1. 2 Chapter 10 Chemical Quantities or 3 How you measure how much? How you measure how much? n You...

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Chapter 10 Chapter 10

Chemical QuantitiesChemical Quantitiesoror

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How you measure how much?How you measure how much? You can measure mass, You can measure mass, or volume,or volume, or you can count pieces.or you can count pieces. We measure mass in grams.We measure mass in grams. We measure volume in liters.We measure volume in liters.

We count pieces in We count pieces in MOLES.MOLES.

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MolesMoles 1 1 molemole is is 6.02 x 106.02 x 102323 particles. particles. Particles can be atoms, molecules, Particles can be atoms, molecules,

formula units.formula units. 6.02 x 106.02 x 102323 is called is called Avogadro's Avogadro's

number.number.

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Representative particlesRepresentative particles The smallest pieces of a substance.The smallest pieces of a substance. For an element it is an For an element it is an atomatom..

– Unless it is diatomicUnless it is diatomic For a molecular compound it is a For a molecular compound it is a

moleculemolecule.. For an ionic compound it is a For an ionic compound it is a

formula unitformula unit..

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Conversion factorsConversion factors Used to change units.Used to change units. Three questionsThree questions

– What unit do you want to get rid of?What unit do you want to get rid of?

– Where does it go to cancel out?Where does it go to cancel out?

– What can you change it into?What can you change it into?

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Calculation questionCalculation question How many molecules of COHow many molecules of CO22 are the in are the in

4.56 moles of CO4.56 moles of CO22 ? ?

4.56 moles CO4.56 moles CO22 x x 6.02 x 106.02 x 102323 MoleculesMolecules = = 2.75 x 102.75 x 102424

Molecules COMolecules CO22

1 Mole CO1 Mole CO22

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Calculation questionCalculation question How many moles of water is 5.87 x 10How many moles of water is 5.87 x 102222

molecules?molecules?

5.87 x 105.87 x 102222 molecules H molecules H22O x O x 1 Mole H1 Mole H22O____________O____________ = = .098 .098 Moles HMoles H22OO

6.02 x 106.02 x 102323 Molecules HMolecules H22OO

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Calculation questionCalculation question How many atoms of carbon are there How many atoms of carbon are there

in 1.23 moles of Cin 1.23 moles of C66HH1212OO66 ? ?

– 1.23 moles C1.23 moles C66HH1212OO6 6 x x 6 moles C Atoms_____ 6 moles C Atoms_____ x 6.02 x 1023 Atoms C

1 mole C1 mole C66HH1212OO6 6 Molecules 1 mole C atomsMolecules 1 mole C atoms

= = 4.44 x 10 4.44 x 10 2424 atoms Catoms C

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Measuring MolesMeasuring Moles The decimal number on the periodic The decimal number on the periodic

table istable is

– The mass of the average atom in The mass of the average atom in amuamu

– The mass of 1 mole of those atoms The mass of 1 mole of those atoms in gramsin grams..

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Gram Atomic MassGram Atomic Mass Gram Atomic Mass is the mass of 1 Gram Atomic Mass is the mass of 1

mole of an element in grams.mole of an element in grams. We can write this as We can write this as

12.01 g C = 1 mole 12.01 g C = 1 mole We can count things by weighing We can count things by weighing

them.them.

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

carbon weigh?carbon weigh?

2.34 moles C x 2.34 moles C x 12 g C__12 g C__ = = 28.1 g C28.1 g C

1 mole C1 mole C

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ExamplesExamples How many moles of magnesium in How many moles of magnesium in

4.61 g of Mg?4.61 g of Mg?

4.61 g Mg x 4.61 g Mg x 1 mole Mg 1 mole Mg = = 0.19 moles Mg0.19 moles Mg

24.3 g Mg24.3 g Mg

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Examples How much would 3.45 x 10How much would 3.45 x 102222 atoms atoms

of U weigh?of U weigh?

