chapter 10 the mole “making measurements in chemistry” t. witherup 2006

Chapter 10 The MoleChapter 10 The Mole

““Making Measurements in Making Measurements in Chemistry”Chemistry”

T. Witherup 2006

Chapt. 10 OBJECTIVESChapt. 10 OBJECTIVES

Define the “mole” & describe its importance.Define the “mole” & describe its importance. Identify & use Avogadro’s number.Identify & use Avogadro’s number. Define “molar mass” & explain how it relates the Define “molar mass” & explain how it relates the

mass of a substance to its number of particles.mass of a substance to its number of particles. Convert among the number of particles, moles & Convert among the number of particles, moles &

mass of a substance.mass of a substance. Describe “molar volume” & use it to solve Describe “molar volume” & use it to solve

problems.problems. Find the percentage composition of a formula.Find the percentage composition of a formula. Use percentage composition to find the formula Use percentage composition to find the formula

of an unknown sample.of an unknown sample. Find empirical & molecular formulas.Find empirical & molecular formulas.

10-1 Chemical Measurements10-1 Chemical Measurements Atomic MassAtomic Mass: the mass of an atom expressed : the mass of an atom expressed

relative to the mass assigned to the carbon-12 relative to the mass assigned to the carbon-12 isotope, in isotope, in amuamu ( (aatomic tomic mmass ass uunits).nits).

1 1 amuamu = 1/12 of the mass of Carbon-12. = 1/12 of the mass of Carbon-12. Since C-12 has 6p + 6n = 12 particles, its mass = Since C-12 has 6p + 6n = 12 particles, its mass =

12 12 amu.amu. Since C-13 has 6p + 7n = 13 particles, its mass = Since C-13 has 6p + 7n = 13 particles, its mass =

13 13 amuamu.. And C-14 has a mass of ??? And C-14 has a mass of ??? amuamu?? (Why don’t we consider the mass of the electrons in (Why don’t we consider the mass of the electrons in

these atoms?)these atoms?)

Average Atomic MassAverage Atomic Mass

Since “natural” carbon has 1.1% the C-13 Since “natural” carbon has 1.1% the C-13 isotope and only a trace amount of the C-isotope and only a trace amount of the C-14 isotope, its 14 isotope, its average atomic massaverage atomic mass is is dominated by C-12, or 12.011 dominated by C-12, or 12.011 amuamu..

The average atomic mass accounts for all The average atomic mass accounts for all (natural) isotopes of an element.(natural) isotopes of an element.

The average atomic mass of each element The average atomic mass of each element may be found on the Periodic Table.may be found on the Periodic Table.

NOTE:NOTE: In our course, when using average atomic In our course, when using average atomic masses, always round to the ‘hundredths’ place.masses, always round to the ‘hundredths’ place.

Average Atomic Mass of ElementsAverage Atomic Mass of Elements

ELEMENTELEMENT AVERAGE ATOMIC AVERAGE ATOMIC MASS (MASS (amuamu))

AVERAGE ATOMIC AVERAGE ATOMIC MASS, rounded MASS, rounded ((amuamu))

HydrogenHydrogen 1.007941.00794 1.011.01

CarbonCarbon 12.011112.0111 12.0112.01

OxygenOxygen 15.999415.9994 16.0016.00

ChlorineChlorine 35.45335.453 35.4535.45

IronIron 55.84755.847 55.8555.85

Formula MassFormula Mass

Formula MassFormula Mass: the sum of the atomic masses : the sum of the atomic masses of of allall atoms in a compound. atoms in a compound. Water (HWater (H22O) has 2 Hydrogen atoms and 1 O) has 2 Hydrogen atoms and 1

Oxygen atom.Oxygen atom. Formula Mass of HFormula Mass of H22O isO is 2 X (1.01) 2 X (1.01) amuamu + 16.00 + 16.00 amuamu = 18.02 = 18.02 amuamu.. What is the formula mass of methane, CHWhat is the formula mass of methane, CH44?? Of NaCl? (For ionic compounds we refer to a Of NaCl? (For ionic compounds we refer to a

“formula unit”“formula unit” rather than a “formula mass,” but rather than a “formula mass,” but it is essentially the same idea.)it is essentially the same idea.)

Of ammonia, NHOf ammonia, NH33?? Of glucose, COf glucose, C66HH1212OO66??

