The Mathematics of Chemical Equations Chapter 11

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STOICHIOMETRYaka USING THE REACTION EQUATION

LIKE A RECIPE

Nearly everything we use is manufactured from chemicals.Soaps, shampoos, conditioners,

cd’s, cosmetics, medications, clothes, etc.

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.

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

USING EQUATIONS

Equations are a chemist’s recipe.Equations tell chemists what amounts

of reactants to mix and what amounts of products to expect.

When you know the quantity of one substance in a rxn, you can calculate the quantity of any other substance consumed or created in the rxn.Quantity meaning the amount of a

substance in grams, liters, molecules, or moles.

USING EQUATIONS

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

22.22.4 L4 L

22.22.4 L4 L

22.22.4 L4 L

22.22.4 L4 L

22.22.4 L4 L

22.22.4 L4 L

N2 + 3H2 2NH3

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 more cookies than the yield of the recipe, the amounts of ingredients can be doubled or tripled.

USING EQUATIONS

Chocolate Chip Cookies!!1 cup butter

1/2 cup white sugar

1 cup packed brown sugar

1 teaspoon vanilla extract

2 eggs

2 1/2 cups all-purpose flour

1 teaspoon baking soda

1 teaspoon salt

2 cups semisweet chocolate chips

Makes 3 dozen

How many eggs are needed to make 3 dozen cookies?

How much butter is needed for the amount of chocolate chips used?

How many eggs would we need to make 9 dozen cookies?

How much brown sugar would I need if I had 1 ½ cups white sugar?

Cookies and Chemistry…Huh!?!?

Just like chocolate chip cookies have recipes, chemists have recipes as well

Instead of calling them recipes, we call them chemical equations

Furthermore, instead of using cups and teaspoons, we use moles

Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds & elements as ingredients

Chemistry Recipes

An equation tells us how much of something you need to react with something else to get a product (like the cookie recipe)

Be sure you have a balanced reaction before you start!

Example: 2 Na + Cl2 2 NaCl This reaction tells us that by mixing 2 moles of

sodium with 1 mole of chlorine we will get 2 moles of sodium chloride

What if we wanted 4 moles of NaCl? 10 moles? 50 moles?

Conversion Factors Conversion Factors from a Chemical from a Chemical

EquationEquation2 C6H6 (l) + 15 O2 (g) 12 CO2 (g) + 6 H2O (g)

In this equation there are In this equation there are 22 molecules of molecules of benzene reacting with benzene reacting with 1515 molecules of oxygen molecules of oxygen to produce to produce 1212 molecules of carbon dioxide molecules of carbon dioxide and and 66 molecules of water molecules of water.

This equation could also be read as This equation could also be read as 22 moles moles of benzene reacts with of benzene reacts with 1515 moles of oxygen to moles of oxygen to produce produce 1212 moles of carbon dioxide and moles of carbon dioxide and 66 moles of water.moles of water.

Since 6.02 x 1023 is the common factor between the relationship between the actual number of molecules and the number of moles present we can use the ratio of the # moles of one substance in the equation to find the # of moles another substance in the equation.

This ratio, # moles of unknown

# moles of given

is known as the MOLE RATIO

Conversion Factors from a Conversion Factors from a Chemical EquationChemical Equation

2 C6H6 (l) + 15 O2 (g) 12 CO2 (g) + 6 H2O (g)

The The MOLE RATIO,MOLE RATIO, from the balanced equation, for oxygen and carbon from the balanced equation, for oxygen and carbon dioxide is 15 : 12. dioxide is 15 : 12. It can be written as:It can be written as:

12 moles CO2 or as 15 moles of O2

15 moles O2 12 moles of CO2

NOTE: The MOLE RATIO is used for converting moles of one substance into moles of another substance. The numbers that go in front of “moles of given” & “moles of unknown” are the coefficients from the balanced equation!

IT IS ALWAYS MOLES OVER MOLES!

Conversion Factors from a Conversion Factors from a Chemical EquationChemical Equation

Moles of unknownMoles of Given

What was that again?!

# Moles Unknown

# Moles Given

The #’s in the numerator & denominator MUST come from the balanced chemical equation

I I Stoichiometry !!!Stoichiometry !!!

Consider: 4NHConsider: 4NH33 + 5O + 5O22 6H 6H22O + 4NOO + 4NO

Is this equation balanced?Yes!!You must always start stoich. problems with a balanced equation!

Write all of the possible mole ratios that can be formed from this equation. YES! NOW!!!

