CHAPTER 10

CHEMICAL QUANTITIES

What is a Mole? In chemistry you will do calculations

using a measurement called a mole. The mole, the SI unit that measures the

amount of substances, is a unit just like the dozen.

The mole can be related to: the number of particles(ion,atoms, etc.) the mass (in grams)volume of an element (STP conditions)

The Number of Particles in a Mole

Because atoms, molecules and ions are very small, the number of individual particles in an object is very large. Instead of counting these particles individually, you can count them using a term that represents a specific number of particles.

Here’s the number: 602 214 199 000 000 000 000 000.

A mole is the Avogadro’s number of particles…. rather like a dozen is twelve, the mole is just that, an Avogadro’s number of particles (The number above was named for Amadeo Avogadro, an Italian chemist who worked on gases in the nineteenth century.)

Representive particles- species present in the substance (usually atoms, molecules, or formula units)

Examples: How many moles of a

magnesium is 1.25 X 1023

atoms of magnesium? The answer: 0.21 moles How many moles is 2.80 X

1024 atoms of silicon? The answer: 4.65 moles When given # of particles,

divide by 6.02 x 1023

Mass of a Mole of a Compound Gram atomic mass- the atomic mass of an element expressed in

grams (use the Periodic Table) The gram atomic mass of any element contains 1 mole of atoms of that

element. 1 atom Carbon = 12 amu…atomic mass units 1 mole Carbon = 12 grams (6.02 x 1023 atoms)

Examples: The formula of hydrogen is H2. What is the gram atomic mass? Note: The subscript = the # of atoms

Pg. 179

Finding Formula Mass

Element # of Atoms x Atomic Mass = Total Mass

Subscript use periodic table

Examples: Find gram formula mass for: Aluminum Chloride Lithium Sulfide Calcium Hydroxide

Relating Moles and Grams

Moles = grams given Formula Mass

Use Periodic table to determine the formula mass of substance

The Molar Mass of a Substance

Molar Mass- the mass in grams of a mole of any substance Mol = gr FmExamples: 1. How many grams are in 9.45 mol of dinitrogen trioxide

(N2O3)?2. Find the number of moles are in 92.2 g of iron (III) oxide

(Fe2O3)3. Find the number of grams for 2.5 moles of Calcium Bromide.

1. 718.2 gr. 2. .578 moles 3. 500grpg. 182

The Volume of a Mole of Gas

The volume of gas changes when temperature or pressure changes.

Because of the variation gas is measured in Standard Temperature and Pressure(STP)

Standard temperature = 0o C (273 0Kelvin)(32 0F) Standard pressure is 101.3 kPa (KiloPascals)

- Molar Volume of gas= 22.4 L and is measured at STP

- 1 mole of any gas at STP will occupy 22.4 liters of volume

Calculate the number of liters if you have16 grams of Sulfur Dioxide@STP

% Composition with Formulas

Use the atomic masses on Periodic Table Find the % composition, by mass, for the

formulas of : H2O Na2SO4

Calculating the Percent Composition of a Compound

The percent composition of a compound has as many percent values as there are different elements in a compound

grams of element A grams of compound

Example: An 8.20-g piece of magnesium combines completely with 5.40 g of oxygen to form a compound. What is the percent composition of this compound?

pg. 305

Percent Composition- percent by mass of each element of a compound

X 100% mass of element A=

Using Percent as a Conversion Factor

You can use percent composition to calculate the number of grams of an element contained in a specific amount of a compound.

Example: calculate the mass of a carbon in 82.0 g of propane (C3H8)

Calculating Empirical Formulas

Empirical Formula- gives the lowest whole-number ratio of atoms (of the element) in a compound.

Empirical formula for H2O2 is HO Empirical formula for CO2 is CO2

Examples-Find the Empirical Formula if given:

--6 gr of Magnesium and 4 gr of Oxygen--79.8 gr of Carbon and 20.2 gr of Hydrogen --67.6 gr.Mercury (Hg), 10.8 gr. Sulfur, 21.6 gr.

Oxygenpg.

309

Molecular Formulas Molecular Compounds- gives the actual

ratio of atoms in a compound. This can be calculated once you have found

the Empirical formula. Example: CO2 Same molecular & Empirical C6H12O6 = molecular CH2O = empirical Example: Calculate the molecular formula

of the compound whose molar mass is 60.0 g and empirical formula is CH4N

Calculating Molecular Formulas

Steps to Solve: Find the Formula Mass of the Empirical

Formula

Divide total grams givenFormula Mass of Empirical

Multiply this answer by the Empirical Formula, this will equal the Molecular Formula

STOICHIOMETRY

USING THE REACTION EQUATION LIKE A

RECIPE

USING EQUATIONS Nearly everything we use

is manufactured from chemicals.Soaps, shampoos, conditioners, cd’s, cosmetics, medications, and clothes.