3.45 x 103.45 x 102222 atoms U x atoms U x 1 mole U_________ 1 mole U_________ x x 238g U238g U = = 13.6 g U13.6 g U

6.02 x 106.02 x 102323 atoms U 1 mole U atoms U 1 mole U

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What about compounds?What about compounds? In 1 mole of HIn 1 mole of H22O molecules there are two O molecules there are two

moles of H atoms and 1 mole of O atoms.moles of H atoms and 1 mole of O atoms. ((Be carefulBe careful))

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 elements Determine the moles of the elements

contained in the compound.contained in the compound.– Find out how much each would weighFind out how much each would weigh– add them upadd them up

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What about compounds?What about compounds? Molar MassMolar Mass is the mass of one mole of is the mass of one mole of

any compoundany compound. . Example: Example: What is the What is the Molar mass of CHMolar mass of CH44??

» 1 mole of C = 12.01 g1 mole of C = 12.01 g

» 4 mole of H = 4 x 1.01 g = 4.04g4 mole of H = 4 x 1.01 g = 4.04g

» 1 mole CH1 mole CH44 = 12.01 + 4.04 = = 12.01 + 4.04 = 16.05g/mol16.05g/mol

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Molar MassMolar Mass What is the molar mass of FeWhat is the molar mass of Fe22OO33??

» 2 moles of Fe = 2 x 55.85 g = 111.70 g2 moles of Fe = 2 x 55.85 g = 111.70 g

» 3 moles of O = 3 x 16.00 g = 48.00 g3 moles of O = 3 x 16.00 g = 48.00 g

» The Molar mass = 111.70 g + 48.00 g = The Molar mass = 111.70 g + 48.00 g = 159.70g/mol159.70g/mol

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Using Molar MassUsing Molar Mass

Finding moles of compoundsFinding moles of compounds

Counting pieces by weighingCounting pieces by weighing

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Molar MassMolar Mass The number of grams in 1 mole of The number of grams in 1 mole of

atoms, formula units, or molecules.atoms, formula units, or molecules. We can make conversion factors We can make conversion factors

from these.from these. To change grams of a compound to To change grams of a compound to

moles of a compound.moles of a compound. Or moles to gramsOr moles to grams

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For exampleFor example How many moles is 5.69 g of NaOH?How many moles is 5.69 g of NaOH?

– 5.69 g NaOH x 5.69 g NaOH x 1 mole NaOH 1 mole NaOH Molar mass of NaOHMolar mass of NaOH

Molar Mass NaOH = Na + O + H = 23.0g+16g + 1.0g = 40g/moleMolar Mass NaOH = Na + O + H = 23.0g+16g + 1.0g = 40g/mole

Substitute 40g/mole into the equation above and get:Substitute 40g/mole into the equation above and get:

– 5.69 g NaOH x 5.69 g NaOH x 1 mole NaOH1 mole NaOH = = 0.14 moles NaOH0.14 moles NaOH 40 g NaOH40 g NaOH

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Gases and the MoleGases and the Mole

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GasesGases Many of the chemicals we deal with Many of the chemicals we deal with

are gases.are gases. They are difficult to weigh, so we’ll They are difficult to weigh, so we’ll

measure volumemeasure volume Need to know how many moles of gas Need to know how many moles of gas

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

–Temperature and pressureTemperature and pressure–Compare at the same temp. and Compare at the same temp. and

pressure.pressure.

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Standard Temperature and Standard Temperature and PressurePressure

Avogadro's HypothesisAvogadro's Hypothesis - at the same - at the same temperature and pressure equal temperature and pressure equal volumes of gas have the same number volumes of gas have the same number of particles.of particles.

0ºC and 1 atmosphere pressure 0ºC and 1 atmosphere pressure Abbreviated atmAbbreviated atm 273 K and 101.3 kPa273 K and 101.3 kPa kPa is kiloPascalkPa is kiloPascal

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AtAt S Standardtandard T Temperatureemperature andand PPressureressure

abbreviated STPabbreviated STP At STP At STP 1 mole of gas occupies 22.4 L1 mole of gas occupies 22.4 L Called the Called the molar volumemolar volume Used for conversion factorsUsed for conversion factors Moles to Liter (x 22.4), and L to mol Moles to Liter (x 22.4), and L to mol

(divide by 22.4)(divide by 22.4)

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ExamplesExamplesWhat is the volume of 4.59 What is the volume of 4.59

mole of COmole of CO22 gas at STP? gas at STP?

