AP Chemistry

A Review of Analytical

Chemistry

AP Chemistry

Ch 1

(Prentice Hall)

What Temperature Do You Read? • A measurement always has

some amount of

uncertainty

• To indicate the uncertainty

of a single measurement

scientists use a system

called significant figures

• What temperature do you

read?

• Are you certain there

is a 30?

• Are you certain about

32?

• Are you UNcertain

about the decimal

point?

Uncertainty in Measurement

Consider:

• 32 degrees

• 32.5 degrees

• 32.3 degrees

• 32. 7 degrees

• It is important to know which digits of the reported number are uncertain

• The last number may be different from person to person and is a visual estimate

• Therefore, the third digit is an uncertain number

Reporting Measurements

• Unless stated otherwise, uncertainty is in the last

digit 1

• For example, 32.5 0.1, means the temperature

ranges from 32.4 to 32.6 degrees

• The numbers recorded in a measurement (all the

certain numbers plus the first uncertain number)

are called significant figures

• Uncertainty comes from limitations of the techniques used for comparison (the limitation is due to the instrument)

• Let’s weigh the same item using three different scales…

• Let’s write down these number & we sill use them for our practice calculations

Cu = Cu = Cu =

Rules For Counting Significant

Figures

1. Nonzero integers are always significant:

– For example:

1234 has four sig figs

2. Dealing With Zeros. There are three classes:

– Leading zeros never significant, for example:

0.0025 has 2 sig figs

– Captive zeros are significant, for example:

1.008 has 4 sig figs

– Trailing zeros are significant if the number has a decimal point, for example:

100 has 1 sig fig

100. Has 3 sig figs

3. Exact Numbers are numbers known with

certainty and were not obtained by measuring

devices: (They have an unlimited number of

significant figures)

• They arise from definitions (the periodic table)

– Amu (___g/1 mole)

– 22.4 L/1 mole

• They arise from counting numbers

– Number of moles

– Number of coefficients

4. Exponents do not count as sig. figs.

For example:

6.02 x 1023 = 3 sig. figs

6.02214199 x 1023 = 9 sig. figs

Rules for Rounding Off

If the number is less than 5, the preceding digit stays the same, for example:

1.33 rounds to 1.3

If the number is equal to or greater than 5, the preceding digit is increased by 1, for example:

1.36 rounds to 1.4

(worked problems in the text show the correct number of sig figs in each step)

Calculations Using Significant Figures

• Calculators do not know about sig. Figs

– For example, try this on your calculator:

8.315 = 0.027902685

298

• Answers to calculations must be rounded to the

proper number of significant figures (answers

must reflect the measuring device)

Multiplication and Division Using Sig Figs

• The answer has the same number of sig figs as

the number in the problem with the least number

of sig figs

• For example: 4.56 x 1.4 = 6.384 Final answer 6.4

• Now try this rule using our three measuring

devices, convert to moles:

Adding and Subtracting Using Sig Figs

• Go with the least number of decimals

• For example:

12.11

18.0 least number of decimals

+ 1.013

31.123 Final answer 31.1

Now let’s add our measured masses together…

Accuracy vs. Precision

• Accuracy refers to how close a measurement is

to some accepted value, we’ll use % error

% error = exp value – accepted value x 100%

accepted value

• Precision refers to the repeatability of a

measurement

– Using sig figs

– Using Absolute Deviation

Remember Scientific Notation:

• …is a technique used to express very large

or very small numbers

• …is based on the power of 10

• To use numbers written in scientific

notation (calculator):

– Use the EE or Exp button, this means x 10

– Never hit x 10 or you’ll be off by 10 times!

Let's practice scientific notation...

You will need to memorize the following common SI Prefixes

& conversions: nm m

1,000,000,000 nm = 1 m

or

1.0 x 109 nm = 1 m

Let's do a few...

1. 500 nm to m (color of blue)

2. 3.4 m to nm (FM radiowave)

Required way to show your work

• You have two jobs in this class:

1. To be able to perform the conversions

2. To be able to prove that you know why

the answer is correct

3. In short, SHOW YOUR WORK!!!

4. Note the 2006 AP Exam Scoring

Guidelines…

• Note the grading rubric & how units

are given, even for multiplying…

• Note how this answer

earned all points…

• Note how this answer lost

points due to significant

figures

AP Chem

Ch 4-1

(Prentice Hall)

1,000,000,000 nm = 1 m

• This conversion comes in handy when dealing with wavelengths

with atomic spectra chemistry:

Frequency (v) = c = speed of light (3 x 108m/s)

wavelength

What’s the frequency of a wavelength of light of 500 nm?