4 moles NH3

5 moles O2

5 moles O2

4 moles NH3

5 moles O2

6 moles H2O6 moles H2O5 moles O2

4 moles NH3

6 moles H2O6 moles H2O4 moles NH3

6 moles H2O4 moles NO

4 moles NO6 moles H2O

ETC.

Using the coefficients of balanced equations and our knowledge of mole conversions we can perform powerful calculations! A.K.A. stoichiometry.

A balanced equation 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 equation.

I I Stoichiometry !!! Stoichiometry !!!

Practice

Write the balanced reaction for hydrogen gas reacting with oxygen gas.

2 H2 + O2 2 H2O How many moles of reactants are needed? What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much hydrogen

would we need to react and how much water would we get?

What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced?

Mole Ratios

These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical

Example: How many moles of chlorine is needed to react with 5 moles of sodium (without any sodium left over)?

2 Na + Cl2 2 NaCl

5 moles Na 1 mol Cl2

2 mol Na= 2.5 moles Cl2

MOLE – MOLE EXAMPLE

The following rxn shows the synthesis of aluminum oxide.

3O2(g) + 4Al(s) 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

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

Solve for the unknown:

3.7 mol Al2O3 2 mol Al2O3

4 mol Al

= 7.4 mol Al

MOLE – MOLE EXAMPLE

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

Mole Ratio

Consider : 4NH3 + 5O2 6H2O + 4NOHow many moles of H2O are produced if

0.176 mol of O2 are used?

How many moles of NO are produced in the reaction if 17 mol of H2O are also produced?

6 mol H2O 5 mol O2

x

Mole - Mole Problems

# mol H2O= 0.176 mol O2 0.211 mol H2O

=

4 mol NO 6 mol H2O

x # mol NO= 17 mol H2O

11 mol NO

=

Mole-Mole Conversions

How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal?

Mole-Mass Conversions

Most of the time in chemistry, the amounts are given in grams instead of moles

We still go through moles and use the mole ratio, but now we also use molar mass to get to grams

Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride?

2 Na + Cl2 2 NaCl

5.00 moles Na 1 mol Cl2 71.0g Cl2

2 mol Na 1 mol Cl2 = 178g Cl2

Consider : 4NH3 + 5O2 6H2O + 4NOHow many grams of H2O are produced if 1.90

mol of NH3 are combined with excess oxygen?

How many grams of O2 are required to produce 0.3 mol of H2O?

6 mol H2O 4 mol NH3

x

Mole-Mass Problems

1.90 mol NH3 51.3 g H2O

= 18.0 g H2O 1 mol H2O

x

5 mol O2 6 mol H2O

x 0.3 mol H2O8 g O2=

32.0 g O2 1 mol O2

x

Practice

Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum.

Mass-Mole We can also start with mass and convert to

moles of product or another reactant We use molar mass and the mole ratio to get

to moles of the compound of interest Calculate the number of moles of ethane (C2H6)

needed to produce 10.0 g of water 2 C2H6 + 7 O2 4 CO2 + 6 H20

10.0 g H2O 1 mol H2O 2 mol C2H6

18.0 g H2O 6 mol H20

= 0.185 mol C2H6

Practice

Calculate how many moles of oxygen are required to make 10.0 g of aluminum oxide

Mass-Mass Conversions

Most often we are given a starting mass and want to find out the mass of a product we will get (called theoretical yield) or how much of another reactant we need to completely react with it (no leftover ingredients!)

Now we must go from grams to moles, mole ratio, and back to grams of compound we are interested in

Mass-Mass Conversion

Ex. Calculate how many grams of ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.

N2 + 3 H2 2 NH3

2.00g N2 1 mol N2 2 mol NH3 17.0g NH3

28.0g N2 1 mol N2 1 mol NH3

= 2.43 g NH3

Consider : 4NH3 + 5O2 6H2O + 4NOHow many grams of NO is produced if 12 g of

O2 is combined with excess ammonia?

4 mol NO 5 mol O2

x

Mass – Mass Problems

12 g O2

9.0 g NO=

30.0 g NO 1 mol NO

x 1 mol O2 32.0 g O2

x

Practice

How many grams of calcium nitride are produced when 2.00 g of calcium reacts with an excess of nitrogen?

Volume – Volume Problems

Ex. What volume of hydrogen gas will be needed to produce 256.7 L of ammonia gas (NH3) at STP? N2 + 3 H2 2 NH3

256.7 L NH3 22.4 L NH3

1 mole NH3

mole NH3

mole H2

2

31 mole H2

22.4 L H2

Mole Ratio= 385.1 L H2

Limiting Reagents

Caution: this stuff is difficult to follow at first.

