# Mole Fraction Question

1. Oct 1, 2009

So I've been thinking about this problem for quite some time now any help will be appreciated.

A sample of $$C_{2}H_{2}(g)$$ has a pressure of 7.4 kPa. After some time a portion of it reacts to form $${C}_6{H}_6 (g)$$. The total pressure of the mixture of gases is then 3.6 kPa. Assume the volume and the temperature do not change. Find the mol fraction of $$C_{2}H_{2}(g)$$ that has undergone reaction.

The mol fraction, X, is given by

$$X = P_{1}/P_{total} = n_{1}/n_{total}$$

where P= pressure and n= number of moles

I think that $$3C_{2}H_{2}(g)->{C}_6{H}_6 (g)$$. So I assumed STP and thought that if i can find the number of moles of $${C}_6{H}_6 (g)$$ then I can use stoichiometry to find the moles of $$C_{2}H_{2}(g)$$ and subtract that from the number of moles found by using the ideal gas equation and then use the mol fraction formula to find it. However, this results in a negative number (3.25858*10-3-4.754*10-3) and I can't seem to figure out what I did wrong. Like I said, any help would be appreciated.

2. Oct 2, 2009

### chemisttree

You shouldn't need to use the ideal gas equation.

Consider that Ntotal = N1 + N2
where N1 = C2H2 and N2 = C6H6

Initially N2 = 0 and later N2 = N1 - Ntotal.

Now we can express N2 in terms of pressure and N1... right?

There are other things to work out but this understanding is critical to solving the problem.

3. Oct 3, 2009

ok, thanks i'll try to work it out from there,
also, am i on the right track with 3C2H2-->C6H6 or is this chemical equation irrelevant to solving the problem?

4. Oct 3, 2009

### Staff: Mentor

It is very important - combined with Avogadro's hypothesis it explains why pressure went down. Think about stochiometry of the process.

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5. Oct 3, 2009

ok, so so far i have
3C2H2-->C6H6
ntotal=nC2H2 final + nC6H6
$$\frac{P_{C_{2}H_{2 initial}}}{n_{C_{2}H_{2 initial}}}=\frac{P_{C_{2}H_{2 final}}}{n_{C_{2}H_{2 final}}}$$ from avogadro's law
PC2H2 final=Ptotal-PC6H6
nC6H6=nC2H2 final-ntotal=nC2H2 final-(PtotalnC2H2 final)/PC2H2 final
nC2H2 initial - x moles of C2H2 = nC2H2 final
x moles of C2H2/3 = nC6H6 final
nC2H2 initial-3*nC6H6=nC2H2 final

i'm not sure how to proceed from here, is there a way to find the change in moles based on the change of pressure and use that to find the mol fraction?

6. Oct 3, 2009

### Staff: Mentor

If you have no other ideas, assume some volume - say 1L - and do caculations on real numbers. Or assume volume V and do calculations on symbols - in the end V will cancel out. The simplest approach is to use pV=nRT and realize it can be written as n=kp, where k is some constant - that in the end will cancel out as well.

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7. Oct 3, 2009

Ok, so I used your way and did P=nk which means that P/n = k, so P1/n1=P2/n2=k. After substitution that gives me 7.4/n1=3.6/n2. I then found the ratio of n1 to n2 to be n1/n2= 7.4/3.6 = 37/18 so I interpreted that to mean that we started out with 37 mols of C2H2 initially and ended up with a mixture of C2H2 and C6H6 that totaled in 18 mols.

so Let w = mols of C2H2
y = mols of C6H6

37 - x mols = w
x mols/3 = y from the stoichiometric ratio
w+y = 18
so that means that
37-3y = w
37-3(18-w) = w
w = mols of C2H2 = 17/2
since the total amount of C2H2 and C6H6 is 18 the mol fraction should be (17/2)/18 = .47222...since this is online homework, I get instant feedback which resulted in a wrong answer. Where did I go wrong?