Chapter 6, Problem 6.56E

A sample of gas weighs $0.176\text{g}$ and has a volume of $114.0\text{mL}$ at a pressure and temperature of $640.\text{torr}$ and $20.°\text{C}$. Determine the molecular weight of the gas, and identify it as $\text{CO}$, ${\text{CO}}_2$, or ${\text{O}}_2$.

Interpretation Introduction

Interpretation:

The molecular weight of the given gas is to be calculated. Whether the given sample of gas is $\text{CO}$, ${\text{CO}}_2$, or ${\text{O}}_2$ 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 $0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}$.

Answer to Problem 6.56E

The molecular mass of gas is $44.1\text{g}/\text{mol}$. The given sample of the given gas is ${\text{CO}}_{\text{2}}$.

Explanation of Solution

The pressure of the given gas is $640\text{torr}$.

The temperature of the given gas is $20°\text{C}$.

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

$\begin{array}{rll}T& =& \left(273+20°\text{C}\right)\text{K}\\ & =& 293\text{K}\end{array}$

It is given that a sample of the given gas has a mass of $0.176\text{g}$.

The given volume of the given gas is $114.0\text{mL}$.

The molar mass of $\text{CO}$ is $28.01\text{g}/\text{mol}$.

The molar mass of ${\text{CO}}_2$ is $44.01\text{g}/\text{mol}$.

The molar mass of ${\text{O}}_2$ is $32.00\text{g}/\text{mol}$.

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 $0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}$.

Rearrange the above equation for the value of $n$.

$n=\frac{PV}{RT}$

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

$\begin{array}{rll}n& =& \frac{\left(114.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(640\text{torr}\right)\left(\frac{1\text{atm}}{760\text{torr}}\right)}{\left(0.08206\text{L}\cdot \text{atm}/\text{K}\cdot \text{mol}\right)\left(293\text{K}\right)}\\ & =& 3.9927\times {10}^{-3}\text{mol}\end{array}$

Therefore, the number of moles of the given gas is $3.9927\times {10}^{-3}\text{mol}$.

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

$M=\frac{m}{n}$ …(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).

$\begin{array}{rll}M& =& \frac{0.176\text{g}}{3.9927\times {10}^{-3}\text{mol}}\\ & =& 44.08\text{g}/\text{mol}\\ & \approx & 44.1\text{g}/\text{mol}\end{array}$

The molar mass of given gas is $44.1\text{g}/\text{mol}$ that is approximately equal to the molar mass of ${\text{CO}}_{\text{2}}$. Therefore, the given sample of the given gas is ${\text{CO}}_{\text{2}}$.

Therefore, the molecular mass of gas is $44.1\text{g}/\text{mol}$.

Conclusion

The molecular mass of gas is $44.1\text{g}/\text{mol}$. The given sample of the given gas is ${\text{CO}}_{\text{2}}$.