But what is a “mole”?But what is a “mole”?

That furry creature who burrows in the That furry creature who burrows in the yard?yard?

The dark pigmentation on our skin?The dark pigmentation on our skin? A massive stone structure used as a A massive stone structure used as a

breakwater or pier?breakwater or pier? An undercover agent?An undercover agent?

NO! NO! A mole is a special chemical term A mole is a special chemical term used to count atoms!used to count atoms!

The MOLE The MOLE (mol)(mol) A A molemole of any element is defined as the amount of of any element is defined as the amount of

the element that contains as many atoms as there the element that contains as many atoms as there are in exactly 12 g of the carbon-12 isotope.are in exactly 12 g of the carbon-12 isotope.

A mole is found experimentally to be equal to A mole is found experimentally to be equal to 6.022 X 106.022 X 102323 atoms of C-12, which is called atoms of C-12, which is called Avogadro’s Number (Avogadro’s Number (NNAA).). 12 g Carbon-12 = 1 mol Carbon 12 atoms = 6.022 X 1012 g Carbon-12 = 1 mol Carbon 12 atoms = 6.022 X 102323

Carbon 12 atomsCarbon 12 atoms

The molar mass of any substance is the mass of one mole of that substance. Molar mass is numerically equal to the atomic mass,

molecular mass, or formula mass of the substance. A mole of A mole of anyany substance contains Avogadro’s number of substance contains Avogadro’s number of

units of that substance (6.022 X 10units of that substance (6.022 X 102323 units). units).

What’s in a Mole?What’s in a Mole?

AtomsAtoms – one mole of an element contains 6.022 X – one mole of an element contains 6.022 X 10102323 atoms of the element. atoms of the element.

MoleculesMolecules – one mole of a molecular (covalent) – one mole of a molecular (covalent) compound contains 6.022 X 10compound contains 6.022 X 102323 molecules. molecules.

FormulaFormula UnitsUnits – one mole of an ionic compound – one mole of an ionic compound contains 6.022 X 10contains 6.022 X 102323 formula units. formula units.

GizmosGizmos – one mole of gizmos contains 6.022 X 10 – one mole of gizmos contains 6.022 X 102323 gizmos.gizmos.

AnythingAnything – one mole of – one mole of anythinganything contains 6.022 X contains 6.022 X 10102323 anythings!anythings!

NNAA (6.022 X 10 (6.022 X 102323) is very practical for counting small ) is very practical for counting small particles, especially things like atoms, ions and particles, especially things like atoms, ions and molecules.molecules.

EXAMPLESEXAMPLES How many pens in 1 mole of pens?How many pens in 1 mole of pens? How many atoms in 63.546g of Cu?How many atoms in 63.546g of Cu? How many atoms in 6.3546g of Cu?How many atoms in 6.3546g of Cu? How many molecules in 1 mole of sugar How many molecules in 1 mole of sugar

(C(C66HH1212OO66)?)? How many molecules in 10 moles of sugar?How many molecules in 10 moles of sugar? How many carbon atoms in 1 mole of sugar? How How many carbon atoms in 1 mole of sugar? How

many oxygen atoms?many oxygen atoms? How many formula units in 1 mole of CaClHow many formula units in 1 mole of CaCl22?? How many calcium ions in 1 mole of CaClHow many calcium ions in 1 mole of CaCl22?? How many chloride ions in 1 mole of CaClHow many chloride ions in 1 mole of CaCl22?? How many chloride ions in 0.1 mole of CaClHow many chloride ions in 0.1 mole of CaCl22? ?

10-2 Mole Conversions by Factor Label 10-2 Mole Conversions by Factor Label Method (1)Method (1)

Mass and MolesMass and Moles Use the molar mass of the substance.Use the molar mass of the substance.

1 Mole = 1 Mole = nn grams of the substance, so grams of the substance, so 1 = 1 = nn grams/Mole, but also grams/Mole, but also 1 = 1 = Mole/Mole/nn grams of the substance. grams of the substance.

Use these conversions by setting up Use these conversions by setting up Factor Factor LabelsLabels to cancel the units! to cancel the units!

See examples on next slide.See examples on next slide.

10-2 Mole Conversions by Factor Label 10-2 Mole Conversions by Factor Label Method (1) ExamplesMethod (1) Examples

Example A: How many moles in 75.0 g of iron?How many moles in 75.0 g of iron?

