Answer to Question #96471 in General Chemistry for Angel

Question

A gaseous sample containing a mixture of only propane and methane, occupies a volume of 2.95L at 30°C.
The partial pressure of C3H8 is 0.395 atm and that of CH4 is 0.485 atm.
What is the total pressure of the mixture?


What is the mole fraction of C3H8 in the mixture?


How many moles of C3H8 are in the sample?


What is the total mass of the sample?

Answer 1

1)According to the Dalton’s Law the total pressure is the sum of the partial pressures thus

$$p = {p_{{C_3}{H_8}}} + {p_{C{H_4}}} = 0.395[{\rm{atm}}] + 0.485[{\rm{atm}}] = 0.88[{\rm{atm}}]$$

2)The partial pressure of the i-th component can be found as (it follows from the ideal gas law) as product of the mole fraction and total pressure

$${p_i} = {\omega _i}p$$

thus the mole fraction of ${{C_3}{H_8}}$ is

$${\omega _{{C_3}{H_8}}} = {{{p_{{C_3}{H_8}}}} \over p} = {{0.395[{\rm{atm}}]} \over {0.88[{\rm{atm}}]}} \approx 0.449$$

3)We can find the number of moles of ${{C_3}{H_8}}$ using the ideal gas law for it

$${p_{{C_3}{H_8}}}V = {\nu _{{C_3}{H_8}}}RT$$

thus

$${\nu _{{C_3}{H_8}}} = {{{p_{{C_3}{H_8}}}V} \over {RT}}$$

and we can do calculations

$${\nu _{{C_3}{H_8}}} = {{0.395 \cdot 101325[{\rm{Pa}}] \cdot 2.95 \cdot {{10}^{ - 3}}[{{\rm{m}}^3}]} \over {8.314[{{\rm{J}} \over {{\rm{K}} \cdot {\rm{mol}}}}] \cdot (30 + 273.15)[{\rm{K}}]}} \approx 0.0468[{\rm{mol}}]$$

4)We can calculate the number of moles of $C{H_4}$ using the same way

$${\nu _{C{H_4}}} = {{0.395 \cdot 101325[{\rm{Pa}}] \cdot 2.95 \cdot {{10}^{ - 3}}[{{\rm{m}}^3}]} \over {8.314[{{\rm{J}} \over {{\rm{K}} \cdot {\rm{mol}}}}] \cdot (30 + 273.15)[{\rm{K}}]}} \approx 0.0575[{\rm{mol}}]$$

and then calculate the mass of the sample

$$m = {\nu _{{C_3}{H_8}}}{M_{{C_3}{H_8}}} + {\nu _{C{H_4}}}{M_{C{H_4}}}$$

$$m = 0.0468[{\rm{mol}}] \cdot 44.097[{{\rm{g}} \over {{\rm{mol}}}}] + 0.0575[{\rm{mol}}] \cdot 16.043[{{\rm{g}} \over {{\rm{mol}}}}] = 2.986[{\rm{g}}]$$