Question

What is the empirical formula of a compound that contains 50.0% carbon, 6.7% hydrogen, and 43.3% oxygen by mass?

Answer 1

Assume 100 g of unknown compound.

50.0⋅g 12.011⋅g⋅mol−1 = 4.163 mol⋅C.

6.7⋅g 1.00794⋅g⋅mol−1 = 6.65 mol⋅H.

43.3⋅g 15.999⋅g⋅mol−1 = 2.71 mol⋅O.

divide thru by the lowest molar quantity, that of oxygen to get a formula of C1.54H2.45O1, double to give a whole number EMPIRICAL formula of C3H5O2

Answer 2

C₃H₅O₂

Step-by-step explanation:

Mass of C = 50 g

Molar mass of C = 12 g/mol

Moles of C = 50 g / 12 g/mol = 4.16

Mass of H = 6.7g

Molar mass of H = 1 g/mol

Moles of H = 6.7g / 1 g/mol = 6.7 mol

Mass of O = 43.3

Molar mass of O = 16 g/mol

Moles of O = 43.3 g / 16 g/mol = 2.7 mol

Dividing by the smallest mole:

Molar ratio C = 4.16 / 2.7 = 1.5

Molar ratio H = 6.7/ 2.7 = 2.5

Molar ratio O = 2.7 / 2.7 = 1

Getting a whole number ratio:

C = 1.5 x 2 = 3, H = 2.5 X 2 = 5, O = 1 X 2 =2

Therefore the empirical formula is :

C₃H₅O₂