The Mole

The MoleThe Mole

Dimensional Analysis ReviewDimensional Analysis Review

How many seconds are in 5.0 hours?How many seconds are in 5.0 hours?5.0 hr5.0 hr5.0 5.0 hr hr x x 60 60 minmin x x 60 sec60 sec = 18000 sec = 18000 sec

1 1 hrhr 1 1 minmin

Dimensional Analysis ReviewDimensional Analysis Review

Calculate the number of inches in 26 yardsCalculate the number of inches in 26 yards26 yards26 yards26 26 yards yards x x 3 3 ftft x x 12 inches 12 inches = 940 inches= 940 inches

1 1 ydyd 1 1 ftft

StoichiometryStoichiometryStoichiometryStoichiometry is just a long word for changing is just a long word for changing units in chemistry units in chemistry Just remember to ALWAYS start with your Just remember to ALWAYS start with your given!given!If you can do Dimensional Analysis, you can do If you can do Dimensional Analysis, you can do stoichiometrystoichiometry

StepsSteps

1.1. Start with your givenStart with your given

2.2. Use conversion factors and cross out Use conversion factors and cross out until you get what you wanteduntil you get what you wanted

3.3. Check sig. figsCheck sig. figs

The MoleThe Mole

Chemists need a convenient method for Chemists need a convenient method for counting accurately the number of atoms, counting accurately the number of atoms, molecules, or formula units in a sample of molecules, or formula units in a sample of a substance.a substance.

The The molemole, commonly abbreviated mol, is , commonly abbreviated mol, is the SI base unit used to measure the the SI base unit used to measure the amount of a substance.amount of a substance.

The MoleThe Mole

A mole of anything contains 6.02 x 10A mole of anything contains 6.02 x 102323 representative particles. representative particles.

A representative particle is any kind of A representative particle is any kind of particle such as atoms, molecules, formula particle such as atoms, molecules, formula units, electrons, or ions. units, electrons, or ions.

6.02 x 106.02 x 102323 is called Avogadro’s number is called Avogadro’s number

Conversion Factor #1Conversion Factor #1

Representative ParticlesRepresentative Particles

Anything - Representative particlesAnything - Representative particlesElements – AtomsElements – AtomsCovalent Compounds – MoleculesCovalent Compounds – MoleculesIonic Compounds – Formula UnitsIonic Compounds – Formula UnitsIons - IonsIons - Ions

Mole – Representative Particle Mole – Representative Particle CalculationsCalculations

Calculate the number of atoms in 3.50 Calculate the number of atoms in 3.50 moles of coppermoles of copperStart with your givenStart with your given3.50 mol Cu3.50 mol CuDraw your lineDraw your line3.50 mol Cu x _________3.50 mol Cu x _________


Place the conversion factorsPlace the conversion factors3.50 mol Cu x 3.50 mol Cu x 6.02 x 106.02 x 102323 atoms atoms

1 mol Cu 1 mol Cu Work the problemWork the problem2.11 x 10 2.11 x 10 24 24 atoms Cuatoms Cu

Another ExampleAnother Example

Calculate the number of molecules in 2.6 Calculate the number of molecules in 2.6 moles of Hmoles of H22OO

2.6 mol H2.6 mol H22O x O x 6.02 x 106.02 x 102323 molec molec = 1.6 x 10 = 1.6 x 102424

1 mol H1 mol H22O molec HO molec H22OO


How many moles of MgO are in 9.72 x 10 How many moles of MgO are in 9.72 x 10 23 23 molecules of MgO?molecules of MgO?1.61 mol MgO1.61 mol MgO


How many moles are in 4.50 x 10How many moles are in 4.50 x 1024 24 atoms atoms of Zinc?of Zinc?7.48 mol Zn7.48 mol Zn

Mass & the MoleMass & the Mole

The mass in grams of 1 mole of a The mass in grams of 1 mole of a substance is called the molar masssubstance is called the molar massIt can also be called molecular mass, It can also be called molecular mass, molecular weight, and formula massmolecular weight, and formula massTo get the molecular weight you just add To get the molecular weight you just add up all of the masses of all of the elements up all of the masses of all of the elements in a compoundin a compound

Molecular WeightMolecular Weight

Calculate the molecular weight of the Calculate the molecular weight of the following:following:CaCa40.08 g/mol40.08 g/molNaNa22.99 g/mol22.99 g/mol

Molecular WeightMolecular WeightCalculate the molecular weight of the following:Calculate the molecular weight of the following:MgOMgO40.31 g/mol40.31 g/molNaClNaCl58.44 g/mol58.44 g/molHH22OO18.02 g/mol18.02 g/molFeFe22OO33

159.70 g/mol159.70 g/mol

Conversion Factor # 2Conversion Factor # 2

1 mole1 mole Molecular weight (g)Molecular weight (g)

The molecular mass comes The molecular mass comes from the periodic table!from the periodic table!

