Question

Aspartame, an artificial sweetener used in soft drinks, has the following percent composition: 57.14 percent C, 6.16 percent H, 9.52 percent N, and 27.18 percent O. What is the empirical formula of aspartame?

C28H36N4O10

C18H2N3O4

C7HNO

C14H18N2O5

Answer 1

wooahh

Answer 2

C -> 57.14 / 12 = 4.75
H -> 6.16 / 1 = 6.16
N -> 9.52 / 14 = 0.68
O -> 27.18 / 16 = 1.69

divided by the smalles number of moles :
C = 4.75 / 0.68 = 14
H = 6.16/ 0.68 = 18
N = 0.68/ 0.68 = 2
O = 1.69/ 0.68 = 5

You know the answer :)

Answer 3

To find an empirical formula of elements p and q.
Let the amount of p be x% in the compound.
And that of q be y%.

P( no.of moles )=x/atomic weight of p=t

Q(no.of moles )=y/atomic weight of q.=k

The empirical formula=p_tq_k

Answer 4

@NvidiaIntely
Your recent posts on chemistry are all very interesting. However, all these chemistry questions would be better posted in the chemistry section. The reason the chemistry section is not as active as it should be is because relevant questions are posted elsewhere. You are free to post a \(link\) to your post in the chemistry anywhere you want.

Answer 5

@mathmate lol... yeah XD