Unit 6Unit 6Chapter 12Chapter 12Chemical Chemical QuantitiesQuantities

oror

Stoichiometry

stoichiometry—using stoichiometry—using balanced chemical equations balanced chemical equations to obtain infoto obtain info

12.1 Counting Particles of Matter

Particles of matter are too small and numerous to count

SI unit of chemical quantity = the mole (abbreviated mol)

The mole is a counting unit

How do you measure how How do you measure how much?much?

You can measure mass = gramsYou can measure mass = grams

or volume = litersor volume = liters

or or you can count pieces = you can count pieces = MOLESMOLES

MolesMoles

1 mole = 6.02 x 101 mole = 6.02 x 102323 representative representative particlesparticles

602 000 000 000 000 000 000 000602 000 000 000 000 000 000 000

Treat it like a very large dozenTreat it like a very large dozen

6.02 x 106.02 x 102323 is called Avogadro's is called Avogadro's numbernumber

Representative particlesRepresentative particles

= The smallest pieces of a = The smallest pieces of a substancesubstance

representative particles = representative particles = ATOMS ATOMS IONSIONS MOLECULESMOLECULES FORMULA UNITSFORMULA UNITS

Representative particlesRepresentative particles

For a molecular (covalent)For a molecular (covalent)

compounds it is a molecule compounds it is a molecule

compound with all nonmetals compound with all nonmetals

CO, BFCO, BF33 , Cl , Cl22

1 mole = 6.02 x 101 mole = 6.02 x 102323 molecules molecules

Representative particlesRepresentative particles

For an element it is an atomFor an element it is an atom– Unless it is diatomicUnless it is diatomic

one symbol, no charge: Br,Cs

1 mole = 6.02 x 101 mole = 6.02 x 102323 atoms atoms

Representative particlesRepresentative particles

For an ionic compound it isFor an ionic compound it is

a formula unita formula unit

compound with metal and nonmetal - KI, Na2SO4

1 mole = 6.02 x 101 mole = 6.02 x 102323 formula units formula units

Representative particlesRepresentative particles

For ions it is 1 mol ionsFor ions it is 1 mol ions

one symbol with charge (monatomic) or more than one symbol with charge

(polyatomic: Na+ , N3- , (C2H3O2) –

1 mole ions = 6.02 x 101 mole ions = 6.02 x 102323 ions ions

Molar MassMolar Mass

= The mass of 1 mole of an element = The mass of 1 mole of an element in grams.in grams.

We can make conversion factors We can make conversion factors from these.from these.– Example - We can write this as Example - We can write this as

12.01 g C = 12.01 g C = 1 mol1 mol

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

Molar MassesMolar Masses the atomic masses on the periodic the atomic masses on the periodic

tabletable– have a unit of amu (atomic mass have a unit of amu (atomic mass

unit)unit)

GAM = GAM = gram atomic massgram atomic mass

= = the atomic mass (listed on the the atomic mass (listed on the periodic table) written in gramsperiodic table) written in grams

1 atom Xe = 131.30 u 1 atom Xe = 131.30 u GAM of Xe = 131.301 gGAM of Xe = 131.301 g

molar mass—molar mass—the mass, in g, of 1 mole the mass, in g, of 1 mole of a substanceof a substance

Add up the gram atomic mass of all Add up the gram atomic mass of all elements in compound to calculate elements in compound to calculate molar mass of a substancemolar mass of a substance

Find the molar mass of methane, CHFind the molar mass of methane, CH44..

CHCH44 = 1(12.0) + 4(1.0) = 12.0 + 4.0 = = 1(12.0) + 4(1.0) = 12.0 + 4.0 = 16.0 g16.0 g

Find the molar mass of calcium Find the molar mass of calcium hydroxide, Ca(OH)hydroxide, Ca(OH)22..

Ca(OH)Ca(OH)22 = 1(40.1) + 2(16.0) + 2(1.0) = 1(40.1) + 2(16.0) + 2(1.0) = 74.1 g= 74.1 g

1. What is the molar mass of 1 mole

of phosphorus?

2. What is the molar mass of a

molecule of H2?

