A 5.00 g sample of compound contains 2.43 g of carbon, 0.41 g of hydrogen, and the remainder of the mass is oxygen. Its molecular mass is 74.09 g/mol. Answer the following questions:

Question

a. What is the empirical formula of the compound?

b. Show mathematical proof along with explaining why the empirical formula and the molecular formula are the same?

c. Draw a possible, valid Lewis Structure for the molecular formula of the compound.

Answer 1

To find the empirical formula, convert grams of each element to moles.

moles C = 2.43 g C x 1 mol C/12 g = 0.2025 moles C

moles H = 0.41 g H x 1 mol H/1 g = 0.41 moles H

moles O = 5.00 g - 2.43 g - 0.41 g = 2.16 g O x 1 mol O/16 g = 0.135 moles O

Now, try to get whole numbers by dividing all by 0.135...

moles C = 0.2025/0.135 = 1.5 moles C

moles H = 0.41/0.135 = 3.0

moles O = 0.135/0.135 = 1

Now just multiply all by 2 to get all whole numbers...

moles C = 3

moles H = 6

moles O = 2

Empirical formula = C₃H₆O₂

Molar mass of this formula = 3x12 + 6x1 + 2x16 = 36 + 6 + 32 = 74 g/mol

b). Since the molar mass of the empirical formula is the same as that of the molecular formula, the two must be the same

c). Unfortunately, I cannot draw Lewis structures on the platform.