4.59 mole CO4.59 mole CO22 x x 22.4L CO22.4L CO22 = = 102.8 L CO102.8 L CO22

1 mole CO1 mole CO22

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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 the mass To find the density we need the mass

and the volume.and the volume. If you assume you have 1 mole than If you assume you have 1 mole than

the mass is the molar mass (PT)the mass is the molar mass (PT) At STP the volume is 22.4 L.At STP the volume is 22.4 L.

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ExamplesExamples Find the density of COFind the density of CO22 at STP. at STP.

NoteNote: At STP the volume is 22.4 L: At STP the volume is 22.4 L

D =m/v = D =m/v = Molar mass COMolar mass CO22 = = ((12g + (16g x 2g))((12g + (16g x 2g)) = = 1.96g/l1.96g/l

22.4 L 22.4L22.4 L 22.4L

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

D=m/v D=m/v ((12.0g +(1.0g x 4))((12.0g +(1.0g x 4)) = = 2.14 g/l2.14 g/l

22.4L22.4L

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The other wayThe other way Given the density, we can find the Given the density, we can find the

molar mass of the gas.molar mass of the gas. Again, pretend you have a mole at Again, pretend you have a mole at

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

have 22.4 L of the stuff.have 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? M=D x V = density of 1.964 g/L? M=D x V = 1.964g/l x 22.4L = 1.964g/l x 22.4L = 44g44g

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All the things we can changeAll the things we can change

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Volume

IonsAtoms

Representative Particles

MassPT

Moles

6.02 x 1023

22.4 L

Count

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Percent CompositionPercent Composition Like all percentsLike all percents Part x 100 %Part x 100 %

whole whole Find the mass of each component,Find the mass of each component, divide by the total mass.divide by the total mass.

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ExampleExample Calculate the percent composition of a Calculate the percent composition of a

compound that is 29.0 g of Ag with 4.30 g compound that is 29.0 g of Ag with 4.30 g of S. of S. PartPart x 100 x 100WholeWhole– Total weight = 29.0g Ag + 4.30g S = 33.3 gTotal weight = 29.0g Ag + 4.30g S = 33.3 g

%Ag = %Ag = 29.0g Ag29.0g Ag x 100 = x 100 = 87.1% Ag87.1% Ag 33.3 g33.3 g% S = % S = 4.30g S4.30g S x 100 = x 100 = 12.9% S12.9% S 33.3g33.3g

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

have 1 mole.have 1 mole. Then you know the pieces and the Then you know the pieces and the

whole.whole.

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ExamplesExamples Calculate the percent composition of Calculate the percent composition of

CC22HH44??– Assume 1 mole Assume 1 mole – 1 mole C1 mole C22HH44 = (12g x 2) +(1.0g x 4) = = (12g x 2) +(1.0g x 4) =

28g/mole 28g/mole total masstotal mass

% C = % C = (12g x 2)(12g x 2) x 100 = x 100 = 85.7% C85.7% C 28 g28 g% H = % H = (1.0g x 4)(1.0g x 4) x 100 = x 100 = 14.3% H14.3% H 28 g28 g

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ExamplesExamples What is the percent composition of What is the percent composition of

Aluminum carbonate.Aluminum carbonate.

– Write the formula: AlWrite the formula: Al22(CO)(CO)3 3 (ionic Formula writing) (ionic Formula writing)– Assume 1 moleAssume 1 mole– Calculate Total Mass: (Al x 2) + (C x 3) + (O x 3) = (27g x 2) + Calculate Total Mass: (Al x 2) + (C x 3) + (O x 3) = (27g x 2) +

(12g x 3) + (16g x 3) = 54g + 36g + 48g = (12g x 3) + (16g x 3) = 54g + 36g + 48g = 138g138g

» % Al = % Al = (27g x 2)(27g x 2) x 100 = x 100 = 39.1%Al39.1%Al 138g138g % C = % C = (12g x 3)(12g x 3) x 100 = x 100 = 26.1% C26.1% C 138g138g % O = % O = (16g x 3(16g x 3) x 100 = ) x 100 = 34.8% O34.8% O 138g138g

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Percent to MassPercent to Mass Multiply % by the total mass to find Multiply % by the total mass to find

the mass of that component.the mass of that component. How much aluminum in 450 g of How much aluminum in 450 g of

aluminum carbonate?aluminum carbonate?– Write the formula: AlWrite the formula: Al22(CO)(CO)3 3 (ionic Formula writing) (ionic Formula writing)

