Answer:
The answer to your question is C₄H₅N₂O
Explanation:
Process
1.- Calculate the percent of oxygen in the sample
Percent of oxygen = 100 - 49.49 - 5.15 - 28.87
Percent of oxygen = 16.49 %
2.- Write the percents as grams
C = 49.49 g
H = 5.15 g
N = 28.87 g
O = 16.47 g
3.- Convert the grams to moles
C 12 g ------------------- 1 mol
49.49 g ---------------- x
x = (49.49 x 1) 12
x = 4.12 moles
H 1 g ------------------- 1 mol
5.15 g ---------------- x
x = (5.15 x 1)/ 1
x = 5.15 moles
N 14 g --------------- 1 mol
28.87 g ---------- x
x = (28.87 x 1) / 14
x = 2 mol
O 16 g ---------------- 1 mol
16.49 g ----------- x
x = (16.49 x 1) / 16
x = 1.03 moles
4.- Divide by the lowest number of moles
C 4.12 / 1.03 = 4
H 5.15 / 1.03 = 5
N 2 / 1.03 = 1.9 ≈ 2
O 1.03 / 1.03 = 1
5.- Write the empirical formula
C₄H₅N₂O
