Chemistry for Today: General, Organic, and Biochemistry
Chemistry for Today: General, Organic, and Biochemistry
9th Edition
ISBN: 9781305960060
Author: Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher: Cengage Learning
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Chapter 6, Problem 6.56E

A sample of gas weighs 0.176 g and has a volume of 114.0 mL at a pressure and temperature of 640. torr and 20. ° C . Determine the molecular weight of the gas, and identify it as CO , CO 2 , or O 2 .

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The molecular weight of the given gas is to be calculated. Whether the given sample of gas is CO, CO2, or O2 is to be identified.

Concept introduction:

The ideal gas equation is used to represent the relation between the volume, pressure, temperature and number of moles of an ideal gas. The ideal gas equation is given as,

PV=nRT

Where,

V represents the volume occupied by the ideal gas.

P represents the pressure of the ideal gas.

n represents the number of moles of the ideal gas.

T represents the temperature of the ideal gas.

R represents the ideal gas constant with value 0.08206Latm/Kmol.

Answer to Problem 6.56E

The molecular mass of gas is 44.1g/mol. The given sample of the given gas is CO2.

Explanation of Solution

The pressure of the given gas is 640torr.

The temperature of the given gas is 20°C.

The temperature of the given gas in Kelvin is represented as,

T=(273+20°C)K=293K

It is given that a sample of the given gas has a mass of 0.176g.

The given volume of the given gas is 114.0mL.

The molar mass of CO is 28.01g/mol.

The molar mass of CO2 is 44.01g/mol.

The molar mass of O2 is 32.00g/mol.

The ideal gas equation is given as,

PV=nRT …(1)

Where,

V represents the volume occupied by the ideal gas.

P represents the pressure of the ideal gas.

n represents the number of moles of the ideal gas.

T represents the temperature of the ideal gas.

R represents the ideal gas constant with value 0.08206Latm/Kmol.

Rearrange the above equation for the value of n.

n=PVRT

Substitute the value of P, V, T and R in the equation (1).

n=(114.0mL)(1L1000mL)(640torr)(1atm760torr)(0.08206Latm/Kmol)(293K)=3.9927×103mol

Therefore, the number of moles of the given gas is 3.9927×103mol.

The relation between mass and number of moles of a substance is given as,

M=mn …(2)

Where,

m represents the mass of the substance.

M represents the molar mass of the substance.

n represents the number of moles of the substance.

Substitute the value of mass and number of moles of the given gas in the equation (2).

M=0.176g3.9927×103mol=44.08g/mol44.1g/mol

The molar mass of given gas is 44.1g/mol that is approximately equal to the molar mass of CO2. Therefore, the given sample of the given gas is CO2.

Therefore, the molecular mass of gas is 44.1g/mol.

Conclusion

The molecular mass of gas is 44.1g/mol. The given sample of the given gas is CO2.

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Chapter 6 Solutions

Chemistry for Today: General, Organic, and Biochemistry

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