Answer:
The answer to your question is: C₁₈ H₂₇ N O₃
Explanation:
Data
Carbon = 70.79 g
Hydrogen = 8.91 g
Nitrogen = 4.58 g
Oxygen = 15.72 g
Process
AT C = 12 g
AT H = 1 g
AT N = 14 g
AT O = 16 g
Carbon
12 g ------------------------ 1 mol
70.79 g ------------------------- x
x = (70.79 x 1) / 12
x = 5.9 mol of C
Hydrogen
1 g ----------------------- 1 mol
8.91 g --------------------- x
x = (8.91 x 1) / 1
x = 8.91 mol of H
Nitrogen
14 g ---------------------- 1 mol
4.58 g ------------------- x
x = (4.58 x 1) / 14
x = 0.33 mol
Oxygen
16 g ------------------------ 1 mol
15.72 g -------------------- x
x = (15.72 x 1)/16
x = 0.98
Divide by the lowest number of moles
Carbon 5.9 / 0.33 = 17.9 ≈ 18
Hydrogen 8.91 / 0.33 = 27
Nitrogen 0.33 / 0.33 = 1
Oxygen 0.98 / 0.33 = 2.9 ≈ 3
C₁₈ H₂₇ N O₃
