Chemistry Forum

I've been working for hours on a problem about molar fractions.

The problem is: You have a solution of ethanol and wather. You know that X(eth)=0,90 and you want to calculate its % M/M (masse/masse pourcentage) in the current solution.

How do you solve this?

I've worked with the molar fraction equation: X(eth)= n(eth) / n(eth)+n(w) and found:
m(eth)=9n(eth)m(w) / n(w)
and
m(w)=n(w)m(eth) / 9n(eth)

but that doesn't help me much...

If I use these 2 answers in my % M/M formula, I end up with:

% M/M(eth) = 9n(eth) / n(w)m(eth) X 100

wich is, in fact, = 1/m(w)

And I'm still far from any kind of answer :S

Killy wrote:I've been working for hours on a problem about molar fractions.

The problem is: You have a solution of ethanol and wather. You know that X(eth)=0,90 and you want to calculate its % M/M (masse/masse pourcentage) in the current solution.

How do you solve this?

I've worked with the molar fraction equation: X(eth)= n(eth) / n(eth)+n(w) and found:
m(eth)=9n(eth)m(w) / n(w)
and
m(w)=n(w)m(eth) / 9n(eth)

but that doesn't help me much...

If I use these 2 answers in my % M/M formula, I end up with:

% M/M(eth) = 9n(eth) / n(w)m(eth) X 100

wich is, in fact, = 1/m(w)

And I'm still far from any kind of answer :S

Assume you have just 1 mole of the mixture:
m(total) = M(eth) * 0.9 + M(w) * 0.1
m(eth) = M(eth) * 0.9
where M - stands for molar weight and m - for actual mass

You need to find mass percent, that is:
m(eth)/m(total)*100% = (M(eth) * 0.9) / (M(eth) * 0.9 + M(w) * 0.1) * 100%
That's it!
Just substitue molar weights and do the math.