Question

question is attached. Please Help! Fan and Medal!!

Answer 1

so, the 2 peaks observed are 44 amu and 42 amu, the % abundance of 42 is about twice that of 44.

so, \(A_{42}= 100\%\; and\; A_{44}=50\%\;\rightarrow A_{42}=\dfrac{2}{3}\; and \;A_{44}=\dfrac{1}{3}\)

we know that: \(A_{average\;atomic\;weight} = 40.5929 amu = A_{42}*A_{42}\%+A_{44}*A_{44}\%\)

But, \(A_{44}*(\dfrac{2}{3})+A_{42}*(\dfrac{1}{3})=42.6667 amu\)

So the calculated and observed average atomic values are not the same.

Can you take it from here?

Answer 2

first of all how did u get A_42 is 2/3 and A_44 is 1/3??

Answer 3

I used the height of the peaks which is their relative abundance.
for example, if you have A = 100% and B=50%, in total of 150 %.
In terms of absolute abundance
A=100/150=2/3
B=50/150=1/3

Answer 4

btw, i just noticed i wrote "\(A_{44}∗(2/3)+A_{42}∗(13)\)=42.6667amu", wrong.

I switched the abundance around, BUT when i plugged into the calc i did it right.

it should read:

" \(A_{42}∗(\dfrac{2}{3})+A_{44}∗(\dfrac{1}{3})\)=42.6667 amu "

Answer 5

ok, now after that what do i need to do??

Answer 6

It says in the question, "Sketch a best guess for what the spectrum should be".

Answer 7

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