Determine the empirical formula of a compound containing 48.38 grams of carbon, 8.12 grams of hydrogen, and 53.5 grams of oxygen. In an experiment, the molar mass of the compound was determined to be 180.15 g/mol. What is the molecular formula of the compound?

For both questions, show your work or explain how you determined the formulas by giving specific values used in calculations. (10 points)

Empirical formula = CH2O

Molecular formula = C6H12O6

Explanation:

According to this question, a compound is said to contain the following: 48.38g Carbon, 8.12g H, and 53.5g O.

First, we divide each mass value by its molar mass to convert to a mole value;

C = 48.38g ÷ 12g/mol = 4.03mol

H = 8.12g ÷ 1g/mol = 8.12mol

O = 53.5g ÷ 16g/mol = 3.34mol

Next, we divide each mole value by the lowest value (3.34mol)

C = 4.03mol ÷ 3.34 = 1.21

H = 8.12mol ÷ 3.34 = 2.43

O = 3.34mol ÷ 3.34 = 1

Approximately, the empirical ratio of CHO = 1:2:1, hence, the empirical formula is CH2O.

2. If the molar mass of the compound was determined to be 180.15 g/mol.

C = 12, H = 1, O = 16

(CH2O)n = 180.15

{12 + 1(2) + 16}n = 180.15

(12 + 2 + 16)n = 180.15

(30)n = 180.15

n = 180.15 ÷ 30

n = 6.005

Hence, (CH2O)n = (CH2O)6

Molecular formula = C6H12O6

1.solid

2.carbon dioxide

3.helium