empirical formulas(68)

7/28/2019 Empirical Formulas(68)

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Timberlake LecturePLUS 1

Chapter 9

Empirical Formulas


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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

CH2

O C5

H10

O5

ribose


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Empirical Formulas

Write your own one-sentence definition for

each of the following:

Empirical formula

Molecular formula


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An empirical formula represents the

simplest who le number rat io o f the

atoms in a compound.

The molecular formulais the t rue or

actual rat io o f the atoms in a

compound .


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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. What is a molecular formula for CH2O?

1) CH2O 2) C2H4O2 3) C3H6O3


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Solution EF-1

A. What is the empirical formula for C4H8?

2) CH2

B. What is the empirical formula for C8H14?

1) C4H7C. What is a molecular formula for CH2O?

1) CH2O 2) C2H4O2 3) C3H6O3


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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


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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 alsobe 4 atoms of S.


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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 massmolecular formula =

2 x empirical formula

molecular formula = or > empirical formula


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Empirical

Formula

Empirical

Mass

Molecular

Formula

Molecular

Mass


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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/EF176.0 g = 2.00

88.0


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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


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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


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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 isknown 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


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2. Calculate the number of moles of each

element.

71.65 g Cl x 1 mol Cl = 2.02 mol Cl35.5 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


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Why moles?

Why do you need the number of moles

of each element in the compound?


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3. Find the smallest who le number rat ioby

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 H2.02

4. Write the simplest or empirical formula

CH2Cl


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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


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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 gH, and 35.5 g O.


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Solution EF-5

60.0 g C x ___________= ______ mol C

4.5 g H x ___________= _______mol H

35.5 g O x ___________= _______mol O


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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


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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?_____


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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?_____


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Finding Subscripts

A fraction between 0.1 and 0.9 must not

be rounded. Multiply all results by an

integer to give whole numbers forsubscripts.

(1/2) 0.5 x2 = 1

(1/3) 0.333 x 3 = 1(1/4) 0.25 x4 = 1

(3/4) 0.75 x 4 = 3


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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


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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?


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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

empirical formulas(68)

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