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Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 1

Introductory Chemistry:

A Foundation

FOURTH EDITION

by Steven S. Zumdahl

University of Illinois Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 2

Chemical Composition

Chapter 8

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 3

Atomic Masses

• Balanced equation tells us the relative numbers of molecules of reactants and products

C + O2→ CO2

1 atom of C reacts with 1 molecule of O2 to make 1 molecule of CO2

• If I want to know how many O2molecules I will need or how many CO2 molecules I can make, I will need to know how many C atoms are in the sample of carbon I am starting with

Copyright©2000 by Houghton

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Atomic Masses

• Dalton used the percentages of elements in compounds and the chemical formulas to deduce the relative masses of atoms

• Unit is the amu.

– atomic mass unit – 1 amu = 1.66 x 10-24g

• We define the masses of atoms in terms of atomic mass units

– 1 Carbon atom = 12.01 amu, – 1 Oxygen atom = 16.00 amu

– 1 O2molecule = 2(16.00 amu) = 32.00 amu

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Atomic Masses

• Atomic masses allow us to convert weights into numbers of atoms

If our sample of carbon weighs 3.00 x 1020amu we

will have 2.50 x 1019atoms of carbon

atoms C 10 x 2.50 amu 12.01 atom C 1 x amu 10 x 3.00 20 = 19

Since our equation tells us that 1 C atom reacts with 1 O2molecule, if I have 2.50 x 1019C

atoms, I will need 2.50 x 1019molecules of O 2

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Example #1

• Determine the mass of 1 Al atom 1 atom of Al = 26.98 amu • Use the relationship as a conversion factor

amu 2024 atom Al 1 amu 26.98 x atoms Al 75 =

(2)

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Chemical Packages - Moles

• We use a package for atoms and molecules called a mole

• A mole is the number of particles equal to the number of Carbon atoms in 12 g of C-12 • One mole = 6.022 x 1023units

• The number of particles in 1 mole is called

Avogadro’s Number

• 1 mole of C atoms weighs 12.01 g and has 6.02 x 1023atoms

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Mifflin Company. All rights reserved. 8

Copyright©2000 by Houghton

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The Mole

• When two chemicals react with each other, we want to know how many atoms or molecules of each are used do that a formula of the product can be established.

• This means we need a way to count atoms, but HOW?

• The solution to this problem is to define a convenient unit of matter that contains a know number of particles.

• 1 moles = 6.022136736×1023particles

– This value is commonly known has Avogadro’s number.

• Remember that one mole always contains the same number of particles, no matter what the substance.

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Molar Mass

• The mass of one mole of atoms of any element is the molar mass, which is numerically equal to the atomic mass, but in grams.

• Example:

– 1 mole of sodium weighs 22.99 g

• This not only true for atoms, but also for molecules, but we use the molecular mass instead of the atomic mass.

– To determine the molecular mass of a molecule, we add up the atomic masses of the elements that make up the molecule.

• Example:

– CCl4

» We have 1 carbon and 4 chlorines » 12.001 g/mol + 4(35.4527 g/mol) = 153.81 g/mol » Therefore, if we have 1 mole of CCl4it would weigh 153.81 g.

Molecules, Compounds, and The

Mole

• Thinking about moles with respect to molecules and compounds is not all that different than with atoms.

• If I have 1 mole of CH4:

1. How many CH4molecules do I have? • I have 6.022 ×1023molecules of CH

4

2. How much does it weigh?

• 16.032 g

3. How many carbon atoms do I have?

• 6.022 ×1023

4. How many hydrogen atoms do I have?

• 4 × 6.022 ×1023= 2.409×1024

MOLES are the CENTER

# of moles of X # of particles Vol. of an “ideal” gas @ STP Mass a b c d e f a) Multiply by Avogadro’s Number b) Divide by Avogadro’s Number c) Multiply by the molecular weight of X d) Divide by the molecular

weight of X

e) Multiply by 22.4 L / mol f) Divide by 22.4 L / mol

(3)

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EXAMPLES

• What is the mass, in grams, of 1.5 mols of silicon?

42.1283 g • How many moles are

represented by 454 g of sulfur? How many atoms?

14.2 mols 8.53×1024atoms

• Calculate the molar mass of CaCO3(limestone).

100.08 g/mol • If you have 454 g of CaCO3,

how many moles do you have? How many molecules? How many atoms of oxygen?

4.54 mols 2.73×1023molecules

8.20×1024atom of O

At first these concepts are difficult to understand, but working problems is the best way to become comfortable with these concepts.

