The atomic mass of lead obtained from its four isotopes with atomic masses and relative abundance of 203.973 amu(1.4%), 205.974 amu(24.1%), 206.976 amu(22.1%), and 207.977 amu(52.4%), is 207.217 amu.
The atomic mass of lead is given by:
[tex] A = m_{1}\%_{1} + m_{2}\%_{2} + m_{3}\%_{3} + m_{4}\%_{4} [/tex] (1)
Where:
m: is the atomic mass of the isotopes
%: is the abundance percent of the isotopes
After entering the given values into equation (1) we have:
[tex] A = 203.973 \:amu*1.4\% + 205.974 \: amu*24.1\% + 206.976\: amu*22.1\% + 207.977\: amu*52.4\% [/tex]
To calculate the atomic mass we need to change the percent values to decimal ones
[tex] A = 203.973 \:amu*0.014 + 205.974 \: amu*0.241 + 206.976\: amu*0.221 + 207.977\: amu*0.524 = 207.217 amu [/tex]
Therefore, the atomic mass of lead is 207.217 amu.
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