Question

A gaseous mixture of helium and oxygen is found to have a density of $0.518gd{m^{ - 3}}$ at ${25^ \circ }C$ and $720torr$ . What is the percent by mass of helium in this mixture?

Answer 1

Hint: Mass percent is among one of the ways of representing concentration of an element in a compound. Mass percent is mass of solute dissolved per $100$ ml of solution. Density of a substance is defined as mass per unit volume of that substance.
Formula used: $P{m_{av}} = dRT$
Mole fraction$ = \dfrac{n}{N} = \dfrac{m}{{M \times N}}$
Average molecular weight$ = \dfrac{{{M_T}}}{N}$
Mass percent of helium$ = \dfrac{{{m_{He}}}}{{{M_T}}} \times 100$
Where, $P = $ pressure (in $atm$ ), ${m_{av}} = $ average molecular weight (in $gmo{l^{ - 1}}$ ), $d = $ density (in $gd{m^{ - 3}}$ ), $R = $ gas constant (in $atmLmo{l^{ - 1}}{K^{ - 1}}$ ), $T = $ temperature (in $K$ ), $n = $ number of moles of substance, $N = $ total moles, $m = $ given mass of substance, $M = $ molecular mass of substance, ${M_T} = $ total mass of components, ${m_{He}} = $ mass of helium, ${m_{{O_2}}}$ is mass of oxygen, ${M_{He}} = $ molecular mass of helium, ${M_{{O_2}}} = $ molecular mass of oxygen.

Complete step by step solution:
Average molecular weight is defined as total weight of components divided by total number of molecules.
Mole fraction of a substance is defined as the number of moles of that substance divided by the total number of molecules present. Sum of mole fraction of all the substances present in a mixture is $1$ .
We know, $Pm = dRT$
Here pressure is in the atmosphere. As $1torr = \dfrac{1}{{760}}atm$
                                                                $720torr = \dfrac{{720}}{{760}}atm$
Hence $P = \dfrac{{720}}{{760}}atm$
Temperature is in Kelvin. As ${0^ \circ }C = 273K$
                                                   ${25^ \circ }C = 273 + 25 = 298K$
Hence $T = 298K$
$R = 0.0821atmLmo{l^{ - 1}}{K^{ - 1}}$
$d = 0.518gd{m^{ - 3}}$ (given)
Put these values in above formula,
$
  \dfrac{{720}}{{760}} \times {m_{av}} = 0.518 \times 0.0821 \times 298 \\
  {m_{av}} = \dfrac{{760 \times 0.0821 \times 0.518 \times 298}}{{720}} \\
  {m_{av}} = 13.37gmo{l^{ - 1}} \\
 $
Mole fraction of total components present in a mixture is one. Components present in this mixture are oxygen and helium. Therefore the sum of mole fraction of oxygen and helium is one.
Let mole fraction of helium be $\alpha $
Then mole fraction of oxygen is $1 - \alpha $
We know Average molecular weight$ = \dfrac{M}{N}$
Average molecular weight$ = \dfrac{{{m_{He}}}}{N} + \dfrac{{{m_{{O_2}}}}}{N}$ where ${m_{He}}$ is mass of helium and ${m_{{O_2}}}$ is mass of oxygen
Also Mole fraction$\left( x \right) = \dfrac{n}{N} = \dfrac{m}{{M \times N}}$
Therefore $\dfrac{m}{N} = x \times M$
And average molecular weight${m_{av}}$$ = \left( {{x_{He}} \times {M_{He}}} \right) + \left( {{x_{{O_2}}} \times {M_{{O_2}}}} \right)$ (from above equation)
$
  13.37 = \left( {\alpha \times 4} \right) + \left( {\left( {1 - \alpha } \right)32} \right) \\
  13.37 = 32 - 28\alpha \\
  28\alpha = 32 - 13.37 \\
  \alpha = 0.66 \\
 $
Mole fraction of helium is $0.66$
Mole fraction of oxygen is $1 - 0.66 = 0.34$
Mass percent of helium$ = \dfrac{{{m_{He}}}}{{{M_T}}} \times 100$
${m_{He}} = \alpha \times {M_{He}} \times N = 0.66 \times 4 \times N = 2.64N$
$
  {M_T} = {m_{He}} + {m_{{O_2}}} = \left( {\left( {1 - \alpha } \right){M_{{O_2}}} \times N} \right) + \left( {\alpha \times {M_{He}}} \right) \\
  {M_T} = \left( {1 - \alpha } \right)32N + \alpha \times 4N \\
  {M_T} = 0.34 \times 32N + 0.66 \times 4N \\
  {M_T} = 13.52N \\
 $
Hence mass percent of helium$ = \dfrac{{2.64N}}{{13.52N}} \times 100 = 19.53\% $
So answer is $19.53\% $

Note: Do not get confused with average molecular weight and mole fraction. Mole fraction is number of moles of substance divided by total number of moles and average molecular weight is mass of mixture divided by total number of molecules.