Naturally occurring lead exists as four stable isotopes: 204Pb with a mass of 203.973 amu (1.480%); 206Pb, 205.9745 amu (23.60%); 207Pb, 206.9759 amu (22.60%); and 208Pb, 207.9766 amu (52.30%). Calculate the average atomic mass of lead.
2007-07-20 19:40:47 · 3 answers · asked by lulu 1 in Science & Mathematics ➔ Chemistry
Multiply the amu of each isotope with its weighted percentage and add the numbers.
(203.973 x 1.48 + 205.9745 x 23.6 + 206.9759 x 22.6 + 207.9766 x 52.3) / 100 = 207.1770976 = 207.1771
2007-07-20 19:58:18 · answer #1 · answered by Swamy 7 · 0⤊ 0⤋
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2016-11-10 01:03:31 · answer #2 · answered by ? 4 · 0⤊ 0⤋
2nd way:
(203.973*0.0148)+(205.9745*0.2360)+(206.9759*0.2260)+(207.9766*0.5230) = 207.1770976
2007-07-20 22:46:28 · answer #3 · answered by MO. NO 1 · 0⤊ 0⤋