Mole – Mass CalculationsMole – Mass Calculations

What is the mass of 4.21 moles of iron (III) What is the mass of 4.21 moles of iron (III) oxide?oxide?Start with your given:Start with your given:4.21 mol Fe4.21 mol Fe22OO33

Draw your lineDraw your line4.21 mol Fe4.21 mol Fe22OO3 3 x _____________x _____________


Place conversion factorsPlace conversion factors4.21 mol Fe4.21 mol Fe22OO3 3 x x 159.70 g Fe159.70 g Fe22OO33

1 mol Fe1 mol Fe22OO33

Cross out units & work the problemCross out units & work the problem672 g Fe672 g Fe22OO33


Calculate the mass of 1.630 moles of NaCalculate the mass of 1.630 moles of Na37.47 g Na37.47 g Na


How many moles of Ca(OH)How many moles of Ca(OH)22 are in 325 are in 325 grams?grams?4.39 moles Ca(OH)4.39 moles Ca(OH)22

Mass – Particle ConversionsMass – Particle Conversions

How many atoms of gold are in 25.0 g of How many atoms of gold are in 25.0 g of gold?gold?

25.0 g Au x 25.0 g Au x 1 mol Au1 mol Au x x 6.02 x 106.02 x 102323 atoms Au atoms Au 196.79 g Au 1 mol Au196.79 g Au 1 mol Au

7.65 x 107.65 x 102222 atoms of Au atoms of Au

Mass – Particle ConversionsMass – Particle Conversions

How many grams of He are in 5.50 x 10How many grams of He are in 5.50 x 102222 atoms of He?atoms of He?0.366 g He0.366 g He

Moles of CompoundsMoles of Compounds

Freon has the formula CClFreon has the formula CCl22FF22

This means thatThis means that1 mol CCl1 mol CCl22FF22 = 1 mol C = 1 mol C

1 mol CCl1 mol CCl22FF22 = 2 mol Cl = 2 mol Cl

1 mol CCl1 mol CCl22FF22 = 2 mol F = 2 mol F

This now gives us new conversion factorsThis now gives us new conversion factors

ExampleExample

How many moles of F are in 5.50 mol of How many moles of F are in 5.50 mol of CClCCl22FF22??

5.50 mol CCl5.50 mol CCl22FF2 2 x x 2 mol F2 mol F = 10.0 mol F = 10.0 mol F

1 mol CCl1 mol CCl22FF22

ExampleExample

How many moles of Al are in 1.25 mol of How many moles of Al are in 1.25 mol of AlAl22OO33??

2.50 mol Al2.50 mol Al22OO33

ExampleExample

How many ClHow many Cl-- ions are in 35.6 g of AlCl ions are in 35.6 g of AlCl33??

4.82 x 10 4.82 x 10 2323 ions Cl ions Cl--

How many Al How many Al +3+3 ions? ions?1.61 x 10 1.61 x 10 2323 ions Al ions Al+3+3

% Composition, Empirical % Composition, Empirical Formulas, & Molecular FormulasFormulas, & Molecular Formulas

% Composition% Composition

% = (part / whole ) x 100% = (part / whole ) x 100

When calculating the % composition, you When calculating the % composition, you are calculating the % of each element in a are calculating the % of each element in a compoundcompound


Calculate the % Composition of MgOCalculate the % Composition of MgO

Mg = 24.31g

O = 16.00g

Total = 40.31g

/ 40.31

/ 40.31

X 100

X 100

= 60.31 % Mg

= 39.69% O


Calculate the % Composition of iron (III) Calculate the % Composition of iron (III) oxideoxide% Fe = 69.94%% Fe = 69.94%% O = 30.06%% O = 30.06%

Empirical & Molecular FormulasEmpirical & Molecular Formulas

Empirical formulaEmpirical formula – the smallest whole – the smallest whole number ratio of elementsnumber ratio of elements

Molecular formulaMolecular formula – the true number of – the true number of elements in a compoundelements in a compound

Empirical FormulaEmpirical Formula

What is the empirical formula for HWhat is the empirical formula for H22OO22??

HOHOWhat is the empirical formula for CWhat is the empirical formula for C66HH1212OO66??