Be patient.

Limiting Reactants Available IngredientsAvailable Ingredients

4 slices of bread 1 jar of peanut butter 1/2 jar of jelly

Limiting ReactantLimiting Reactant bread

Excess ReactantsExcess Reactants peanut butter and jelly

Grilled Cheese Sandwich

Bread + Cheese ‘Cheese Melt’

2 B + C B2C

100 bread 30 slices ? sandwiches

Limiting Reactant: Cookies1 cup butter

1/2 cup white sugar

1 cup packed brown sugar

1 teaspoon vanilla extract

2 eggs

2 1/2 cups all-purpose flour

1 teaspoon baking soda

1 teaspoon salt

2 cups semisweet chocolate chips

Makes 3 dozen

If we had the specified amount of all ingredients listed, could we make 4 dozen cookies?

What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies?

What if we only had one egg, could we make 3 dozen cookies?

Limiting Reactant

Most of the time in chemistry we have more of one reactant than we need to completely use up another reactant.

That reactant is said to be in excess (there is too much).

The other reactant limits how much product we get. Once it runs out, the reaction s. This is called the limiting reactant.

Limiting Reactants

Limiting ReactantLimiting Reactant used up in a reaction determines the amount of product

Excess ReactantExcess Reactant More than enough to react with the limiting

reagent – some left over! added to ensure that the other reactant is

completely used up cheaper & easier to recycle

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants

1. Write a balanced equation.

2. For each reactant, calculate the

amount of product formed.

3. Smaller answer indicates: limiting reactant, and amount of product formed

Real-World Stoichiometry:Limiting Reactants

LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 366

IdealStoichiometry

LimitingReactants

Real-World Stoichiometry:Limiting Reactants

LeMay Jr, Beall, Robblee, Brower, Chemistry Connections to Our Changing World , 1996, page 366

IdealStoichiometry

LimitingReactants

Fe + S FeS

S =

Fe =

Limiting Reactants

aluminum + chlorine gas aluminum chloride

Al(s) + Cl2(g) AlCl3

2 Al(s) + 3 Cl2(g) 2 AlCl3

100 g 100 g ? g

A. 200 g B. 125 g C. 667 g D. ???

Limiting Reactant To find the correct answer, we have to try all of

the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one.

The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it!

Remember! You can’t compare to see which is greater and which is lower unless the product is the same!

Limiting Reactant: Example

10.0g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting

2 Al + 3 Cl2 2 AlCl3 Start with Al:

Now Cl2:

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3

27.0 g Al 2 mol Al 1 mol AlCl3

= 49.4g AlCl3

35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3

71.0 g Cl2 3 mol Cl2 1 mol AlCl3

= 43.9g AlCl3

LimitingLimitingReactantReactant

LR Example Continued

We get 49.4g of aluminum chloride from the given amount of aluminum, but only 43.9g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction comes to a complete .

Limiting Reactant Practice

15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made.

Finding the Amount of Excess

By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

Can we find the amount of excess potassium in the previous problem?

Finding Excess Practice 15.0 g of potassium reacts with 15.0 g of iodine.

2 K + I2 2 KI We found that Iodine is the limiting reactant, and

19.6 g of potassium iodide are produced.

15.0 g I2 1 mol I2 2 mol K 39.1 g K

254 g I2 1 mol I2 1 mol K= 4.62 g K USED!

15.0 g K – 4.62 g K = 10.38 g K EXCESS

Given amount of excess reactant

Amount of excess reactant actually used

Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

Excess Reactant

2 Na + Cl2 2 NaCl50 g 50 g x g

“Have”

/ 23 g/mol / 71 g/mol

2.17 mol 0.70 mol

“Need” 1.40 mol

EXCESSEXCESS LIMITING

1.40 mol

x 58.5 g/mol

81.9 g NaCl

1 : 2coefficients

Excess Reactant (continued)

2 Na + Cl2 2 NaCl50 g 50 g x g81.9 g NaCl

All the chlorine is used up…

81.9 g NaCl-50.0 g Cl2

31.9 g Na is consumed in reaction.

How much Na is unreacted?

