Molar Ratio Problem: Find Ethane to Methane

[ƒ(x)]

Homework Statement

Unknown amounts of methane and ethane are mixed at STP. The resulting mixture has a density of 1.18 g/L. Find the ratio of ethane to ethane.

The Attempt at a Solution

Since it's a ratio, we can convert it to 1 mol methane : X mol ethane

1 L methane at STP = \frac{RT}{P} = \frac{.08206 * 273}{1} = 22.402 mol

X L ethane at STP = 22.402X mol

d = \frac{m}{V}
1.18 = \frac{1.18(22.402 + 22.402*X)}{2*22.402}
1 = X

So it's a 1:1 ratio.

I was talking to someone who's better at chemistry than I am, and he said it got a 1:4 ratio. Plus, this method doesn't seem right since I didn't use the masses of CH4 or C2H6 (which would mean that there would always be a 1:1 ratio, regardless of the compounds). How do I do this?

 

[Borek]

You are wrong from the start - V is not RT/P, 1L is not 22.4 moles. Quite the opposite - 1 mole is 22.4 liters.

Hint: for ideal gas there is a simple dependence between molar mass and density. And "molar mass" in the case of a gas mixture can be expressed as a weighted (by molar fractions) average of molar masses of components.

 

[ƒ(x)]

You are wrong from the start - V is not RT/P, 1L is not 22.4 moles. Quite the opposite - 1 mole is 22.4 liters.

Hint: for ideal gas there is a simple dependence between molar mass and density. And "molar mass" in the case of a gas mixture can be expressed as a weighted (by molar fractions) average of molar masses of components.

So...

n = PV/RT, setting V = 1 gets .044 mol of CH4 and .044*X mol C2H6.

1.18 = (.044*16+.044*X*30)/(2)

X = 1.22...so it's 1:1.22 which is about 4:5?

My teacher gave me this question btw, the choices for the answer were:

A) 1:2
B) 1:4
C) 1:1
D) 4:1
E) none of the above

 

[ƒ(x)]

You are wrong from the start - V is not RT/P, 1L is not 22.4 moles. Quite the opposite - 1 mole is 22.4 liters.

Hint: for ideal gas there is a simple dependence between molar mass and density. And "molar mass" in the case of a gas mixture can be expressed as a weighted (by molar fractions) average of molar masses of components.

I know that d = PM/RT, where M is the molar mass.

I solved for it and got M = 26.434 g/L, but don't know where to go from there.

 

[Borek]

Imagine you have 1 mole of mixture. Assume molar fraction of one gas is x. That means you have x moles of one gas and 1-x of another, and total mass is x*molar mass of the first gas + (1-x)* molar mass of the second gas. Total mass of one mole is just a molar mass. Can you calculate amounts of both gases in the mixture?

 

[ƒ(x)]

Imagine you have 1 mole of mixture. Assume molar fraction of one gas is x. That means you have x moles of one gas and 1-x of another, and total mass is x*molar mass of the first gas + (1-x)* molar mass of the second gas. Total mass of one mole is just a molar mass. Can you calculate amounts of both gases in the mixture?

x mol CH4
1-x mol C2H6

16*x + 30*(1-x) = 26.434
-14*x = -3.566
x ~ .25

So it's a 1:4 ratio.

 

[ƒ(x)]

Why is the way I did it in my second post incorrect though?

Edit: Nevermind

 

[Borek]

x ~ .25

So it's a 1:4 ratio.

If x is 0.25, it is 1:3, or 1 in 4, but not 1:4. For 1:4 you would need x=0.2.

Yeah, I know - there is 1:3 answer. But that's not my fault.

 

[ƒ(x)]

If x is 0.25, it is 1:3, or 1 in 4, but not 1:4. For 1:4 you would need x=0.2.

Yeah, I know - there is 1:3 answer. But that's not my fault.

Eh...I'm not a fan of my chem professor.

Thanks for walking me through this.

 

[Borek]

Oops, I meant "there is no 1:3 answer". But you probably got it.