Answer:
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