The frequency of a red light is 4.74 x 1014 calculate its wavelength.

sec

AP Chemistry

Solution Chemistry

Ch 15.1-15.4

(Prentice Hall)

• Solution: is a homogeneous mixture, where the components are uniformly intermingled.

– Solvent: the substance present in the largest amount (usually the component doing the dissolving).

– Solute: the substance present in the smaller amount

(usually the component being dissolved).

• Solutions can be categorized by their physical state:

– Solid solutions

– Gas solutions

– Liquid solutions

• Solid solutions: contains two or more metals and are called alloys

• The more abundant element = solvent

• The less abundant element = solute

• For example steel = carbon + iron

What is the solvent?

What is the solute?

– Gaseous solutions: contain two or more gases that do not react with each other

• For example: Air:

• 78% N2

• 21% O2

• 1% other gases

– Liquid solutions: have a liquid solvent and the solute can consist of either a gas, liquid, or solid

• Aqueous solutions are special solutions that have water as the liquid solvent

You will need to memorize the following common SI

Prefixes & conversions:

mL L 1000 mL = 1 L

Let's practice a few, use the "picket fence method"

1. 459 L to mL

2. 0.0032 mL to L

You will need to memorize the following common

SI Prefixes & conversions:

mL L

1000 mL = 1 L

• For example, this conversion comes in handy when dealing with

solution chemistry:

Molarity (M) = moles of solute = mol

Liters of solution L

8.320g NiCl2 added to 250 mL of water, what’s the concentration?

How many grams of KCl are needed to prepare 0.750 L of a 1.50 M

1000 g = 1 kg

Molality (m) = moles of solute = mol

Kg of solvent kg

What is the molality of a solution containing 75.2 g of

AgClO4 dissolved in 885 g of benzene?

What is the molality of a solid solution of 0.125 g Cr &

81.3 g of Fe?

• Mass Percent: is one way of describing a

solution’s composition, sometimes called

weight percent, (%w/w) which is the mass of a

solute in a given mass of solution.

• Mass Percent = mass of solute x 100%

mass of solution

or Mass Percent = g of solute x 100%

g of solute + g of solvent

Does this look familiar? part/whole x 100%

• Milk is a solution it is composed of a dissolved

sugar called lactose.

• Cow’s milk typically contains 4.5% by mass of

lactose.

• Calculate the mass of lactose present in 175 g of

milk. Use the formula below…

Mass Percent = mass of solute x 100%

mass of solution

• Dissolved sugar called lactose = solute

• Mass % = 4.5%

• Solution mass = 175 g of milk

4.5% = mass of solute (sugar) x 100%

175 g

7.9 g = mass of solute (sugar lactose)

Switch To Clickers

• A sample of brass contains 68 g copper and 7 g zinc.

• What is the mass percent of this solid solution, (alloy)?

(remember to first identify the solute and solvent)

• A sample of brass contains 68 g copper and 7 g zinc.

% Cu = 68 g x 100%

68 g + 7 g

= 68 g x 100%

75 g

% Cu = 90.67%

10-1

Calculating AMU

• Next, let’s determine the mass of each atom.

• As we learned, there are different types of

atoms known as elements.

• Each element has a different mass and it is

called it’s amu

• atomic mass units

• Another name for amu is molar mass

• Molar mass is obtained by summing the masses of the component atoms.

For example: NH3 has the following amu:

N = 1 atom x 14.01 amu = 14.01 amu

H = 3 atoms x 1.01 amu = 3.03 amu

total amu = 17.04 amu

or 17.04 molar mass

Percent Composition

Ch 10-3

• If we wanted to know the percent boys in the class currently, how could we do this?…

• How about the percent girls in the class?

• Correct, count the number of boys/girls (part)

• Count total number in the class (whole)

• Then divide the part over the whole and multiply by 100 %

Percent composition = Part x 100%

whole

Does this formula make sense?

