Question

calculate the average atomic mass for oxygen given the following: isotope %abundance relative mass mass contribution o-16 99.7572 15.9949146 amu ( 1 )(15.9949146 amu) = 2 amu o-17 0.0381 16.9991317 amu ( 3 )(16.9991317 amu) = 4 amu o-18 0.20514 17.9991611 amu ( 5 )(17.9991611 amu) = 6 amu 7 amu first we need to convert the % abundance into abundance by dividing by 100. thus, 99.7572% becomes: 99.7572 / 100 = 8 a. 15.9995 b. 15.9994791 c. 0.0020514 d. 0.000381 e. 0.997572 f. 99.7572. g. 0.03692348 h. 0.0064767 i. 15.956079 j 1.5956079

Explanation:

Step1: Convert O - 16 % abundance to abundance

Divide 99.7572 by 100: $\frac{99.7572}{100}=0.997572$. So the value for blank 1 and 8 is 0.997572.

Step2: Calculate O - 16 mass contribution

Multiply the abundance of O - 16 by its relative mass: $0.997572\times15.9949146\ amu\approx15.994791\ amu$. So the value for blank 2 is 15.994791.

Step3: Convert O - 17 % abundance to abundance

Divide 0.0381 by 100: $\frac{0.0381}{100}=0.000381$. So the value for blank 3 is 0.000381.

Step4: Calculate O - 17 mass contribution

Multiply the abundance of O - 17 by its relative mass: $0.000381\times16.9991317\ amu\approx0.0064767\ amu$. So the value for blank 4 is 0.0064767.

Step5: Convert O - 18 % abundance to abundance

Divide 0.20514 by 100: $\frac{0.20514}{100}=0.0020514$. So the value for blank 5 is 0.0020514.

Step6: Calculate O - 18 mass contribution

Multiply the abundance of O - 18 by its relative mass: $0.0020514\times17.9991611\ amu\approx0.03692348\ amu$. So the value for blank 6 is 0.03692348.

Step7: Calculate average atomic mass

Sum up the mass - contributions of all isotopes: $15.994791 + 0.0064767+0.03692348 = 15.99959118\approx15.9995\ amu$. So the value for blank 7 is 15.9995.

Answer:

1. E. 0.997572
2. B. 15.9994791
3. D. 0.000381
4. H. 0.0064767
5. C. 0.0020514
6. G. 0.03692348
7. A. 15.9995