The elemental mass percent composition of malonic acid is 34.63% C, 3.87% H, and 61.50% O. Part A Determine the empirical formula of malonic acid. Express your answer as a chemical formula.

The empirical formula is C3H4O4

Explanation:

Step 1: Data given

Suppose we have 100 grams of malonic acid

Malonic acid contains:

34.63 % C = 34.63 grams C

3.87 % H = 3.87 grams H

61.50 % O = 61.50 grams O

Molar mass of C = 12.01 g/mol

Molar mass of H = 1.01 g/mol

Molar mass of O = 16.00 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = 34.63 grams / 12.01 g/mol

Moles C = 2.883 moles

Moles H = 3.87 grams / 1.01 g/mol

Moles H = 3.83 moles

Moles O = 61.50 grams / 16.00 g/mol

Moles O = 3.844 moles

Step 3: Calculate mol ratio

We divide by the smallest amount of moles

C: 2.883 / 2.883 = 1

H: 3.83/ 2.883 = 1.33

O: 3.844 / 2.883 = 1.33

The empirical formula is C3H4O4

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.