Calculating Empirical and Molecular Formulas

What is an Empirical Formula?•The formula of a compound expressed as the smallest possible whole-number ratio of subscripts of the elements in the formula.

What is a Molecular Formula?

It is the formula of a compound in which the subscripts give the actual number of each element in the formula.

Calculating Empirical Formula• What if a percent composition is given. How do we find the empirical

formula?

• Example: Chemical analysis of a liquid shows that it is 60.00% C, 13.40% H, and 26.60% O by mass. Calculate the empirical formula of this substance.

– Step 1: Convert to grams

• Assume you have a 100.00 g sample, and convert percentages to grams.

– 60.00% Carbon is the same as 60.00 g of C

– 13.40% of Hydrogen is the same as 13.40 g of H

– 26.60% of Oxygen is the same as 26.60 g of O

Step 2: Convert grams to moles

Step 3: Divide each by the smallest decimal (in moles) to get whole numbers. * These numbers will be the subscripts.

1 mole C

13.40 g H

26.60 g O

60.00 g C

1 mole H

1 mole O

12.01 g C

1.01 g H

16.00 g O

= 5.00 mole C

= 13.27mole H

= 1.66 mole O

÷ 1.66

÷ 1.66

÷ 1.66

= 3.01 mol C

= 7.99 mol H

= 1.00 mol C

Of the three calculated numbers, this is the smallest decimal in moles, so you divide all the other decimals and itself by that number.

Round those answers to the nearest whole number and use them as the Subscripts on the final Empirical Formula.

THE EMPIRICAL FORMULA WOULD BE:

C3H8O1 C3H8O

Example 2 for Empirical Formulas• A compound consists of 72.2% magnesium and 27.8% nitrogen by mass.

What is the empirical formula?

• Step 1: Convert to grams : 72.2 % Mg = 72.2 g Mg 27.8 % N = 27.8 g N• Step 2: Convert grams to moles

72.2 g Mg

24.31 g Mg

27.9 g N

14.01 g N

1 mole Mg

1 mole N

= 2.97 mole Mg

= 1.99 mole N

= 1.49 mole Mg

÷ 1.99

Step 3: Divide each by the smallest decimal (in moles) to get whole numbers.

• Still not whole numbers. We have to get rid of the 1.49. If we multiply by 1 we get the same number of course. So, lets multiply by 2 and see what happens. Remember whatever we do to one side we do it to the other.

1.00 mole N x 2 = 2 mole N

THE EMPIRICAL FORMULA WOULD BE:Mg3N2

How do we determine the molecular formula from the empirical formula?• Example: The empirical formula for a compound is P2O5.

Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound.

• Step 1: Find the molar mass of the empirical formula.• 2 mol P x 30.97 g = 61.94 g P• 5 mol O x 16.00g = + 80.00 g O

141.94 g/mol

Step 2: Molar mass of a compound divided by the molar mass of empirical formula.

Experimental molar mass of compound

Molar mass of empirical formula

284 g/mol

141.94 g/ mol= 2.00

YOUR MOLECULAR FORMULA IS:

P4O10

Assignment

Determine the Empirical Formula for each of the following. Use the calculated empirical formula to calculate the molecular formula.

1. A sample compound with a molar mass of 34.00g/mol is found to consist of 0.44g H and 6.92g O. Calculate both empirical and molecular formulas.

2. A compound has a molar mass of 456.18 g/mol and consists of 3.0 g of Fe and 4.81g of S. Calculate both empirical and molecular formulas

3. A compound consists of 36.48% Na, 25.41% S, and 38.11% O. It

has a molar mass of 252.10 g/mol. Calculate both empirical and

molecular formulas.