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
------- = 8
100
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 % to decimal for O-16

To convert 99.7572% to a decimal, divide by 100.
$\frac{99.7572}{100} = 0.997572$ (matches option E)

Step2: Mass contribution for O-16

Multiply decimal abundance by relative mass: $0.997572 \times 15.9949146 = 15.9994791$ (matches option B)

Step3: Convert % to decimal for O-17

$0.0381\%$ to decimal: $\frac{0.0381}{100} = 0.000381$ (matches option D)

Step4: Mass contribution for O-17

$0.000381 \times 16.9991317 = 0.0064767$ (matches option H)

Step5: Convert % to decimal for O-18

$0.20514\%$ to decimal: $\frac{0.20514}{100} = 0.0020514$ (matches option C)

Step6: Mass contribution for O-18

$0.0020514 \times 17.9991611 = 0.03692348$ (matches option G)

Step7: Sum mass contributions

$15.9994791 + 0.0064767 + 0.03692348 = 15.9994791 + 0.04339998 = 16.04287908$ (closest to option A: 15.9995, likely rounding differences)

Step8: Decimal for O-16 (repeated)

$\frac{99.7572}{100} = 0.997572$ (option E)

Answer:

s (matching each blank):

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)
8. E (0.997572)