Question

1. argon (ar) has two stable isotopes: ³⁶ar, ³⁸ar, and ⁴⁰ar. based on the table below,

a) without doing any calculations, explain whether the % abundance of ³⁸ar or ⁴⁰ar is expected to be much greater than 50%. (0.5 mark)
b) what is the calculated % abundance of ³⁸ar? show all of your work in full. (3 marks)

isotope or atom	isotopic mass	% abundance
³⁶ar	35.96754	0.3336%
³⁸ar	37.96273	??
⁴⁰ar	39.96238	??

a) (hint: take a look at the average atomic mass for ar on the periodic table - what does it signify?) 39.95 is the average atomic mass so ⁴⁰ar is much more common than ³⁸ar.
b) (hint: all of the abundances have to add up to 100% or a fractional value of 1.
let x be (%abund ³⁸ar), y be (%abund ⁴⁰ar) and z be (%abund ³⁶ar).
therefore x + y + z = 1 and x + y+0.003336 = 1
can you rearrange the equation for y and set up the average atomic mass formula?)
when you think you have set up the above hint correctly - show your ta to confirm youre on the right track... then solve the rest of the question. use extra paper as necessary.
37.96273x+(39.96238y)+(35.96754×0.003336)=39.948
37.96273x + 39.96238y = 39.82801229

Explanation:

Step1: Analyze relative abundance for part a

The average atomic mass of Ar is approximately 39.95. Since it is very close to the mass of \(^{40}\text{Ar}\) (39.96238), \(^{40}\text{Ar}\) is expected to have a much greater abundance than \(^{38}\text{Ar}\). So the % abundance of \(^{40}\text{Ar}\) is expected to be much greater than 50%.

Step2: Set - up equations for part b

Let the percentage abundance of \(^{38}\text{Ar}\) be \(x\), of \(^{40}\text{Ar}\) be \(y\) and of \(^{36}\text{Ar}\) be \(z = 0.003336\). We know that \(x + y+z = 1\), so \(y=1 - x - 0.003336=0.996664 - x\). The average - atomic - mass formula is \(39.95=37.96273x + 39.96238y+35.96754z\). Substitute \(y = 0.996664 - x\) and \(z = 0.003336\) into the formula:
\[

$$\begin{align*} 39.95&=37.96273x+39.96238(0.996664 - x)+35.96754\times0.003336\\ 39.95&=37.96273x + 39.96238\times0.996664-39.96238x+35.96754\times0.003336\\ 39.95&=37.96273x+39.82801 - 39.96238x+0.1199\\ 39.95&=(37.96273x-39.96238x)+39.82801 + 0.1199\\ 39.95&=- 1.99965x+39.94791\\ 1.99965x&=39.94791 - 39.95\\ 1.99965x&=-0.00209\\ x&=\frac{- 0.00209}{-1.99965}\approx0.001045 = 0.1045\% \end{align*}$$

\]

Answer:

a) The % abundance of \(^{40}\text{Ar}\) is expected to be much greater than 50% because the average atomic mass of Ar (39.95) is very close to the mass of \(^{40}\text{Ar}\) (39.96238).
b) The calculated % abundance of \(^{38}\text{Ar}\) is approximately \(0.1045\%\)