Empirical and Molecular Formulas
Empirical Formula Ł� a formula with the lowest whole number ratio of elements in a compound.
" used for ionic compounds NaCl, Al2O3
Molecular Formula Ł� the actual number of each atom in a stable molecule
Name Molecular Formula Empirical Formula
Glucose C6H12O6 CH2O
Vinegar CH3COOH CH2O
Octane C8H18 C4H9
Benzene C6H6 CH
Determination of the Empirical Formula from Percent Composition
Steps:
1. Find the number of moles of each element for 100g of the compound (use molar mass).
2. Divide the number of moles of each element by the smallest number of moles.
3. Multiply all values to get a whole number.
example #20 d.
In 100g Convert to Moles Divide by the Smallest Get Whole #
17.6% Na Ł� 17.6g Na x 1 mol Na = 0.765550& mol Na � 0.76346& = 1 mol Na Ł� 2 mol Na
22.99g Na ę!do not round off!
39.7% Cr Ł� 39.7g Cr x 1 mol Cr = 0.76346& mol Cr � 0.76346& = 1 mol Cr Ł� 2 mol Cr
52.00g Cr ę!smallest
42.7% O Ł� 42.7g O x 1 mol O = 2.66875 & mol O � 0.76346& = 3.5 mol O Ł� 7 mol O
16.00g O
Empirical Formula Ł� Na2Cr2O7 sodium dichromate
Check (optional) Ł� find percent composition of the empirical formula
Na2Cr2O7 2 x Na = 2 (22.99g Na) = 45.98g Na � 262.0g Na2Cr2O7 = 17.549618& % Na
2 x Cr = 2 (52.00g Cr) = 104.0g Cr � 262.0g Na2Cr2O7 = 39.694656 & % Cr
7 x O = 7 (16.00g O) = 112.0g O � 262.0g Na2Cr2O7 = 42.74809 & % O
Total = 262.0g Na2Cr2O7 = 100 %
Determine Molecular Formula
Method 1: Divide Molecular Mass by the Empirical Mass
1. Determine the Empirical Formula
2. Calculate Empirical Mass Ł� multiply number of each element in the empirical formula by its atomic mass
3. Divide the Molecular Mass (MM) by the Empirical Mass (EM)
4. Multiply each subscript by this ratio
example Given: Empirical Formula: CH4N Molecular Mass = 64g
Calculate Empirical Mass = 1C + 4H + 1N = 32.05g MM � EM = 64g � 32.05g = 2
4" Molecular Formula Ł� C2H8N2 (Molecular Mass = 64.10g)
Method 2: Multiply Percent Compositions by Molecular Mass instead of 100g