Calculate the empirical formula for each stimulant based on its elemental mass percent composition. a. nicotine (found in tobacco leaves): c 74.03%, h 8.70%, n 17.27% b. caffeine (found in coffee beans): c 49.48%, h 5.19%, n 28.85%, o 16.48%

A.

B.

Explanation:

Hello,

a. In this case, based on the given percentages, one computes the moles of C, H and N as follows:

Now, we divide each element's moles by the smallest amount of moles, that is nitrogen's moles:

So the empirical formula is:

b. In this case, based on the given percentages, one computes the moles of C, H, N and O as follows:

Now, we divide each element's moles by the smallest amount of moles, that is oxugen's moles:

So the empirical formula is:

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the correct answer is b. a strong acid completely dissociates to its ions when dissolved in water. completely dissociates to and ions. a is incorrect because does not fully ionize to and ions and some remain as the compound . c and d are incorrect because, and do not fall into the category of bronsted lowry acids. a bronsted lowry acid is a proton donor and since both molecules do not have a proton to donate they are not bronsted lowry acids. likewise,   if we use the lewis acid definition, they cannot be acids. a lewis acid is an electron pair acceptor. both and have a complete octet so can not accept any electrons which disqualifies them as acids.