# Cl-35 has an atomic weight of "34.9689 and Cl-37 has an atomic weight of "36.9695. What is the percent abundance of Cl-35?

Nov 6, 2014

The percent abundance of Cl-35 = 76.0%.

#### Explanation:

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:

$x = \text{abundance of Cl-35}$

$1 - x = \text{abundance of Cl-37}$

SOLUTION:

1. Set up the equation as follows:

$\left(34.9689\right) x + \left(36.9695\right) \left(1 - x\right) = 35.45$

2. Carry out multiplication.

$34.9689 x + 36.9695 - 36.9695 x = 35.45$

3. Combine the x values.

$- 2.0006 x + 36.9695 = 35.45$

4. Subtract 36.9695 from both sides.

$- 2.0006 x = - 1.52$

(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 figures 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.

$\text{% abundance of Cl-35}$ is 0.760xx100% = 76.0%

$\text{% abundance of Cl-37}$ is 0.24xx100% = 24.0%