50.0 g - 31.9 g = 18.1 g Natotal used “excess”

Conservation of Mass is Obeyed

2 Na + Cl2 2 NaCl50 g 50 g x g81.9 g NaCl

2 Na + Cl2 2 NaCl50 g 50 g x g81.9 g NaCl

31.9 g + 18.1 g

18.1 g

+ Na

100 g reactant

100 g product

Limiting Reactant: Recap

1. You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT.

2. Convert ALL of the reactants to the SAME product (pick any product you choose.)

3. The reactant that gave you the lowest answer is the LIMITING REACTANT.

4. The other reactant is in EXCESS.5. To find the amount of excess left over, subtract the amount

used from the given amount.6. If you have to find more than one product, be sure to start

with the limiting reactant. You don’t have to determine which is the LR over and over again!

Percent Yield

100yield ltheoretica

yield actualyield %

calculated on paper

measured in lab

% yield

Experimental yield

Theoretical yieldX 100 = % yield

Sample problemQ - What is the % yield of H2O if 138 g H2O is

produced from 16 g H2 and excess O2?

Step 1: write the balanced chemical equation2H2 + O2 2H2O

Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:

2 mol H2O 2 mol H2

x # g H2O= 16 g H2 143 g= 18.02 g H2O1 mol H2O

x 1 mol H2

2.02 g H2

x

Step 3: Calculate % yield138 g H2O 143 g H2O

= % yield = x 100% 96.7%= actualtheoretical x 100%

Practice problemQ - What is the % yield of NH3 if 40.5 g NH3 is

produced from 20.0 mol H2 and excess N2?

Step 1: write the balanced chemical equationN2 + 3H2 2NH3

Step 2: determine actual and theoretical yield. Actual is given, theoretical is calculated:

2 mol NH3 3 mol H2

x # g NH3= 20.0 mol H2 227 g= 17.04 g NH3

1 mol NH3

x

Step 3: Calculate % yield40.5 g NH3 227 g NH3

= % yield = x 100% 17.8%= actualtheoretical x 100%

Percent Yield

When 45.8 g of K2CO3 react with excess HCl,

46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl.

K2CO3 + 2HCl 2KCl + H2O + CO2

45.8 g ? g

actual: 46.3 g

Percent Yield

45.8 gK2CO3

1 molK2CO3

138.21 gK2CO3

= 49.4g KCl

2 molKCl

1 molK2CO3

74.55g KCl

1 molKCl

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g ? g

actual: 46.3 g

Theoretical Yield:

Percent Yield

Theoretical Yield = 49.4 g KCl

% Yield =46.3 g

49.4 g 100 = 93.7%

K2CO3 + 2HCl 2KCl + H2O + CO2 45.8 g 49.4 g

actual: 46.3 g

Challenging question

2H2 + O2 2H2O

What is the % yield of H2O if 58 g H2O are produced by combining 60 g O2 and 7.0 g H2?

Hint: determine limiting reagent first

2 mol H2O 2 mol H2

x # g H2O= 7.0 g H2 62.4 g= 18.02 g H2O1 mol H2O

x 1 mol H2

2.02 g H2

x

58 g H2O 62.4 g H2O

= % yield = x 100% 92.9%= actualtheoretical x 100%

2 mol H2O 1 mol O2

x # g H2O= 60 g O2 68 g= 18.02 g H2O1 mol H2O

x 1 mol O2

32 g O2

x

Air Bag Design

Exact quantity of nitrogen gas must be produced in an instant.

Use a catalyst to speed up the reaction

2 NaN3(s) 2 Na(s) + 3 N2(g)

6 Na(s) + Fe2O3(s) 3 Na2O(s) + 2 Fe (s)

Airbag Design

2 NaN3(s) 2 Na(s) + 3 N2(g)

6 Na(s) + Fe2O3(s) 3 Na2O(s) + 2 Fe(s)

Assume that 65.1 L of N2 gas are needed to inflate an air bag to the proper size. How many grams of NaN3 must be included in the gas generant to generate this amount of N2? (Hint: The density of N2 gas at this temperature is about 0.916 g/L).

How much Fe2O3 must be added to the gas generant for this amount of NaN3?

65.1 L N2

x 0.916 g/L N2

59.6 g N2

X g NaN3 = 59.6 g N2 (1 mol N2) (2 mol NaN3) (65 g NaN3)(28 g N2) (3 mol N2) (1 mol NaN3)

X = 92.2 g NaN3

X g Fe2O3 = 92.2 g NaN3 (1 mol NaN3) (2 mol Na) (1 mol Fe2O3) (159.6 g Fe2O3)

(65 g NaN3) (2 mol NaN3) (6 mol Na) (1 mol Fe2O3)

X = 37.7 g Fe2O3

Water from a Camel

Camels store the fat tristearin (C57H110O6) in the hump. As well as being a source of energy, the fat is a source of water, because when it is used the reaction

takes place. What mass of water can be made from 1.0 kg of fat?

