Chapter 10 Chemical Quantities Hingham High School Mr. Clune

Chapter 10Chapter 10Chemical QuantitiesChemical Quantities

Hingham High SchoolHingham High School

Mr. CluneMr. Clune

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..

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 particles. 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 Avogadro’s number.number.

Representative particlesRepresentative particles

The smallest pieces of a The smallest pieces of a substance.substance.– For a molecular compound: it is the For a molecular compound: it is the

molecule.molecule.– For an ionic compound: it is the For an ionic compound: it is the

formula unit (ions).formula unit (ions).– For an element: it is the atom.For an element: it is the atom.

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

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

551717

3322

55

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

Molar Mass Molar Mass 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 12.01 grams of C has the same number of pieces as 1.008 grams number of pieces as 1.008 grams of H and 55.85 grams of iron.of H and 55.85 grams of iron.

We can write this as We can write this as 12.01 g C = 1 mole C 12.01 g C = 1 mole C

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

ExamplesExamples

How much would 2.34 moles of How much would 2.34 moles of carbon weigh?carbon weigh?

1 mole of C is 12.01g or 1 mole of C is 12.01g or

12.01 g/mole12.01 g/mole

2.34 mole2.34 mole 12.01 g12.01 g

molemoleXX == 28.10g28.10g

ExamplesExamplesHow many moles of How many moles of magnesium is 24.31 g of Mg?magnesium is 24.31 g of Mg?

1 mole of Mg is 24.31g or 1 mole of Mg is 24.31g or

24.31 g/mole24.31 g/mole

24.31g24.31g XX1 mole1 mole24.31 g24.31 g == 1 mole1 mole

ExamplesExamples

How many atoms of lithium is How many atoms of lithium is 1.00 g of Li?1.00 g of Li?

6.94g/mole6.94g/mole 6.02x106.02x102323 atoms atoms

molemole

1.00g1.00gXX

1 mole1 mole

6.94g6.94g

6.02x106.02x102323 atoms atoms

molemoleXX

8.67x108.67x102222atomsatoms

6.02x106.02x102323 atoms atoms

molemole

XX

238.03g/mole238.03g/mole

1 mole1 mole

6.02x106.02x102323atmatm

XX

13.6g13.6g

How much would How much would

3.45 x 103.45 x 102222 atoms of U weigh? atoms of U weigh?

3.45x103.45x102222atmatm 238.03g238.03g

molemole

What about compounds?What about compounds?

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 haveFind out how much they would Find out how much they would

weighweighadd them upadd them up


1 mole of H1 mole of H22O molecules has O molecules has two moles of H atoms and 1 two moles of H atoms and 1 mole of O atomsmole of O atoms

HH22O O


H:2 moles X 1.08g/mole = 2.16gH:2 moles X 1.08g/mole = 2.16g

HH22O O

O:1 mole X 16.00g/mole= 32.00gO:1 mole X 16.00g/mole= 32.00g+

1 mole of H1 mole of H220 = 34.16g0 = 34.16g


What is the mass of one What is the mass of one mole of CHmole of CH44??

1 mole CH1 mole CH44 = 16.05g = 16.05g

1 mole of C x 12.01 g = 12.02g1 mole of C x 12.01 g = 12.02g4 mole of H x 1.01 g = 4.04g4 mole of H x 1.01 g = 4.04g

+


The The Molar Mass Molar Mass of CHof CH44 is is 16.05g16.05g

– this is the mass of one mole of a this is the mass of one mole of a molecular compound.molecular compound.

Moles to Mass ConversionsMoles to Mass ConversionsHow many grams of Na are there in How many grams of Na are there in

1.23 moles?1.23 moles?

1. How many grams of Na are there 1. How many grams of Na are there in 1 mole?in 1 mole?

22.99g/mol22.99g/mol2. How many grams of Na are there 2. How many grams of Na are there

in 1.23 moles?in 1.23 moles?1.23 mol1.23 mol 22.99 g22.99 g

mol molXX == 28.28g28.28g

Moles to Mass ConversionsMoles to Mass Conversions

How many grams of CHow many grams of C66HH1212OO66 are are

there in 0.56 moles?there in 0.56 moles?1. How many grams of C1. How many grams of C66HH1212OO66 are are

there in 1 mole?there in 1 mole?

