

Empirical Formula vs Molecular Formula
Presentation
•
Chemistry
•
10th - 12th Grade
•
Hard
Joseph Anderson
FREE Resource
12 Slides • 0 Questions
1
Empirical and Molecular Formula
Determine the chemical formula from the percentage composition or the molar mass.

2
Follow the Rhyme to get it every time
Change percent(%) to mass (g)
Convert mass to moles (g --> mol)
Divide by the smallest
Multiply til whole.
3
Sample Problem
An unknown compound is analyzed and found to contain 24.3% carbon, 4.1% hydrogen, and 71.6% chlorine. If the molecular mass of the compound is 98.8g/mol, what is the compounds empirical formula and molecular formula?
4
Change percent to grams
24.3g C
4.1g H
71.6g Cl
5
Convert grams to moles
24.3g C x 1 mol/12.01g = 2.02
4.1g H x 1 mol/1.01g = 4.06
71.6g Cl x 1 mol/35.45g = 2.02
6
Divide by the smallest
24.3g C x 1 mol/12.01g = 2.02/2.02 = 1
4.1g H x 1 mol/1.01g = 4.06/2.02 = 2
71.6g Cl x 1 mol/35.45g = 2.02/2.02 = 1
7
Multiply til whole (if necessary)
24.3g C x 1 mol/12.01g = 2.02/2.02 = 1
4.1g H x 1 mol/1.01g = 4.06/2.02 = 2
71.6g Cl x 1 mol/35.45g = 2.02/2.02 = 1
C1H2Cl1 --> CH2Cl is the empirical formula (Ef)
8
Find the molar mass of the empirical formula
CH2Cl
C - 1 x 12.01 amu = 12.01
H - 2 x 1.01 amu = 2.02
Cl- 1 x 35.45 amu = 35.45
__________________
49.48 amu
9
Divide the masses Mf/Ef
98.8 amu/49.48 amu = 2
10
Multiply the integer by the empirical to get the molecular formula
(CH2Cl)2 = C2H4Cl2 is the molecular formula (Mf)
11
Try this one on your own!
Naphthalene, commonly known as moth balls, is composed of 93.7% carbon and 6.3% hydrogen. The molecular mass is 128g/mol. Determine the empirical and molecular formulas for naphthalene.
12
Answers
Ef = C5H4
Mf = C10H8
Empirical and Molecular Formula
Determine the chemical formula from the percentage composition or the molar mass.

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