Example B: How many grams in 0.250 mol Na?How many grams in 0.250 mol Na?

75.0g Fe X1 mol Fe

55.85g Fe= 1.34 mol Fe

0.250 mol Na X 22.99g Na

1 mol Na= 5.75g Na

Solving ‘Mole Problems” (1)Solving ‘Mole Problems” (1)

MOLES

MASS

Molar mass

mol/g

g/mol


Particles and MolesParticles and Moles Use Avogadro’s Number (6.022 X 10Use Avogadro’s Number (6.022 X 102323) of ) of

particles.particles.

(6.022 X 10(6.022 X 102323)particles/mol )particles/mol 1 mol/(6.022 X 1 mol/(6.022 X 10102323)particles )particles

Again, set up Again, set up Factor LabelsFactor Labels to cancel to cancel units!units!

See examples on next slide.See examples on next slide.

10-2 Mole Conversions by Factor Label 10-2 Mole Conversions by Factor Label Method (2) ExamplesMethod (2) Examples

= 0.697 mol = 0.697 mol COCO22

Example C: How many atoms in 0.25 mol Na?How many atoms in 0.25 mol Na?

0.250 mol Na X0.250 mol Na X 6.022 X106.022 X102323 atoms Na atoms Na

1 mol Namol Na= 1.51 X 10= 1.51 X 102323 atoms Na atoms Na

Example DExample D: How many moles in 4.20 X 10: How many moles in 4.20 X 102323 molecules of CO molecules of CO22??

4.20 X 104.20 X 102323 Molecules CO Molecules CO22 X X1 mol CO2

6.022 X 106.022 X 102323 molecules CO molecules CO22


MOLES

PARTICLES

Number of Particles in1 mole(6.02 X 10(6.02 X 102323))


Gases and MolesGases and Moles Avogadro proposed that equal volumes of Avogadro proposed that equal volumes of

gases contain the same number of gas gases contain the same number of gas particles at a given temperature & pressure.particles at a given temperature & pressure.

Therefore one mole of gas #1 would have the Therefore one mole of gas #1 would have the same volume as one mole of gas #2.same volume as one mole of gas #2.

It is observed that one mole of It is observed that one mole of anyany gas gas occupies 22.4 liters @ STP (occupies 22.4 liters @ STP (molarmolar volumevolume).).

STPSTP = Standard Temperature and = Standard Temperature and Pressure, 0° C and 1 atmosphere.Pressure, 0° C and 1 atmosphere.

Once more, set up Once more, set up Factor LabelsFactor Labels to cancel to cancel the units!the units!

10-2 Mole Conversions by Factor Label 10-2 Mole Conversions by Factor Label Method (3) Examples (Gases)Method (3) Examples (Gases)

Example E: What is the volume of 13.0 moles of Example E: What is the volume of 13.0 moles of hydrogen gas at STP?hydrogen gas at STP?

13.0 mol H13.0 mol H22 X X22.4 L H22.4 L H22

1 mol H1 mol H22

= 291 L H= 291 L H22

Example F: How many moles are in 250. mL of oxygenExample F: How many moles are in 250. mL of oxygen at STP?at STP?

250. mL O250. mL O2 2 XX1 L O1 L O22

1000 mL O1000 mL O22

1 mol O1 mol O22

22.4 L O2

= 0.0112 mol O= 0.0112 mol O22X


MOLES

VOLUME of gas @ STP

Molar volume(22.4L/mol @ STP)

10-2 Mole Conversions by Factor Label 10-2 Mole Conversions by Factor Label Method (Summary) (See p 330.)Method (Summary) (See p 330.)

Mass and MolesMass and Moles Use the molar mass of the substance.Use the molar mass of the substance.

Particles and MolesParticles and Moles Use Avogadro’s Number (6.02 X 10Use Avogadro’s Number (6.02 X 102323) of particles.) of particles.

Gases and MolesGases and Moles Use the Molar Volume (22.4 liters @ STP).Use the Molar Volume (22.4 liters @ STP). STP = Standard Temperature and Pressure, 0° C and 1 STP = Standard Temperature and Pressure, 0° C and 1

atmosphere.atmosphere. Set up Set up Factor LabelsFactor Labels to cancel units! to cancel units!

