Class 11 Chemistry : Determining Empirical and Molecular Formula of Compounds

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Molecular formula

It describes the exact number and type of atoms in a single molecule of a compound. The constituent elements are represented by their chemical symbols, and the number of atoms of each element present in each molecule is shown as a subscript following that element’s symbol.

Empirical formula

The empirical formula of a compound is defined as the formula that shows the ratio of elements present in the compound, but not the actual numbers of atoms found in the molecule. The ratios are denoted by subscripts next to the element symbols.

The empirical formula is also known as the simplest formula because the subscripts are the smallest whole numbers that indicate the ratio of elements.

Steps for determining an Empirical Formula

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

If percentages are given, assume that the total mass is 100 grams so that

the mass of each element = the percent given.

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.
If the number is too far to be round of then multiply each solution by the same factor to get the lowest whole number multiple.

e.g. If one solution is 1.5, then multiply each solution in the problem by 2 to get 3.

e.g. If one solution is 1.25, then multiply each solution in the problem by 4 to get 5.

Once the empirical formula is found, the molecular formula for a compound can be determined if the molar mass of the compound is known. Simply calculate the mass of the empirical formula and divide the molar mass of the compound by the mass of the empirical formula to find the ratio between the molecular formula and the empirical formula. Multiply all the atoms (subscripts) by this ratio to find the molecular formula.

Example 1: A compound is analyzed and calculated to consist of 13.5 g Ca, 10.8 g O, and 0.675 g H. Find the empirical formula of the compound.

Solution: Start by converting the mass of each element into moles by looking up the atomic numbers from the periodic table. The atomic masses of the elements are 40.1 g/mol for Ca, 16.0 g/mol for O, and 1.01 g/mol for H.

13.5 g Ca x (1 molCa / 40.1 g Ca) = 0.337 molCa

10.8 g O x (1 mol O / 16.0 g O) = 0.675 mol O

0.675 g H x (1 mol H / 1.01 g H) = 0.668 mol H

Next, divide each mole amount by the smallest number or moles (which is 0.337 for calcium) and round to the nearest whole number:

0.337 mol Calcium / 0.337 = 1.00 mol Calcium

0.675 mol Oxygen / 0.337 = 2.00 mol Oxygen

0.668 mol Hydrogen / 0.337 = 1.98 mol Hydrogen which rounds up to 2.00

Now we have the subscripts for the atoms in the empirical formula: CaO2H2

Finally, apply the rules of writing formulas to present the formula correctly. The cation of the compound is written first, followed by the anion. The empirical formula is properly written as Ca(OH)2

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