For a manufacturer to make a profit the cost of making any of these items can’t be more than the money paid for them.

USING EQUATIONS Chemical processes carried

out in industry must be economical, this is where balanced equations help.

Equations are a chemist’s recipe.Eqs tell chemists what amounts of reactants to mix and what amounts of products to expect.

USING EQUATIONS When you know the

quantity of one substance in a reaction, you can calculate the quantity of any other substance consumed or created in the reaction.

Quantity meaning the amount of a substance in grams, liters, molecules, or moles.

USING EQUATIONS The calculation of

quantities in chemical reactions is called stoichiometry.

When you bake cookies you probably use a recipe.

A cookie recipe tells you the amounts of ingredients to mix together to make a certain number of cookies.

If you need a larger number of cookies than the yield of the recipe, the amounts of ingredients can be doubled or tripled.

In a way, a cookie recipe provides the same kind of information that a balanced chemical equ. providesIngredients are the reactantsCookies are the products.

Imagine you are in charge of manufacturing for Rugged Rider Bicycle Company.

The business plan for Rugged Rider requires production of 128 custom-made bikes each day.

One of your responsibilities is to be sure that there are enough parts available at the start of each day.

Assume that the major components of the bike are the frame (F), the seat (S), the wheels (W), the handlebars (H), and the pedals (P).

The finished bike has a “formula” of FSW2HP2.

The balanced equation for the production of 1 bike is.F +S+2W+H+2P FSW2HP2

Now in a 5 day workweek, Rugged Riders is scheduled to make 640 bikes. How many wheels should be in the plant on Monday morning to make these bikes?

What do we know?Number of bikes = 640 bikes

1 FSW2HP2=2W (balanced eqn)

What is unknown?# of wheels = ? wheels

The connection between wheels and bikes is 2 wheels per bike. We can use this information as a conversion factor to do the calculation.640 FSW2HP2

1 FSW2HP2

2 W

= 1280 wheels

We can derive the same kind of information from a normal chemical reaction equation.

For instance the synthesis reaction of ammonia:N2(g) + 3H2(g) 2NH3(g)

• What kinds of information can we glean from this equation?

• Well for starters…

PARTICLES SUMMARY 1 molecule of N2 reacts with 3

molecules of H2 to produce 2 molecules of ammonia.

N2 and H2 will always react to form ammonia in this 1:3:2 ratio of molecules.

So if you started with 10 molecules of N2 it would take 30 molecules of H2 and would produce 20 molecules of NH3

PARTICLES SUMMARY It isn’t possible to count such

small numbers of molecules and allow them to react.

You could react Avogadro’s number of N2 molecules and make them react with 3 times Avogadro’s number of H2 molecules forming 2 times Avogadro’s number of NH3 molecules.

N2 + 3H2 2NH3

2 atoms N + 6 atoms H 2 atoms N & 6 atoms H

1 molecule N2 + 3 molecule H2 2 molecule NH3

10molecules N2 + 30molecules H2 20 molecules

NH3

6.02X1023

molecules N2

+6.02X1023

molecules H2

6.02X1023

molecules NH3

1X 3X 2X

MOLES SUMMARY We have recently learned

that Avogadro’s number of particles is the same as a mole of a substance.

On the basis of the particle interpretation we just discussed, the equation also tells you the number of moles of reactants and products.

MOLES SUMMARY 1 mole of N2 molecules

reacts with 3 moles of H2 molecules to make 2 moles of NH3 molecules.

The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical reaction.

MOLES SUMMARY This is the most important

information that a reaction equation provides.

Using this information, you can calculate the amounts of reactants and products.

MASS SUMMARY A balanced chemical equation

must also obey the law of conservation of mass.Mass can be neither created nor destroyed in ordinary chemical or physical processes.

Remember that mass is related to the number of atoms in a compound through the mole.

MASS SUMMARY The mass of 1 mol of N2

molecules is 28 g; the mass of 3 mols of H2 molecules is 6 g for a total mass of reactnts of 34 g.

The mass of 2 moles of NH3 molecules is 2 * 17g or 34 g.

As you can see the reactants mass is equal to the mass of the products.

VOLUME SUMMARY Remember that 1 mole of

any gas at STP occupies 22.4 L of space.

It follows that 22.4 L of N2 reacts with 67.2 L of H2 to form 44.8 L of Ammonia gas.

1 mol N2 + 3 mol N2 2 mol NH3

28 g N2 + 3 (2 g H2) 2 (17 g NH3)

34 g reactants 34 g products

+

22.4 L N2 67.2 L H2 44.8 L NH3

22.4 L

22.4 L

22.4 L

22.4 L

22.4 L

22.4 L

N2 + 3H2 2NH3

USING EQUATIONS You can see how much

information is stored in a simple balanced reaction eqn

We can combine this information with our knowledge of mole conversions to perform important common stoichiometric calculations.