180.18g/mol180.18g/mol

C: 6 x 12.01g = 72.06gC: 6 x 12.01g = 72.06g

H: 12 x 1.01g = 12.12gH: 12 x 1.01g = 12.12g

O: 6 x 16.00g = 96.00gO: 6 x 16.00g = 96.00g+

Moles to Mass ConversionsMoles to Mass Conversions

2. How many grams of C2. How many grams of C66HH1212OO66 are are

there in 0.56 moles?there in 0.56 moles?

0.56mol0.56mol 180.18 g180.18 g mol mol

XX == 28.28g28.28g

Representative Representative ParticlesParticlesHow many moles of water is 5.87 x How many moles of water is 5.87 x

10102222 molecules? molecules?

MolesMoles

1 mole1 mole

6.02 x 106.02 x 102323moleculesmolecules5.87 x 105.87 x 102222moleculesmolecules X =

0.098 moles or 9.8 x 100.098 moles or 9.8 x 10-2-2 moles moles

How many molecules of COHow many molecules of CO22 are are there in 4.56 moles of COthere in 4.56 moles of CO22 ? ?

4.56 mol4.56 mol 6.02 x 106.02 x 102323moleculesmolecules

1 mol1 molX =

2.75 x 102.75 x 102424 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 ? ?

1.23 mol1.23 mol 6.02 x 106.02 x 102323moleculesmolecules

1 mol1 molX =

7.40 x 107.40 x 102323moleculemolecule 6 atoms of C6 atoms of C

1 molecule1 moleculeX =

4.44 x 104.44 x 102424 atoms of C atoms of C

How many moles is 7.78 x 10How many moles is 7.78 x 102424 formula units of MgClformula units of MgCl22??

1 mole1 mole


12.92 moles12.92 moles

What is the weight of 3.01 x 10What is the weight of 3.01 x 102424 molecules of NaOH?molecules of NaOH?

1 mole1 mole


5.00 moles5.00 moles

1. Find the number of moles.1. Find the number of moles.

2. Find the mass of one mole 2. Find the mass of one mole of NaOH.of NaOH.

Na : 1 x 22.99 = 22.99gNa : 1 x 22.99 = 22.99g

O : 1 x 16.00 = 16.00gO : 1 x 16.00 = 16.00g

H : 1 x 1.08 = 1.08gH : 1 x 1.08 = 1.08g ++

40.07g40.07g

3. Find the mass of 5 moles3. Find the mass of 5 molesof NaOH.of NaOH.

5 mol5 mol 40.07g40.07g

1 mol1 molX = 200.35g200.35g

Homework 10-1Homework 10-1

Practice Problems Worksheet1-16

Due: 01/10/05

Molar MassMolar Mass

Molar massMolar mass is the generic term is the generic term for the mass of one mole of any for the mass of one mole of any substance (in grams) substance (in grams)

The same as: 1) gram molecular The same as: 1) gram molecular mass, 2) gram formula mass, and mass, 2) gram formula mass, and 3) gram atomic mass- just a 3) gram atomic mass- just a much broader term.much broader term.

ExamplesExamples

Calculate the molar mass of the Calculate the molar mass of the following and tell what type it is:following and tell what type it is:

NaNa22SS

NN22OO44

CC

Ca(NOCa(NO33))22

CC66HH1212OO66

(NH(NH44))33POPO44

Molar MassMolar Mass

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 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.

For exampleFor example

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



5 69. g



5 69. g mole

g

need to change grams to moles



5 69. g mole

g

need to change grams to moles for NaOH



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



5 69. g mole

g


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



5 69. g 1 mole

40.00 g





5 69. g 1 mole

40.00 = 0.142 mol NaOH

g



GasesGases

Many of the chemicals we deal with Many of the chemicals we deal with are gases.are gases.–They are difficult to They are difficult to weighweigh..