• Let’s apply this formula

% Composition = part x 100%

whole

by using the formula of the compound…

What is the % N to % H of

Windex, which is ammonia,

(NH3)

• First let’s determine the number of elements

in the formula NH3:

• N:1 atom

• H:3 atoms

Now do part x 100%

whole

N = 1 atom x 14.01 amu = 14.01 x 100%

17.03

= 82.21 % N

H = 3 atoms x 1.01 amu = 3.03 x 100%

17.03

= 17.79 % H

• The percentages may not always total to 100% due to rounding,

for example if you go to “1 decimal spot”

82.2% N

+ 17.7% H

99.9 % total

• But if you go to “2 decimal spots” you get closer to 100%

82.21% N

+ 17.79% H

100.00 % total

• Overall your numbers should

add up close to 100%

Switch To Clickers

You and your lab partner find the %Cl in

this formula (CCl2F2).

• You should have calculated

• 58.64 % Cl

Calculating % Composition By

Given Masses

• http://videos.howstuffworks.com/science-

channel/29291-100-greatest-discoveries-

atomic-weight-video.htm

• Dalton’s discovery of relative weights…

• According to the Law of Constant

Composition, any sample of a pure

compound always consists of the same

elements combined in the same proportion

by mass.

Percent composition can be determined of each

element in a compound by its mass:

% Composition = part mass x 100%

whole mass

This formula can be applied by using the

formula of the compound or by experimental

mass analysis of the compound

A sample of butane (C4H10)--lighter fluid--

contains 288 g carbon and 60 g hydrogen.

Find %C and %H in butane

First find total mass of sample (“whole”)

288 g C + 60 g O = 348 g

Next find the percent of each element in the formula… Use the following equation:

Percent composition = Part x 100%

whole

288 g C + 60 g O = 348 g

Part Part whole

Can you figure it out?…

288 g C (C part) x 100 % = 82.8 % C

348 g (whole)

60 g O (O part) x 100% = 17.2 % O

348 g (whole) + ---------------

100.0 %

Percents should add up close to 100%

(It may be a little over or under, that’s ok…)

Switch To Clickers

• Now you and your lab partner try this one…

• What is the percentage composition of a

carbon and oxygen compound, that contains

40.8 g of carbon and 54.4 g of oxygen.

• The total mass of the compound is 95.2 g.

% C = ? And % O = ?

• You should have calculated…

42.86 % C

57.14 % O

Ch 10-1

The Mole

Molar Mass

• The mass of one mole of atoms of any element is the molar mass which is numerically equal to the atomic mass unit (amu), but in grams…

• Molar mass = __g = 1 amu

1 mole

• Therefore CO2 has an amu = 44.01

or 44.01 g

1 mol

Using The Mole Map

Converting

from grams to moles

EXAMPLE: A student weighs out 88 grams

of solid CO2 (dry ice), how many moles

does the student have?

First, find the amu of CO2 :

C = 1 atom x 12.01 amu = 12.01 amu

O = 2 atoms x 16.00 amu = 32.00 amu

CO2 total amu = 44.01 amu

CO2 = 44g

1 mol

• Then convert from grams to moles

• Use your mole map…

Using The Mole Map

(What’s Given) x 1 mole

(molar mass)

88 g CO2 x 1 mole = 2 moles CO2

44 g

• Now you try to find the moles of various

compounds…

ALWAYS

SHOW YOUR WORK!!!

Converting To Particles

• A particle can

be defined as:

– an atom,

– a molecule,

– or formula unit

(ionic

compounds)

• Next, convert from mol to molecules:

• Use your mole map…

(moles calculated) x (Avogadro’s #)

1 mole

2 mole CO2 x 6.02 x 1023 molecules

1 mole

= 1.204 x 1024 molecules CO2

Empirical Formulas

and

Molecular Formulas

Ch 10-3

• Empirical Formulas are the simplest

(lowest) whole number ratio of atoms in a

molecule or ionic compound

• For example:

• C6H6 = CH

• H2O2 = HO

• C6H12O6 = CH2O

Empirical Formulas

…can be determined from % composition,

here is the “process:”

1. % is the same as grams

2. Convert from grams to moles

3. Next divide by the smallest # of moles

4. …this gives the empirical formula

Empirical Formula of Eugenol a

Component of Clove Oil?