3. What is the molar mass of a

formula unit of MgCl2?1. 30.97 g2. 2.016 g3. 94 g

Molar Volume: volume-to-mole and mole-to-volume conversions

At STP, all gases occupy the same amount of space:

MOLAR VOLUME of any gas at STP:

22.4 L = 1 mol

dimensional analysis = using the units (dimensions) to solve problems

steps for success:1) identify unknown (read carefully)2) identify known (read carefully)“Play checkers” with the units, moving them diagonally, canceling when appropriate. All units should cancel except those of the desired answer.3) plan solution4) calculate5) check (sig.figs., units, and math)

CONVERSION FACTOR SUMMARY:

6.02 x 10^23 representative particles

1 MOLE

&

1 MOLE 6.02 x 10^23 representative particles

CONVERSION FACTOR SUMMARY:

MOLAR MASS (g) & 1 MOLE1 MOLE MOLAR MASS

(g)

for a gas at STP:22.4 L & 1 MOLE 1 MOLE 22.4 L

Calculation questionCalculation question

How many molecules of COHow many molecules of CO22 are the in 4.56 moles of COare the in 4.56 moles of CO22 ? ?

4.56 moles CO2 X 6.02 x 1023 molecules CO2

1 1 mole CO2

= 27.5 x 1023 molecules CO2

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

NaOH?NaOH?

5.69 g NaOH x 1 mole= 0.142 mol NaOH

1 40.0 gNaOH

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

1 mole NaOH = 40.00 g

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

carbon weigh?carbon weigh?

2.34 mol C X 12.01 g C 1 1mol C

= 28.10 g C

Calculation questionCalculation question

How many moles of salt is 5.87 x How many moles of salt is 5.87 x 10102222 formula units? formula units?

5.87 X 1022 formula units NaCl X 1 mole NaCl_ 1 6.02 x 1023 f.u. NaCl

= 0.0975 moles NaCl

ExamplesExamples How many moles of How many moles of

magnesium in 4.61 g of Mg?magnesium in 4.61 g of Mg?

4.61 g Mg X _1 mol Mg 1 24.3 g Mg

= 0.1897 mol Mg

exampleWhat is the volume, in L, of 0.495 mol of NO2 gas at STP?

0.495 mol NO2 x 22.4 L NO2 = 11.1 L NO2

1 1 mol NO2

How many moles are found in 84 L of neon gas at STP?

84 L Ne x 1 mol Ne = 3.8 mol Ne 1 22.4 L Ne

There are many types of mole problems:

1 step: r.p. mol & mol r.p.

mass mol & mol mass

2 step: mass r.p. & r.p. mass

mass volume & volume mass

r.p. volume & volume r.p.

Moles of Compounds

• 1 mole of a compound contains as many moles of each element as are indicated by the subscripts in the formula for the compound.

• Example:

1 mole of ammonia (NH3) has

1 mole of nitrogen atoms and

3 moles of hydrogen atoms.

Percent composition • composition of a compound is the

percent by mass of each element in the compound.

Find the mass of each Find the mass of each elementelement

divide by the total mass divide by the total mass of compound.of compound.

• Determine the percent composition of calcium chloride (CaCl2).

• Step 1: determine the molar masses of each element in compound and the compound

Ca = 40.078 g

Cl2 = 2 x 35.453 g = + 70.906 g

CaCl2 = 110.984 g

Step 2: divide the molar mass of each element by the molar mass of the compound and times by 100

****make sure they add up to 100

Empirical FormulaEmpirical Formula

The lowest whole number ratio The lowest whole number ratio of elements in a compound.of elements in a compound.

