Empirical and molecular formula in chemistry

Empirical formula

• a formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms
• In chemistry, the empirical formula of a chemical compound is the simplest positive integer ratio of atoms present in a compound.

For example, CH₃COOH has two carbons, four hydrogens and two oxygens. So we could write the formula like this: C₂H₄O₂ and so it reduces to CH₂O.

molecular formula:

• the molecular formula is a multiple of 6 times the empirical formula:

C_{(1 x 6)} H_{(2 x 6)} O_{(1 x 6)} which becomes C₆H₁₂O₆

• the formula of a compound in which the subscripts give the actual number of each element in the formula

Molecular formula	Empirical formula
H2O	H2O
CH3COOH	CH2O
CH2O	CH2O
C6H12O6	CH2O

Notice two things:1. The molecular formula and the empirical formula can be identical. 2. You scale up from the empirical formula to the molecular formula by a whole number factor.

• When teaching the method for converting percentage composition to an empirical formula, I have devised the following rhyme:

1. Percent to mass

1. Mass to mole

1. Divide by small

1. Multiply 'til whole

Step 1: Obtain the mass of each element present in gramsElement % = mass in g = mStep 2: Determine the number of moles of each type of atom presentm/atomic mass = Molar amount (M)Step 3: Divide the number of moles of each element by the smallest number of molesM / least M value = Atomic Ratio (R)Step 4: Convert numbers to whole numbers. This set of whole numbers are the subscripts in the empirical formula.R * whole number = Empirical Formula

• Here's an example of how it works. A compound consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula?

(1) Percent to mass:

Assume 100 g of the substance, then 72.2 g magnesium and 27.8 g nitrogen.

(2) Mass to moles:

for Mg: 72.2 g Mg x (1 mol Mg/24.3 g Mg) = 2.97 mol Mg
for N: 27.8 g N x (1 mol N/14.0 g N) = 1.99 mol N

(3) Divide by small:

for Mg: 2.97 mol / l.99 mol = 1.49
for N: 1.99 mol / l.99 mol = 1.00

(4) Multiply 'til whole:

for Mg: 2 x 1.49 = 2.98 (i.e., 3)
for N: 2 x 1.00 = 2.00

and the formula of the compound is Mg₃N₂.

Example

An oxide of aluminum is formed by the reaction of 4.151 g of aluminum and 3.692 g of oxygen. Calculate the empirical formula for this compound.

What do we know?

• The compound contains 4.151 g of aluminum and 3.692 g of oxygen
• We also know the atomic masses by looking them up on the periodic table. Aluminum (26.98 g/mol) and oxygen (16.00 g/mol).

Let's go through the steps to solve this:

Step 1: Determine the masses

We have these: 4.151 g of Al and 3.692 g of O

Step 2: Determine the number of moles by dividing the grams by the atomic mass

So let's do that now:

4.151 g Al x (1 mol Al / 26.98 g Al) = 0.1539 mol Al atoms

3.692 g O x (1 mol O / 16.00 g O) = 0.2398 mol O atoms

Step 3: Divide the number of moles of each element by the smallest number of moles

0.1539 mol Al / 0.1539 = 1.000 mol Al atoms

0.2398 mol O / 0.1539 = 1.500 mol O atoms

Step 4: Convert numbers to whole numbers

1.000 Al * 2 = 2.000 Al atoms and 1.500 O atoms * 2 = 3.000 O atoms

The compound contains 2 Al atoms for every 3 O atoms

Empirical Formula = Al2O3

Example Problem 

A compound was analyzed and found to contain 13.5 g Ca, 10.8 g O, and 0.675 g H.  What is the empirical formula of the compound?

Start with the number of grams of each element, given in the problem.

Convert the mass of each element to moles using the molar mass from the periodic table.

Divide each mole value by the smallest number of moles calculated.  Round to the nearest whole number.

This is the mole ratio of the elements and is represented by subscripts in the empirical formula.