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.

1. The empirical formula is CH2O

2. The molecular formula is C6H12O6

Explanation:

Data obtained from the question include:

Carbon (C) = 48.38g

Hydrogen (H) = 8.12g

Oxygen (O) = 53.5g

Molar Mass of the compound = 180.15 g/mol

1. The empirical formula can be obtained as follow:

C = 48.38g

H = 8.12g

O = 53.5g

Divide each by their molar mass

C = 48.38/12 = 4.032

H = 8.12/1 = 8.12

O = 53.5/16 = 3.344

Next, divide by the smallest

C = 4.032/3.344 = 1

H = 8.12/3.344 = 2

O = 3.344/3.344 = 1

Therefore, the empirical formula is CH2O.

2. The molecular formula is mostly a multiple of the empirical i. e

Molecular formula = > [CH2O]n

Now, we need to find the value of 'n' in order to obtain the molecular formula.

From the question given, we were told that the molar mass of the compound is 180.15 g/mol. With this information, the molecular formula can be obtained as follow:

[CH2O]n = 180.15

[12 + (2x1) + 16]n = 180.15

[12 + 2 + 16]n = 180.15

30n = 180.15

Divide both side by the coefficient of n i. e 30

n = 180.15/30

n = 6

Therefore, the Molecular formula is

=> [CH2O]n

=> [CH2O]6

=> C6H12O6