CHCH22

HH22OO

You can determine the empirical formula from percent composition and mole ratios*** percent means “parts per hundred”

assume you have 100 g of the compoundStep 1: change the percent sign to g (grams)

Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N

Step 2:calculate the number of moles for each element by converting grams to moles using

1 mole = molar mass

38.67 g C x 1mol C = 3.220 mole C 1 12.01 g C

16.22 g H x 1mol H = 16.1 mole H 1 1.01 g H

45.11 g N x 1mol N = 3.220 mole N 1 14.01 g N

Step 3: divide each number of moles by the smallest number of moles

3.220 mole C = 1 mol C 3.22016.1 mole H = 5 mol

H 3.220

3.220 mole N = 1 mol N 3.220

So empirical formula is CH5N

Determine the empirical formula of a compound with percent composition of 38.4 % Mn, 16.8 % C, 44.7 % O

38.4 g Mn X 1mol Mn = 0.699 mole Mn 1 55 g Mn

16.8 g C X 1mol C = 1.339 mole C 1 12 g C

45.7 g O X 1mol O = 2.798 mole O 1 16 g O

0.699 mole Mn = 1 mol Mn 0.699

1.339 mole C = 2 mol C 0.699

2.798 mole O = 4 mol O 0.699

So formula is MnC2O4

For many compounds, the For many compounds, the empirical formula is not the empirical formula is not the true formula true formula

A A molecular formulamolecular formula tells tells the exact number of atoms of the exact number of atoms of each element in a molecule each element in a molecule

Example: acetic acid Example: acetic acid

molecular formula = Cmolecular formula = C22HH44OO22

empirical formula = CHempirical formula = CH22OO

The molecular formula for a The molecular formula for a compound is always a whole-compound is always a whole-number multiple of the empirical number multiple of the empirical formula. formula.

To determine molecular formula you To determine molecular formula you need to know the molar mass of the need to know the molar mass of the compoundcompound

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

Multiply the empirical formula by Multiply the empirical formula by this number.this number.

molar mass of compoundmolar mass of compoundmolar mass of empirical formulamolar mass of empirical formula

Example: A compound has an empirical formula of ClCH2 and a molar mass of 98.96 g/mol. What is its molecular formula?

molar mass of ClCH2 = 49.0 g/mol

98.96 = 2.01 49.0 2 X (ClCH2)

Empirical formula = ClCH2

Molecular formula = Cl2C2H4

ExampleExample A compound has an empirical

formula of CH2O and a molar mass of 180.0 g/mol. What is its molecular formula?

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?

Unit 6Unit 6Chemical Chemical Quantities Quantities ContinuedContinued

remember our friend the remember our friend the mole?mole?

Recall that the mole should be Recall that the mole should be treated like a dozentreated like a dozen

1 dozen = 12 pieces &1 dozen = 12 pieces & 1 mole = 602 000 000 000 000 000 1 mole = 602 000 000 000 000 000

000 000 pieces (or 6.02 X 10000 000 pieces (or 6.02 X 102323 r.p) r.p)

remember our friend the remember our friend the mole?mole?

You still need to know how and when You still need to know how and when to use:to use:

Avogadro’s number (6.02 x 1023), Avogadro’s number (6.02 x 1023), representative particlesrepresentative particles

Molar mass, grams in one mole, Molar mass, grams in one mole,

Molar volume of a gas at STP (22.4 Molar volume of a gas at STP (22.4 L)L)

Conversion factor reviewConversion factor review

1 mole = 6.02 x 1023 r.p1 mole = 6.02 x 1023 r.p

1 mole = molar mass (g) 1 mole = molar mass (g)

1 mole = 22.4 L (gas @STP)1 mole = 22.4 L (gas @STP)

So what about this silly mole?So what about this silly mole?

Our useful friend the mole allows us Our useful friend the mole allows us to do calculations called to do calculations called

stoichiometrystoichiometry—using balanced —using balanced chemical equations to obtain infochemical equations to obtain info

ex: 2C + Oex: 2C + O22 2CO 2CO

Mole - Mole (MOL – MOL) Mole - Mole (MOL – MOL) ConversionsConversions

new conversion factor – new conversion factor –

# of mol A = # of mol B# of mol A = # of mol B

# = coefficients in balanced # = coefficients in balanced equationequation

I.Mole - Mole (MOL – MOL) I.Mole - Mole (MOL – MOL) ConversionsConversions

A.A.the most important, most basic the most important, most basic stoich calculationstoich calculation

BB. . uses the coefficients of a uses the coefficients of a balanced equation to compare balanced equation to compare the amounts of reactants and the amounts of reactants and productsproducts

