1 Chapter 10 “Chemical Quantities” Chemistry Pioneer High School Mr. David Norton

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Chapter 10Chapter 10

“Chemical Quantities”“Chemical Quantities”

ChemistryChemistryPioneer High SchoolPioneer High School

Mr. David NortonMr. David Norton

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Section 10.1Section 10.1The Mole: A Measurement of The Mole: A Measurement of

MatterMatter OBJECTIVES:OBJECTIVES:

–DescribeDescribe methods of methods of measuring the amount of measuring the amount of something.something.

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–DefineDefine Avogadro’s Avogadro’s number as it relates to a number as it relates to a mole of a substance.mole of a substance.

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–DistinguishDistinguish between the between the atomic mass of an atomic mass of an element and its molar element and its molar mass.mass.

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–DescribeDescribe how the mass how the mass of a mole of a compound of a mole of a compound is calculated.is calculated.

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What is a Mole?What is a Mole? You can measure You can measure massmass, , or or volumevolume,, or you can or you can count piecescount pieces..

We measure mass in We measure mass in gramsgrams.. We measure volume in We measure volume in litersliters.. We count pieces in We count pieces in MOLESMOLES..

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Moles (abbreviated: mol)Moles (abbreviated: mol) Defined as the number of carbon Defined as the number of carbon

atoms in exactly 12 grams of atoms in exactly 12 grams of carbon-12.carbon-12.

1 mole is 6.02 x 101 mole is 6.02 x 102323 representative particles.representative particles.

Treat it like a very large dozenTreat it like a very large dozen 6.02 x 106.02 x 102323 is called: is called:

Avogadro’s numberAvogadro’s number..

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Representative particlesRepresentative particles The smallest pieces of a The smallest pieces of a

substance:substance:

–For a molecular compound: it is For a molecular compound: it is the the moleculemolecule..

–For an ionic compound: it is the For an ionic compound: it is the formula unitformula unit (made of ions). (made of ions).

–For an element: it is the For an element: it is the atomatom..

»Remember the 7 diatomic Remember the 7 diatomic elements (made of molecules)elements (made of molecules)

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Types of questionsTypes of questions How many oxygen atoms in the How many oxygen atoms in the

following?following?–CaCOCaCO33

–AlAl22(SO(SO44))33

How many ions in the following?How many ions in the following?–CaClCaCl22–NaOHNaOH–AlAl22(SO(SO44))33

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Some practice problems:Some practice problems: How many molecules of COHow many molecules of CO22 are are

there in 4.56 moles of COthere in 4.56 moles of CO22 ? ? How many moles of water is 5.87 How many moles of water is 5.87

x 10x 102222 molecules? molecules? How many atoms of carbon are How many atoms of carbon are

there in 1.23 moles of Cthere in 1.23 moles of C66HH1212OO66 ? ? How many moles is 7.78 x 10How many moles is 7.78 x 102424

formula units of MgClformula units of MgCl22??

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Measuring MolesMeasuring Moles Remember relative atomic mass?Remember relative atomic mass? The amu was one twelfth the The amu was one twelfth the

mass of a carbon-12 atom.mass of a carbon-12 atom. Since the mole is the number of Since the mole is the number of

atoms in 12 grams of carbon-12,atoms in 12 grams of carbon-12, the decimal number on the the decimal number on the

periodic table is also the mass of 1 periodic table is also the mass of 1 mole of those atoms in grams.mole of those atoms in grams.

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Gram Atomic Mass (gam)Gram Atomic Mass (gam) Equals the mass of 1 mole of an Equals the mass of 1 mole of an

element in gramselement in grams 12.01 grams of C has the same number 12.01 grams of C has the same number

of pieces as 1.008 grams of H and of pieces as 1.008 grams of H and 55.85 grams of iron.55.85 grams of iron.

We can write this as: 12.01 g C = We can write this as: 12.01 g C = 1 mole C (this is also the 1 mole C (this is also the molar massmolar mass))

We can count things by weighing them.We can count things by weighing them.

