Percent composition and empirical formulas

Question

A sample of lidocaine contains 3.588 g of C, 0.473 g of H, 0.598 g of N, and 0.342 g of O. What’s the perfect composition and empirical formula of lidocaine

Answer 1

See explanation in previous question/answer.

3.588 g C x 1 mol/12 g = 0.299 mol C

0.473 g H x 1 mol/1 g = 0.473 mol H

0.598 g N x 1 mol/14 g = 0.0427 mol N

0.342 g O/16 g = 0.0214 mol O

Dividing all by 0.0214 we get...

14 mol C

22 mol H

2 mol N

1 mol O

Empirical formula = C₁₄H₂₂N₂O