CHCH22OO

Steps for Calculating the Empirical Steps for Calculating the Empirical FormulaFormula

1.1. List your givensList your givens2.2. Change % to gramsChange % to grams3.3. Change grams to molesChange grams to moles4.4. Divide everything by the smallest number Divide everything by the smallest number

of molesof moles5.5. Write your formulaWrite your formula

Empirical Formula ProblemEmpirical Formula Problem

Calculate the empirical formula of a Calculate the empirical formula of a compound containing 40.05 % S and compound containing 40.05 % S and 59.95 % O.59.95 % O.

40.05 g S x

59.95 g O x

1 mol S =32.07 g S

1 mol O =16.00 g O

1.249 mol

3.747 mol

/

/

1.249 mol = 1

3.747 mol = 3

SO3

Empirical Formula ProblemEmpirical Formula Problem

Calculate the empirical formula for a Calculate the empirical formula for a compound containing 48.64 g C, 8.16 g compound containing 48.64 g C, 8.16 g H, and 43.20 g O.H, and 43.20 g O.

CC33HH66OO22

Steps for Calculating Molecular Steps for Calculating Molecular FormulaFormula

1.1. Calculate the empirical formulaCalculate the empirical formula2.2. Get the molecular mass of the empirical Get the molecular mass of the empirical

formula that you just determinedformula that you just determined3.3. Divide the experimentally determined molecular Divide the experimentally determined molecular

mass (given) by the molecular mass of the mass (given) by the molecular mass of the empirical formulaempirical formula

4.4. You will get a whole numberYou will get a whole number5.5. Multiply everything in the empirical formula by Multiply everything in the empirical formula by

this numberthis number

Molecular Formula ProblemMolecular Formula Problem

Calculate the molecular formula of a Calculate the molecular formula of a compound containing 40.68%C, 5.08%H, compound containing 40.68%C, 5.08%H, and 54.25%O with an experimentally and 54.25%O with an experimentally determined molecular weight of 118.1 determined molecular weight of 118.1 g/molg/mol


40.68g C x 40.68g C x 1 mol C1 mol C = 3.387 / 3.387 = (1)2 = 2 = 3.387 / 3.387 = (1)2 = 2 12.01 g C12.01 g C5.08g H x 5.08g H x 1 mol H1 mol H = 5.04 / 3.387 = (1.5)2= 3 = 5.04 / 3.387 = (1.5)2= 3 1.01 g H 1.01 g H 54.25 g Ox 54.25 g Ox 1 mol O1 mol O = 3.390 / 3.387 = (1)2 = 2 = 3.390 / 3.387 = (1)2 = 2 16.00 g O16.00 g O

CC22HH33OO22


CC22HH33OO22

Molecular Mass = 59.04Molecular Mass = 59.04EDMM / EFMM = 118.1 / 59.04 = 2EDMM / EFMM = 118.1 / 59.04 = 2Molecular Formula Molecular Formula CC44HH66OO44


Calculate the molecular formula of a Calculate the molecular formula of a compound containing 57.84 g C, 3.64 g H, compound containing 57.84 g C, 3.64 g H, and 38.52 g O with an experimentally and 38.52 g O with an experimentally determined molecular mass of 249.21 determined molecular mass of 249.21 g/molg/molCC1212HH99OO66

StoichiometryStoichiometry

StoichiometryStoichiometry

Using the methods of Using the methods of stoichiometrystoichiometry, we , we can measure the amounts of substances can measure the amounts of substances involved in chemical reactions and relate involved in chemical reactions and relate them to one another.them to one another.

StepsSteps

1.1. Write the chemical equationWrite the chemical equation2.2. Balance the chemical equationBalance the chemical equation3.3. Start with your givenStart with your given4.4. Cross out until you get what you wantCross out until you get what you want5.5. Check sig. figs, units, and circle your Check sig. figs, units, and circle your

answeranswer

Conversion Factor Conversion Factor

# Moles A# Moles A# Moles B# Moles B

The #’s in from if A & B MUST come from The #’s in from if A & B MUST come from the balanced chemical equationthe balanced chemical equation

Mole – Mole RelationshipMole – Mole Relationship

4 Fe + 3O4 Fe + 3O22 2Fe 2Fe22OO33

4 mol Fe / 3 mol O4 mol Fe / 3 mol O22

4 mol Fe / 2 mol Fe4 mol Fe / 2 mol Fe22OO33

3mol O3mol O22 / 2 mol Fe / 2 mol Fe22OO33

Mole – Mole RelationshipMole – Mole Relationship

How many moles of FeHow many moles of Fe22OO33 will I form from will I form from 5.0 mol of Fe?5.0 mol of Fe?5.0 mole Fe x 5.0 mole Fe x 2 mol Fe2 mol Fe22OO33