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ExamplesExamples How much would 2.34 moles of How much would 2.34 moles of

carbon weigh?carbon weigh? How many moles of magnesium is How many moles of magnesium is

24.31 g of Mg?24.31 g of Mg? How many atoms of lithium is 1.00 How many atoms of lithium is 1.00

g of Li?g of Li? How much would 3.45 x 10How much would 3.45 x 102222

atoms of U weigh?atoms of U weigh?

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What about compounds?What about compounds? in 1 mole of Hin 1 mole of H22O molecules there are O molecules there are

two moles of H atoms and 1 mole of two moles of H atoms and 1 mole of O atomsO atoms

To find the mass of one mole of a To find the mass of one mole of a compound compound –determine the moles of the determine the moles of the

elements they haveelements they have–Find out how much they would Find out how much they would

weigh (from the periodic table)weigh (from the periodic table)–add them upadd them up

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Calculating Formula MassCalculating Formula Mass

Calculate the formula mass of Calculate the formula mass of magnesium carbonate, MgCOmagnesium carbonate, MgCO33..

24.31 g + 12.01 g + 3(16.00 g) =24.31 g + 12.01 g + 3(16.00 g) = 84.32 g84.32 g

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Section 10.2Section 10.2Mole-Mass and Mole-Volume Mole-Mass and Mole-Volume

RelationshipsRelationships OBJECTIVES:OBJECTIVES:

–DescribeDescribe how to convert how to convert the mass of a substance to the mass of a substance to the number of moles of a the number of moles of a substance, and moles to substance, and moles to mass.mass.

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Section 10.2Section 10.2Mole-Mass and Mole-Volume Mole-Mass and Mole-Volume

RelationshipsRelationships OBJECTIVES:OBJECTIVES:

–IdentifyIdentify the volume of a the volume of a quantity of gas at STP.quantity of gas at STP.

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Molar MassMolar MassMolar mass is the generic term is the generic term

for the mass of one mole of for the mass of one mole of anyany substance substance (expressed in grams/mol)(expressed in grams/mol)

The same as: The same as: 1) 1) GGram ram MMolecular olecular MMass, ass, 2) 2) GGram ram FFormula ormula MMass, and ass, and 3) 3) GGram ram AAtomic tomic MMassass

just a much broader term.just a much broader term.

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ExamplesExamples Calculate the molar mass of the Calculate the molar mass of the

following following andand tell what type it is: tell what type it is: NaNa22SS

NN22OO44

CC Ca(NOCa(NO33))22

CC66HH1212OO66

(NH(NH44))33POPO44

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Molar Mass is…Molar Mass is… The number of grams of 1 mole The number of grams of 1 mole

of atoms, ions, or molecules.of atoms, ions, or molecules. We can make We can make conversion conversion

factorsfactors from these. from these.

–To change grams of a To change grams of a compound to moles of a compound to moles of a compound.compound.

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For exampleFor example

How many moles is 5.69 g How many moles is 5.69 g of NaOH?of NaOH?

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How many moles is 5.69 g of NaOH?How many moles is 5.69 g of NaOH?

5 69. g

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5 69. g mole

g

need to change grams to moles

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5 69. g mole

g

need to change grams to moles for NaOH

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5 69. g mole

g

need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g

1 mole of H = 1.01 g

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5 69. g mole

g


1 mole of H = 1.01 g 1 mole NaOH = 40.00 g

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5 69. g 1 mole

40.00 g



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5 69. g 1 mole

40.00 = 0.142 mol NaOH

g



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The Mole-Volume RelationshipThe Mole-Volume Relationship Many of the chemicals we deal with are Many of the chemicals we deal with are

gases.gases.

–They are difficult to They are difficult to weigh (or mass)weigh (or mass).. Need to know how many moles of gas Need to know how many moles of gas

we have.we have. Two things effect the volume of a gasTwo things effect the volume of a gas

–Temperature and pressureTemperature and pressure We need to compare them at the same We need to compare them at the same

temperature and pressure.temperature and pressure.

