The average molar mass of a natural gas is 17.4g/mol consisting of CH4 and C2H6,what is the molar fraction of each gas?
NG*16+BT*30=17.4
but NG+BT=1.00
(1-BT)16+BT*30=17.4
BT*14=1.4
BT=.0875 or 8.75 percent
NG=.9125 or 91.25 percent
To find the molar fraction of each gas, we need to determine the number of moles of each gas and then divide it by the total number of moles of all the gases.
Let's assume x represents the molar fraction of CH4 and y represents the molar fraction of C2H6.
From the given information, we know that the average molar mass of the natural gas is 17.4 g/mol.
The molecular weight of CH4 (methane) is 16 g/mol, and the molecular weight of C2H6 (ethane) is 30 g/mol.
Now we can set up two equations based on the molar fraction and the molar mass information:
(1) x + y = 1 (the sum of the molar fractions is 1)
(2) x(16) + y(30) = 17.4 (the weighted average of the molar masses is 17.4 g/mol)
Now we can solve these equations to find the values of x and y.
Rearrange equation (1) to solve for x:
x = 1 - y
Substitute this value of x into equation (2):
(1 - y)(16) + y(30) = 17.4
Let's simplify this equation:
16 - 16y + 30y = 17.4
14y = 1.4
y = 1.4 / 14
y = 0.1
Substitute this value of y back into equation (1) to find x:
x = 1 - 0.1
x = 0.9
So, the molar fraction of CH4 (methane) is 0.9, and the molar fraction of C2H6 (ethane) is 0.1.
To determine the molar fraction of each gas in the natural gas sample, we need to compare the number of moles of each gas component.
Let's assume we have x moles of CH4 and y moles of C2H6 in the natural gas sample.
We can set up a system of equations based on the molar mass and molar fraction of the gases:
1. Molar mass of CH4 = 16 g/mol
Molar mass of C2H6 = 30 g/mol
2. Average molar mass of natural gas = (x * Molar mass of CH4 + y * Molar mass of C2H6) / (x + y)
17.4 g/mol = (x * 16 g/mol + y * 30 g/mol) / (x + y)
Now, we can solve this system of equations to find the values of x and y.
Let's multiply both sides of equation 2 by (x + y) to eliminate the denominator:
17.4 g/mol * (x + y) = x * 16 g/mol + y * 30 g/mol
Expanding and rearranging the equation:
17.4x + 17.4y = 16x + 30y
Rearranging again:
17.4x - 16x = 30y - 17.4y
Combining like terms:
1.4x = 12.6y
Now, we can find the molar fraction of each gas:
Molar fraction of CH4 = x / (x + y)
Molar fraction of C2H6 = y / (x + y)
Dividing the equation we derived above by y, we get:
1.4x/y = 12.6
Simplifying the equation:
x/y = 12.6/1.4
x/y = 9
Substituting this value of x/y in the equation for the molar fraction of CH4:
Molar fraction of CH4 = 9 / (9 + 1)
Molar fraction of CH4 = 9/10 = 0.9
To find the molar fraction of C2H6, we can subtract the molar fraction of CH4 from 1:
Molar fraction of C2H6 = 1 - 0.9
Molar fraction of C2H6 = 0.1
Therefore, the molar fraction of CH4 is 0.9 and the molar fraction of C2H6 is 0.1 in the natural gas sample.