Chemical Formulas Lecture-6

8/8/2019 Chemical Formulas Lecture-6

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Empirical and Molecular

formulae

Dr. ABDALLA AHMED ELBASHIR

ASSOCIATE PROFESSOR

CHEMISTRY DEPARTMENT

UNIVERSITYofKHARTOUM


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Determining formulae

The percentage composition of a compoundleads directly to its empirical formula.

Recall: An empirical formula for a compound is

the formula of a substance written with the

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smallest integer subscripts.

Eg. Consider hydrogen peroxide:

Molecular formula = H2O2

Empirical formula = HO


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Types of Formulas

Empirical Molecular (true) Name

CH C H acet lene

The formulas for compounds can be expressedas an empirical formula and as a molecular

(true) formula.

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CH C6H6 benzene

CO2 CO2 carbon dioxide

CH2O C5H10O5 ribose


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Compounds with different molecular

formulae can have the same empiricalformula, and such substances will have

the same percentage composition.

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g. ace y ene = 2 2benzene = C6H6

both have the empirical formula = ?


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Empirical Formulas

Write your own one-sentence

definition for each of the

5

o ow ng:

Empirical formula

Molecular formula


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An empirical formula representsthe simplest whole number ratio

of the atoms in a compound.

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The molecular formula is the

true or actual ratio of 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) CH

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B. What is the empirical formula for C8H14?

1) C4H7 2) C6H12 3) C8H14


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Solution EF-1A. What is the empirical formula for C4H8?

2) CH2

B. What is the empirical formula for

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C8H14?

1) C4H7


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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.

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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.

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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 = em irical mass

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molecular formula = empirical formula

n = 2 molar mass = 2 x empirical mass

molecular formula = 2 x empirical formula

molecular formula = or > empirical formula


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Empirical

Formula

Molecular

Formula

12

Empirical

Mass

Molecular

Mass


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Empirical formula from CompositionConsider the following flow-diagram:

Percent composition

Mass Composition

Number of moles of

13

Divide by smallest number of

moles to find the molar ratios

Multiply by appropriate number to

get whole number subscripts


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A compound has a formula mass of 176.0and an empirical formula of C3H4O3. What

is the molecular formula ? (C=12, O=16,

H=1)

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3 4 3

2) C6H8O6

3) C9H12O9


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A compound has a formula mass of 176.0

and an empirical formula of C3H4O3. Whatis the molecular formula?

2) C6H8O6

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C3H4O3 = 88.0 g/EF

176.0 g = 2.00

88.0


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If there are 192.0 g of O in themolecular formula, what is the true

formula if the EF is C7H6O4?

1) C7H6O4

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2) C14H12O8

3) C21H18O12


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If there are 192.0 g of O in the

molecular formula, what is the trueformula if the EF is C7H6O4?

3) C21H18O12

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4 7 6 4

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 is

known to be 99.0 g/mol. (Cl = 35.5, C=12,

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H=1.01)

1. State mass percents as grams in a100.00-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 Cl

35.5 g Cl

=

19

. .

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

20

compoun


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3. Find the smallest whole number ratio by

dividing each mole value by the smallest

mole values:

Cl: 2.02 = 1 Cl

2.02

=

21

.

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

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.

E M 49.5 g/EM

7.Molecular formula

(CH2Cl)2 = C2H4Cl2


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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.

23


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60.0 g C x ___________= ______ mol C

4.5 g H x ___________= _______mol H

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35.5 g O x ___________= _______mol O


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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

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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 = ________________

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______

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__

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.

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 for

subscripts.

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(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


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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

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: . mo x = mo

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

C9H8O4


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A compound is 27.4% S, 12.0%

N and 60.6 % Cl. If the compoundhas a molar mass of 351 g/mol,

what is the molecular formula?

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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

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Mol. Mass/ Empirical mass 351/117.1 = 3

Molecular formula = S3N3Cl6


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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.

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For ex: H2O H-O-H

C2H6 CH3 H-C- C- H

H H


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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.

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o ecu ar ormu a

The actual numbers of atoms in a molecule.

Structural Formula

Show the relative arrangements of atoms in amolecule


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Timberlake LecturePLUS 34


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HYDRATESSolids which are found in combinedform with water in definite proportionare called as HYDRATES. When

hydrates are heated, H2O evaporates,

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amorphous .

(w/o a certain geometric structure,

generally in powdered form. H2Omolecules surround ionic substanceswith certain amounts.


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WATER OF HYDRATION : Water molecules of a hydrate.

Na2CO3.10H2O Na2CO3(s) + 10H2O(g) DEHYDRATION: Evaporation of water of hydration.

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2 3. 2 ,

CaSO4.2H2O,

CuSO4

.5H2

O


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Chemical Formulas Lecture-6