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Standard Temperature and PressureStandard Temperature and Pressure0ºC and 1 atm pressure0ºC and 1 atm pressureabbreviated “abbreviated “STP”STP”At STP 1 mole of any gas At STP 1 mole of any gas

occupies 22.4 Loccupies 22.4 L

–Called the Called the molar volumemolar volume1 mole = 22.4 L of any gas at 1 mole = 22.4 L of any gas at

STPSTP

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Practice ExamplesPractice Examples

What is the volume of 4.59 What is the volume of 4.59 mole of COmole of CO22 gas at STP? gas at STP?

How many moles is 5.67 L of How many moles is 5.67 L of OO22 at STP?at STP?

What is the volume of 8.8 g of What is the volume of 8.8 g of CHCH44 gas at STP? gas at STP?

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Density of a gasDensity of a gas D = m / VD = m / V

–for a gas the units will be g / Lfor a gas the units will be g / L We can determine the density of any We can determine the density of any

gas at STP if we know its formula.gas at STP if we know its formula. To find the density we need: 1) To find the density we need: 1) massmass

and 2) and 2) volumevolume.. If you assume you have 1 mole, then If you assume you have 1 mole, then

the mass is the the mass is the molar massmolar mass And, at STP the volume is 22.4 L.And, at STP the volume is 22.4 L.

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Practice ExamplesPractice Examples

Find the density of COFind the density of CO2 2 at STP.at STP.

Find the density of CHFind the density of CH44 at STP. at STP.

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Another way:Another way: Given the density, we can find the molar Given the density, we can find the molar

mass of the gas.mass of the gas. Again, pretend you have 1 mole at STP, Again, pretend you have 1 mole at STP,

so V = 22.4 L.so V = 22.4 L. m = D x Vm = D x V m is the mass of 1 mole, since you have m is the mass of 1 mole, since you have

22.4 L of the stuff.22.4 L of the stuff. What is the molar mass of a gas with a What is the molar mass of a gas with a

density of 1.964 g/L?density of 1.964 g/L? How about a density of 2.86 g/L?How about a density of 2.86 g/L?

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SummarySummary These four items are all equal:These four items are all equal:

a) 1 molea) 1 mole

b) molar mass (in grams/mol)b) molar mass (in grams/mol)

c) 6.02 x 10c) 6.02 x 102323 representative representative particlesparticles

d) 22.4 L of gas at STPd) 22.4 L of gas at STP

Thus, we can make Thus, we can make conversion conversion factorsfactors from them. from them.

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Section 10.3Section 10.3Percent Composition and Percent Composition and

Chemical FormulasChemical Formulas OBJECTIVES:OBJECTIVES:

–DescribeDescribe how to calculate how to calculate the percent by mass of an the percent by mass of an element in a compound.element in a compound.

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–InterpretInterpret an empirical an empirical formula.formula.

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–DistinguishDistinguish between between empirical and molecular empirical and molecular formulas.formulas.

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Calculating Percent Composition of Calculating Percent Composition of a Compounda Compound

Like all percent problems:Like all percent problems: Part Part

wholewhole Find the mass of each of the Find the mass of each of the

components (the elements),components (the elements), Next, divide by the total mass Next, divide by the total mass

of the compound, then x 100of the compound, then x 100

x 100 %x 100 %

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ExampleExample

Calculate the percent Calculate the percent composition of a composition of a compound that is 29.0 g compound that is 29.0 g of Ag with 4.30 g of S.of Ag with 4.30 g of S.

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Getting it from the formulaGetting it from the formula If we know the formula, If we know the formula,

assume you have 1 mole.assume you have 1 mole.Then you know the mass Then you know the mass

of the pieces and the of the pieces and the wholewhole (these values come from (these values come from the periodic table).the periodic table).

