Timberlake LecturePLUS 1

Determining formulae

The percentage composition of a compound leads directly to its empirical formula.

Recall: An empirical formula for a compound is the formula of a substance written with the smallest integer subscripts.

Eg. Consider hydrogen peroxide:

Molecular formula = H2O2

Empirical formula = HO

Timberlake LecturePLUS 2

Types of Formulas

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

Empirical Molecular (true) Name

CH C2H2 acetylene

CH C6H6 benzene

CO2 CO2 carbon dioxide

CH2O C5H10O5 ribose

Timberlake LecturePLUS 3

Compounds with different molecular formulae can have the same empirical formula, and such substances will have the same percentage composition.

Eg. acetylene = C2H2

benzene = C6H6

both have the empirical formula = ?

Timberlake LecturePLUS 4

Empirical Formulas

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

Empirical formula

Molecular formula

Timberlake LecturePLUS 5

• 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 6

Learning Check EF-1

A. What is the empirical formula for C4H8?

1) C2H4 2) CH2 3) CH

B. What is the empirical formula for C8H14?

1) C4H7 2) C6H12 3) C8H14

C. What is a molecular formula for CH2O?

1) CH2O 2) C2H4O2 3) C3H6O3

Timberlake LecturePLUS 7

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

Timberlake LecturePLUS 8

Learning Check EF-2

If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.

1) SN

2) SN4

3) S4N4

Timberlake LecturePLUS 9

Solution EF-2

If the molecular formula has 4 atoms of N, what is the molecular formula if SN is the empirical formula? Explain.

3) S4N4

If the actual formula has 4 atoms of N, and S is related 1:1, then there must also be 4 atoms of S.

Timberlake LecturePLUS 10

Empirical and Molecular Formulas

molar mass = a whole number = n

simplest mass

n = 1 molar mass = empirical mass

molecular formula = empirical formula

n = 2 molar mass = 2 x empirical mass

molecular formula =

2 x empirical formula

molecular formula = or > empirical formula

Timberlake LecturePLUS 11

Empirical Formula

Empirical Mass

Molecular Formula

MolecularMass

Timberlake LecturePLUS 12

Empirical formula from CompositionConsider the following flow-diagram:

Percent composition

Mass Composition

Number of moles of each element

Divide by smallest number of moles to find the molar ratios

Multiply by appropriate number to get whole number subscripts

Timberlake LecturePLUS 13

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 14

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 15

Learning Check EF-4

If there are 192.0 g of O in the molecular

formula, what is the true formula if the EF

is C7H6O4?

1) C7H6O4

2) C14H12O8

3) C21H18O12

Timberlake LecturePLUS 16

Solution EF-4

If there are 192.0 g of O in the molecular

formula, what is the true formula if the EF

is C7H6O4?

3) C21H18O12

192 g O = 3 x O4 or 3 x C7H6O4

64.0 g O in EF

Timberlake LecturePLUS 17

Finding the Molecular Formula

A compound is Cl 71.65%, C 24.27%, and H 4.07%. What are the empirical and molecular formulas? The molar mass is known to be 99.0 g/mol.

1. State mass percents as grams in a 100.00-g sample of the compound.

Cl 71.65 g C 24.27 g H 4.07 g

Timberlake LecturePLUS 18

2. Calculate the number of moles of each element.

71.65 g Cl x 1 mol Cl = 2.02 mol Cl

35.5 g Cl

24.27 g C x 1 mol C = 2.02 mol C

12.0 g C

4.07 g H x 1 mol H = 4.04 mol H

1.01 g H

Timberlake LecturePLUS 19

Why moles?

Why do you need the number of moles of each element in the compound?

Timberlake LecturePLUS 20

3. Find the smallest whole number ratio by dividing each mole value by the smallest mole values:

Cl: 2.02 = 1 Cl 2.02

C: 2.02 = 1 C 2.02

H: 4.04 = 2 H 2.024. Write the simplest or empirical formula

CH2Cl

Timberlake LecturePLUS 21

5. EM (empirical mass)

= 1(C) + 2(H) + 1(Cl) = 49.5

6. n = molar mass/empirical mass

Molar mass = 99.0 g/mol = n = 2 E M 49.5 g/EM

7.Molecular formula

(CH2Cl)2 = C2H4Cl2

Timberlake LecturePLUS 22

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 23

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 24

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 25

Divide by the smallest # of moles.

5.00 mol C = ________________

______ mol O

4.5 mol H = ________________

______ mol O

2.22 mol O = ________________

______ mol O

Are are the results whole numbers?_____

Timberlake LecturePLUS 26

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 O

Are are the results whole numbers?_____

Timberlake LecturePLUS 27

Finding Subscripts

A fraction between 0.1 and 0.9 must not be rounded. Multiply all results by an integer to give whole numbers for subscripts.(1/2) 0.5 x 2 = 1

(1/3) 0.333 x 3 = 1

(1/4) 0.25 x 4 = 1

(3/4) 0.75 x 4 = 3

Timberlake LecturePLUS 28

Multiply everything x 4

C: 2.25 mol C x 4 = 9 mol C

H: 2.0 mol H x 4 = 8 mol H

O: 1.00 mol O x 4 = 4 mol O

Use the whole numbers of mols as the subscripts in the simplest formula

C9H8O4

Timberlake LecturePLUS 29

Learning Check EF-6

A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g/mol, what is the molecular formula?

Timberlake LecturePLUS 30

Solution EF 6

0.853 mol S /0.853 = 1 S

0.857 mol N /0.853 = 1 N

1.71 mol Cl /0.853 = 2 Cl

Empirical formula = SNCl2 = 117.1 g/EF

Mol. Mass/ Empirical mass 351/117.1 = 3

Molecular formula = S3N3Cl6

Timberlake LecturePLUS 31

STRUCTURAL FORMULAThe atoms in a molecule are connected or chemically bonded in a precise way. A SF.

Shows how the atoms in a molecule are arranged.

For ex: H2O H-O-H

C2H6 CH3 H-C- C- H

H H

Timberlake LecturePLUS 32

Empirical formula•The simplest whole number ratio of atoms of elements in a compound, described with the use of subscripts.•Ionic compounds are always shown as empirical formulas.

Molecular FormulaThe actual numbers of atoms in a molecule.

Structural FormulaShow the relative arrangements of atoms in a molecule

Timberlake LecturePLUS 33

Timberlake LecturePLUS 34

HYDRATESSolids which are found in combined form with water in definite proportion are called as HYDRATES. When hydrates are heated, H2O evaporates, and only solid is obtained in amorphous .(w/o a certain geometric structure, generally in powdered form. H2O molecules surround ionic substances with certain amounts.

Timberlake LecturePLUS 35

WATER OF HYDRATION : Water molecules of a hydrate.

• Na2CO3.10H2O Na2CO3(s) + 10H2O(g)

• DEHYDRATION: Evaporation of water of hydration.

• Na2CO3.10H2O,

• CaSO4.2H2O,

• CuSO4.5H2O

Timberlake LecturePLUS 36