4 mol Fe4 mol Fe2.5 mol Fe2.5 mol Fe22OO33

Mass – Mole RelationshipMass – Mole Relationship

How many g of NaCl will be produced from How many g of NaCl will be produced from 1.25 mol of chlorine gas reacting with 1.25 mol of chlorine gas reacting with sodium?sodium?Write the reactionWrite the reactionNa + ClNa + Cl22 NaCl NaCl

Balance the reactionBalance the reaction2Na + Cl2Na + Cl22 2NaCl 2NaCl

Mass – Mole RelationshipMass – Mole Relationship

Work the problemWork the problem1.25 mol Cl1.25 mol Cl22 x x 2 mol NaCl2 mol NaCl x x 58.44 g NaCl58.44 g NaCl

1 mol Cl2 1 mol NaCl1 mol Cl2 1 mol NaCl146 g NaCl146 g NaCl

Mass – Mass RelationshipsMass – Mass Relationships

Ammonium nitrate decomposes into Ammonium nitrate decomposes into dinitrogen monoxide gas and water. dinitrogen monoxide gas and water. Determine that amount of water produced Determine that amount of water produced if 25.0 g of ammonium nitrate if 25.0 g of ammonium nitrate decomposes. decomposes. NHNH44NONO33 N N22O + HO + H22OO

NHNH44NONO33 N N22O + 2 HO + 2 H22OO

Mass – Mass RelationshipsMass – Mass Relationships

25.0 g NH25.0 g NH44NONO33 x x 1 mol NH1 mol NH44NONO33 x 2 x 2 mol Hmol H22OO x x 18.02 g H18.02 g H22OO

80.04 g80.04 g NHNH44NONO3 3 1 mol NH1 mol NH44NONO3 3 1 mol H 1 mol H22O O

11.2 g H11.2 g H22OO


Determine the number of moles of carbon Determine the number of moles of carbon dioxide that can be formed from the dioxide that can be formed from the combustion of 10.0 moles of Ccombustion of 10.0 moles of C33HH88..

CC33HH88 + O + O22 CO CO22 + H + H22OO

CC33HH88 + 5 O + 5 O22 3 CO 3 CO22 + 4 H + 4 H22OO

10.0 mol C10.0 mol C33HH88 x x 3 mol CO3 mol CO22 = 30.0 mol CO = 30.0 mol CO22

1 mol C1 mol C33HH88

Conversion FactorsConversion Factors

1 mol 1 mol 6.02 x 10 6.02 x 10 2323 particles particles

1 mol1 molmw in gramsmw in grams

# mol A# mol A# mol B# mol B

Limiting Reactants, Theoretical Limiting Reactants, Theoretical Yield, and % YieldYield, and % Yield

Limiting ReactantsLimiting Reactants

Limiting reactantsLimiting reactants are reactants that limit are reactants that limit the amount of products that can be formedthe amount of products that can be formed– They are consumed first in the reactionThey are consumed first in the reaction

Excess reactants Excess reactants are reactants that are are reactants that are not completely consumednot completely consumed– They are left overThey are left over

Limiting Reactant Problems

A limiting reactant problem can be easily identified because you have TWO givens

You basically are just going to do 2 stoichiometry problems

Steps to determine the limiting Steps to determine the limiting reactantreactant

1.1. Balance the equationBalance the equation2.2. Calculate the number of moles of each Calculate the number of moles of each

reactantreactant3.3. Convert one given to moles of 1 product Convert one given to moles of 1 product

(you can pick any product)(you can pick any product)4.4. Repeat for second givenRepeat for second given5.5. The reactant that produces the smallest The reactant that produces the smallest

amount (moles) of product is the LRamount (moles) of product is the LR6.6. Use the LR in any calculations Use the LR in any calculations

ExampleExample

If 18.1 g or NHIf 18.1 g or NH33 are reacted with 90.4 g of are reacted with 90.4 g of CuO, which is the limiting reactant and CuO, which is the limiting reactant and how many grams of Nhow many grams of N22 will be formed? will be formed?