X g H2O = 1 kg ‘fat” (1000 g ‘fat’) (1 mol “fat”) (110 mol H2O) (18 g H2O)

(1 kg ‘fat’) (890 g ‘fat’) (2 mol ‘fat’) (1 mol H2O)

X = 1112 g H2O

or 1.112 liters water

2 C57H110O6(s) + 163 O2(g) 114 CO2(g) + 110 H2O(l)

Rocket Fuel

The compound diborane (B2H6) was at one time considered for use as a rocket fuel. How many grams of liquid oxygen would a rocket have to carry to burn 10 kg of diborane completely? (The products are B2O3 and H2O).

B2H6 + 3 O2 B2O3 + 3 H2O

B2H6 + O2 B2O3 + H2OChemical equation

Balanced chemical equation

X g O2 = 10 kg B2H6 (1000 g B2H6) (1 mol B2H6) (3 mol O2) (32 g O2)

(1 kg B2H6) (28 g B2H6) (1 mol B2H6) (1 mol O2)

X = 34,286 g O2

10 kg X g

Water in Space

In the space shuttle, the CO2 that the crew exhales is removed from the air by a reaction within canisters of lithium hydroxide. On average, each astronaut exhales about 20.0 mol of CO2 daily. What volume of water will be produced when this amount of CO2 reacts with an excess of LiOH? (Hint: The density of water is about 1.00 g/mL.)

CO2(g) + 2 LiOH(s) Li2CO3(aq) + H2O(l)

X mL H2O = 20.0 mol CO2 (1 mol H2O) (18 g H2O) (1 mL H2O)(1 mol CO2) (1 mol H2O) (1 g H2O)

excess20.0 mol x g

X = 360 mL H2O

Real Life Problem Solving

Determine the amount of LiOH required for a seven-day mission in spacefor three astronauts and one ‘happy’ chimpanzee. Assume each passengerexpels 20 mol of CO2 per day.

(4 passengers) x (10 days) x (20 mol/day) = 800 mol CO2

Plan for a delay

CO2(g) + 2 LiOH(s) Li2CO3(aq) + H2O(l)

800 mol X g

Note: The lithium hydroxide scrubbers are only 85% efficient.

CO2(g) + 2 LiOH(s) Li2CO3(aq) + H2O(l)

800 mol X g

X g LiOH = 800 mol CO2 (2 mol LiOH) (1 mol CO2 )

(23.9 g LiOH)( 1 mol LiOH)

= 38,240 g LiOH

Needed(actual yield)

% Yield = Actual YieldTheoretical Yield

0.85 = 38,240 g LiOH X g LiOH

X = 44,988 g LiOH

800 mol 1600 mol1:2

x 23.9 g/mol

Note: The lithium hydroxide scrubbers are only 85% efficient.

Amount of LiOH tobe taken into space

Limiting Reactants

79.1 g of zinc react with 0.90 L of 2.5M HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP?

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

Limiting Reactants

79.1g Zn

1 molZn

65.39g Zn

= 27.1 L H2

1 molH2

1 molZn

22.4 LH2

1 molH2

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants

22.4L H2

1 molH2

0.90L

2.5 molHCl

1 L= 25 L

H2

1 molH2

2 molHCl

Zn + 2HCl ZnCl2 + H2 79.1 g ? L0.90 L

2.5M

Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

Limiting Reactants

Zn: 27.1 L H2 HCl: 25 L H2

Limiting reactant: HCl

Excess reactant: Zn

Product Formed: 25 L H2

left over zincCourtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

% error

Experimental yield – Theoretical yield

Theoretical yieldX 100 = % error

What YOU got in the experiment

What you predicted you would get (use stoichiometry)

Review: Molar Mass of Compounds

The molar mass (MM)- add up all the atomic masses for the molecule (or compound) Ex. Molar mass of CaCl2 Avg. Atomic mass of Calcium = 40.1g Avg. Atomic mass of Chlorine = 35.5g Molar Mass of calcium chloride =

40.1 g/mol Ca + (2 X 35. 5) g/mol Cl 111.1 g/mol CaCl2

20

Ca  40.08 17

Cl 35.45

Review: Flowchart

Atoms or Molecules

Moles

Mass (grams)

Divide by 6.02 X 1023

Multiply by 6.02 X 1023

Multiply by atomic/molar mass from periodic table

Divide by atomic/molar mass from periodic table