Question

isotope mass (amu) natural abundance
b-10 10.013 19.78% → 0.1978×10.013 = 1.9806
b-11 11.009 80.22% → 0.8022×11.009 = 8.8314
\t\t\t\t\t\t\t\t\t\t\t\t\t\t\t 10.812 amu

1. calculate the average atomic mass of boron, showing all work and using proper significant figures.

10.81 amu

Explanation:

Step1: Convert % to decimals

B-10: $19.78\% = 0.1978$; B-11: $80.22\% = 0.8022$

Step2: Calculate weighted mass for B-10

$0.1978 \times 10.013 = 1.9805714$

Step3: Calculate weighted mass for B-11

$0.8022 \times 11.009 = 8.8314198$

Step4: Sum weighted masses

$1.9805714 + 8.8314198 = 10.8119912$

Step5: Round to sig figs

Round to 4 significant figures (from isotope masses)

Answer:

10.81 amu