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Chapter 4Chemical Composition

• Mole Quantities

• Moles, Masses, and Particles

• Determining Empirical and Molecular Formulas

• Chemical Composition of Solutions

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Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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The Mole• The unit that acts as a bridge between the

microscopic world and the macroscopic world

• Contains 6.022 x 1023 particles (molecules, atoms, ions, formula units, etc.)

– This number is called Avogadro’s number.

• The amount of substance that contains as many basic particles (atoms, molecules,

or formula units) as there are atoms in exactly 12 g of carbon-12

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Moles of Various Elements and Compounds

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Figure 4.8

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Molar Mass• Describes the mass of 1 mole of a substance• We obtain the Molar Mass (MM) from the

periodic table by assigning different units to the atomic mass. – Instead of assigning the atomic mass units of amu,

we assign the atomic mass units of grams per 1 mole.

• Molar mass is the conversion factor between mass and moles.

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Practice - Molar Mass• Complete the table.

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NaCl

Molar Mass

H2O

Cl

Na

O

H

Atomic MassElement or Compound

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Practice Solutions – Molar Mass

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58.44 g/mol58.44 amuNaCl

18.02 g/mol

35.45 g/mol

22.99 g/mol

16.00 g/mol

1.008 g/mol

Molar Mass

18.02 amuH2O

35.45 amuCl

22.99 amuNa

16.00 amuO

1.008 amuH

Atomic MassElement or Compound

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Percent Composition by Mass

• An expression of the

portion of the total mass contributed by

each element

• To find the percent composition of E (E

is any element):

100% x sample of mass

E of mass E % =

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Practice - Percent Composition

1. What are the percent iron and the percent sulfur in an 8.33-g sample of

chalcopyrite that contains 2.54 g Fe and 2.91 g S?

2. A 4.55-g sample of limestone (CaCO3)

contains 1.82 g of calcium. What is the percent Ca in limestone?

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Practice Solutions - Percent Composition

1. What are the percent iron and the percent

sulfur in an 8.33-g sample of chalcopyrite that contains 2.54 g Fe and 2.91 g S?

samplein S 34.9% 100% x sample g 8.33

S g 2.91 S % ==

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samplein Fe 30.5% 100% x sample g 8.33

Fe g 2.54 Fe % ==

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Practice Solutions - Percent Composition

2. A 4.55-g sample of limestone (CaCO3) contains 1.82 g of calcium. What is the

percent Ca in limestone?

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limestonein Ca 40.0% 100% x limestone g 4.55

Ca g 1.82 Ca % ==

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Practice – Conversions with Molar Mass

1. How many moles of sulfur are in the

1.28 g of sulfur (S) found in a

sample of chalcopyrite?

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Practice Solutions – Conversions with

Molar Mass

1. How many moles of sulfur are in the 1.28 g of sulfur (S) found in a sample of chalcopyrite?

The conversion factor needed is the MM of sulfur: 1 mole S = 32.07 g S

We can write this equality as a ratio as well:1 mole S or 32.07 g S32.07 g S 1 mole S

To solve, we need to cancel out the grams of S. Therefore:

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S mol 0.0399 S g 32.07

S mol 1 x S g 1.28 =

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Extra Practice – Conversions with

Molar Mass1. How many moles of aspartame (C14H18N2O5)

are found in 40. mg of aspartame? How many molecules of aspartame are found in this

mass?

2. If one aspirin tablet contains 0.324 g of acetylsalicylic acid (C9H8O4), then how many molecules of acetylsalicylic acid are in 2

aspirin tablets?

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Extra Practice Solutions – Conversions with Molar Mass

1. How many moles of aspartame (C14H18N2O5) are found in 40. mg of aspartame? How many molecules of aspartame are found in this mass?

1. We first need to convert from mg to g:

2. Next, we need to find the MM of C14H18N2O5:(14 moles C x 12.01 g C) + (18 moles H x 1.01 g H) +

1 mol C 1 mol H(2 moles N x 14.01 g N) + (5 moles O x 16.00 g O)

1 mol N 1 mol O

= 294.34 g/mol C14H18N2O5

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ONHC g 0.040 ONHC mg 1000

ONHC g 1 x ONHC mg 40. 521814

521814

521814521814 =

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Extra Practice Solutions Continued – Conversions with Molar Mass

1. How many moles of aspartame (C14H18N2O5) are found in 40. mg of aspartame? How many

molecules of aspartame are found in this mass?

