesofgas1andgas2Letgas1=HeliumM1=4.003Heliumeffusesthroughaporus3.20timesasfastasthecompounddoesSo,3.20=(M24.003)1/2M2=(3.20)2×4.003=10.24×4.003=41.0g/mol

  • To determine: the empirical formula mass

Empirical formula mass =41.0g/mol

EmpiricalformulaisC2H3N.TheempiricalformulamassofC2H3 2×12+ 3×1+1×14 = 41.0g/mol

  • To divide: the molar mass by empirical formula

nMolarmassEmpiricalformula=1

n=MolarmassEmpiricalformulaMolar mass of the given compound is=41.0g/molEmpirical formula mass=41.0g/moln=41.0g/mol41.0g/mol=1

  • To multiply: the empirical formula by n

By multiplying the empirical formula by n,

molecularformulaofthecompound=C2H3N

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

n=1andempiricalofthecompound=C2H3NC2H3N×1=C2H3NSo,themolecularformulaofthecompoundisC2H3N

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

Ch. 8 - Prob. 3ALQCh. 8 - Prob. 4ALQCh. 8 - Prob. 6ALQCh. 8 - Prob. 8ALQCh. 8 - Prob. 11ALQCh. 8 - Prob. 12ALQCh. 8 - Prob. 15ALQCh. 8 - Prob. 16ALQCh. 8 - Draw molecular-level views that show the...Ch. 8 - Prob. 20QCh. 8 - Prob. 21QCh. 8 - Prob. 22QCh. 8 - Prob. 23QCh. 8 - Prob. 24QCh. 8 - Prob. 25QCh. 8 - Consider two different containers, each filled...Ch. 8 - Prob. 27QCh. 8 - Prob. 28QCh. 8 - Prob. 29QCh. 8 - Prob. 30QCh. 8 - Prob. 31QCh. 8 - Prob. 32QCh. 8 - Prob. 33QCh. 8 - Prob. 34QCh. 8 - Prob. 35QCh. 8 - Prob. 36QCh. 8 - Prob. 37ECh. 8 - Prob. 38ECh. 8 - A sealed-tube manometer (as shown below) can be...Ch. 8 - Prob. 40ECh. 8 - A diagram for an open-tube manometer is shown...Ch. 8 - Prob. 42ECh. 8 - Prob. 43ECh. 8 - Prob. 44ECh. 8 - Prob. 45ECh. 8 - Prob. 46ECh. 8 - Prob. 47ECh. 8 - Prob. 48ECh. 8 - Prob. 49ECh. 8 - Prob. 50ECh. 8 - The Steel reaction vessel of a bomb calorimeter,...Ch. 8 - A 5.0-L flask contains 0.60 g O2 at a temperature...Ch. 8 - Prob. 53ECh. 8 - A person accidentally swallows a drop of liquid...Ch. 8 - A gas sample containing 1.50 moles at 25C exerts a...Ch. 8 - Prob. 56ECh. 8 - Prob. 57ECh. 8 - What will be the effect on the volume of an ideal...Ch. 8 - Prob. 59ECh. 8 - Prob. 60ECh. 8 - An ideal gas is contained in a cylinder with a...Ch. 8 - Prob. 62ECh. 8 - A sealed balloon is filled with 1.00 L helium at...Ch. 8 - Prob. 64ECh. 8 - Consider the following reaction:...Ch. 8 - A student adds 4.00 g of dry ice (solid CO2) to an...Ch. 8 - Air bags are activated when a severe impact causes...Ch. 8 - Concentrated hydrogen peroxide solutions are...Ch. 8 - In 1897 the Swedish explorer Andre tried to reach...Ch. 8 - Sulfur trioxide, SO3, is produced in enormous...Ch. 8 - A 15.0-L rigid container was charged with 0.500...Ch. 8 - An important process for the production of...Ch. 8 - Consider the reaction between 50.0 mL liquid...Ch. 8 - Urea (H2NCONH2) is used extensively as a nitrogen...Ch. 8 - Prob. 75ECh. 8 - Prob. 76ECh. 8 - Prob. 77ECh. 8 - A compound has the empirical formula CHCl. A...Ch. 8 - Prob. 79ECh. 8 - Given that a sample of air is made up of nitrogen,...Ch. 8 - Prob. 81ECh. 8 - Prob. 82ECh. 8 - Prob. 83ECh. 8 - Prob. 84ECh. 8 - Consider the