C. C. coefficients are mole ratioscoefficients are mole ratios

D. D. the way to go from substance A to the way to go from substance A to substance Bsubstance B

E. mol – mol is the only time the mole E. mol – mol is the only time the mole number in the conversion is not number in the conversion is not automatically 1. (Avogadro’s #, automatically 1. (Avogadro’s #, molar mass, and 22.4 L are all = to 1 molar mass, and 22.4 L are all = to 1 mol)mol)

MOL – MOL MOL – MOL : : # mol A# mol A # mol B # mol B

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

# = coefficients# = coefficients

MOL – MOL ConversionsMOL – MOL Conversions

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

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

ex: How many moles of carbon ex: How many moles of carbon monoxide are produced when 0.750 monoxide are produced when 0.750 mol of oxygen reacts with carbon?mol of oxygen reacts with carbon?

2C + O2C + O22 2CO 2CO

***always start with a balanced ***always start with a balanced chemical equationchemical equation

How many moles of carbon monoxide are How many moles of carbon monoxide are produced when 0.750 mol of oxygen reacts produced when 0.750 mol of oxygen reacts with carbon?with carbon?

2C + O 2C + O22 2CO 2CO

Step 1; determine given and Step 1; determine given and unknownunknown

given = 0.750 mol Ogiven = 0.750 mol O22

unknown = ? mol COunknown = ? mol CO

How many moles of carbon monoxide are How many moles of carbon monoxide are produced when 0.750 mol of oxygen reacts produced when 0.750 mol of oxygen reacts with carbon?with carbon?

2C + O 2C + O22 2CO 2CO

Step 2; setup conversion using mole Step 2; setup conversion using mole ratios (coefficients) in balanced ratios (coefficients) in balanced equationequation

0.750 mol O0.750 mol O2 2 X X 2 mol CO2 mol CO = 1.5 mol = 1.5 mol COCO

1 1 mol O1 1 mol O22

Find the number of moles of the reactants, given 0.661 mol of product is formed.

4Al + 3O4Al + 3O22 2Al 2Al22OO33

Step 1; determine given and Step 1; determine given and unknownunknown

given = 0.661 mol Algiven = 0.661 mol Al22OO33

unknown = ? mol Alunknown = ? mol Al

= ? mol O= ? mol O22

Step 2; setup conversion using mole Step 2; setup conversion using mole ratios (coefficients) in balanced ratios (coefficients) in balanced equationequation

0.661 mol Al0.661 mol Al22OO3 3 X X 4 mol Al4 mol Al = 1.32 mol = 1.32 mol AlAl

1 2 mol Al1 2 mol Al22OO33

0.661 mol Al0.661 mol Al22OO3 3 X X 3 mol O3 mol O22 = 0.992 mol = 0.992 mol OO22

1 2 mol Al1 2 mol Al22OO33

4Al + 3O4Al + 3O22 2Al 2Al22OO33

II. MASS – MASS Conversions II. MASS – MASS Conversions

– – Using molar mass in stoich problems to Using molar mass in stoich problems to predict masses of reactants and/or predict masses of reactants and/or productsproducts

A.A.a balanced chemical equation can be used a balanced chemical equation can be used to compare masses of reactants and to compare masses of reactants and productsproducts

B. B. mass – mass cannot change which mass – mass cannot change which substance you are dealing with; substance you are dealing with; only only mol – mol – mol can do thatmol can do that

MASS – MASS Conversions MASS – MASS Conversions

GIVEN g A GIVEN g A X X 1 mol A1 mol A X X # mol B # mol B X X PT g PT g BB

1 PT g A # mol A 1 mol B1 PT g A # mol A 1 mol B

PT = periodic table, molar mass PT = periodic table, molar mass

# = coefficients, chemical equation# = coefficients, chemical equation

How many grams of hydrochloric acid How many grams of hydrochloric acid are made from the reaction of are made from the reaction of 0.500 g of hydrogen gas with 0.500 g of hydrogen gas with excess chlorine gas? excess chlorine gas?

