   # An organic compound containing only C, H, and N yields the following data. i. Complete combustion of 35.0 mg of the compound produced 33.5 mg CO 2 and 41.1 mg H 2 O. ii. A 65.2-mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129), giving 35.6 mL of dry N 2 at 740. torr and 25°C. iii. The effusion rate of the compound as a gas was measured and found to be 24.6 mL/min. The effusion rate of argon gas, under identical conditions, is 26.4 mL/min. What is the molecular 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 8, Problem 130AE
Textbook Problem
706 views

## An organic compound containing only C, H, and N yields the following data.i. Complete combustion of 35.0 mg of the compound produced 33.5 mg CO2 and 41.1 mg H2O.ii. A 65.2-mg sample of the compound was analyzed for nitrogen by the Dumas method (see Exercise 129), giving 35.6 mL of dry N2 at 740. torr and 25°C.iii. The effusion rate of the compound as a gas was measured and found to be 24.6 mL/min. The effusion rate of argon gas, under identical conditions, is 26.4 mL/min.What is the molecular formula of the compound?

Interpretation Introduction

Interpretation: From the given data, 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
• Mass%ofanelement=MassofthatelementinthecompoundMolarmassofthecompound×100

### Explanation of Solution

Explanation

• To determine: The mass percent from the given data

26.1%Carbon13.1%Hydrogen61.0%Nitrogen

Mass % of carbon

Since,completecombustionof35.0mgofthecompoundproduced33.5mgCO2and41.1mgH2ONumber ofmolesofcarbon dioxide =GivenmassMolecularmassSince,35.0 mgof thecompoundis combusted to produce33.5gCO2Givenmass=33.5mgMolecularmassofCO2=44.01mgNumber ofmolesofcarbon dioxide = 33.5mgCO244.01mgCO2=0.761molEverymoleofCO2willcontains1moleofcarbonand2molesofoxygen.Since,everymoleofCO2containsonemoleofcarbonNumberofmolesofcarbon=0.761molMassofcarboninthecompound=Number ofmolesofcarbon×atomic massofcarbonMassofcarboninthecompound=0.761mol×12.01mgC=9.14mgCMass%ofanelement=MassofthatelementinthecompoundMolarmassofthecompound×100Since,35.0 mgof thecompoundis combusted to produce33.5gCO2Molarmassofthecompound=35.0mgMass%ofcarbon=9.14mgC35.0mg×100=26.1%C

Mass % of Hydrogen

Since,completecombustionof35.0mgofthecompoundproduced33.5mgCO2and41.1mgH2ONumber ofmolesofwater =GivenmassMolecularmassSince,35.0 mgof thecompoundis combusted to produce41.1mgH2OGivenmass=41.1mgMolecularmassofH2O= 18.02 mgNumber ofmolesofwater = 41.1mg18.02mg=2.27molEverymoleofH2Ocontains2molesofHydrogenand1moleofoxygen.NumberofmolesofHydrogen= 2×2.284.54molMassofHydrogeninthecompound=Number ofmolesofHydrogen×AtomicmassofHydrogenMassofHydrogeninthecompound=4.54mol×1.008mgH=4.60mgHMass%ofanelement=MassofthatelementinthecompoundMolarmassofthecompound×100Since,35.0 mgof thecompoundis combusted to produce41.1mgH2OMolarmassofthecompound=35.0gMass%ofHydrogen=4.60mgH35.0g×100=13.1%H

Mass % of Nitrogen

The Partial pressure of N2 is 740torr=740torr×1atm760torr=0.973atm .

The volume of N2 gas is given as 35.6mL=35.6mL×1L1000mL=35.6×103L .

The temperature of N2 gas is given as 25oC=(25+273)K=298K .

The equation for finding number of moles of a substance,

According to ideal gas equation,

Numberofmoles=Pressure×VolumeR×Temperature

Numberofmoles=0.973atm×35.6×103L0.08206Latm/Kmol×298K =1.42×10-3mol

Numberofmolesof Nitroogen=1.42×10-3molMassof Nitrogenin thecompound=Number ofmolesofNitrogen×MolecularmassofnitrogenMassofNitrogeninthecompound=1.42×10-3mol×28.02g=3

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