flasks in the following diagram. What...Ch. 8 - Consider the flask apparatus in Exercise 85, which...Ch. 8 - Prob. 87ECh. 8 - At 0C a 1.0-L flask contains 5.0 102 mole of N2,...Ch. 8 - Prob. 89ECh. 8 - A tank contains a mixture of 52.5 g oxygen gas and...Ch. 8 - Prob. 91ECh. 8 - Helium is collected over water at 25C and 1.00 atm...Ch. 8 - At elevated temperatures, sodium chlorate...Ch. 8 - Xenon and fluorine will react to form binary...Ch. 8 - Methanol (CH3OH) can be produced by the following...Ch. 8 - In the Mthode Champenoise, grape juice is...Ch. 8 - Hydrogen azide, HN3, decomposes on heating by the...Ch. 8 - Prob. 98ECh. 8 - Some very effective rocket fuels are composed of...Ch. 8 - The oxides of Group 2A metals (symbolized by M...Ch. 8 - Prob. 101ECh. 8 - Prob. 102ECh. 8 - Prob. 103ECh. 8 - Prob. 104ECh. 8 - Prob. 105ECh. 8 - Prob. 106ECh. 8 - Prob. 107ECh. 8 - Prob. 108ECh. 8 - Prob. 109ECh. 8 - Prob. 110ECh. 8 - Prob. 111ECh. 8 - Prob. 112ECh. 8 - Prob. 113ECh. 8 - Prob. 114ECh. 8 - Prob. 115ECh. 8 - Prob. 116ECh. 8 - Prob. 117ECh. 8 - Prob. 118ECh. 8 - Prob. 119ECh. 8 - Prob. 120ECh. 8 - Prob. 121ECh. 8 - Prob. 122ECh. 8 - Prob. 123AECh. 8 - At STP, 1.0 L Br2 reacts completely with 3.0 L F2,...Ch. 8 - Prob. 125AECh. 8 - A 2.747g sample of manganese metal is reacted with...Ch. 8 - Prob. 127AECh. 8 - Cyclopropane, a gas that when mixed with oxygen is...Ch. 8 - The nitrogen content of organic compounds can be...Ch. 8 - Prob. 130AECh. 8 - A 15.0L tank is filled with H2 to a pressure of...Ch. 8 - A spherical glass container of unknown volume...Ch. 8 - Prob. 133AECh. 8 - A 20.0L stainless steel container at 25C was...Ch. 8 - Metallic molybdenum can be produced from the...Ch. 8 - Prob. 136AECh. 8 - Prob. 137AECh. 8 - One of the chemical controversies of the...Ch. 8 - An organic compound contains C, H, N, and O....Ch. 8 - Prob. 140AECh. 8 - The total volume of hydrogen gas needed to fill...Ch. 8 - Prob. 142AECh. 8 - Prob. 143CWPCh. 8 - Prob. 144CWPCh. 8 - A certain flexible weather balloon contains helium...Ch. 8 - Prob. 146CWPCh. 8 - A 20.0L nickel container was charged with 0.859...Ch. 8 - Consider the unbalanced chemical equation below:...Ch. 8 - Prob. 149CWPCh. 8 - Which of the following statements is(are) true? a....Ch. 8 - A chemist weighed out 5.14 g of a mixture...Ch. 8 - A mixture of chromium and zinc weighing 0.362 g...Ch. 8 - Prob. 153CPCh. 8 - You have an equimolar mixture of the gases SO2 and...Ch. 8 - Methane (CH4) gas flows into a combustion chamber...Ch. 8 - Prob. 156CPCh. 8 - Prob. 157CPCh. 8 - Prob. 158CPCh. 8 - You have a helium balloon at 1.00 atm and 25C. You...Ch. 8 - Prob. 160CPCh. 8 - You are given an unknown gaseous binary compound...Ch. 8 - Prob. 162CPCh. 8 - Calculate w and E when 1 mole of a liquid is...Ch. 8 - The preparation of NO2(g) from N2(g) and O2(g) is...Ch. 8 - In the presence of nitric acid, UO2+ undergoes a...Ch. 8 - Silane, SiH4, is the silicon analogue of methane,...Ch. 8 - Prob. 167IPCh. 8 - Prob. 168IPCh. 8 - Prob. 169MP
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