HH22 + Cl + Cl22 2HCl 2HCl

Known = 0.500g HKnown = 0.500g H22

Unknown = ? g HCl Unknown = ? g HCl

0.500 g H0.500 g H22 X X 1 mol H1 mol H22 X X 2 mol HCl 2 mol HCl X X 36.5 36.5 gHClgHCl

1 2.0 g H1 2.0 g H22 1 mol H 1 mol H22 1 mol HCl 1 mol HCl

= 18 g HCl= 18 g HCl

H2 + Cl2 2HCl

GIVEN g A GIVEN g A X X 1 mol A1 mol A X X # mol B # mol B X X PT g PT g BB

1 PT g A # mol A 1 mol B1 PT g A # mol A 1 mol B

Calculate the numbers of grams of oxygen Calculate the numbers of grams of oxygen formed when 25.0 g of sodium nitrate formed when 25.0 g of sodium nitrate decomposes into sodium nitrite and oxygen.decomposes into sodium nitrite and oxygen.

2NaNO2NaNO33 2NaNO 2NaNO22 + O + O22

25.0 g 25.0 g NaNONaNO33 X X 1 mol 1 mol NaNONaNO33 X X 1 mol O1 mol O22 X X 32.0 g32.0 g OO22

1 85.0g 1 85.0g NaNONaNO33 2 mol 2 mol NaNONaNO33 1 mol O 1 mol O22

= 20.3 g O= 20.3 g O22

““mole – mass” (mass – mole) mole – mass” (mass – mole) calculationscalculations

““MASS – MOLE”:MASS – MOLE”:

GIVEN g A GIVEN g A X X 1 mol A 1 mol A X X # mol B# mol B

PT g A # mol APT g A # mol A

““MOLE – MASS”:MOLE – MASS”:

GIVEN mol A GIVEN mol A X X # mol B # mol B X X PT g BPT g B

# mol A 1 mol B# mol A 1 mol B

PT = periodic table, molar mass PT = periodic table, molar mass

# = coefficients# = coefficients

How many g of water are produced How many g of water are produced from the complete combustion of from the complete combustion of 0.6829 mol of C0.6829 mol of C22HH22??

2C2C22HH22 + 5O + 5O22 4CO 4CO22 + 2H + 2H22OO

0.6829 mol C0.6829 mol C22HH22 X X 2 mol H2 mol H22O O X X 18.0 g H18.0 g H22OO

1 2 mol C1 2 mol C22HH22 1 mol H 1 mol H22OO

= 12.3 g H= 12.3 g H22OO

Using the equation Using the equation

2C2C22HH22 + 5O + 5O22 4CO 4CO22 + 2H + 2H22O O

how many moles of Ohow many moles of O22 would be would be

needed to produce 56.09 g of COneeded to produce 56.09 g of CO22??

56.09 g CO56.09 g CO22 X X 1 mol CO1 mol CO22 X X 5 mol O5 mol O22

1 1 44.0 g CO44.0 g CO22 4 mol CO4 mol CO22

= 1.59 mol O= 1.59 mol O22

““mass – volume” (volume – mass) mass – volume” (volume – mass) calculations – calculations –

““MASS – VOLUME”: (gases @ STP)MASS – VOLUME”: (gases @ STP)

GIVEN g A GIVEN g A X X 1 mol A 1 mol A X X # mol B # mol B X X 22.4 L B22.4 L B

1 PT g A # mol A 1 mol B1 PT g A # mol A 1 mol B

““VOLUME – MASS”: (gases @ STP)VOLUME – MASS”: (gases @ STP)

GIVEN L A GIVEN L A X X 1 mol A 1 mol A X X # mol B # mol B X X PT g BPT g B

1 22.4 L A 1 22.4 L A # mol A 1 mol B# mol A 1 mol B

PT = periodic table, molar mass PT = periodic table, molar mass

# =# = coefficients coefficients

How many L of hydrogen are How many L of hydrogen are produced from the decomposition of produced from the decomposition of 3.50 g of water at STP?3.50 g of water at STP?

