Find mass from mole frac and solution mass

[jesuslovesu]

1. 75g sample of a gaseous mixture contains .51 mole fraction propane. The remainder is butane. What are the masses of propane and butane?



2. X = mol A / mol total

The Attempt at a Solution

P = mol propane
B = mol butane
I tried doing .49P + .49B = X mol propane and .51P + .51B = Y mol butane however that doesn't seem to be solvable. I'm not quite sure how I should manipulate the 75 g of solution.

 

[Borek]

0.49B + 0.51P = n

second equation must contain masses of both gases - what is a mass of 0.49B?

 

[Blueshift5]

First, let's define our variables:
P = mass of propane (in grams)
B = mass of butane (in grams)
X = mole fraction of propane (in decimal form)
Y = mole fraction of butane (in decimal form)

To find the masses of propane and butane, we need to use the equation:

mass = mole fraction * solution mass

We know that the solution mass is 75 g, so we can set up two equations using this equation:

P = X * 75 g
B = Y * 75 g

Now, we also know that the mole fractions must add up to 1, so we can set up another equation:

X + Y = 1

We can solve this system of equations by substituting the value of Y from the third equation into the first two equations:

P = X * 75 g
B = (1 - X) * 75 g

We can then solve for P and B by plugging in the given value of X (0.51):

P = 0.51 * 75 g = 38.25 g
B = (1 - 0.51) * 75 g = 36.75 g

Therefore, the masses of propane and butane in the 75 g sample are 38.25 g and 36.75 g, respectively.