Need to know how many moles of Need to know how many moles of gas we have.gas we have.Two things effect the volume of a Two things effect the volume of a gasgas–Temperature and pressureTemperature and pressure

We need to compare them at the We need to compare them at the same temperature and pressure.same temperature and pressure.

Standard Temperature and Standard Temperature and PressurePressure0ºC and 1 atm pressure0ºC and 1 atm pressure

abbreviated abbreviated STPSTP

At STP 1 mole of gas occupies At STP 1 mole of gas occupies 22.4 L22.4 L

Called the Called the molar volumemolar volume

1 mole = 22.4 L of any gas at 1 mole = 22.4 L of any gas at STPSTP

ExamplesExamples

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?

4.59 mol4.59 mol 22.4 L22.4 L

1 mol1 molX =

102.81 L102.81 L

ExamplesExamples

How many moles is 5.67 L of How many moles is 5.67 L of OO2 2 at STP?at STP?

5.67 L5.67 L 1 mol 1 mol

22.4 L22.4 LX =

0.25 mol0.25 mol

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?

1. Find the molar mass of CH1. Find the molar mass of CH44..C: 1 X 12.01 = 12.01gC: 1 X 12.01 = 12.01g

H: 4 X 1.01 = 4.04gH: 4 X 1.01 = 4.04g++16.05g16.05g

2. Find the number of moles.2. Find the number of moles.

8.8g8.8gXX

1 mol1 mol

16.05g16.05g== 0.55 mol0.55 mol

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?

3. Find the volume of CH3. Find the volume of CH44..

22.4 L22.4 L

1 mol1 mol0.55 mol0.55 mol XX ==

12.32 L12.32 L

Mole

Volume

Mass Particle1 m

ol

molar m

ass

molar m

ass

1 mol

22.4

L1

mol

1 mol

22.4L

1 mol

6.02X10 23part.6.02X10 23part.

1 mol

HomeworkHomework

Finish worksheet.Finish worksheet.

Due: 1/11/05Due: 1/11/05

Quiz onQuiz on

ConversionsConversions

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, then If you assume you have 1 mole, then the mass is the molar mass (from PT)the mass is the molar mass (from PT)

At STP the volume is 22.4 L.At STP the volume is 22.4 L.

ExamplesExamples

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.

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 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 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?density of 1.964 g/L?

2.86 g/L?2.86 g/L?

SummarySummary

These four items are all equal:These four items are all equal:a) 1 molea) 1 mole

b) molar mass (in grams)b) molar mass (in grams)

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

d) 22.4 L at STPd) 22.4 L at STP

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

Section 10.3Section 10.3Percent Composition and Percent Composition and

Chemical FormulasChemical Formulas

OBJECTIVES:OBJECTIVES:– Calculate the percent composition Calculate the percent composition

of a substance from its chemical of a substance from its chemical formula or experimental data.formula or experimental data.

Calculating Percent Composition Calculating Percent Composition of a Compoundof a Compound

Like all percent problems:Like all percent problems:

Part Part X 100%X 100%

wholewhole

Find the mass of each Find the mass of each component,component,

Then divide by the total mass.Then divide by the total mass.

ExampleExampleCalculate the percent composition Calculate the percent composition

of a compound that is 29.00 g of of a compound that is 29.00 g of Ag with 4.30 g of S.Ag with 4.30 g of S.

1. Calculate the total mass.1. Calculate the total mass.

Ag + S = 29.00g + 4.30g = 33.30gAg + S = 29.00g + 4.30g = 33.30g

ExampleExample

2. Calculate % of each element.2. Calculate % of each element.

29.00g29.00g

33.3g33.3g X 100% = 87%X 100% = 87%AgAg

4.30g4.30g

33.3g33.3g X 100% = 13%X 100% = 13%SS

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 of Then you know the mass of the pieces and the whole.the pieces and the whole.