2H2H22O 2HO 2H22 + O + O22

3.50 g H3.50 g H22O O X X 1 mol H1 mol H22O O X X 2 mol H2 mol H22 X X 22.4 L 22.4 L HH22

1 18.0 g H1 18.0 g H22O 2 mol HO 2 mol H22O 1 mol HO 1 mol H22

= 4.36 L H= 4.36 L H22

Using the equation Using the equation

2C2C22HH22 + 5O + 5O22 4CO 4CO22 + 2H + 2H22O O

how many liters of water vapor are how many liters of water vapor are produced when 5.02 g of Cproduced when 5.02 g of C22HH22 undergoes complete combustion?undergoes complete combustion?

5.02 g C5.02 g C22HH22 x x 1 mol C1 mol C22HH22 x x 2 mol H2 mol H22O O x x 22.4 L H22.4 L H22OO

1 26.0 g C1 26.0 g C22HH22 2 mol C 2 mol C22HH22 1 mol H 1 mol H22OO

= 4.32 L H= 4.32 L H22OO

““volume – volume” calculationsvolume – volume” calculations

GIVEN L A GIVEN L A x x 1 mol A 1 mol A x x # mol B # mol B x x 22.4 L B22.4 L B

1 22.4 L A 1 22.4 L A # mol A 1 mol B# mol A 1 mol B

# = coefficients (SHORT CUT: # = coefficients (SHORT CUT: compare coefficients!)compare coefficients!)

How many L of sulfur trioxide are How many L of sulfur trioxide are produced from the reaction of 36.1 L of produced from the reaction of 36.1 L of oxygen with sulfur dioxide at STP?oxygen with sulfur dioxide at STP?

2SO2SO22 + O + O22 2SO 2SO33

36.1 L O36.1 L O22 x x 1 mol O1 mol O22 x x 2 mol SO2 mol SO33 x x 22.4 L SO22.4 L SO33

1 22.4 L O1 22.4 L O22 1 mol O 1 mol O22 1 mol SO 1 mol SO3 3

= 72.2 L SO= 72.2 L SO3 3

SHORTCUT: coefficient of OSHORTCUT: coefficient of O22 = 1 = 1 coefficient of SOcoefficient of SO3 3 = 2 so 36.1 L x 2 = = 2 so 36.1 L x 2 = 72.2 L SO72.2 L SO33

How many liters of COHow many liters of CO2 2 are produced from are produced from 0.252 L of HCl reacting with NaHCO0.252 L of HCl reacting with NaHCO33??

NaHCONaHCO33 + HCl NaCl + CO + HCl NaCl + CO22 + H + H22OO

0.252 L HCl 0.252 L HCl x x 1 mol HCl 1 mol HCl x x 1 mol CO1 mol CO22 x x 22.4 L CO22.4 L CO22

1 22.4 L HCl 1 mol HCl 1 mol CO1 22.4 L HCl 1 mol HCl 1 mol CO22

= 0.252 L CO= 0.252 L CO22

SHORTCUT: coefficients = 1 mol HCl to 1 mol SHORTCUT: coefficients = 1 mol HCl to 1 mol COCO22. .

so 0.252 L HCl = 0.252 L COso 0.252 L HCl = 0.252 L CO22

mass –particle (particle – mass) calculationsmass –particle (particle – mass) calculations

““MASS – PARTICLE”:MASS – PARTICLE”:GIVEN g A GIVEN g A x x 1 mol A 1 mol A x x # mol B # mol B x x 6.02 x 106.02 x 102323 r.p. B r.p. B

1 PT g A # mol A 1 mol B1 PT g A # mol A 1 mol B

““PARTICLE – MASS”:PARTICLE – MASS”:GIVEN r.p. A GIVEN r.p. A x x 1 mol A 1 mol A x x # mol B # mol B x x PT g BPT g B

1 6.02 x 101 6.02 x 102323 r.p. A r.p. A # mol A 1 mol B# mol A 1 mol B

PT = periodic table, molar mass PT = periodic table, molar mass

# = coefficients# = coefficients

How many molecules of NHHow many molecules of NH33 are are produced from reacting 2.07 g of produced from reacting 2.07 g of HH22 with excess N with excess N22??

