Question

1. given the relative abundance of the following naturally occurring isotopes of oxygen, calculate the average atomic mass of oxygen. show your work.

oxygen - 16: 99.76%
oxygen - 17: 0.037%
oxygen - 18: 0.204%

1. calculate the atomic mass of chlorine. isotope cl - 35 has an atomic mass of 34.97 amu and a relative abundance of 75.77%. isotope cl - 37 has an atomic mass of 37.00 amu and a relative abundance of 24.23%. show your work.

Explanation:

Question 2:

Step1: Convert percentages to decimals

For oxygen - 16: \(99.76\%=0.9976\), for oxygen - 17: \(0.037\% = 0.00037\), for oxygen - 18: \(0.204\%=0.00204\)

Step2: Multiply mass of each isotope by its relative abundance (in decimal)

Oxygen - 16: \(16\times0.9976 = 15.9616\)
Oxygen - 17: \(17\times0.00037=0.00629\)
Oxygen - 18: \(18\times0.00204 = 0.03672\)

Step3: Sum the results

\(15.9616+0.00629 + 0.03672=16.00461\approx16.00\)

Step1: Convert percentages to decimals

For Cl - 35: \(75.77\%=0.7577\), for Cl - 37: \(24.23\%=0.2423\)

Step2: Multiply mass of each isotope by its relative abundance (in decimal)

Cl - 35: \(34.97\times0.7577=34.97\times(0.75 + 0.0077)=34.97\times0.75+34.97\times0.0077=26.2275+0.269269 = 26.496769\)
Cl - 37: \(37\times0.2423 = 8.9651\)

Step3: Sum the results

\(26.496769+8.9651=35.461869\approx35.46\)

Answer:

The average atomic mass of oxygen is approximately \(16.00\) amu.

Question 3: