Let X = percent bromide
and Y = percent chloride
Then 1.4368(X) + 1.3785(Y) = mixture*100
You can do it another way.
1.4368-1.3785 = 0.05830 difference between the two pure materials.
So how far up the curve have we moved in going from 1.3785 (pure chloride) to 1.4194 for the mixture.
1.4194-1.3785 = 0.04090.
[0.04090/0.05830]*100 = 70.15% bromide and if I solve the two equations you posted I get 70.15% Br.
OR you can do it a third way.
Just graph it on the x axis moving left to right 100 mole percent Cl (0%Br) to 0%Cl (100 mole % Br).
This is a lab question...I don't understand how to do it.
A solution consisting of isobutyl bromide and isobutyl chloride is found to have a refractive index of 1.4194 at 20 degrees celsius. The refractive indices at 20 degrees celsius of isobutyl bromide and isobutyl chloride are 1.4368 and 1.3785, respectively. Determine the molar composition (in percent) of the mixture by assuming a linear relation between the refractive index and the molar composition of the mixture.
From poking around the internet I got that I should be using this equation:
(1.4194)(100) = 1.4368x + 1.3785y
and
x + y = 100
Obviously x + y = 100 is used because we are looking at % composition and it has to add up to 100. But I don't really understand the first equation.
2 answers
25% bromide and 75% chloride
3 answers