NN22 + 3H + 3H22 2NH 2NH33

2.07 g H2.07 g H22 x x 1 mol H1 mol H22 x x 2 mol NH2 mol NH33 x x 6.02 x 106.02 x 102323 NHNH33

1 2.0 g H1 2.0 g H22 3 mol H 3 mol H22 1 mol NH 1 mol NH33

= 4.2 x 10= 4.2 x 102323 molecules NH molecules NH33

Limiting Reactants

• Rarely are the reactants in a chemical reaction present in the exact mole ratios specified in the balanced equation.

• Usually, one or more of the reactants are present in excess, and the reaction proceeds until all of one reactant is used up.

• The left-over reactants are called excess reactants.

• The reactant that is used up is called the limiting reactant.

Limiting ReactantLimiting Reactant

To figure out the limiting reactant:To figure out the limiting reactant:

You must do the stoichiometric You must do the stoichiometric calculation with all reactants.calculation with all reactants.

The one that produces the least The one that produces the least amount of product is the limiting amount of product is the limiting reactant.reactant.

Ex:Ex: 4Al + 3O4Al + 3O2 2 2Al 2Al22OO33

What is the limiting reactant when 35 g of What is the limiting reactant when 35 g of aluminum reacts with 35 g of oxygen? How aluminum reacts with 35 g of oxygen? How much aluminum oxide is formed in this reaction?much aluminum oxide is formed in this reaction?

35 g Al 35 g Al X X 1 mol Al 1 mol Al X X 2 mol Al2 mol Al22OO33 ==

1 27 g Al 4 mol Al 1 27 g Al 4 mol Al

35 g O35 g O22 X X 1 mol O1 mol O22 X X 2 mol Al2 mol Al22OO33 = 1.1 mol = 1.1 mol AlAl22OO33

1 16 g O1 16 g O22 4 mol O 4 mol O22Aluminum is limiting reactant0.65 mol Al0.65 mol Al22OO33 X X 162 g Al162 g Al22OO33 = = 105.3 g 105.3 g AlAl22OO33

1 1 mol Al1 1 mol Al22OO33

0.65 mol 0.65 mol AlAl22OO33

Percent YieldPercent Yield

theoretical yield theoretical yield ——amount of amount of product predicted by the math product predicted by the math (theory)(theory)

actual yield actual yield ——amount of product amount of product obtained in labobtained in lab

percent yield percent yield ——percentage of percentage of product recovered; comparison product recovered; comparison of actual andof actual and

theoretical yieldstheoretical yields

% YIELD = % YIELD = ACTUAL YIELD ACTUAL YIELD X 100 X 100

THEORETICAL YIELDTHEORETICAL YIELD

35.0 g of product should be 35.0 g of product should be recovered from an experiment. A recovered from an experiment. A student collects 22.9 g at the end student collects 22.9 g at the end of the lab. What is the percent of the lab. What is the percent yield?yield?

22.9 g 22.9 g X 100 = X 100 = 65.4%65.4%

35.0 g35.0 g

What is the percent yield if 2.89 g of NaCl is What is the percent yield if 2.89 g of NaCl is produced when 1.99 g of HCl reacts with excess produced when 1.99 g of HCl reacts with excess NaOH? Water is the other product.NaOH? Water is the other product.

HCl + NaOH NaCl + HHCl + NaOH NaCl + H22OO

Actual yield = 2.89 g NaClActual yield = 2.89 g NaCl

Theoretical yield = ?Theoretical yield = ?

1.99 g HCl 1.99 g HCl x x 1 mol HCl 1 mol HCl x x 1 mol NaCl 1 mol NaCl x x 58.5 g 58.5 g NaClNaCl

1 36.5 g HCl 1 mol HCl 1 mol NaCl1 36.5 g HCl 1 mol HCl 1 mol NaCl

= THEORETICAL YIELD = 3.19 g NaCl= THEORETICAL YIELD = 3.19 g NaCl

% YIELD = % YIELD = 2.89 g NaCl 2.89 g NaCl X 100 = 90.6% X 100 = 90.6%

3.19 g NaCl3.19 g NaCl

The EndThe End