# Question 750e4

Apr 15, 2015

Let's assume that $m$ represents the mass of each gas added to the container.

The total mass of gas in the container will be

${m}_{\text{total}} = m + m + m = 3 m$

Because oxygen, hydrogen, and methane have different molar masses, you won't get the same number of moles of each gas in the container.

Moreover, the number of moles of each gas present in the container will be proportional to the respective gas' molar mass.

So, the molar masses for the gases are

M_("M hydrogen") = "1.00794 g/mol" ~= "1 g/mol"

M_("M oxygen") = "32 g/mol"

M_("M methane") = "16.04 g/mol" ~= "16 g/mol"

You can determine how many moles of each gas you have in the container by using

$n = \frac{m}{M} _ M$

${n}_{\text{hydrogen" = (mcancel("g"))/(1cancel("g")/"mol") = "m moles}}$

${n}_{\text{oxygen" = (mcancel("g"))/(32cancel("g")/"mol") = m/32" moles}}$

${n}_{\text{methane" = (mcancel("g"))/(16cancel("g")/"mol") = m/16" moles}}$

Now just divide the number of moles of one gas to the number of moles of another gas to get their ratio

n_"hydrogen"/n_"oxygen" = cancel("m")/(cancel("m")/32) = color(green)(32)

n_"hydrogen"/n_"methane" = cancel("m")/(cancel("m")/16) = color(green)(16)

n_"oxygen"/n_"methane" = (cancel("m")/32)/(cancel("m")/16) = color(green)(1/2)

If you're looking for the mole fraction, which represents the ration between the number of moles of a gas and the total number of moles present, simply add the number of moles of each gas to get the total number of moles present

${n}_{\text{total}} = m + \frac{m}{16} + \frac{m}{32} = \frac{35}{32} m$

Therefore

chi_"hydrogen" = n_"hydrogen"/n_"total" = (cancel("m"))/(35/32cancel("m")) = color(green)(32/35)

chi_"oxygen" = n_"oxygen"/n_"total" = (cancel("m")/32)/(35/32cancel("m")) = color(green)(1/35)

chi_"methane" = n_"methane"/n_"total" = (cancel("m")/16)/(35/32cancel("m")) = color(green)(2/35)#