Question

I NEED HELP PLEASE! :)

Answer 1

Answer:

$C_{21} H_{23} NO_{5}$

Explanation:

First thing is we have assume all the percents are grams so we have

68.279g C, 6.2760g H, 3.7898g N, and 21.656g O

Now convert each gram to moles by dividing the the molar mass of each element

68.279g/12.01g= 5.685 moles of C

6.2760g/1.01g= 6.214 moles of H

3.7898g N/14.01g= 0.271 moles  of N

21.656g O/ 16.00g= 1.354 moles of O

Now to find the lowest ratios divide all the moles by the smallest number of moles you found, in our case, the smallest moles is 0.271 moles of N so divide everything by that....

5.685 moles/0.271 moles ------> ~21 C

6.214 moles/0.271 moles --------> ~23 H

0.271 moles  / 0.271 moles  ---------> 1 N

1.354 moles/ 0.271 moles ----------> ~5 O

So the empirical formula is C21H23NO5 $C_{21} H_{23} NO_{5}$