# Chemistry - Using a H NMR spectrum to determine the structure of a protein supplement

Let me try to explain the point about the 10 atoms and the integral.

The area under the peaks and the ratios of these areas correspond to the ratios of the hydrogen atoms. The blue line adds all these areas up as it goes along. That means at the first peak on the left you simply get the area of that peak. But at the second peak you still have the amount from the first peak plus the new area, which is why you have to look at the difference. If we denote the three peak areas as $$A_1$$, $$A_2$$, and $$A_3$$, the height of the blue line goes from $$A_1$$ to $$A_1 + A_2$$, then lastly $$A_1 + A_2 + A_3$$.

With your values we have, \begin{aligned} A_1 &= \pu{2.7 cm} \\ A_1+A_2 &= \pu{4.75 cm} \\ A_1+A_2+A_3 &= \pu{6.9 cm} \end{aligned} Which allows us to easily determine $$A_2 = \pu{2.05 cm}$$ and $$A_3 = \pu{2.15 cm}$$. (Alternatively you could have simply read off the difference of the blue line between two peaks directly when you measured it).

This gives us the ratios of $$\pu{2.7 cm} : \pu{2.05 cm} : \pu{2.15 cm}$$. If we assume that the signal at $$\pu{1.25 ppm}$$ (i.e. $$A_3$$) is a methyl group with $$3$$ hydrogen atoms, we can divide all values by $$\pu{2.15 cm}$$ and multiply by 3 to normalize the ratio. We get approximately hydrogen atom ratios of $$3.76 : 2.86 : 3.00$$. Rounding the result gives a ratio of $$4:3:3$$ which agrees with $$10$$ atoms in total.