3. Next, convert from grams to moles:

• Finally, we convert from moles to molecules:

5218144-

521814

521814521814 ONHC mol 10 x 1.4

ONHC g 294.34

ONHC mol 1 x ONHC g 0.040 =

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52181419

521814

52181423

5218144- ONHC molecules 10 x 8.2

ONHC mol 1

ONHC molecules 10 x 6.022 x ONHC mol 10 x 1.4 =

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Extra Practice Solutions –Conversions with Molar Mass

2. If one aspirin tablet contains 0.324 g of acetylsalicylic acid (C9H8O4), then how many molecules of acetylsalicylic acid are in 2 aspirin tablets?

= 1.08 x 1021 molecules C9H8O4

1 aspirin tablet

= 2.16 x 1021 molecules C9H8O4

OHC mol 1

OHC molecules 10 x 6.022 x

OHC g 180.17

OHC mol 1 x

bletaspirin ta 1

OHC g 0.324

489

48923

489

489489

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bletaspirin ta 1

OHC molecules 10 x 1.08 x bletsaspirin ta 2

48921

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Summary – Conversions with Molar

Mass and Avogadro’s Number

• To convert from moles to grams or from grams to moles, use a molar mass (MM) as your conversion factor.

• To convert from moles to particles (molecules, atoms, ions, or formula units) or from particles to moles, use Avogadro’s number as your conversion factor.

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Determining Empirical and

Molecular Formulas• Empirical formula

– Expresses the simplest ratios of atoms in a

compound

– Written with the smallest whole-number subscripts

• Molecular formula

– Expresses the actual number of atoms in a

compound

– Can have the same subscripts as the empirical formula or some multiple of them

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Determining Empirical and

Molecular Formulas

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Practice – Empirical and Molecular Formulas

• For which of these substances is the empirical formula the same as the molecular formula?

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Empirical or Molecular Formulas

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CH2C6H12cyclohexane

CaCl2

This compound does

not have a molecular formula

calcium chloride

H2SH2Shydrogen

sulfide

CH2C2H4ethylene

CH2C5H10cyclopentane

Empirical FormulaMolecular FormulaSubstance

Table 4.1 Some Empirical and Molecular Formulas

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Empirical Formulas from Percent Composition

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Finding Empirical Formulas• To find the empirical formula:

1. If starting with a percent composition, find

the mass of the element by assigning the percent composition (which has no units

but a % instead) the units of grams.

– If starting with another set of units, then convert the units to masses if necessary.

2. Convert from mass to moles using the MM

of the element.

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Finding Empirical Formulas• To find the empirical formula cont’d:

3. Repeat for all elements in the compound.

4. Find whole number subscripts by:

1. Dividing the moles of the each element by the smallest number of moles. The

quotients will give whole numbers which are now the subscripts for the empirical formula.

2. If #1 does not give whole numbers, then multiply all numbers by a multiplier that

will resolve the quotients into whole numbers.

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Practice – Finding Empirical Formulas

1. Determine the empirical formula for the mineral covellite, which has the percent

composition 66.5% Cu and 33.5% S.

2. Shattuckite is a fairly rare copper mineral.

It has the composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31%

hydrogen. Calculate the empirical formula of shattuckite.

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Practice Solutions – Finding Empirical

Formulas1. Determine the empirical formula for the mineral

covellite, which has the percent composition 66.5% Cu and 33.5% S.

First, reassign the percentages units of grams: 66.5 g Cu and 33.5 g S.

Then, convert to moles and divide both numbers by the lowest number.

66.5 g Cu x 1 mol Cu = 1.05 mol Cu 1.05 mol Cu = 1

63.55 g Cu 1.05 mol

33.5 g S x 1 mol S = 1.05 mol S 1.05 mol S = 1

32.07 g S 1.05 mol

The whole numbers in purple then become our subscripts.

The empirical formula is therefore: CuS

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Practice Solutions Continued – Finding

Empirical Formulas2. Shattuckite is a fairly rare copper mineral. It has the

composition 48.43% copper, 17.12% silicon, 34.14% oxygen, and 0.31% hydrogen. Calculate the empirical formula of shattuckite.

