How do I solve this problem on relative abundance?

Question

"The atomic mass of copper is 63.540 amu. There are two isotopes, Cu-63 and Cu-65 having atomic masses of 62.930 amu and 64.928 amu respectivlery. What is the relative abudance of these isotopes in naturallal occuring samples of copper?"

Normally I would multiply 63 and 65 (the atomic masses) by a percentage but there is not percentage given. How do I solve this equation?

Answer 1

Take 100 g of natural Cu, then if x is the mass of the isotope with atomic mass 62.930, the mass of the isotope with atomic mass 64.928 must be (100 - x) g. So we have the equation:

62.930*x/100 + 64.928*(100 - x)/100 = 63.540

Solving for x:

x = 69.45% of the isotope with atomic mass = 62.930, and:

100 - x = 100 - 69.45 = 30.55% of the isotope with atomic mass = 64.928

Answer 2

In this case you need to calculate the percentages. So you do a reverse calculation.
The fraction (percentages) will be x and y where y = 1 - x.

The average atomic weight will be given by:

63.540 = 62.930 * x + 64.928 (1 - x).

Solve for x and y = 1-x

Multiply x and y by 100 to get percentages.

Answer 3

you will go in reverse and set algebraic equation:
let aboundance of isotop 1 be x and the other (1-x):
63x + 65(1-x) = 63.540.
dont forget that x is fraction and you need to multiply it x by 100 for it to be %