Calculating Approximate Atomic Mass of Lead | Isotopes of Pb and Percent by Mass

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SUMMARY

The discussion focuses on calculating the approximate atomic mass of lead (Pb) using its isotopes and their percent by mass. The isotopes provided are 204Pb, 206Pb, 207Pb, and 208Pb, with respective abundances of 1.37%, 26.25%, 20.82%, and 51.55%. The initial calculation mistakenly used mass numbers instead of atomic masses, leading to an incorrect approximate atomic mass of 232.3 amu. The correct calculation, using atomic masses, yields an approximate atomic mass of 207.1912 amu, emphasizing that the result is approximate due to the unknown composition of less than 0.01% of lead isotopes.

PREREQUISITES
  • Understanding of isotopes and their significance in atomic mass calculations.
  • Familiarity with atomic mass versus mass number.
  • Basic proficiency in weighted averages and percentage calculations.
  • Knowledge of atomic masses of specific isotopes, particularly lead isotopes.
NEXT STEPS
  • Research the atomic masses of lead isotopes using reliable sources like Wolfram Alpha or scientific databases.
  • Learn about the concept of mass defect and its impact on atomic mass calculations.
  • Explore the differences between mass number and atomic mass in greater detail.
  • Study the methods for calculating weighted averages in chemistry.
USEFUL FOR

Chemistry students, educators, and anyone involved in nuclear physics or isotopic studies will benefit from this discussion, particularly those interested in atomic mass calculations and isotopic abundance.

tigerwoods99
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Homework Statement


1. The Table below lists the isotopes of Pb and their percent by mass:

Isotope

% Mass

82p, 122n 1.37%

82p, 124n 26.25%

82p, 125n 20.82%

82p, 126n 51.55%(a) Using these data, calculate the approximate atomic mass of lead;

(b) Why is this number in (a) only approximate?

Show all work!

Homework Equations



Look below, i was able to solve A, however I am wondering why this number is only an approximate? Thanks!

The Attempt at a Solution



82p, 122n --> 204 mass number (.1370)
82p, 124n --> 206 mass number (.2625)
82p, 125n --> 207 mass number (.2082)
82p, 126n --> 208 mass number (.5155)

(204*.1370) + (206*.2625) + (207*.2082) + (208*.5155) = approximate atomic mass of lead
(27.95) + (54.08) + (43.10) + (107.2) = approximate atomic mass of lead
(a) 232.3 amu = approximate atomic mass of lead
 
Last edited:
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100.00%?

No need for crying, we can read standard font no problems.
 
Borek said:
100.00%?

No need for crying, we can read standard font no problems.


Im just trying to figure out the answer to the part that says like why is the average weight using the % natural occurrence only an approximate.
 
And I have gave you a hint. Do you know 100.00% composition of the mixture?
 
Borek said:
And I have gave you a hint. Do you know 100.00% composition of the mixture?

im not sure, not really understanding what you are talking about
 
tigerwoods99 said:

Homework Statement


The Attempt at a Solution



82p, 122n --> 204 mass number (.1370)
82p, 124n --> 206 mass number (.2625)
82p, 125n --> 207 mass number (.2082)
82p, 126n --> 208 mass number (.5155)

(204*.1370) + (206*.2625) + (207*.2082) + (208*.5155) = approximate atomic mass of lead
(27.95) + (54.08) + (43.10) + (107.2) = approximate atomic mass of lead
(a) 232.3 amu = approximate atomic mass of lead

Somewhere this is wrong, If you have all the individual masses in the range 204-208, how can the average be 232?

As to why it's approximate, what is the relative atomic mass of 204Pb?
 
You have assumed that fractions given sum to 100%, but have you checked it?

What if I will tell you that lead contains 26.25% of ^{124}_{82}Pb and 20.82% ^{125}_{82}Pb? Will you be able to calculate average atoms mass?

