The study of the The study of the quantitativequantitative relationships relationships between reactants and products in a reactionbetween reactants and products in a reaction It is used to answer questions like; If I have this much It is used to answer questions like; If I have this much

reactant, how much product can I make?reactant, how much product can I make? If I want this much product, how much reactant do I If I want this much product, how much reactant do I

need?need? These questions have real life application, These questions have real life application,

particularly in manufacturingparticularly in manufacturing. . It allows us to convert the mass of a substance It allows us to convert the mass of a substance

to the number of particles (atoms, ions or to the number of particles (atoms, ions or molecules) it contains.molecules) it contains.

These numbers can be really large, so they are These numbers can be really large, so they are counted in groupscounted in groups

Much like when we count a lot of pennies we stack them Much like when we count a lot of pennies we stack them in 10’s and count by 10in 10’s and count by 10

Atoms are very tiny, so small Atoms are very tiny, so small that the grouping we use to count that the grouping we use to count them must be very largethem must be very large MOLE;MOLE; the group (unit of measure) used to the group (unit of measure) used to

count atoms, molecules, formula units or count atoms, molecules, formula units or ions of a substanceions of a substance

1 mole of a substance has a particular 1 mole of a substance has a particular number of particles in it! number of particles in it! Much like 1 dozen always means 12; whether it is Much like 1 dozen always means 12; whether it is 12 eggs 12 eggs 12 oranges or 12 12 oranges or 12

gold barsgold bars

The number of particles The number of particles in a mole in a mole = 6.02 x = 6.02 x 10 10 2323 or or 602,000,000,000,000,000,000,602,000,000,000,000,000,000,000 !000 !

This is known as This is known as Avogadro’s NumberAvogadro’s Number

Using this, We can easily count the number Using this, We can easily count the number of particles in all kinds of things !of particles in all kinds of things !

There are 6.02 x 10 There are 6.02 x 10 23 23 Carbon atoms Carbon atoms in a mole of in a mole of carbon carbon

There are 6.02 x 10 There are 6.02 x 10 2323 COCO22 molecules molecules in a mole of COin a mole of CO2 2

There are 6.02 x 10 There are 6.02 x 10 2323 sodium ions sodium ions in a mole of sodium in a mole of sodium There are 6.02 x 10 There are 6.02 x 10 2323 marblesmarbles in a mole of marbles in a mole of marbles

That’s a lot of marbles!That’s a lot of marbles!

The The Size of a mole of a substance changesSize of a mole of a substance changes, the bigger , the bigger the substance the more space a mole of the the substance the more space a mole of the substance takes up, substance takes up, but the number of particles in a but the number of particles in a mole is always the samemole is always the same! !

Chemicals do not come bundled in moles, like a Chemicals do not come bundled in moles, like a dozen eggs comes in a 1 dozen or 1 ½ dozen dozen eggs comes in a 1 dozen or 1 ½ dozen package so we use the mole as a grouping unit.package so we use the mole as a grouping unit.

The mass of 1 mole of a pure substance called The mass of 1 mole of a pure substance called it’s it’s molar massmolar mass

If I want to produce 500g of methanol using If I want to produce 500g of methanol using the following equation, CO2 +3H2 the following equation, CO2 +3H2 CH3OH + CH3OH + H20 how many grams of CO2 and H2 do I H20 how many grams of CO2 and H2 do I need?need?

These are the questions stoichiometry These are the questions stoichiometry answersanswers!

If I want to produce 500g of methanol using If I want to produce 500g of methanol using the following equation;the following equation;

COCO22 +3H +3H22 CH CH33OH + HOH + H220 0 How many grams of COHow many grams of CO22 and H and H22 do I need? do I need?

This equation relates the molecules of reactants This equation relates the molecules of reactants and products, NOT THEIR MASSES!and products, NOT THEIR MASSES! 1 molecule of CO1 molecule of CO22 and 3 molecules of H and 3 molecules of H2 2 will make 1 will make 1

molecule of CHmolecule of CH33OHOHWe need to relate the We need to relate the massesmasses to the to the number of number of

moleculesmolecules..

Remember; Remember; The average The average atomic masses of the elements are atomic masses of the elements are found on the Periodic Table!found on the Periodic Table!

We can use the atomic masses on We can use the atomic masses on the PT to relate the mass of the the PT to relate the mass of the compound to the mass of a mole!compound to the mass of a mole!

