[leena]

There are 3 parts to this question.The 1st two parts were not related to Mass spectrometry,but the gas produced(Y) in those reactions was used in this part of the question(I don't know if that's relevant here but I just thought I'd mention it)

(Q) c) The mass spectrum of Y gave a major peak at m/e=64 and a minor one at m/e=66 ,in the ratio of approximately 23:1.The common isotopes of the elements which maybe involved are given below

isotope        abundances (%)

¹H      99.985
²H      0.01200
¹⁶O     99.801
¹⁸O      0.2000
²³Na    100.0
³²S       95.00
³³S       0.800
³⁴S       4.200
³⁵Cl       72.50
³⁷Cl       24.50

1) which element is responsible for the mass spectrum containing in the ratio of approximately 23:1 ?
2) Suggest a formula,including isotopic masses of the species responsible for the peak at m/e = 64

I know I'm supposed to post my attempt at solving the problem,but I really really don't know where to start.
I have looked at my class notes and countless websites,but they didn't help.We've done questions where we had to find the RAM of an element by looking at the peaks shown in the graphs given,but that was a bit different from this(I think).

First of all,I hope someone can tell me what the 23:1 ratio indicates?

Thank you

[sjb]

There are 3 parts to this question.The 1st two parts were not related to Mass spectrometry,but the gas produced(Y) in those reactions was used in this part of the question(I don't know if that's relevant here but I just thought I'd mention it)

(Q) c) The mass spectrum of Y gave a major peak at m/e=64 and a minor one at m/e=66 ,in the ratio of approximately 23:1.The common isotopes of the elements which maybe involved are given below

isotope        abundances (%)

¹H      99.985
²H      0.01200
¹⁶O     99.801
¹⁸O      0.2000
²³Na    100.0
³²S       95.00
³³S       0.800
³⁴S       4.200
³⁵Cl       72.50
³⁷Cl       24.50

1) which element is responsible for the mass spectrum containing in the ratio of approximately 23:1 ?
2) Suggest a formula,including isotopic masses of the species responsible for the peak at m/e = 64
...
First of all,I hope someone can tell me what the 23:1 ratio indicates?

Thank you

Well, you have an individual molecular weight based on one isotope being 64, and another being 66. Can you see if there are any individual atoms in your list that have two (or more) isotopes differing by 2 mass units? Once you have done that, can you make an estimation of their relative abundances (relative to the isotope with the highest mass)?

[leena]

Thanks for responding,sjb!

Can you see if there are any individual atoms in your list that have two (or more) isotopes differing by 2 mass units?

Oh yes!
Those atoms would be
¹⁶O , ¹⁸O
³²S , ³⁴S  and
³⁵Cl , ³⁷Cl  ?

can you make an estimation of their relative abundances (relative to the isotope with the highest mass)?

I'm not quite sure what you mean here though
¹⁸O has almost 100%
¹⁶O  about 1%,so % adundancy of O-16 is approximately a 100 times more

About the same for ³²S and ³⁴S ?
and the % abundance for³⁷Cl is approx. 3 times more than that for ³⁵ Cl ?
But I don't really know what I'm supposed to do with these values

[sjb]

...
I'm not quite sure what you mean here though
¹⁸O has almost 100%
¹⁶O  about 1%,so % adundancy of O-16 is approximately a 100 times more

About the other way round, actually. But yes, ¹⁶O is approx 100 times more abundant than ¹⁸O.

About the same for ³²S and ³⁴S ?

Are they? What is 95 / 4.2?

and the % abundance for³⁷Cl is approx. 3 times more than that for ³⁵ Cl ?
But I don't really know what I'm supposed to do with these values

Again, the other way around...

[leena]

and the % abundance for³⁷Cl is approx. 3 times more than that for ³⁵ Cl ?
But I don't really know what I'm supposed to do with these values

Again, the other way around...

Yes.so ³⁵ Cl is approximately 3 times more abundant than ³⁷Cl

What is 95 / 4.2?

23. That is ³²S is approximately 23 times more abundant than ³⁴S.

Now I think I know what they are asking for in the 1st question.
So the element responsible for the mass spectrum in the 23:1 ration would be S ?

Now for part 2,
Inorder to get an m/e of 64,there are only 2 possible compounds (I think?)
¹⁶O  ³²S ¹⁶O (i.e. SO₂)
and
³²S ³²S.
The 2nd one doesn't seem like a likely compound to form,so would SO₂ be the the answer?

Thanks in advance!

[sjb]

23. That is ³²S is approximately 23 times more abundant than ³⁴S.

Now I think I know what they are asking for in the 1st question.
So the element responsible for the mass spectrum in the 23:1 ration would be S ?

Sounds good here - note that there are other isotopes that also contribute ^12/14C, and similar, but the largest contribution is from the sulfur.

Now for part 2,
Inorder to get an m/e of 64,there are only 2 possible compounds (I think?)
¹⁶O  ³²S ¹⁶O (i.e. SO₂)
and
³²S ³²S.
The 2nd one doesn't seem like a likely compound to form,so would SO₂ be the the answer?

Thanks in advance!

Well, I don't know what the earlier parts were about, but in the absence of any further information, I'd also note that in the case of ³²S=³²S you'd probably get effects like in chlorine - have you covered that? SO₂ is probably the most likely, in my opinion.

[leena]

Well,in part b) of the question,we were asked to react Na₂SO₃ with HCl which after reacting gives out a gas Y(which is obviously SO₂?),
which we then use in the mass spectrometer,so I think the answer's infact SO₂.

I'd also note that in the case of 32S=32S you'd probably get effects like in chlorine - have you covered that?

What effects are there,I'm just curious to know.If it's not too much trouble can you expand on this?

Already very grateful for everything.
Thank you very much,sjb!

[sjb]

Have a look through http://www.chemguide.co.uk/analysis/masspec/elements.html , and http://www.chemguide.co.uk/analysis/masspec/mplus2.html .

[leena]

Wow!those links were just amazingly easy to understand!Everythings so crystal clear to me now(I wish I had found them earlier but)Thank you so much for sharing them!

Now for part 2 of the question I've posted,the chemical formula would be
 (¹⁶O=³²S=¹⁶O )⁺ ion,right ?

[sjb]

Probably, at your level, yes.