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ExamplesExamples Calculate the percent composition Calculate the percent composition

of Cof C22HH44?? How about Aluminum carbonate?How about Aluminum carbonate? Sample Problem 10.10, p.307Sample Problem 10.10, p.307 We can also use the percent as a We can also use the percent as a

conversion factorconversion factor Sample Problem page 308Sample Problem page 308

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FormulasFormulas

• molecular formula = (empirical molecular formula = (empirical formula)formula)nn [ [nn = integer] = integer]

• molecular formula = Cmolecular formula = C66HH66 = (CH) = (CH)66

• empirical formula = CHempirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.

Molecular formula: the true number of atoms of each element in the formula of a compound.

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FormulasFormulas (continued)(continued)

Formulas for Formulas for ionic compoundsionic compounds are are ALWAYSALWAYS empirical (lowest whole empirical (lowest whole number ratio).number ratio).

Examples:Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

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FormulasFormulas (continued)(continued)

Formulas for Formulas for molecular compoundsmolecular compounds MIGHTMIGHT be empirical (lowest whole be empirical (lowest whole number ratio).number ratio).

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

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Calculating EmpiricalCalculating Empirical Just find the lowest whole number ratioJust find the lowest whole number ratio

–CC66HH1212OO66

–CHCH44NN It is not just the ratio of It is not just the ratio of atomsatoms, it is also , it is also

the ratio of the ratio of molesmoles of atoms. of atoms. In 1 mole of COIn 1 mole of CO22 there is 1 mole of there is 1 mole of

carbon and 2 moles of oxygen.carbon and 2 moles of oxygen. In one molecule of COIn one molecule of CO22 there is 1 atom there is 1 atom

of C and 2 atoms of O. of C and 2 atoms of O.

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Calculating EmpiricalCalculating EmpiricalWe can get a ratio from the We can get a ratio from the

percent composition.percent composition.Assume you have a 100 g.Assume you have a 100 g.

–The percentages become The percentages become grams.grams.

Convert grams to moles. Convert grams to moles. Find lowest whole number ratio by Find lowest whole number ratio by

dividing by the smallest value.dividing by the smallest value.

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ExampleExample Calculate the empirical formula of Calculate the empirical formula of

a compound composed of 38.67 % a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.C, 16.22 % H, and 45.11 %N.

Assume 100 g soAssume 100 g so 38.67 g C x 1mol C = 3.220 mole C 38.67 g C x 1mol C = 3.220 mole C

12.01 gC 12.01 gC 16.22 g H x 1mol H = 16.09 mole H 16.22 g H x 1mol H = 16.09 mole H

1.01 gH1.01 gH 45.11 g N x 1mol N = 3.219 mole N 45.11 g N x 1mol N = 3.219 mole N

14.01 gN14.01 gN

Now divide each value by the smallest value

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ExampleExample The ratio is 3.220 mol C = 1 mol CThe ratio is 3.220 mol C = 1 mol C

3.219 mol N 1 mol N 3.219 mol N 1 mol N The ratio is 16.09 mol H = 5 mol HThe ratio is 16.09 mol H = 5 mol H

3.219 mol N 1 mol N 3.219 mol N 1 mol N

= = CC11HH55NN11

A compound is 43.64 % P and 56.36 % A compound is 43.64 % P and 56.36 % O. What is the empirical formula?O. What is the empirical formula?

Caffeine is 49.48% C, 5.15% H, 28.87% Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical N and 16.49% O. What is its empirical formula?formula?

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Empirical to molecularEmpirical to molecular Since the empirical formula is the Since the empirical formula is the

lowest ratio, the actual molecule lowest ratio, the actual molecule would weigh more.would weigh more.–By a whole number multiple.By a whole number multiple.

DivideDivide the the actual molar massactual molar mass by the by the empirical formula massempirical formula mass – you get a – you get a whole number to whole number to increase each increase each coefficientcoefficient in the empirical formula in the empirical formula

Caffeine has a molar mass of 194 g. Caffeine has a molar mass of 194 g. what is its molecular formula?what is its molecular formula?

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Documents

1 Chapter 10 “Chemical Quantities” Chemistry Pioneer High School Mr. David Norton