NHNH33 + CuO + CuO N N22 + Cu + H + Cu + H22OO

Balance the equationBalance the equation2NH2NH33 + 3CuO + 3CuO N N22 + 3Cu + 3H + 3Cu + 3H22OO

2NH2NH33 + 3CuO + 3CuO N N22 + 3Cu + 3H + 3Cu + 3H22OO

Determine the number of molesDetermine the number of moles18.1 g NH18.1 g NH33 x x 1 mol NH1 mol NH33 x x 1 mol N1 mol N22 = 0.53 mol N = 0.53 mol N22

17.03 g NH17.03 g NH3 3 2 mol NH 2 mol NH3 3

90.4 g CuO x 90.4 g CuO x 1 mol CuO 1 mol CuO x x 1 mol N1 mol N22 = 0.38 mol N = 0.38 mol N22

79.55 g CuO 3 mol CuO79.55 g CuO 3 mol CuO

CuO produces less product, so it is the LRCuO produces less product, so it is the LR

ExampleExample

CuO is what we use to calculate the amount of CuO is what we use to calculate the amount of NN22 produced produced

0.38 mol N0.38 mol N22 x x 28.0 g N28.0 g N22 = 10.6 g = 10.6 g NN22

1 mol1 mol N N22

ExampleExample

10.6 g NN22


If 25.0 kg of nitrogen are reacted with 5.00 If 25.0 kg of nitrogen are reacted with 5.00 kg of hydrogen to form ammonia, how kg of hydrogen to form ammonia, how much ammonia will be produced?much ammonia will be produced?Balanced equationBalanced equationNN22 + 3H + 3H22 2NH 2NH33

28.0 kg NH28.0 kg NH33

Theoretical YieldTheoretical Yield

The The theoretical yield theoretical yield is the amount of is the amount of product that would be produced if all of the product that would be produced if all of the limiting reactant were completely limiting reactant were completely consumedconsumed– The maximum amount of product that can be The maximum amount of product that can be

produced with the given quantities of produced with the given quantities of reactantsreactants

Percent YieldPercent Yield

The The percent yield percent yield is the actual yield of product, is the actual yield of product, expressed as a percentexpressed as a percent Actual yield Actual yield x 100% = percent yield x 100% = percent yield

Theoretical yieldTheoretical yieldWhen performing an experiment, things do not always go exactly perfectSome product may get spilled, some may get sneezed on, or the reaction may not have gone to completion

ExampleExample

When 68.5 kg CO react with 8.60 kg HWhen 68.5 kg CO react with 8.60 kg H22, , what is the percent yield if 35.7 kg of what is the percent yield if 35.7 kg of methanol (CH3OH)are formed?methanol (CH3OH)are formed?Balanced equationBalanced equation2H2H22 + CO + CO CH CH33OHOH

ExampleExample

Find the LRFind the LR68.5 kg CO x 68.5 kg CO x 1000 g CO 1000 g CO x x 1 mol CO 1 mol CO x x 1 mol CH1 mol CH33OH OH = 2440 mol CH = 2440 mol CH33OHOH

1 kg CO 28 g CO 1 mol CO1 kg CO 28 g CO 1 mol CO8.60 kg H8.60 kg H22 x x 1000 g H1000 g H22 x x 1 mol H1 mol H22 x x 1 mol CH1 mol CH33OHOH = 2135 mol CH = 2135 mol CH33OHOH

1 kg H1 kg H22 2 g H 2 g H22 2 mol H 2 mol H22

HH2 2 produces less CHproduces less CH33OH, so it is the LROH, so it is the LR

ExampleExample

If all of the HIf all of the H22 is consumed, 2135 mol is consumed, 2135 mol CHCH33OH will be producedOH will be produced2135 mol CH2135 mol CH33OH x OH x 32 g CH32 g CH33OH OH x x 1 kg CH 1 kg CH33OH OH = 68.6 kg CH = 68.6 kg CH33OHOH

1 mol CH1 mol CH33OH 1000 g CHOH 1000 g CH33OHOH

theoretical yieldtheoretical yield

ExampleExample

Calculate the %yieldCalculate the %yield35.7 kg CH35.7 kg CH33OH OH x 100% = 52 % x 100% = 52 %

68.6 kg CH68.6 kg CH33OHOH

Another exampleAnother example

CC77HH66OO33 + C + C44HH66OO33 C C99HH88OO44 + HC + HC22HH33OO22

When 1.50 g CWhen 1.50 g C77HH66OO33 with 2.00 g C with 2.00 g C44HH66OO33 , , 1.50 g C1.50 g C99HH88OO44 was produced. What is the was produced. What is the %yield for this experiment?%yield for this experiment?76.5%76.5%

The Mole

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