A solution consisting of isobutyl bromide and isobutyl chloride is found to have a

refractive index of 1.3931 at 20oC. The refractive indices at 20oC 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.

How do I find this assuming that the total is 100%?
PLEASE EXPLAIN!
THANK YOU!

Let's call mole fraction isobutylchloride = C.

and mole fraction isobutylbroomide = B

Now if we draw a straight line from left to right
|____________________________________|
and label the left index as 0 mole fraction B (meaning that is pure C and no B) and the right index as 1.0 mole fraction B (meaning pure B and no C).
Now the left index is marked as 1.3789 refractive index (pure C) and the right index is marked as 1.4368 (pure B). Check me out on that.
What we want to do is to find the fraction that 1.3931 represents. The difference from left to right is
1.4368-1.3785- ??.
The difference from 1.3931(the sample) and the left mark is -1.3785 = ??.
Now take the ratio of the two and that should be the mole fraction of B in the mixture. That times 100 should convert to percent. Check my thinking.

Thank you!

What is the percent by mass of Ca3N2?

To find the molar composition of the mixture, we can use the linear relation between the refractive index and the molar composition. The equation for this linear relation is:

n = x1 * nb + x2 * nc

Where:
n is the refractive index of the mixture
x1 and x2 are the molar compositions of isobutyl bromide and isobutyl chloride respectively
nb and nc are the refractive indices of isobutyl bromide and isobutyl chloride respectively

Given:
n = 1.3931
nb = 1.4368
nc = 1.3785

Let's substitute these values into the equation:

1.3931 = x1 * 1.4368 + x2 * 1.3785

Now, since the total molar composition is assumed to be 100%, we have:

x1 + x2 = 1

We can solve these two simultaneous equations to find the values of x1 and x2.

Let's rearrange the equation by isolating one of the variables. We can isolate x1 by subtracting x2 from both sides:

x1 = 1 - x2

Now substitute this expression into the first equation:

1.3931 = (1 - x2) * 1.4368 + x2 * 1.3785

Simplify the equation:

1.3931 = 1.4368 - 1.4368 * x2 + 1.3785 * x2

Combine like terms:

1.3931 = 1.4368 - 0.0583 * x2

Subtract 1.4368 from both sides:

1.3931 - 1.4368 = -0.0583 * x2

Simplify:

-0.0437 = -0.0583 * x2

Now, divide both sides by -0.0583 to solve for x2:

x2 = -0.0437 / -0.0583

x2 ≈ 0.7495

Since x1 + x2 = 1, substitute the value of x2 to find x1:

x1 + 0.7495 = 1

x1 ≈ 0.2505

The molar composition of the mixture is approximately 25.05% isobutyl bromide and 74.95% isobutyl chloride when the total is assumed to be 100%.