Bell Work: % Comp Review**Turn in late Folder Checks**A 100 gram sample of carbon dioxide is 27.3% carbon.1. How many grams of C are in the sample?2. How many grams of O are in the sample?3. What’s the molar mass of CO2?

Empirical and Molecular Formulas

Section 11.4

Empirical formulas are like reduced fractions.

can be reduced to . . . . . . 56

Mg2O2 can be reduced to . . . . . . MgO

C6H12O6 can be reduced to . . . . . .CH2O

Timberlake LecturePLUS 4

Types of FormulasThe formulas for compounds can be expressed as an empirical formula and as a molecular(true) formula.

Empirical Molecular (true) Name CH C2H2 acetyleneCH C6H6 benzene

CO2 CO2 carbon dioxide

CH2O C5H10O5 ribose

Timberlake LecturePLUS 5

Empirical Formulas

Write your own one-sentence definition for each of the following: Empirical formula

Molecular formula

Timberlake LecturePLUS 6

An empirical formula represents the simplest whole number ratio of the atoms in a compound.

The molecular formula is the true or actual ratio of the atoms in a compound.

Timberlake LecturePLUS 7

Learning Check EF-1

A. What is the empirical formula for C4H8?

1) C2H4 2) CH2 3) CHB. What is the empirical formula for C8H14?

1) C4H7 2) C6H12 3) C8H14

C. Which are possible molecular formulas for CH2O?

1) CH2O 2) C2H4O2 3) C3H6O3

Timberlake LecturePLUS 8

Solution EF-1

A. What is the empirical formula for C4H8?

2) CH2

B. What is the empirical formula for C8H14?

1) C4H7

C. What is a molecular formula for CH2O?

1) CH2O 2) C2H4O2 3) C3H6O3

Empirical FormulasWhat if you don’t know the chemical formula?

We can use the percent composition to find the empirical formula.

An empirical formula represents the simplest whole number ratio of the atoms in a compound.

HF5

Empirical FormulaThe empirical formula may or may not be the same as the

actual molecular formula.Molecular Formula: the actual number of atoms of each

element in one molecule or formula unit of a substance.

Example: glucose Molecular formula = C6H12O6

Empirical formula = CH2O

What is their common element ratio?Ratio of one C to 2 H to 1 O

What is the scaling factor?6

Empirical Formula

If you have the percent composition given to you, you can determine the empirical formula.

You need to assume a few things.The total mass of the compound is 100.0g.The percent composition of the element is equal to the mass

in grams of the element.

Example 1: An oxide of sulfur has a percent composition of 40.05% S and 59.95% O.

In 100 g of the compound, 40.05 g are S and 59.95 g are O.

Next, find the amount of mol for each element.

Finding Empirical Formulas from %Composition

OK, Now what?

These numbers will help us determine the subscripts for the empirical formula.

You cannot use the exact numbers that we just found because they are not whole numbers.

Empirical Formula

Next, find the amount of mol for each element.

1 mol S40.05g S x 1.249 mol S32.07g S

1 mol O59.95g O x 3.747 mol O.16.00g O

Divide all numbers by the smallest number.

So, S has a subscript of one:1.249 mol S / 1.249 = 1 mol S

Then, divide the mol O by the same number to find its subscript in the empirical formula.3.747 mol O/ 1.249 = 3 mol O

Then write your empirical formula using your subscripts: SO3

Empirical Formula

Pg. 333: 46

Example Problem 2

N: 2.63 / 2.63 = 1 molO: 3.95 / 2.63 = 1.5 mol

Have to double moles to get whole # subscripts.N2O3

Bell Work: Empirical

1. __________ formulas show the actual ratio of elements found in a compound in nature.

2. _________ formulas show the simplified ratio of elements in a compound.

3. Name three compounds that have identical molecular and empirical formulas.

4. Percent to ____, mass to ___, _____ by small, multiply ‘til _____.

Bell Work: Empirical vs. Molecular

1. Draw a Venn diagram to compare and contrast Empirical and Molecular formulas.

2. Draw a Venn diagram to compare and contrast moles and grams.

3. Draw a Venn diagram to compare and contrast formula units and molecules.

The molecular formula needs to be found by going one step further

Molecular formula = (Empirical formula) x scaling factor

To find the scaling factor1) Determine the empirical mass (total mass of all elements in empirical formula)2) Divide molecular mass by empirical mass

Molecular Formula

Chemical analysis of succinic acid indicates it is composed of 40.68% C, 5.08% H, and 54.24 % O, and has a molar mass of 118.1 g/mol. Determine the empirical and molecular formulas for succinic acid.

1)Convert the percent for each element into moles (use the percent given as the amount in grams for each element in 100 g of the compound)

40.68 g C x (1 mol C/12.0 g C) = 3.39 mol C5.08 g H x (1 mol H/1.0 g H) = 5.08 mol H54.24 g O x (1 mol O/16.0 g O) = 3.39 mol O

Finding a Molecular FormulaExample 3:

2)Next, divide each mol amount by the smallest mol amount.

3.39 mol C/ 3.39 = 1 mol C5.08 mol H/ 3.39 = 1.5 mol H3.39 mol O/ 3.39 = 1 mol O

Ratio of C : H : O = 1 : 1.5 : 1

3) Write the Empirical Formula:You can’t have half-moles, so multiply everything by 2.

Empirical Formula: C2H3O2

Molecular Formula

Molecular Formula

4) We need to find the empirical mass using the masses of each element.

2 mol C x (12.0 g C/1 mol C) = 24.0 g C.3 mol H x (1.0 g H/1 mol H) = 3.0 g H.2 mol O x (16.0 g O/1 mol O) = 32.0 g O.

Empirical Mass: 59.0 g/ mol C2H3O2

Molecular Formula5) Now, divide the molar mass by the empirical mass to

determine the scaling factor.

118.1 / 59.0 = 2.00

Multiply the subscripts of the empirical formula by 2 to find the molecular formula.

Molecular Formula: C4H6O4

Timberlake LecturePLUS 23

Learning Check EF-3

A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?

1) C3H4O3

2) C6H8O6

3) C9H12O9

Timberlake LecturePLUS 24

Solution EF-3

A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?2) C6H8O6

C3H4O3 = 88.0 g/EF

176.0 g = 2.00 88.0

Timberlake LecturePLUS 25

Learning Check EF-5

Aspirin is 60.0% C, 4.5 % H and 35.5 O. Calculate its simplest formula. In 100 g of aspirin, there are 60.0 g C, 4.5 g H, and 35.5 g O.

Timberlake LecturePLUS 26

Solution EF-5

60.0 g C x ___________= ______ mol C

4.5 g H x ___________ = _______mol H

35.5 g O x ___________ = _______mol O

Timberlake LecturePLUS 27

Solution EF-5

60.0 g C x 1 mol C = 5.00 mol C 12.0 g C

4.5 g H x 1 mol H = 4.5 mol H 1.01 g H

35.5 g O x 1mol O = 2.22 mol O 16.0 g O

Timberlake LecturePLUS 28

Divide by the smallest # of moles.5.00 mol C = ________________

______ mol O

4.5 mol H = ______________________ mol O

2.22 mol O = ______________________ mol OAre are the results whole numbers?_____

Timberlake LecturePLUS 29

Divide by the smallest # of moles.5.00 mol C = ___2.25__

2.22 mol O

4.5 mol H = ___2.00__2.22 mol O

2.22 mol O = ___1.00__2.22 mol OAre are the results whole numbers?_____

Pg 335: 52

Practice Problems

A Handy Flowchart