MOLE – MOLE CALCULATIONS

A balanced rxn eqn is essential for all calculations involving amounts of reactants and products.

If you know the number of moles of 1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn eqn.

MOLE – MOLE CALCULATIONS Let’s go back to our synthesis

of ammonia rxn.

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

• The MOST important interpretation of this rxn is that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

MOLE – MOLE CALCULATIONS These connections of the

coefficients allows us to set up conversion factors called mole ratios.

The mole ratios are used to calculate the connections in moles of compounds in our reaction equation.

We can start calculating…

MOLE – MOLE CALCULATIONS Sample Mole – Mole

problem:How many moles of ammonia are produced

when .60 moles of N2 are reacted with H2?

Given: .60 moles of N2

Uknown: ____ moles of NH3

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

MOLE – MOLE CALCULATIONS According to the reaction

equation, for every 1 mole of N2 reacted we form 2 mols of NH3.

To determine the number of moles of NH3, the given quantity of N2 is multiplied by the mole ratio from the rxn eqn in such a way that the units of “mol N2” cancel

MOLE – MOLE CALCULATIONS Solve for the unknown:

.6 mol N21 mol N2

2 mol NH3

= 1.2 mol NH3

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

MOLE – MOLE EXAMPLE #2 This equation shows the

formation of aluminum oxide.4Al(s) + 3O2(g) 2Al2O3(s)• How many moles of aluminum are needed to form 3.7 mol Al2O3?

Given: 3.7 moles of Al2O3

Uknown: ____ moles of Al

MOLE – MOLE EXAMPLE 2 This equation shows the

formation of aluminum oxide.4Al(s) + 3O2(g) 2Al2O3(s)• How many moles of aluminum are needed to form 3.7 mol Al2O3?

Given: 3.7 moles of Al2O3

Uknown: ____ moles of Al

MOLE – MOLE CALCULATIONS Solve for the unknown:

3.7 mol Al2O32 mol Al2O3

4 mol Al

= 7.4 mol Al

Coefficients in the

balanced equation

Coefficients in the

balanced equation

4Al(s) + 3O2(g) 2Al2O3(s)

MASS – MASS CALCULATIONS No lab balance measures

moles directly, instead the mass of a substance is usually measured in grams.

From the mass of a reactant or product, the mass of any other reactant or product in a given chemical equation can be calculated.

MASS – MASS CALCULATIONS The mole – mole connection

is still vital to do these calcs.1. If the given sample is

measured in grams, the mass can be converted to moles by using the molar mass.

2. Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown.

MASS – MASS CALCULATIONS3.If it is mass of the

unknown that needs to be determined, the number of moles of the unknown can be multiplied by the molar mass of the desired compound.

As in mole-mole calcs, the unknow can be either a reactant or a product.

MASS – MASS CALCULATIONS Again back to our synthesis

of ammonia rxn.

N2(g) + 3H2(g) 2NH3(g)• Calculate the number of grams of NH3 produced by the reaction of 5.4 g of H2 with an excess of N2.

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

MASS – MASS CALCULAT’NS What do we know?

Mass of H2 = 5.4 g H23 mol H2 = 2 mol NH3 (from balanced equation - AKA mole ratio)

Molar mass of H2 = 2.0 g H2Molar mass of NH3=17.0g NH3

What are we asked for?Mass of ammonia produced

MASS – MASS CALCULAT’NS

Step 1: convert mass of given to moles of given using MM of G

5.4 g H22.0 g H2

1 mol H2

= 2.7 mol H2

Coefficients in the

balanced equation

Molar mass of

H2

MASS – MASS CALCULAT’NS

Step 2: convert mols of G to mols of U using the mole ratio

2.7 mol H23 mol H2

2 mol NH3

= 1.8 mol NH3

Coefficients in the

balanced equation

Coefficients from the balanced equation

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

MASS – MASS CALCULAT’NS

Step 3: convert moles of desired compound to mass using MM of

U

1.8 mol NH31 mol NH3

17.0 g NH3

= 31.0 g NH3

Coefficients in the

balanced equation

Molar mass of

NH3

MASS – MASS CALCULAT’NS Mass to mass calculations

always follow those same three steps.

It uses the mole math that we have had lots of practice with (mass to moles and moles to mass)

The only difference is the new middle step where we use our newly acquired mole ratio

MASS – MASS CALCULAT’NS

Let’s do another one:Acetylene gas (C2H2) is

produced by adding water to calcium carbide (CaC2).