…is 73.14% C, 7.37% H and 19.49 g O

Remember, % is the same as grams (g)

Empirical Formula of Eugenol,

continued…

Next, convert to the central unit, the mole

73.14 g C x 1 mole = 6.09 mol C

12.01 g

7.37 g H x 1 mole H = 7.31 moles H

1.0079 g

19.49 g O x 1 mole O = 1.22 moles O

15.9994 g

Empirical Formula of Eugenol,

continued…

Finally divide by the smallest # of moles

6.09 mol C 7.31 moles H 1.22 moles O

1.22 mol 1.22 mol 1.22 moles

C: 4.99 H: 5.99 O: 1.00

Or C: 5 atoms H: 6 atoms O: 1.00 atoms

Therefore C5H6O

is Eugenol’s empirical formula

Eugenol

Count the number of

carbon atoms,

hydrogen atoms and

oxygen atoms, does

this fit the empirical

formula that we just

derived? No, because

we did not find the

molecular formula…

Molecular Formulas

Molecular formulas are also known as the

“true formula” of a molecule.

To derive this use amu:

Molecular Formula = True amu

empirical amu

True Formulas

• The molar mass of Eugenol is 164.2 g/mol,

what’s the molecular formula of Eugenol?

• Use: True amu

empirical amu

164.2 g/mol = 164.2 g/mol = 2

C5H6O 82 g/mol

Therefore 2(C5H6O) = C10H12O2

Hydrated Compounds

Ch 7-3

&

Ch 14-3

Hydrated Compounds

• If ionic compounds

are prepared water

solutions and then

isolated as solids, the

crystals often have

molecules of water

trapped in the lattice,

for example CuSO4

CuSO4 • 5 H2O

Should you add the water when figuring the molar mass?

Yes, therefore

CuSO4 • 5 H2O has an amu of 249.7 g/mol

Notice the color difference of the dehydrated crystals

Hydrates • A compound that is hydrated is called a hydrate,

they form solids that includes water in their crystal structure

• Water can be driven from a hydrate to leave an anhydrous compound

•

Naming Hydrates

• To name hydrates:

1. Name the compound

2. Plus the word hydrate—use prefixes to indicate how

many waters are associated with the compound

3. Example: Copper (II) Sulfate pentahydrate

4. To write their formulas

Write: the name of the compound • number of H2O

CuSO4 • 5 H2O

Units of Hydration • A student heats hydrated crystals of CuSO4, how

many moles of water are associated with the

crystals?

• Step 1: Find the mass of the crystals:

1.023 g of CuSO4 • x H2O

• Step 2: Subtract the dehydrated crystal mass from

the initial crystal mass = mass of water

1.023 g of CuSO4 • x H2O – 0.654 g of CuSO4

= 0.369 g water

Units of Hydration Continued…

• Step 3: Determine the number of moles

0.369 g H2O x 1 mol

18.02 g = 0.0205 mol H2O

0.654 g CuSO4 x 1 mol/159.6 g = 0.00410 mol CuSO4

• Step 4: Determine the molar ratio (see above)

0.0205 mol H2O 0.00410 mol CuSO4

0.00410 mol CuSO4 0.00410 mol CuSO4

1 CuSO4 • 5 H2O

Ch 11-1

Mole-Mole Relationships

Using Mole Ratios

Ammonia (NH3) is used in huge quantities as a fertilizer. It is manufactured by combining nitrogen and hydrogen according to the following equation:

N2 + 3 H2 2 NH3

How many moles of NH3 can be made from 1.30 mol H2?

N2 + 3 H2 2 NH3

1.30mol ? mol

• Write what’s given first:

1.30 mol H2

• Next look at the balanced equation to

convert from moles of one thing to moles of

another:

1.30 mol H2 x 2 mol NH3 = 0.867 mol NH3

3 mol H2

http://spaceflight.nasa.gov/gallery/video/shuttle/sts-107/html/fd11.html

What if you had the

task of figuring out

how much LiOH was

needed on board a

space shuttle flight?

Solid LiOH is used to

take out CO2 from the

shuttle’s environment.

SAMPLE PROBLEM

• Solid lithium hydroxide is used in space

vehicles to remove exhaled carbon dioxide

from the living environment.

STEP 1 Balance the equation for the reaction:

LiOH + CO2 Li2CO3 + H2O

• STEP 1 Balance the equation for the

reaction:

2 LiOH + CO2 Li2CO3 + H2O

STEP 1 Balance the equation for the reaction:

2 LiOH + CO2 Li2CO3 + H2O

You need 41.8 mol LiOH in order to maintain the

atmosphere of the space shuttle for the

astronauts

How many moles of CO2 will be taken out of the

space shuttle?