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Example #2

‘ Use the Periodic Table to determine the mass of 1 mole of Al

1 mole Al = 26.98 g ’ Use this as a conversion factor for

grams-to-moles Al mol 0.371 g 26.98 Al mol 1 x Al g 10.0 =

Compute the number of moles and number of atoms in 10.0 g of Al

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 15 “ Use Avogadro’s Number to determine the

number of atoms in 1 mole

1 mole Al = 6.02 x 1023atoms

” Use this as a conversion factor for moles-to-atoms atoms Al 10 x 2.23 Al mol 1 atoms 10 x 6.02 x Al mol 0.371 23 = 23

Example #2

Compute the number of moles and number of atoms in 10.0 g of Al

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 16 ‘ Use Avogadro’s Number to determine the

number of atoms in 1 mole

1 mole Al = 6.02 x 1023atoms

’ Use this as a conversion factor for atoms-to-moles Al mol 0.370 atoms 10 x 6.02 Al mol 1 x atoms Al 10 x 2.23 23 23 =

Compute the number of moles and mass of 2.23 x 1023atoms of Al

Example #3

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 17 “ Use the Periodic Table to determine the

mass of 1 mole of Al

1 mole Al = 26.98 g ” Use this as a conversion factor for

moles-to-grams Al g 9.99 Al mol 1 g 26.98 x Al mol 0.370 =

Compute the number of moles and mass of 2.23 x 1023atoms of Al

Example #3

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Mifflin Company. All rights reserved. 18

Molar Mass

• The molar mass is the mass in grams of one mole of a compound

• The relative weights of molecules can be calculated from atomic masses

water = H2O = 2(1.008 amu) + 16.00 amu = 18.02 amu

• 1 mole of H2O will weigh 18.02 g, therefore

the molar mass of H2O is 18.02 g

• 1 mole of H2O will contain 16.00 g of oxygen and 2.02 g of hydrogen

(4)

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 19 • Percentage of each element in a compound

– By mass

• Can be determined from ° the formula of the compound or ± the experimental mass analysis of the

compound

• The percentages may not always total to 100% due to rounding

100%

whole

part

Percentage

=

×

Copyright©2000 by Houghton

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Percent Composition

• Percent Composition

– The percentage of the mass of a compound represented by each of its constituent elements.

• According to the law of constant composition, any sample of a pure compound always consists of the same elements combined in the same proportion by mass.

• Example: What is the percent composition of N and H in ammonia (NH3)? 82.27% 100 NH g 17.03 N g 14.01 NH of mol 1 of Mass NH of mol 1 in N of Mass NH in N Percent Mass 3 3 3 3= = × = 17.76% 100 NH g 17.03 H g 3(1.008) NH of mol 1 of Mass NH of mol 1 in H of Mass NH in H Percent Mass 3 3 3 3= = × = Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 21

Empirical and Molecular Formulas

• We have discussed molecular formulas • Empirical formulas

– A molecular formula showing the simplest possible ratio of atoms in a molecule.

• From percent compositions we can determine empirical and molecular formulas by the following procedure. % A → g A → x mols A % B → g B → x mols B y molB A mol x Find mole

ratio Ratio givesformula

AxBy

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 22 Example

• Eugenol is the active component of oil of cloves. It has a molar mass of 164.2 g/mol and is 73.14% C and 7.37% H; the remainder is oxygen. What are empirical and molecular formulas for eugenol?

• Where to start?!

1. What is the % of oxygen? • 100% - (73.14% + 7.37%) = 19.49% 2. How many grams of C, H, and O?

• Assume you have 100 g of the compound therefore your percentages are the number of grams of C, H, and O.

– 73.14 g of C – 7.37 g of H – 19.49 g of O

3. Change the grams into moles.

• 73.14 g of C × 1 mole of C / 12.011g of C = 6.09 mols of C • 7.37 g of H × 1 mole of H / 1.0079 g of H = 7.31 mols of H • 19.49 g of O × 1 mole of O / 15.9994 g of O = 1.22 mols of O

4. Find the mole ratios (divide the large amount(s) of moles by the smallest amount of moles). • 6.09 mols of C / 1.22 mols of O = 5 mols of C to 1 mol of O

• 7.37 mols of H / 1.22 mols of O = 6 mols of H to 1 mol of O 5. So what does this mean?

• C5H6O is the empirical formula

6. How to get the molecular formula? • Molecular weight of the empirical formula.

– 82 g/mol

• Divide the molecular weight of eugenol by the molecular weight empirical formula.

– 164.2 g/mol / 82 g/mol = 2

• So there are 2 units of the empirical formula.

– (C5H6O)2= C10H12O2

Hydrated Compounds

• If ionic compounds are prepared water solution and then isolated as solids, the crystals often have molecules of water trapped in the lattice.

• Compounds in which molecules of water are associated with the ions of the compound are called hydrated compounds.

• Examples:

– NiCl2• 6H2O

– CaSO4• 2H2O

• But what about the molar mass? (Do I add the water?)

– YES!

