Thursday, September 27, 2012

Getting the Empirical and Molecular Formula

Our professor taught us the right process of getting the Empirical and Molecular Formula of the Compounds. Here is the sample problem that she showed us.

The problem is:
Given the following percent composition 26.50% K, 35.37% Cr and 39.10% O, determine the empirical and molecular formula of the compound if the molar mass of the compound is 294 6/mol.

Step  1: State the given.
% mass of K = 26.50%
% mass of Cr = 35.37%
% mass of O = 39.10%

Step 2: Convert percent to grams by copying the percent value and write grams.
% mass of K = 26.50%  = 26.50 g.
% mass of Cr = 35.37% = 35.37 g.
% mass of O = 39.10% = 39.10 g.

Step 3: Calculate the number of moles by multiplying the atomic weights to the individual masses.

mol of K = 26.50 g x 1 mol/40.00 g = 0.66 mol
mol of Cr = 35.37 g x 1 mol/52.00 g = 0.68 mol
mol of O = 39.10 g x 1 mol/16.00 g = 2.44 mol 

Step 4: Look for the smallest number of mole and use it as a divisor to obtain a whole number ratio.
In the problem, the smallest number of mole is of K which is 0.66 mol.

K = 0.66 mol/ 0.66 mol = 1
Cr = 0.68 mol/0.66 = 1.03
O = 2.44 mol/0.66 mol = 3.69
Empirical Formula is KCrO4

Step 4: Compute for the empirical mass based on the empirical formula, then add the answer.
K = 1 x 40.00 g/mol = 40.00 g/mol
Cr = 1 x 52.00 g/mol = 52.00 g/mol
O = 4 x 16.00 g/mol = 64.00 g/mol
Empirical Mass = 156 g/mol

Step 5: Compute for "n", to solve for Molecular Formula.
n = molar mass/empirical mass
n = 294 g/mol/156 g/mol
n = 1.88
n = 2

Molecular Formula = n(E.F.)
Molecular Formula = 2 (KCrO4)
M. F. = K2Cr2O8

There you go. I hope that I have helped you in some other way in getting the empirical and molecular mass of compounds.

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