Next Step Using this mole ratio, we can calculate

the moles of CO2 needed to react with the given

moles of LiOH:

2 LiOH + CO2 Li2CO3 + H2O

41.8 mol LiOH X 1 MOL CO2 = 20.9 mol CO2

2 mol LiOH

What if we start off with grams?

11-2

Mass-Mass Calculations

Steps for Calculating the Masses of Reactants &

Products in Chemical Reactions

STEP 1 Balance the equation for the reaction.

STEP 2 Convert the masses of reactants or products to

moles

STEP 3 Use the balanced equation to set up the

appropriate

mole ratio(s).

STEP 4 Use the mole ratio(s) to calculate the number

of

moles of the desired reactant or product.

STEP 5 Convert from moles back to mass.

SAMPLE PROBLEM

Solid lithium hydroxide is used in space

vehicles to remove exhaled carbon dioxide

from the living environment.

The products are solid lithium carbonate and

liquid water.

What mass of gaseous carbon dioxide can

1000 g of lithium hydroxide absorb?

http://spaceflight.nasa.gov/gallery/video/shuttle/sts-107/html/fd11.html

What if you had the

task of figuring out

how much LiOH was

needed on board a

space shuttle flight?

Solid LiOH is used to

take out CO2 from the

shuttle’s environment.

What mass of gaseous carbon dioxide

can

1000 g of lithium hydroxide absorb?

2 LiOH + CO2 Li2CO3 + H2O

1000 g ? g

STEP 2 Convert the masses of reactants or products to moles:

1.00 x 10 3 g LiOH x 1 mol LiOH = 41.8 mol LiOH

23.95 g LiOH

STEP 3 The appropriate mole ratio is:

2 LiOH + CO2 Li2CO3 + H2O

1 mol CO2

2 mol LiOH

Step 4 Using this mole ratio, we calculate

the moles of CO2 needed to react with

the given mass of LiOH:

41.8 mol LiOH X 1 MOL CO2 = 20.9 mol CO2

2 mol LiOH

Step 5 We calculate the mass of CO2 by using

its molar mass (44.01 g):

20.9 mol CO2 x 44.01 g = 920. g CO2

1 mol CO2

Thus 1.00x103 g of LiOH(s) can absorb 920. g of CO2

(g).

Now You Try! Phosphorus is placed in a flask of chlorine gas, heat and

light is given off forming phosphorus trichloride:

Step 1: Write and balance the Equation:

P4 + Cl2 PCl3

QuickTime™ and aCinepak decompressor

are needed to see this picture.

Step 1: Write and balance the Equation:

The Balanced Equation

P4 + 6 Cl2 4 PCl3

Now You Try!

How many grams of Cl2 will react with 1.24 g

of P4?

P4 + 6 Cl2 4 PCl3

1.24 g ? g

P4 + 6 Cl2 4 PCl3

1.24 g ? g

g P4 mol P4 mol: mol ratio g Cl2

1.24 g P4 x 1 mol P4 = 0.0100 mol P4

123.88 g

0.0100 mol P4 x 6 mol Cl2 = 0.0601 mol Cl2

1 mol P4

0.0601 mol Cl2 x 70.90 g Cl2 = 4.26 g Cl2

1 mol Cl2

Limiting Reactants

Ch 11-3

• Stoichiometry:

is the process of using a chemical equation

(mole ratio) to calculate the masses of

reactants and products in a reaction

• There are three parts to a reaction when

using stoichiometry…

The reactant that runs out first limits the amount

of product that can be formed and is called the

limiting reactant (or limiting reagent).

The reactant that is left over when the

reactions stops is called the excess of

reactant (or reagent).

A stoichiometric quantity is when reactants are

mixed in exactly correct amounts so that all

reactants “run out” at the same time.

The balanced equation is the correct

“stoichiometric quantity

– nothing is “left over”

– nothing “runs out first”

Let’s Do A Little Activity…

• You will receive a small kit of molecular

models to build.

• We will investigate a simple reaction.

• But we will build by using different mole

relationships to determine the Limiting

Reactant and Excess Reactant.

• Let’s begin…

1. 3 H2 + 1 O2 --> ? H2O

* What is the limiting and excess reactant?

?

1. 3 H2 + 2 O2 --> ? H2O

* What is the limiting and excess reactant?

?