– NiCl2• 6H2O → 237.7 g/mol • 129.6 g/mol from NiCl2 • 108.1 g/mol from 6H2O

– CaSO • 2HO → 172.14 g/mol

Determining the Units of Hydration

• Example: The following reaction takes place. 1.023 g of CuSO4· x H2O + heat → 0.654 g CuSO4+ ? g H2O

• Find the units of hydration. • Step 1: determine the mass of water

1.023 g of CuSO4· x H2O – 0.654 g of CuSO4= 0.369 g of water

• Step 2: determine the number of moles

0.369 g of water × 1 mol of water / 18.02 g of water = 0.0205 mols of water 0.654 g CuSO4× 1 mols CuSO4/ 159.6 g CuSO4= 0.00410 mols CuSO4

• Step 3: determine the molar ratio

• So, we know that we started with 1.023 g of CuSO4· 5 H2O 4 2 4 2 CuSO mol 1.00 O H mol 5.00 CuSO mol 0.00410 O H mol 0.0205 =

(5)

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 25 ‘ Determine the mass of each element in 1

mole of the compound

2 moles C = 2(12.01 g) = 24.02 g 6 moles H = 6(1.008 g) = 6.048 g

1 mol O = 1(16.00 g) = 16.00 g

’ Determine the molar mass of the compound by adding the masses of the elements

1 mole C2H5OH = 46.07 g

Determine the Percent Composition from the Formula C2H5OH

Example #4

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 26 “ Divide the mass of each element by the molar

mass of the compound and multiply by 100% 52.14%C 100% 46.07g 24.02g× = 13.13%H 100% 46.07g 6.048g× = 34.73%O 100% 46.07g 16.00g× =

Determine the Percent Composition from the Formula C2H5OH

Example #4

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 27

Empirical Formulas

• The simplest, whole-number ratio of atoms in a molecule is called the Empirical Formula

– can be determined from percent composition or combining masses

• The Molecular Formula is a multiple of the Empirical Formula % A mass A (g) moles A100g MMA % B mass B (g) moles B100g MMB moles A moles B Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 28 ‘ Find the greatest common factor (GCF) of the

subscripts

factors of 20 = (10 x 2), (5 x 4) factors of 12 = (6 x 2), (4 x 3)

GCF = 4

’ Divide each subscript by the GCF to get the empirical formula

C20H12= (C5H3)4 Empirical Formula = C5H3 Determine the Empirical Formula of

Benzopyrene, C20H12

Example #5

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 29 ‘ Convert the percentages to grams by

assuming you have 100 g of the compound

– Step can be skipped if given masses

47gC 100g 47gC 100g× = 47gO 100g 47gO 100g× = 6.0gH 100g 6.0gH 100g× = Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

Example #6

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 30 ’ Convert the grams to moles

C mol 3.9 12.01g C mol 1 C 47g × = O mol 2.9 16.00g O mol 1 O g 47 × = H mol 6.0 1.008g H mol 1 H g 6.0 × =

Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

(6)

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 31 “ Divide each by the smallest number of moles

1.3 2.9 C mol 3.9 ÷ = 1 2.9 O mol 2.9 ÷ = 2 2.9 H mol 6.0 ÷ =

Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 32 ” If any of the ratios is not a whole number, multiply

all the ratios by a factor to make it a whole number

– If ratio is ?.5 then multiply by 2; if ?.33 or ?.67 then multiply by 3; if ?.25 or ?.75 then multiply by 4

4 3 x 1.3 2.9 C mol 3.9 ÷ = = 3 3 x 1 2.9 O mol 2.9 ÷ = = 6 3 x 2 2.9 H mol 6.0 ÷ = =

Multiply all the Ratios by 3 Because C is 1.3

Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 33 ° Use the ratios as the subscripts in the

empirical formula 4 3 x 1.3 2.9 C mol 3.9 ÷ = = 3 3 x 1 2.9 O mol 2.9 ÷ = = 6 3 x 2 2.9 H mol 6.0 ÷ = = C 4H6O3

Determine the Empirical Formula of Acetic Anhydride if its Percent Composition is 47% Carbon, 47% Oxygen and 6.0% Hydrogen

Example #6

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Molecular Formulas

• The molecular formula is a multiple of the empirical formula

• To determine the molecular formula you need to know the empirical formula and the molar mass of the compound

‘ Determine the empirical formula

• May need to calculate it as previous C5H3

’ Determine the molar mass of the empirical formula

5 C = 60.05 g, 3 H = 3.024 g C5H3= 63.07 g

Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an

empirical formula of C5H3

Example #7

“ Divide the given molar mass of the compound by the molar mass of the empirical formula

– Round to the nearest whole number

4

07

.

63

252

=

g

g

Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an

empirical formula of C5H3

Example #7

(7)

Copyright©2000 by Houghton

Mifflin Company. All rights reserved. 37 ” Multiply the empirical formula by the

calculated factor to give the molecular formula

(C5H3)4= C20H12

Determine the Molecular Formula of Benzopyrene if it has a molar mass of 252 g and an

empirical formula of C5H3

Example #7

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Mifflin Company. All rights reserved. 38

Homework

• Questions and Problems:

• 6, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84

References

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