An empirical formula represents the simplest whole number ratio of various atoms present in a compound whereas the molecular formula shows the exact number of different types of atoms present in a molecule of a compound.
If the mass per cent of various elements present in a compound is known, its empirical formula can be determined. Molecular formula can further be obtained if the molar mass is known. The following example illustrates this sequence.
Problem: A compound contains 4.07 %hydrogen. 24.27 % carbon and 71.65 % chlorine. Its molar mass is 98.96 g. What are its empirical and molecular formulas?
Solution:
Step 1: Conversion of mass per cent to grams.
Since we are having mass per cent, it is convenient. to use 100g of the compound as the starting material. Thus, in the 100 g sample of the above compound. 4.07g hydrogen is present. 24.27g carbon is present and 71.65 g chlorine is present.
Step 2: Convert into number moles of each element
Divide the masses obtained above by respective atomic masses of various elements.
Moles of hydrogen = 4.07 g /1.008g = 4.04
Moles of carbon = 24.27g /12.01g = 2.021
Moles of chlorine = 71.65g /35.453g = 2.021
Step 3: Divide the mole value obtained above by the smallest number
Since 2.021 is smallest value, division by it gives a ratio of 2:1:1 for H:C:CI .
In case the ratios are not whole numbers. then they may be converted into whole number by multiplying by the suit able coefficient.
Step 4: Write empirical formula by mentioning the numbers after writing the symbols of respective elements.
CH2 Cl is, thus, t he empirical formula of the above compound.
Step 5: Writing molecular formula
(a) Determine empirical formula mass: Add the atomic masses of various atoms present in the empirical formula.
For CH2 Cl, empirical formula mass is 12.01 + 2 1.008 + 35.453 = 49.48 g
(b) Multiply empirical formula by n obtained above to get the molecular formula
Empirical formula = CH2Cl, n = 2.
Hence molecular formula is C2 H4 Cl2.