» % Al = % Al = (27g x 2)(27g x 2) x 100 = x 100 = 39.1%Al39.1%Al 138g138g % C = % C = (12g x 3)(12g x 3) x 100 = x 100 = 26.1% C 26.1% C 138g138g % O = % O = (16g x 3(16g x 3) x 100 = ) x 100 = 34.8% O34.8% O 138g138g

– 39.1%Al x1/100 x 450g Al39.1%Al x1/100 x 450g Al22(CO)(CO)33 = = 176g Al176g Al

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Empirical FormulaEmpirical Formula

From percentage to formulaFrom percentage to formula

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The Empirical FormulaThe Empirical Formula The lowest whole number ratio of The lowest whole number ratio of

elements in a compound.elements in a compound. The molecular formula the actual ratio The molecular formula the actual ratio

of elements in a compound.of elements in a compound. The two can be the same. The two can be the same. CHCH22 empirical formula empirical formula CC22HH44 molecular formula molecular formula CC33HH66 molecular formula molecular formula HH22O bothO both

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Finding Empirical FormulasFinding Empirical Formulas Just find the lowest whole number ratioJust find the lowest whole number ratio CC66HH1212OO66

CHCH44NN22

It is not just the ratio of atoms, it is also It is not just the ratio of atoms, it is also the ratio of moles of atoms.the ratio of moles of atoms.

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Calculating Empirical FormulasCalculating Empirical Formulas Means we can get ratio from percent Means we can get ratio from percent

composition.composition. Assume you have a 100 g.Assume you have a 100 g. The percentages become grams.The percentages become grams. Turn grams to moles. Turn grams to moles. Find lowest whole number ratio by Find lowest whole number ratio by

dividing everything by the smallest dividing everything by the smallest moles.moles.

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ExampleExample Calculate the empirical formula of a Calculate the empirical formula of a

compound composed of 38.67 % C, compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.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 = 16.1 mole H 16.22 g H x 1mol H = 16.1 mole H

1.01 gH1.01 gH 45.11 g N x 1mol N = 3.220 mole N 45.11 g N x 1mol N = 3.220 mole N

14.01 gN14.01 gN

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

3.220 molN 1 3.220 molN 1 mol Nmol N

The ratio is 16.1 mol H = 5 mol HThe ratio is 16.1 mol H = 5 mol H 3.220 molN 1 3.220 molN 1

mol Nmol N

CC11HH55NN11

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

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Empirical to molecularEmpirical to molecular Caffeine is 49.48% C, 5.15% H, 28.87% N Caffeine is 49.48% C, 5.15% H, 28.87% N

and 16.49% O. What is its empirical and 16.49% O. What is its empirical formula?formula?

Since the empirical formula is the lowest Since the empirical formula is the lowest ratio the actual molecule would weigh the ratio the actual molecule would weigh the same or more.same or more.

By a whole number multiple.By a whole number multiple. Divide the actual molar mass by the the Divide the actual molar mass by the the

mass of one mole of the empirical formula.mass of one mole of the empirical formula. You will get a whole number.You will get a whole number. Multiply the empirical formula by this.Multiply the empirical formula by this.

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ExampleExample A compound has an empirical

formula of ClCH2 and a molar mass of 98.96 g/mol. What is its molecular formula?

A compound has an empirical formula of CH2O and a molar mass of 180.0 g/mol. What is its molecular formula?

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Percent to molecularPercent to molecular Take the percent x the molar massTake the percent x the molar mass

–This gives you mass in one mole of This gives you mass in one mole of the compoundthe compound

Change this to molesChange this to moles

–You will get whole numbersYou will get whole numbers

–These are the subscriptsThese are the subscripts Caffeine is 49.48% C, 5.15% H, 28.87% N Caffeine is 49.48% C, 5.15% H, 28.87% N

and 16.49% O. It has a molar mass of and 16.49% O. It has a molar mass of 194 g. What is its molecular formula?194 g. What is its molecular formula?

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ExampleExample Ibuprofen is 75.69 % C, 8.80 % H, 15.51

% O, and has a molar mass of about 207 g/mol. What is its molecular formula?