find the percentage composition of 35Cl and 37Cl that amount for atomic masses of Cl 35.45

The percent abundance of chlorine-35 and chlorine-37 must equal 100, or, in decimal form 1.

Given/Known:
atomic weight of Cl-35 = 34.9689
atomic weight of Cl-37 = 36.9695
atomic weight of chlorine = 35.45

Unknown:
x = abundance of Cl-35
1-x = abundance of Cl-37

SOLUTION:

1. Set up the equation as follows:

(34.9689)x + (36.9695)(1-x) = 35.45

2. Carry out multiplication.

34.9689x + 36.9695 - 36.9695x = 35.45

3. Combine the x values.

-2.0006x + 36.9695 = 35.45

4. Subtract 36.9695 from both sides.

-2.0006x = - 1.52 (rounded to two decimal places due to significant figure rounding rules for addition and subtraction)

5. Divide both sides by -2.0006 to calculate abundance.

x = 0.760 (rounded to three significant figure due to rounding rules for multiplication and division)

1 - 0.760 = 0.240 (In this case 1 is an exact number, which has an infinite number of significant figures, so significant figures are dependent upon 0.760, which has three decimal places.)

6. Multiply abundance times 100 to get percent abundance.

0.760 x 100 = 76.0%
0.24 x 100 = 24.0%

Answer:

The percent abundance of Cl-35 = 76.0%.
The percent abundance of Cl-37 = 24.0%.