48.43 g Cu x 1 mol Cu = 0.7621 mol Cu = 2.5 mol Cu x 2 = 5 mol Cu

63.55 g Cu 0.3069

17.12 g Si x 1 mol Si = 0.6095 mol Si = 2 mol Si x 2 = 4 mol Si

28.09 g Si 0.3069

34.14 g O x 1 mol O = 2.134 mol O = 7 mol O x 2 = 14 mol O

16.00 g O 0.3069

0.31 g H x 1 mol H = 0.3069 mol H = 1 mol H x 2 = 2 mol H

1.01 g H 0.3069

Therefore, the empirical formula is: Cu5Si4O14H2

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Molecular Formulas• To determine a molecular

formula, the problem must give a piece of experimental data, such as a molar mass, MM.

• To find the molecular formula:

1. Find the empirical formula first.

2. Divide the empirical formula’s molar mass by the experimental molar mass (which is given).

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Practice - Molecular Formulas

1. Potassium persulfate is a strong

bleaching agent. It has a percent

composition of 28.93% potassium,

23.72% sulfur, and 47.35% oxygen.

The experimental molar mass of

270.0 g/mol. What are the empirical

and molecular formulas of

potassium persulfate?

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Practice Solutions - Molecular Formulas1. Potassium persulfate is a strong bleaching agent. It

has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate?

First, find the empirical formula:

28.93 g K x 1 mol K = 0.7399 mol K = 1 mol K

39.10 g K 0.7396

23.72 g S x 1 mol S = 0.7396 mol S = 1 mol S

32.07 g S 0.7396

47.35 g O x 1 mol O = 2.959 mol O = 4 mol O

16.00 g O 0.7396

Thus, the empirical formula is KSO4.4 -

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Practice Solutions Continued –

Molecular Formulas1. Potassium persulfate is a strong bleaching agent. It

has a percent composition of 28.93% potassium, 23.72% sulfur, and 47.35% oxygen. The experimental molar mass of 270.0 g/mol. What are the empirical and molecular formulas of potassium persulfate?

To find the molecular formula:

Calculate the MM of KSO4 = (1 mol K x 39.10 g/mol K) + (1 mol S x 32.07 g/mol S) + (4 mol O x 16.00 g/mol O) = 135.17 g/mol KSO4

270.0 g/mol (Experimental MM) = 2

135.17 g/mol (Empirical Formula’s MM)

Thus, the molecular formula is K2S2O8.

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Determining Percent Composition Using Molar Mass

compound of Total

compound]in E moles of # x (E) [ E %

MM

MM=

• To determine the percent composition of an element (E) in a compound using molar mass (MM):

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Practice – Percent Composition• Calculate the percent composition of

each element in the following

compounds: [HINT: you must determine the compound formula 1st]

1. iron(II) chloride

2. dinitrogen tetroxide

3. sodium phosphate

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Practice Solutions –

Percent Composition• Calculate the percent composition of each element in the

following compounds:

1. iron(II) chloride – FeCl2Find the MM of the compound 1st:

(1 mol Fe x 55.85 g/mol Fe) + (2 mol Cl x 35.45 g/mol Cl)

= 126.75 g/mol FeCl2

% Fe = (1 mol Fe x 55.85 g/mol Fe) = 44.06% Fe in FeCl2126.75 g/mol FeCl2

% Cl = (2 mol Cl x 35.45 g/mol Cl) = 55.94% Cl in FeCl2126.75 g/mol FeCl2

NOTE: % Cl = 100.00% FeCl2 - 44.06% Fe = 55.94% Cl4 -

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Practice Solutions Continued

Percent Composition• Calculate the percent composition of each element in the

following compounds:

2. Dinitrogen tetroxide – N2O4

Find the MM of the compound 1st:

(2 mol N x 14.01 g/mol N) + (4 mol O x 16.00 g/mol O)

= 92.02 g/mol

% N = (2 mol N x 14.01 g/mol N) = 30.45% N in N2O4

92.02 g/mol N2O4

% O = (4 mol O x 16.00 g/mol O) = 69.55% O in N2O4

92.02 g/mol N2O4

NOTE: % O = 100.00% N2O4 – 30.45% N = 69.55% O4 -

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Practice Solutions Continued

Percent Composition• Calculate the percent composition of each element in the

following compounds:

3. Sodium phosphate – Na2PO4

Find the MM of the compound 1st:

(2 mol Na x 22.99 g/mol Na) + (1 mol P x 30.97 g/mol P) + (4 mol O x

16.00 g/mol O) = 140.95 g/mol Na2PO4

% Na = (2 mol Na x 22.99 g/mol Na) = 32.62% Na in Na2PO4

140.95 g/mol Na2PO4

% P = (1 mol P x 30.97 g/mol P) = 21.97% P in Na2PO4

140.95 g/mol Na2PO4

% O = (4 mol O x 16.00 g/mol O) = 45.41% O in Na2PO4

140.95 g/mol Na2PO4

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Chemical Composition of Solutions

• Solutions

– are any homogeneous mixture at the molecular or ionic scale

– are composed of solutes and solvents

• Solutes

– Are present in a lesser amount

– The substances that are dissolved (can be either wet or dry)

• Solvents

– Are present in the larger amount

– The substances that dissolve

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Making a Solution

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Solution Concentration• Concentration

– Is the relative amounts of solute and solvent in a solution

– When compared with one another, solutions are classified as dilute or concentrated.• Dilute solution

– A solution that contains a relatively small amount of solute

• Concentrated solution– A solution that contains a relatively large

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Concentration of Solutions

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Determining Concentration

• Percent by Mass

– Expresses concentration via percentage

% mass = mass solute x 100%

mass solution

• Molarity (M)

– The moles of solute dissolved in 1 L of solution

– The most common units of concentration

M = moles solute

L solution4 -

Insert diagram at bottom of pg. 138

(or pg. 140)

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Practice – Solution Concentration

1. A solution is prepared from 22.5 g of H2Sdissolved in sufficient water to give 250.0 mL

of solution. What is the molarity of the solution?

2. Bluestone is copper(II) sulfate pentahydrate, CuSO4•5H2O, with a molar mass of 249.7 g/mol. A sample of pond water was found to have a

concentration of 6.2 x 105 M copper(II) sulfate. If the pond has a volume of 1.8 x 107 L, then what mass of bluestone did the farmer add to

the pond?

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Practice Solutions – Solution Concentration1. A solution is prepared from 22.5 g of H2S dissolved in

sufficient water to give 250.0 mL of solution. What is the molarity of the solution?

M = moles of solute

L solution

H2S is the solute, so

Moles of H2S = 22.5 g H2S x 1 mole H2S = 0.660 moles H2S

34.09 g H2S

To find L of solution:

250.0 mL x 1 L = 0.2500 L

1000 mL

M = 0.660 moles H2S = 2.64 M H2S solution

0.2500 L solution

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Practice Solutions Continued –Solution Concentration

2. Bluestone is copper(II) sulfate pentahydrate, CuSO4•5H2O, with a molar mass of 249.7 g/mol. A sample of pond water was found to have a concentration of 6.2 x 105 Mcopper(II) sulfate. If the pond has a volume of 1.8 x 107 L, then what mass of bluestone did the farmer add to the pond?

Start with the volume as it only has one set of units:

1.8 x 107 L x 6.2 x 105 mol x 249.7 g = 2.8 x 1015 g CuSO4•5H2O

1 L 1 mol

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Dilution and the Dilution Equation

• Dilution– The process of adding more solvent to solution

• The dilution equation:

Molescon = MconVcon Moles1 = M1V1

MconVcon = MdilVdil M1V1 = M2V2

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Diluting a More Concentrated Solution

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Practice – Dilution

1. If 42.8 mL of 3.02 M H2SO4 solution

is diluted to a final volume 500.0 mL,

what is the molarity of the diluted

solution of H2SO4?

2. What is the concentration of a

solution prepared by diluting 35.0

mL of 0.150 M KBr to 250.0 mL?

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Practice Solutions – Dilution1. If 42.8 mL of 3.02 M H2SO4 solution is

diluted to a final volume 500.0 mL, what is the molarity of the diluted solution of H2SO4?

MconVcon = MdilVdil

(3.02 M)(42.8 mL) = Mdil(500.0 mL)

Mdil = (3.02 M)(42.8 mL) = 0.259 M

500.0 mL

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Practice Solutions – Dilution

2. What is the concentration of a

solution prepared by diluting 35.0

mL of 0.150 M KBr to 250.0 mL?

MconVcon = MdilVdil

(0.150 M)(35.0 mL) = Mdil(250.0 mL)

Mdil = (0.150 M)(35.0 mL) = 0.0210 M

250.0 mL

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