CaC2 + 2H2O C2H2 + Ca(OH)2

How many grams of acetylene are produced by

adding water to 5.00 g CaC2?

CaC2 + 2H2O C2H2 + Ca(OH)2

MASS – MASS CALCULAT’NS

What do we know?Mass of CaC2 = 5.0 g CaC21 mol CaC2 = 1 mol C2H2 (from balanced equation)

MM of CaC2 = 64.0 g CaC2MM of C2H2 = 26.0g C2H2

What are we asked for?Mass of C2H2 produced

MASS – MASS CALCULAT’NS

Step 1: convert mass of given to moles of given using MM of G

5.0 g CaC264.0 g CaC2

1 mol CaC2

= .078mol CaC2

MASS – MASS CALCULAT’NS

Step 2: convert mols of G to mols of U using the mole ratio

.078mol CaC21 mol CaC2

1 mol C2H2

= .078mol C2H2

CaC2 + 2H2O C2H2 + Ca(OH)2

MASS – MASS CALCULAT’NS

Step 3: convert moles of desired compound to mass

using MM of U

.078 mol C2H2

1 mol C2H2

26.0 g C2H2

= 2.03 g C2H2

MORE MOLE CALCULAT’NS

A balanced reaction equation indicates the relative number of moles of reactants and products.

We can expand our stoichiometric calculations to include any unit of measure-ment that is related to the mole.

MORE MOLE CALCULAT’NS

The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP.

The problems can include mass-volume, volume-volume, and particle-mass calculations.

MORE MOLE CALCULAT’NS

In any of these problems, the given quantity is first converted to moles.

Then the mole ratio from the balanced eqn is used to convert from the moles of given to the number of moles of the unknown

MORE MOLE CALCULAT’NS

Then the moles of the unknown are converted to the units that the problem requests.

The next slide summarizes these steps for all typical stoichiometric problems

MORE MOLE EXAMPLES

How many molecules of O2 are produced when a

sample of 29.2 g of H2O is decomposed by electrolysis according to this balanced

equation:2H2O 2H2 + O2

MORE MOLE EXAMPLES What do we know?

Mass of H2O = 29.2 g H2O 2 mol H2O = 1 mol O2 (from balanced equation)

MM of H2O = 18.0 g H2O 1 mol O2 = 6.02X1023

molecules of O2 What are we asked for?

molecules of O2

Step 1: convert mass of given to moles of given using MM of G

29.2 g H2O18.0 g H2O

1 mol H2O

= 1.62 mol H2O

Step 2: convert mols of G to mols of U using the mole ratio

1.62 mol H2O2 mol H2O

1 mol O2

= .811 mol O2

2H2O 2H2 + O2

Step 3: convert moles of unknown compound to desired

units

.811 mol O21 mol O2

6.02X1023

= 4.88X1023 molecules O2

The last step in the production of nitric acid is the reaction of NO2 with

H2O.3NO2+H2O2HNO3+NO

How many liters of NO2 must react with water to

produce 5.00X1022 molecules of NO?

MORE MOLE EXAMPLES

MORE MOLE EXAMPLES What do we know?

Molecs NO=5.0X1022 molecs NO

1 mol NO = 3 mol NO2 (from balanced equation)

1 mol NO=6.02X1023 molecs NO

1 mol NO2 = 22.4 L NO2 What are we asked for?

Liters of NO2

Step 1: convert molecules of given to moles of given

using Avogadro’s number.

5.0X1022 molecs NO

5.0X1022 molecs NO

6.02X1023

molecs

1 mol NO

= .083 mol NO

Step 2: convert mols of G to mols of U using the mole ratio

.083 mol NO1 mol NO

3 mol NO2

= .249 mol NO2

3NO2+H2O2HNO3+NO

Step 3: convert moles of unknown compound to desired

units

.249 mol NO21 mol NO2

22.4 L NO2

=5.58 L of NO2

Assuming STP, how many liters of oxygen are needed

to produce 19.8 L SO3 according to this balanced

equation?2SO2+O22SO3

MORE MOLE EXAMPLES

MORE MOLE EXAMPLES What do we know?

2 mol SO3 = 1 mol O2 (from balanced equation)

1 mol SO3 = 22.4 L SO31 mol O2 = 22.4 L O2 Volume of SO3 = 19.8 L

What are we asked for? volume of oxygen

Step 1: convert volume of G to moles of G

19.8 L SO322.4 L SO3

1 mol SO3

= .884 mol SO3

Step 2: convert mols of G to mols of U using the mole ratio

.884 mol SO32 mol SO3

1 mol O2

= .442 mol O2

2SO2+O22SO3

MORE MOLE EXAMPLESStep 3: convert moles of

unknown compound to desired units

.442 mol O21 mol O2

22.4 L O2

=9.9 L of O2