This is the part 3 of the calculator post series using my Casio ES991 calculator. This covers how to use the EQN mode of the calculator and using it to find two unknowns simultaneously if you have two equations already. This problem was taken out from Schaum’s Outline for College Chemistry:
Naturally, occurring carbon consists of two isotopes, ^{12}C and ^{13}C. What are the percentage abundances of the two isotopes in a sample of carbon whose atomic weight is 12.01112?
To solve this you must know how to solve for the average atomic weight of an element. The average atomic weight of any element can be computed by summation of the products of the mass of each isotope by its percentage abundance. For example:
It has been found by mass spectrometric analysis that in nature the relative abundances of the various isotopic atoms of silicon are: 92.23% ^{28}Si, 4.67% ^{29}Si, and 3.10% ^{30}Si. Calculate the atomic weight of silicon from this information and from the nuclidic masses.
The solution is just simply: 0.9223(28) + .0467(29) + 0.0310(30) = 28.109 amu. If you add up all the percentage abundances, the sum is 100% or just 1. In our problem, we could say that the abundances of Carbon 12 and Carbon 13 would also add up to 1. So we let x
be the percentage abundance of ^{12}C and y as the percentage abundance of ^{13}C. Therefore, x + y = 1. This will be our equation 1. Equation 2 will come from: x(12) + y(13) = 12.01112.
To solve for the percentage abundances, press MODE à 5 [EQN] à 1 [anx + bny = cn].
Where an, bn and cn are coefficients of your equations. Let me rewrite my equations below as:
1x + 1y = 1
12x + 13y = 12.01112
Input as follows:
a 
b 
c 

1 
1 
1 
1 
2 
12 
13 
12.01112 
Then press enter. The answer should be:
x = 0.98888
y = 0.01112
or 98.89% and 1.11% for ^{12}C and ^{13}C respectively.
Now try:
35Cl and 37Cl are the only naturally occurring chlorine isotopes. What percentage distribution account for the atomic weight, 35.453? Your answer should be 77.35% for ^{35}C and 22.65% for ^{37}C.