Find the Molar Mass.Find the Molar Mass.

ExamplesExamples

Calculate the percent Calculate the percent composition of Ccomposition of C22HH44..

C: 2 x 12.0 = C: 2 x 12.0 = 24.0g24.0gH: 4 x 1.0 = H: 4 x 1.0 = 4.0g 4.0g

Total Mass = 28.0gTotal Mass = 28.0g++

ExampleExample

2. Calculate % of each element.2. Calculate % of each element.

24.0g24.0g

28.0g28.0g X 100% = 86%X 100% = 86%CC

4.0g4.0g

28.0g28.0g X 100% = 14%X 100% = 14%HH

ExamplesExamplesIf a substance which has a mass of If a substance which has a mass of

0.85g is 75% C and 25% 0, how 0.85g is 75% C and 25% 0, how much of each substance is there? much of each substance is there?

C: 0.85g X .75 =0.64gC: 0.85g X .75 =0.64g

O: 0.85g X .25 =0.21gO: 0.85g X .25 =0.21g

HomeworkHomeworkWorksheet 10-3Worksheet 10-3

1-91-9Show all work in Show all work in

Notebook!!!!Notebook!!!!Due: 2/01/06Due: 2/01/06Quiz: 2/2/06Quiz: 2/2/06

The Empirical FormulaThe Empirical Formula

The lowest The lowest whole number whole number ratioratio of elements in a of elements in a compound.compound.The molecular formula = the The molecular formula = the actual actual ratio of elements in a ratio of elements in a compound.compound.The two The two cancan be the same. be the same.

The Empirical FormulaThe Empirical FormulaCHCH22 is an empirical formula is an empirical formula

CC22HH44 is a molecular formula is a molecular formula

CC33HH66 is a molecular formula is a molecular formula

HH22O is both empirical & O is both empirical & molecularmolecular

Calculating EmpiricalCalculating Empirical

Just find the lowest Just find the lowest whole number ratiowhole number ratio

CC66HH1212OO66

CHCH44NN- CH- CH22O O

- CH- CH44NN

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

In 1 mole of COIn 1 mole of CO22 there is 1 there is 1 mole of carbon and 2 moles mole of carbon and 2 moles of oxygen.of oxygen.

In one molecule of COIn one molecule of CO22 there is 1 atom of C and 2 there is 1 atom of C and 2 atoms of O. atoms of O.

Calculating EmpiricalCalculating Empirical

We 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 grams.The percentages become grams.

Convert grams to moles. Convert grams to moles.

Find lowest whole number ratio Find lowest whole number ratio by dividing by the smallest.by dividing by the smallest.

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 so:Assume 100 g so:

C: 38.67g x 1mol = 3.22 mol C C: 38.67g x 1mol = 3.22 mol C

12.0g 12.0g

H 16.22 g x 1mol = 16.09 mol H H 16.22 g x 1mol = 16.09 mol H 1.0g 1.0g

N: 45.11g x 1mol = 3.22 mol N N: 45.11g x 1mol = 3.22 mol N 14.0g 14.0g

The Smallest number is The Smallest number is 3.22mol.3.22mol.

Divide each mole value by Divide each mole value by 3.22mol.3.22mol.

The ratio of C:NThe ratio of C:N

3.22 mol C = 1 mol C3.22 mol C = 1 mol C

3.22 mol N 1 mol N3.22 mol N 1 mol N

The ratio H:NThe ratio H:N

16.09 mol H = 5 mol H 3.22 16.09 mol H = 5 mol H 3.22 mol N 1 mol Nmol N 1 mol N

CC11HH55NN11

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

empirical formula?empirical formula?Assume 100g & find moles.Assume 100g & find moles.

43.64g43.64gXX

1mol1mol

30.97g30.97g== 1.41 mol1.41 mol

P:P:

56.36g56.36gXX

1mol1mol

16.00g16.00g== 3.52 mol3.52 mol

O:O:

Find ratio of moles.Find ratio of moles.

