Lead has four stable isotopes. ^204Pb (1.4%), ^206Pb (24.1%). ^207Pb (22.1%) and ^208Pb (52.4%). Find the atomic mass of ^208 Pb given that ^204Pb has an atomic mass of 203.973 amu, ^206Pb has an atomic mass of 205.974 amu, and ^207Pb has an atomic mass of 206.976 amu.
atomicmass=Sum of massIsotopes*decimalPercent
To find the atomic mass of ^208Pb, we must consider the relative abundance of each isotope and its corresponding atomic mass.
Given the relative abundance of each isotope:
^204Pb = 1.4%
^206Pb = 24.1%
^207Pb = 22.1%
^208Pb = 52.4%
To calculate the atomic mass of ^208Pb, we can use the formula:
Atomic mass = (Mass of isotope 1 x Relative abundance of isotope 1) + (Mass of isotope 2 x Relative abundance of isotope 2) + ... + (Mass of isotope n x Relative abundance of isotope n)
Let's insert the values:
Atomic mass of ^208Pb = (203.973 amu x 1.4%) + (205.974 amu x 24.1%) + (206.976 amu x 22.1%) + (x amu x 52.4%)
Let's solve for x:
Atomic mass of ^208Pb = (203.973 amu x 1.4%) + (205.974 amu x 24.1%) + (206.976 amu x 22.1%) + (x amu x 52.4%)
= (2.855 AMU) + (49.633 AMU) + (45.710 AMU) + (0.524 AMU x)
= 98.722 AMU + (x amu x 52.4%)
Now, since the atomic mass of ^208Pb is the sum of all the terms, we can write:
Atomic mass of ^208Pb = 98.722 AMU + (x amu x 52.4%)
To solve for x, we subtract 98.722 AMU from both sides:
x amu x 52.4% = Atomic mass of ^208Pb - 98.722 AMU
Dividing both sides by 52.4% (or 0.524) gives:
x = (Atomic mass of ^208Pb - 98.722 AMU) / 0.524
Now, let's substitute the known values:
x = (208.980 amu - 98.722 amu) / 0.524
Simplifying the equation:
x = 110.258 amu / 0.524
x ≈ 210.30 amu
Therefore, the atomic mass of ^208Pb is approximately 210.30 amu.