Edit: sjb is right, I missed something important :blushing:
 
Ok yea i found my error, I converted 1.37% to .1370 instead of .0137. So the approximate atomic mass of lead would be 207.1912.

I'm still confused on why this number is an approximate.
 
That sounds better, yes. What is the relative atomic mass of 204Pb?
 
  • #10
sjb-2812 said:
that sounds better, yes. What is the relative atomic mass of 204pb?

204?
 
  • #11
No. Less than that.
 
  • #12
I'm not sure... it says 204.

http://www.wolframalpha.com/entities/isotopes/lead_204/2y/re/x8/
 
Last edited by a moderator:
  • #13
tigerwoods99 said:
I'm not sure... it says 204.

http://www.wolframalpha.com/entities/isotopes/lead_204/2y/re/x8/

No. Read again. You are not looking for a mass number.
 
Last edited by a moderator:
  • #14
82 ?
 
  • #15
What are you calculating?
 
  • #16
is it 204*0.0137? = 2.7948 ?? I'm not really sure what you asking me to do.
 
  • #17
You are calculating AVERAGE ATOMIC MASS, so to get correct result you should use ATOMIC MASSES, not mass numbers.
 
  • #18
Ok, So i got that 207.1912 is the approximate atomic mass of lead. Now i just need to figure out why this number is only an approximate.
 
  • #19
Previous one was approximate, this one (assuming you took atomic masses for all isotopes) should be correct. That is, you have no idea what is 0.01% of the mixture, so there is still some error in calculations.
 
  • #20
Ok so this was my starting data
82p, 122n 1.37%

82p, 124n 26.25%

82p, 125n 20.82%

82p, 126n 51.55%

So the reason this calculation is an approximate is because the atomic mass did not take into consideration the electrons, is that correct? Or why is it an approximate, even with the atomic masses. B/c i know atomic mass = protons + neutrons but the electrons are very minute
 
  • #21
No, I think the RAM takes into account electrons. What does the Wolfram page you have quoted say is the atomic mass of lead-204?
 
  • #23
Alright, but I am still not sure where I am headed with this. I got the right number correct? It's just a matter of explaining why this number is only an approximate.

Thanks everyone
 
  • #24
Question is confusing, that's not your fault.

If you use 204 for atomic mass of the ^{122}_{82}Pb your result will be off for obvious reason - atomic mass of ^{122}_{82}Pb is NOT 204. But if you will use 203.973 your result will be as correct as possible, the only remaining reason for lack of accuracy being fact that you have no idea what isotopes constitute remaining 0.01% of the lead.

--
 
  • #25
Yea that makes sense. How did you get the 203.9... I got it being something around 207.
 
  • #26
According to wolfram page you linked to earlier, 203.9... is a correct atomic mass of ^{122}_{82}Pb.
 
  • #27
Ok so here is what I have right now:

82p, 122n --> 204 mass number (.0137) %
82p, 124n --> 206 mass number (.2625) %
82p, 125n --> 207 mass number (.2082) %
82p, 126n --> 208 mass number (.5155) %

(204*.0137) + (206*.2625) + (207*.2082) + (208*.5155) = approximate atomic mass of lead
(2.795) + (54.08) + (43.10) + (107.2) = approximate atomic mass of lead
(a) 207.2 amu = approximate atomic mass of lead

This is what the directions say (a) Using these data, calculate the approximate atomic mass of lead; (b) Why is this number in (a) only approximate?

I'm still not sure why the number in (a) is only approximate of the atomic mass of lead. Is it because like someone said before, that we don't know what the .01% of the isotope is?
I'm really not sure
 
  • #28
Mass number and atomic mass of the isotope are DIFFERENT. You have calculated average using MASS NUMBERS which are incorrect, so your result must be incorrect.
 
  • #29
Oh. How do I do it using atomic mass?
 
  • #30
Exactly the same way, but using atomic mass instead of mass number.
 

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