Question

Determine the empirical formula for a compound that is 70.79% carbon, 8.91% hydrogen, 4.59% nitrogen, and 15.72% oxygen.

Determine the empirical formula for a compound that is 70.79 carbon, 8.91 hydrogen, 4.59 nitrogen, and 15.72 oxygen.
C17H27NO3
C18H27NO2
C18H27NO3
C17H26NO3

Answer 1

The correct answer for the empirical formula is C18H27NO3.

Answer 2

The answer to your question is: C₁₈ H₂₇ N O₃

Step-by-step 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