Lab Section
Empirical formula: The empirical formula is the formula of the molecule represented with the lowest possible subscripts of the elements. This formula can be obtained from percentage compositions of the elements or from combustion products.
Molecular formula: The molecular formula gives the actual subscripts of the elements in the molecule. It may be the same as the empirical formula or multiplied by a whole-number factor. This formula can be determined from the empirical formula if the molar mass of the molecule is known. In other words, Molecular formula = (Empirical formula) n , where n is a whole number.
The difference in the formulae can be shown using the example of the molecule Benzene. The molecular formula is C6H6 while the empirical formula is CH only.
The significance of an empirical formula, also known as a stoichiometric formula becomes dominant when used to represent ionic (like KBr) or giant covalent molecules (like diamond). A molecular formula is not befitting for such molecules since their network structure can only be represented by (K + Br -)n, n is a large numerical value dependent on the sample size. Secondly, it gives us the names of all the elements in the compound. In addition to this, it also represents the formula mass which is the sum of the atomic masses of all the elements in the compound.
A stepwise approach to finding empirical formula as mentioned by John Kenkel (2011): (a) Find the number of grams of each element in the compound and convert it to moles by dividing it with its molar mass. (b) Divide all quantities with the smallest number of moles. If the result is ± 0.1 to a whole number, then round off. Else, multiply the answer with a whole number until it achieves that condition. This will give the subscripts for the empirical formula.
Next, to find the molecular formula: (a) First find the mass of the empirical formula. (b) Then divide the molecules molar mass with this calculated mass. This will give a scaling number to be multiplied with the subscripts of empirical formula. If the experiment is for a combustion reaction where one of the elements is oxygen, few additional steps are required.
Mass percent: This parameter gives the percentage of mass of each element in the compound. It is determined experimentally by the ratio of the mass of just the element to the molar mass of the entire compound, multiplied by 100.
Example: Let us consider a common pain relief medication called Ibuprofen. It has 75.69% C, 8.80% H, and 15.51% O, by mass. What is its empirical and molecular formula? Molar mass is 206.29 g/mol
Solution: Considering these are the only elements in the compound, the mass percents for Carbon, Hydrogen and Oxygen would add together to give a 100%. Hence, if we consider it to be a 100g sample, then we have 75.69 g C, 8.8 g of H and 15.51 g of O. Using each element’s molar mass, we calculate the moles of carbon nC, hydrogen nH, and oxygen nO, in the sample.
nC = 75.69 g C × 1 mol C /12.01 g C = 6.302 mol C
nH = 8.8 g H × 1 mol H / 1.008 g H = 8.73 mol H
nO = 15.51 g O × 1 mol O / 16.00 g O = 0.9694 mol O
Dividing the moles of each element by the smallest number of moles, which in this case is 0.9694 mol of O, gives the relative number of moles for each element; thus
nC / nO = 6.302 mol C / 0.9694 mol O = 6.5 mol C / 1.00 mol O
nH / nO = 8.73 mol H / 0.9694 mol O = 9 mol H / 1.00 mol O
nO / nO = 0.9694 mol O / 0.9694 mol O = 1.00 mol O / 1.00 mol O
Based on these mole ratios, the empirical formula is C6.5H9O1. Because an empirical formula must have integer subscripts, we multiply the subscripts by 2, giving an empirical formula of C13H18O2.
Given the molar mass of the compound is 206.29 g/mol.
The scaling factor = molar mass of compound / molar mass of empirical formula
= 206.29 / (12.01* 13 + 1.008 * 18 + 16 * 2) = 206.26 / 206.27 = 1
Here, the molecular and empirical formula is observed to be the same.
References
John Kenkel (2011). Basic Chemistry concepts and Exercises, Boca Raton, FL: CRC Press
