[MJF]

A mass spectrum of an unknown compound has a molecular ion peak with a relative intensity of 43.27% and a M+1 peak with a relative intensity of 3.81%. How many C atoms are present in the compound?

a. 8  b. 4  c. 6  d. 2

I recalled from lecture that when dealing with M+1 peak, we used this formula:

intensity of M+1 peak / 0.011 = number of C atoms (important to round to the nearest whole number)

So I dealt with: 0.0381 / 0.011 = 3.46

I decided to go with Choice B since that seemed the closest, however, the correct answer was Choice A and I don't have a clue why. Some help please?

[Archer]

You haven't included the intensity of the ¹²C peak in your calculation.

[MOTOBALL]

It is the ratio of the M peak to the M+1 peak that is used to estimate the C atom count.

Thus, 43.27/3.81 = 11.3; natural abundance of 13C is 1.08%---this is usually rounded to 1.1% to allow for 2H,17O etc.

Then, 11.3/1.1 = 10 and answer A = 8 is the closest value.

[MJF]

It is the ratio of the M peak to the M+1 peak that is used to estimate the C atom count.

Thus, 43.27/3.81 = 11.3; natural abundance of 13C is 1.08%---this is usually rounded to 1.1% to allow for 2H,17O etc.

Then, 11.3/1.1 = 10 and answer A = 8 is the closest value.

Yeah you're right Motoball. I apparently wrote the formula down during lecture but forgot about multiplying the intensity. Won't make that mistake again

[Archer]

It is the ratio of the M peak to the M+1 peak that is used to estimate the C atom count.

Thus, 43.27/3.81 = 11.3; natural abundance of 13C is 1.08%---this is usually rounded to 1.1% to allow for 2H,17O etc.

Then, 11.3/1.1 = 10 and answer A = 8 is the closest value.

No, this is not correct.

3.81(M+1)/43.27(M)*100 = 8.81% ¹³C

8.81/1.1=8 Carbons

[MOTOBALL]

Archer, thanks for spotting my error; I had reversed the ratios.

MJF, take the (M+1)/M ratio, not as I had written.