Molar mass: The mass (in grams)of one mole of a molecule or a formula unit

Molecular mass: mass in atomic mass units of just one molecule

Formula Mass: mass in atomic mass units of one formula unit of an ionic compound

StepsSteps1.1. Find the average Atomic Mass of the Find the average Atomic Mass of the

element on the PT. (state it in grams instead element on the PT. (state it in grams instead of atomic units)of atomic units)a)a) Example: molar mass of Fe = 55.847 gExample: molar mass of Fe = 55.847 gb)b) Example: molar mass of Pt = 195.08 gExample: molar mass of Pt = 195.08 g

2.2. If the element is a molecule, count the number If the element is a molecule, count the number of atoms in the molecule then multiply the of atoms in the molecule then multiply the atomic mass by the number of atoms.atomic mass by the number of atoms.a)a) Example: OExample: O22, the mass of O =16.0g There are 2 , the mass of O =16.0g There are 2

atoms of O in the Oatoms of O in the O22 molecule , 2 atoms X molecule , 2 atoms X 16.0g = 32.00g is the molar mass of the 16.0g = 32.00g is the molar mass of the molecule.molecule.

Calculate the molar mass of each of Calculate the molar mass of each of the following:the following:

1.1. NN22

2.2. ClCl223.3. BrBr22

4.4. II22

5.5. HH22

6.6. FF22

Calculate the molar mass of each of the Calculate the molar mass of each of the following:following:

1.1. NN22 = 14.007g X 2 =28.014 g/mol = 14.007g X 2 =28.014 g/mol2.2. ClCl22 = 35.453g X 2 =70.906 g/mol = 35.453g X 2 =70.906 g/mol3.3. BrBr22 = 79.904g X 2 =159.808 g/mol = 79.904g X 2 =159.808 g/mol 4.4. II22 = 126.904g X 2 =253.808 g/mol = 126.904g X 2 =253.808 g/mol5.5. HH22 = 1.008g X 2 =2.016 g/mol = 1.008g X 2 =2.016 g/mol6.6. FF22 = 18.998g X 2 =37.996 g/mol = 18.998g X 2 =37.996 g/mol

StepsSteps1. Count the number and type of atoms2. Find the Atomic Mass of each atom type,

on the periodic table. Write it in grams.3. Multiply the mass times the # of Atoms.

Then add the totals

1. Count the number and type of atomsEthanol (C2H5OH)

2. Find the Atomic Mass of each atom type, on the periodic table. Write it in grams.

3. Multiply The mass X the # of Atoms. Then add the totals.

Atom type Amount of each atomC 2H 6O 1

Atom type Amount of atom

Ave. Atomic Mass in g

C 2 12.0H 6 1.00O 1 16.0

Atom type

Amount of atom

Ave. Atomic Mass in g

Total

C 2 12.0 =24.0H 6 1.00 =6.0O 1 16.0 =16.0

Molar Mass Of Ethanol (C2H5OH) = 46.0g/mole

Atom Types

Amount of Atoms

Ave. Atomic

Mass in g

Total

Ca 1 40.1 40.1Cl 2 35.5 71.0

Mass of 1 mol of CaCl2 (molar mass) 111.1 g/mole

Example: Calcium Chloride (CaCl2 )

What is the molar mass of each of the following?

1. Fe2 O3 2. H2O3. CO2

4. NaCl5. NH3

6. BaI2

Fe2 O3 = 55.85g X 2= 111.7 g 16.0g X 3 = 48.0g = 159.7 g/mol_______________________________________________ H2O = 1.01g X 2 = 2.02 16.0g X 1 = 16.0 = 18.02 g/mol_______________________________________________ CO2 = 12.01g X 1 = 12.01

16.0g X 2 = 32.0 = 44.01 g/mol________________________________________________NaCl = 22.99 gX1 = 22.99 35.45g X1 = 35.45 = 58.44 g/mol________________________________________________NH3 =14.01g X 1 = 14.01 1.01g X 3 = 3.03 = 17.04 g/mol________________________________________________BaI2 = 137.33g X 1 = 137.33 126.90g X 2 = 253.80 = 391.13 g/mol

If I want to produce 500g of ethanol using If I want to produce 500g of ethanol using the following equation;the following equation;

6CO6CO22 +17H +17H22 3C 3C22HH55OH + 9HOH + 9H220 0 How many grams of COHow many grams of CO22 and H and H22 do I need? do I need?

The Molar Mass Of Ethanol (C2H5OH)= 46.0g/mole Now we need to find the number of atoms

in the sample. How many molecules of ethanol are in 500g?