Ratio of P:ORatio of P:O

1.41 mol of P1.41 mol of P3.52 mol of O3.52 mol of O

2 of P2 of P5 of O5 of O

==

PP22OO55Empirical FormulaEmpirical Formula

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

empirical formula?empirical formula?

ExampleExample

Assume 100g & find moles.Assume 100g & find moles.

28.87g28.87gXX

1mol1mol

14.0g14.0g== 2.06 mol2.06 mol

N:N:

16.49g16.49gXX

1mol1mol

16.0g16.0g== 1.03 mol1.03 mol

O:O:

5.15g5.15gXX

1mol1mol

1.0g1.0g== 5.10 mol5.10 mol

H:H:

49.48g49.48gXX

1mol1mol

12.0g12.0g== 4.12 mol4.12 mol

C:C:

Find ratio of moles.Find ratio of moles.Ratio of C:H:N:ORatio of C:H:N:O

4.12 mol of C4.12 mol of C1.03 mol of O1.03 mol of O

4 of C4 of C1 of O1 of O

==

5.10 mol of H5.10 mol of H1.03 mol of O1.03 mol of O

5 of H5 of H1 of O1 of O

==

2.06 mol of N2.06 mol of N1.03 mol of O1.03 mol of O

2 of N2 of N1 of O1 of O

==

Find ratio of moles.Find ratio of moles.

Ratio of C:H:N:ORatio of C:H:N:O

CC44HH55NN22OO

4:5:2:14:5:2:1

Empirical FormulaEmpirical Formula

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.

Divide the actual molar mass by Divide the actual molar mass by the empirical formula mass.the empirical formula mass.

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

ExampleExample

A compound is known to be A compound is known to be composed of 71.65 % Cl, composed of 71.65 % Cl,

24.27% C and 4.07% H. Its 24.27% C and 4.07% H. Its molar mass is known (from molar mass is known (from gas density) to be 98.96 g. gas density) to be 98.96 g.

What is its molecular formula?What is its molecular formula?

71.65g71.65gXX

1mol1mol

35.45g35.45g== 2.02 mol2.02 mol

Cl:Cl:

4.07g4.07gXX

1mol1mol

1.01g1.01g==4.03 mol4.03 mol

H:H:

24.27g24.27gXX

1mol1mol

12.01g12.01g== 2.02 mol2.02 mol

C:C:

Assume 100g & find moles.Assume 100g & find moles.

Find ratio of moles.Find ratio of moles.Ratio of Cl:C:HRatio of Cl:C:H

2.02 mol of Cl2.02 mol of Cl2.02 mol of C2.02 mol of C

1 of Cl1 of Cl1 of C1 of C

==

4.03 mol of H4.03 mol of H2.02 mol of C2.02 mol of C

2 of H2 of H1 of C1 of C

==

CHCH22ClClEmpirical FormulaEmpirical Formula

Find the mass of the empirical Find the mass of the empirical formula.formula.

CHCH22ClClC: 1 X 12.01g = 12.01gC: 1 X 12.01g = 12.01gH: 2 X 1.01g = 2.02gH: 2 X 1.01g = 2.02g

Cl: 1 X 35.45g = 35.45gCl: 1 X 35.45g = 35.45g++ 49.48g49.48g

Compare to the molar mass of Compare to the molar mass of the molecular formula.the molecular formula.

49.48g for CH49.48g for CH22ClCl

Empirical FormulaEmpirical Formula 98.96g for 98.96g for

Molecular FormulaMolecular Formula

C2H4Cl2

HomeworkHomework

Worksheet 10-3Worksheet 10-3

10 – 2510 – 25

Due: 2/2/06Due: 2/2/06

Quiz: 2/2/06Quiz: 2/2/06

HomeworkHomework

Worksheet 10-3Worksheet 10-3

FinishFinish

Due: 1/18/05Due: 1/18/05

Test: 1/18/05Test: 1/18/05

Documents

Chapter 10 Chemical Quantities Hingham High School Mr. Clune