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Empirical and Molecular Formula

Empirical and Molecular Formula

Assessment

Presentation

Chemistry

10th - 12th Grade

Hard

Created by

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.

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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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