DON’T GET LAZY! Include labels to ensure that DON’T GET LAZY! Include labels to ensure that ALL units cancel correctly.ALL units cancel correctly.

Multi-step conversions are easily done if Multi-step conversions are easily done if you are careful with the labels!you are careful with the labels!

Summary: Solving ‘Mole Problems”Summary: Solving ‘Mole Problems”

MOLES

MASS

Molar mass

PARTICLES

Number of Particles in1 mole(6.02 X 10(6.02 X 102323))

VOLUMEof gas @STP

Molar volume(22.4L/mol @ STP)

10-3 Empirical & Molecular 10-3 Empirical & Molecular FormulasFormulas

Percentage CompositionPercentage Composition The mass of each element in a compound The mass of each element in a compound

compared to the entire mass of the compound compared to the entire mass of the compound and multiplied by 100%.and multiplied by 100%.

Example 1Example 1 2.45 g aluminum oxide decomposes into 1.30 g 2.45 g aluminum oxide decomposes into 1.30 g

aluminum & 1.15 g oxygen. What is the percentage aluminum & 1.15 g oxygen. What is the percentage composition?composition?

%O = (1.15g O/2.45g Aluminum Oxide) X 100% = %O = (1.15g O/2.45g Aluminum Oxide) X 100% = 46.9% Oxygen (O, not O46.9% Oxygen (O, not O22))

%Al = (1.30g Al/2.45g Aluminum Oxide) X 100% = %Al = (1.30g Al/2.45g Aluminum Oxide) X 100% = 53.1% Aluminum53.1% Aluminum

As a check, note that 46.9% + 53.1% = 100.0%.As a check, note that 46.9% + 53.1% = 100.0%.

10-3 Empirical & Molecular Formulas 10-3 Empirical & Molecular Formulas (cont’d)(cont’d)

Percentage Composition (cont’d)Percentage Composition (cont’d) Example 2Example 2

Determine the percent composition of Determine the percent composition of CaCOCaCO33..

Molar Mass = 40.08 + 12.01 +3(16.00) = Molar Mass = 40.08 + 12.01 +3(16.00) = 100.09g/mol100.09g/mol

%Ca = (40.08g Ca/100.09g CaCO%Ca = (40.08g Ca/100.09g CaCO33) X 100% = ) X 100% = 40.04% Ca40.04% Ca

% C = (12.01g C/100.09g CaCO% C = (12.01g C/100.09g CaCO33) X 100% = 12.00% ) X 100% = 12.00% CC

% O = (48.00g O/100.09g CaCO% O = (48.00g O/100.09g CaCO33) X 100% = 47.96% ) X 100% = 47.96% OO

Check: 40.04% + 12.00% + 47.96% = 100.00% Check: 40.04% + 12.00% + 47.96% = 100.00%

Determining Empirical FormulasDetermining Empirical Formulas

Empirical FormulaEmpirical Formula The formula that gives the simplest whole The formula that gives the simplest whole

number ratio of the atoms of the elements in number ratio of the atoms of the elements in the formula.the formula.

Example 1Example 1 What is the empirical formula of a compound containing What is the empirical formula of a compound containing

2.644g of gold and 0.476g of chlorine?2.644g of gold and 0.476g of chlorine?

0.476g Cl X (1 mol Cl/35.45g Cl) = 0.0134 mol Cl0.476g Cl X (1 mol Cl/35.45g Cl) = 0.0134 mol Cl 2.644g of Au X (1 mol Au/196.97g Au) = 0.0134 mol Au2.644g of Au X (1 mol Au/196.97g Au) = 0.0134 mol Au

Empirical formula = AuEmpirical formula = Au0.01340.0134 Cl Cl0.01340.0134 or simply or simply AuClAuCl

Determining Empirical FormulasDetermining Empirical Formulas

Example 2Example 2 What is the empirical formula of a compound with What is the empirical formula of a compound with

5.75 g Na, 3.50 g N & 12.00 g O?5.75 g Na, 3.50 g N & 12.00 g O? First, find the mole amounts.First, find the mole amounts. 5.75g Na X (1 mol Na/22.99g Na) = 0.250 mol Na5.75g Na X (1 mol Na/22.99g Na) = 0.250 mol Na 3.50 g N X (1 mol N/14.01g N) = 0.250 mol N3.50 g N X (1 mol N/14.01g N) = 0.250 mol N 12.00g O X (1 mol O/16.00g O) = 0.750 mol O12.00g O X (1 mol O/16.00g O) = 0.750 mol O Empirical formula = NaEmpirical formula = Na0.2500.250 N N0.2500.250 O O0.7500.750

