kael9610
contestada

Will vote brainliest!! Can someone please help me with the empirical formula of
C:46.66% H:4.48%
N:31.10% O:17.76%
I am getting 3C 4H 2N 1O but this is not one of the options so I don’t know if I’m doing it wrong or the question is wonky because I did all the others right

Respuesta :

Answer:

It’s simple. You’re doing it right, the question is just messed up. If you think about it it’s like H2O

Explanation:

Answer:

The answer is C7H8N4O2.

Same problem as your other question - you rounded in a place where you shouldn't have (see explanation below, exact same process)!

Explanation:

These are percents by mass. Pretend you have a 100g sample. That means you have:

46.66g C

4.48g H

31.10g N

17.76g O

Divide each by molar mass to find the moles of each:

C: 46.66 / 12 = 3.89 mol

H: 4.48 / 1 = 4.48 mol

N: 31.10 / 14 = 2.22 mol

O: 17.76 / 16 = 1.11 mol

Now divide each of these values by the lowest of the three to find the ratio between each element in the compound:

C: 3.89 / 1.11 ≈ 3.5

H: 4.48 / 1.11 ≈ 4

N: 2.22 / 1.11 = 2

O: 1.11 / 1.11 = 1

As with your other question, 3.5 is too different from 3, so you can't round it down. The .5 makes it easy to find the LCD - it's 2. Multiply each of the numbers by 2 to get your final answer:

C: 3.5 x 2 = 7

H: 4 x 2 = 8

N: 2 x 2 = 4

O: 1 x 2 = 2

Again, if you were to take the percent masses of the empirical formula you came up with, you'll see they're quite different from the one the question provided, which is why you can't round in this situation.