How to find abundance of 206, 207 and 208 Pb

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SUMMARY

The discussion focuses on calculating the molar fractions of lead isotopes 206Pb, 207Pb, and 208Pb using given isotopic ratios. Participants confirm that the molar fractions must sum to 1 and can be derived from the ratios 207Pb/206Pb and 208Pb/206Pb. The example provided includes a lead content of 468 ppm, with specific ratios of 0.2515 for 207Pb/206Pb and 0.0581 for 208Pb/206Pb. The solution involves setting up simultaneous equations to solve for the individual molar fractions.

PREREQUISITES
  • Understanding of molar fractions and isotopic ratios
  • Basic algebra skills for solving equations
  • Familiarity with lead isotopes, specifically 206Pb, 207Pb, and 208Pb
  • Knowledge of ppm (parts per million) in chemical analysis
NEXT STEPS
  • Study the concept of isotopic abundance in geochemistry
  • Learn about simultaneous equations in algebra
  • Research the applications of lead isotopes in radiometric dating
  • Explore software tools for isotopic analysis and data visualization
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Chemists, geologists, and students studying isotopic analysis or radiometric dating techniques will benefit from this discussion.

subopolois
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how would i find the abundance of Pb 206, 207 and 208 if I am given the isotopic data of 207Pb/206Pb and 208Pb/206Pb?
 
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OK... Assume that the expression 206Pb means the molar fraction of lead-206. The same is true for 207Pb and 208Pb. You know that the molar fractions sum to 1 as well as the molar ratios of 207Pb/206Pb and 208Pb/206Pb. You can also calculate 207Pb/208Pb. From here it is merely solving for the simultaneous equations.
 
chemisttree said:
OK... Assume that the expression 206Pb means the molar fraction of lead-206. The same is true for 207Pb and 208Pb. You know that the molar fractions sum to 1 as well as the molar ratios of 207Pb/206Pb and 208Pb/206Pb. You can also calculate 207Pb/208Pb. From here it is merely solving for the simultaneous equations.

ok, i think i get it. here's my scenerio,
the Pb content is 468ppm
207Pb/206Pb= 0.2515
208Pb/206Pb= 0.0581
i could just use abgebra to find each right?
206=x
207=0.2515x
208=0.0581x
i just solve for x and put it back into the previous equation?
 
subopolois said:
ok, i think i get it. here's my scenerio,
the Pb content is 468ppm
207Pb/206Pb= 0.2515
208Pb/206Pb= 0.0581
i could just use abgebra to find each right?
206=x
207=0.2515x
208=0.0581x
i just solve for x and put it back into the previous equation?

I would use the definition... 206Pb + 207Pb + 208Pb = 1
 

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