Divide each mole quantity by the smallest to Divide each mole quantity by the smallest to get whole numbers. (0.250 in this case)get whole numbers. (0.250 in this case)

Empirical formula = NaNOEmpirical formula = NaNO33

Determining Molecular FormulasDetermining Molecular Formulas Molecular FormulaMolecular Formula

The formula that gives the The formula that gives the actualactual numbernumber of of atoms of each element in a molecular compound.atoms of each element in a molecular compound.

Example 3Example 3 Hydrogen peroxide has a molar mass of 34.00 g/mol and a Hydrogen peroxide has a molar mass of 34.00 g/mol and a

chemical composition of 5.90% H & 94.1% O. What is its chemical composition of 5.90% H & 94.1% O. What is its molecular formula?molecular formula?

First, find the empirical formula, assuming the percents are First, find the empirical formula, assuming the percents are mass.mass.

For 5.90% H: 5.90gH X (1 mol H/1.01g H) = 5.84 mol HFor 5.90% H: 5.90gH X (1 mol H/1.01g H) = 5.84 mol H For 94.1% O: 94.1g O X (1 mol O/16.00g O) = 5.88 mol OFor 94.1% O: 94.1g O X (1 mol O/16.00g O) = 5.88 mol O Empirical formula = HEmpirical formula = H5.845.84OO5.885.88 or HO. or HO. But molar mass = 34.00g/mol and HO is only 17.01g/mol.But molar mass = 34.00g/mol and HO is only 17.01g/mol. Therefore, Molecular Formula = 2(HO) or HTherefore, Molecular Formula = 2(HO) or H22OO22..

Determining Molecular FormulasDetermining Molecular Formulas

Example 4Example 4 A compound contains 42.56 g Pd and 0.8000 g H. If its A compound contains 42.56 g Pd and 0.8000 g H. If its

molar mass is 216.8 g/mol, find the molecular formula.molar mass is 216.8 g/mol, find the molecular formula. First, find the empirical formula.First, find the empirical formula. 42.56g Pd X (1 mol Pd/106.42 g Pd) = 0.4000 mol Pd42.56g Pd X (1 mol Pd/106.42 g Pd) = 0.4000 mol Pd 0.8000g H X (1 mol H/1.00g H) = 0.800 mol H0.8000g H X (1 mol H/1.00g H) = 0.800 mol H Empirical formula: PdEmpirical formula: Pd0.40000.4000HH0.8000.800 or PdH or PdH22 PdHPdH22 has a mass = 108.44 g/mol. has a mass = 108.44 g/mol.

Since molar mass = 216.8 and the empirical Since molar mass = 216.8 and the empirical mass = 108.4, the Molecular Formula is mass = 108.4, the Molecular Formula is twicetwice the empirical formula [2(PdHthe empirical formula [2(PdH22)] or simply Pd)] or simply Pd22HH44..

Did we cover the OBJECTIVES?Did we cover the OBJECTIVES?

Define the “mole” & describe its importance.Define the “mole” & describe its importance. Identify & use Avogadro’s number.Identify & use Avogadro’s number. Define “molar mass” & explain how it relates the Define “molar mass” & explain how it relates the

mass of a substance to its number of particles.mass of a substance to its number of particles. Convert among the number of particles, moles & Convert among the number of particles, moles &

mass of a substance.mass of a substance. Describe “molar volume” & use it to solve Describe “molar volume” & use it to solve

problems.problems. Find the percentage composition of a formula.Find the percentage composition of a formula. Use percentage composition to find the formula Use percentage composition to find the formula

of an unknown sample.of an unknown sample. Find empirical & molecular formulas.Find empirical & molecular formulas.

How to be successful at solving How to be successful at solving Mole Problems:Mole Problems:

USE FACTOR LABELS!USE FACTOR LABELS! PRACTICE!PRACTICE! PRACTICE!PRACTICE! PRACTICE!PRACTICE!

chapter 10 the mole “making measurements in chemistry” t. witherup 2006

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