Question

1. calculate the atomic mass of lead. the four lead isotopes have atomic masses and relative abundances of 203.973 amu (1.4000%), 205.974 amu (24.1010%), 206.976 amu (22.1000%) and 207.977 amu (52.3990%).

isotopic mass x relative abundance + isotopic mass x relative abundance
overline{qquadqquadqquadqquadqquadqquadqquadqquadqquadqquad 100 qquadqquadqquadqquadqquadqquadqquadqquadqquadqquad} = relative atomic
mass
overline{qquadqquad 7.02 x 8.0 quad + quad 6.02 x 92.0 qquad} = 6.1 atomic mass units
qquadqquadqquadqquadqquad 100 qquadqquadqquadqquadqquad

Explanation:

Step1: Calculate contribution of first isotope

Multiply isotopic mass (203.973 amu) by its relative abundance (1.4000%).
$203.973 \times \frac{1.4000}{100} = 203.973 \times 0.014 = 2.855622$

Step2: Calculate contribution of second isotope

Multiply isotopic mass (205.974 amu) by its relative abundance (24.1010%).
$205.974 \times \frac{24.1010}{100} = 205.974 \times 0.24101 \approx 49.6417$

Step3: Calculate contribution of third isotope

Multiply isotopic mass (206.976 amu) by its relative abundance (22.1000%).
$206.976 \times \frac{22.1000}{100} = 206.976 \times 0.221 \approx 45.7417$

Step4: Calculate contribution of fourth isotope

Multiply isotopic mass (207.977 amu) by its relative abundance (52.3990%).
$207.977 \times \frac{52.3990}{100} = 207.977 \times 0.52399 \approx 108.978$

Step5: Sum all contributions

Add the contributions from all four isotopes.
$2.855622 + 49.6417 + 45.7417 + 108.978 \approx 207.217$

Answer:

The atomic mass of lead is approximately 207.22 amu (rounded to two decimal places).