1. 2 H2 + 1 O2 --> ? H2O

* What is the limiting and excess reactant?

?

Now Balance The Equation (This is the correct Stoichiometric Quantity)

H2 + O2 --> H2O

Are There Any Limiting or

Excess Reactants?

2 H2 + O2 --> 2 H2O

• Using this concept we can complete a

“special mass to mass calculation” to

figure out how much of our reactants

and products we need to get the

“greatest” amount of our product.

What is the correct ratio of Zn to HCl to produce

the maximum amount of H2?

(moles of HCl will be held constant…)

7.00 g 3.27 g 1.31 g 7.00 g 3.27 g 1.31 g

0.107 mol Zn

0.100 mol HCl

0.050 mol Zn

0.100 mol HCl

0.020 mol Zn

0.100 mol HCl

Excess Zinc Reactants match Excess HCl

perfectly; all is

consumed

Steps To Determine The Limiting

1. Compare the moles of reactant 1 to reactant 2

2. This will determine how many moles needed for reactant 2

3. Then determine how many moles you actually have for reactant 2

4. Compare the moles of needed to actual

5. This will help you to determine what is limiting and what is excess

Steps for Solving Stoichiometry Problems

Involving Limiting Reactants

STEP 1 Write and balance the equation for the reaction.

STEP 2 Convert known masses of reactants to moles.

STEP 3 Using the numbers of moles of reactants and the

appropriate

mole ratios, determine which reactant is limiting.

STEP 4 Using the amount of the limiting reactant and the

appropriate mole ratios, compute the number of moles of the

desired product.

STEP 5 Convert from moles of product to grams of product,

using the molar mass (if this is required by the problem).

Calculate the mass of ammonia produced…

25,000 g of nitrogen gas and 5000 g

of hydrogen gas are mixed and reacted to form

ammonia.

Remember, the limiting reactant determines

the amount of product formed…

N2 (g) + 3 H2(g) 2 NH3 (g)

25,000 g 5000 g ?

N2 (g) + 3 H2 (g) 2 NH3 (g) 25,000 g 5000 g

• What is the Limiting & Excess Reactant?

N2 (g) + 3 H2 (g) 2 NH3 (g)

N2 25,000 g N2 x 1 mol N2

28.02 g N2

= 892 mol N2

Mol of N2 available

892 mol N2 x 3 mol H2

1 mol N2

= 2680 mol H2 needed to react

H2

5000 g H2 x 1 mol H2

2.016 g H2

= 2480 mol H2

Mol of H2 available

We have 2480 mol H2 but NEED 2680 mol H2. Therefore H2 will run out first. N2 is excess

25,000 g 5000 g

• Now determine the maximum amount of

product produced if hydrogen is the limiting

reactant…

• Remember 2480 mol H2 Mol of H2 available

N2 (g) + 3 H2 (g) 2 NH3 (g)

Calculate the mass of ammonia produced if

hydrogen is the limiting reactant…

25 g of nitrogen gas and 5 g of hydrogen gas

are mixed and reacted to form ammonia.

Remember, the limiting reactant determines

the amount of product formed…

N2 (g) + 3 H2(g) 2 NH3 (g)

25 g 5 g ?

N2 (g) + 3 H2(g) 2 NH3 (g)

5 g H2 x x 1 mol H2 = 2.48 mol H2

2.016 g H2

2.48 mol H2 x 2 mol NH3 = 1.65 mol NH3

3 mol H2

1.65 mol NH3 x 17.03 g NH3 = 28.1 g NH3

1 mol NH3

Therefore: N2 (g) + 3 H2(g) 2 NH3

25 g 5 g (limiting) ?

25 g 5 g 28.1 g NH3

Percent Yield

Chapter 11-3

Theoretical Yield

• The amount of product formed is controlled

by the limiting reactant—products stop

forming when on reactant runs out.

• The amount of product calculated in this

way is called the theoretical yield.

• This is the amount of product predicted

from the amount of reactants used.

Actual Yield

• However, the amount of product predicted

(the theoretical yield) is seldom obtained.

• One reason for this is the presence of side

reactions (other reactions that consume one

or more of the reactants or products).

• The actual yield of product, is the amount

of product actually obtained.

Percent Yield

• The comparison of the product actually

obtained and theoretically obtained is called

the percent yield:

• Percent Yield = Actual Yield x 100%

Theoretical Yield

Chapter 1 Summary