Steps to finding the number of atoms in a given mass of a sample

1. Use PT to find the molar mass of the substance2. Convert the mass of the substance to

number of moles in the sample (convert using mass of one mole as conversion factor)

3. Use the number of atoms in a mole to find the number of atoms in the sample

4. Solve and check answer by canceling out units

The mass of an iron bar is 16.8g. How many iron(Fe) atoms are in the sample?Step 1: Use PT to find the molar mass of the substance : The molar mass of Fe =55.8g/moleStep 2: Convert the given mass of the substance to number of moles in the sample: Fe =55.8g/mole

(16.8g Fe) (1 mol Fe) (6.022 X 1023 Fe atoms)= 1.81 X 10

23 Fe atoms

(55.8g Fe) (1 mol Fe)Step 3: Use the number of atoms in a mole to find the number of atoms in the sample = 1.18 X 1023

1. 25.0 g silicon, Si

2. 1.29 g chromium, Cr

(25.0 g Si ) ( 1 mol Si ) (6.02 X 1023 Si atoms ) 1 28.1g Si 1 mol Si = 5.36 X1023 atoms Si

(1.29 g Cr ) ( 1 mol Cr ) (6.02 X 1023 Cr atoms ) 1 52.0g Cr 1 mol Cr= 1.49 X1022 atoms Cr

1. 98.3g mercury, Hg2. 45.6g gold, Au3. 10.7g lithium, Li4. 144.6g tungsten, W

1. (98.3 g Hg ) ( 1 mol Hg )(6.02 X 1023 Hg atoms) 1 200.6g Hg 1 mol Hg = 2.95 X1023 atoms Hg

2. (45.6 g Au ) ( 1 mol Au )(6.02 X 1023 Au atoms) 1 197.0g Au 1 mol Au = 1.39 X1023 atoms Au

3. (10.7 g Li ) ( 1 mol Li )(6.02 X 1023 Li atoms) 1 6.94g Li 1 mol Li = 9.28 X1023 atoms Li

4. (144.6 g W ) ( 1 mol W )(6.02 X 1023 W atoms) 1 183.8g W 1 mol W = 4.738 X1023 atoms W

Steps1. Use the PT to calculate the molar mass of

one formula unit2. Convert the given mass of the compound

to the number of molecules in the sample (use the molar mass as the conversion factor)

3. Multiply the moles of the compound by the number of the formula units in a mole (Avagadro’s number) and solve

4. Check by evaluating the units

1. Calculate the molar mass (Fe2O3) 2 Fe atoms 2X 55.8 = 111.6 3 O atoms 3 X 16.0 = +48.0

molar mass 159.6 g/mol(given mass X 1 mole per molar mass X Form Units per 1

mole)

(16.8 g Fe2O3 ) ( 1 mol Fe2O3 )(6.02 X 1023 Fe2O3 Formula units) 1 159.6g Fe2O3 1 mol Fe2O3 = 6.34 X1022 Fe2O3 Formula units

1. 89.0g sodium oxide (Na2O)

2. 10.8g boron triflouride ( BF3)

1. 89.0g sodium oxide (Na2O)Calculate the molar mass (Na2O) 2 Na atoms 2X 23.0 = 46.0 1 O atoms 1 X 16.0 = +16.0

molar mass 62.0 g/mol

(given mass X 1 mole per molar mass X molecules per 1 mole)

(89.0 g Na2O ) ( 1 mol Na2O )(6.02 X 1023 Na2O Form Units) 1 62.0g Na2O 1 mol Na2O = 8.64 X1023 Na2O Formula units

2. 10.8g boron trifloride ( BF3)Calculate the molar mass (Na2O) 1 B atom 1X 10.8 = 10.8 3 F atoms 3 X 19.0 = +57.0

molar mass 67.8 g/mol

( given mass X 1 mole per molar mass X molecules per 1 mole)

(10.8 g BF3 ) ( 1 mol BF3 )(6.02 X 1023 BF3 Form units) 1 67.8g BF3 1 mol BF3 = 9.59 X1022 BF3 Formula units

Steps1. Determine the molar mass2. Change given mass to moles by

using molar mass as the conversion factor. (1/molar mass)

Calculate the number of moles in 6.84g sucrose (C12H22O11)

12 C atoms 12 X 12.0 = 144.0 22 H atoms 22 X 1.0 = 22.0 11 O atoms 11 X 16.0 = +176.0 molar mass 342.0 g/mol

(given mass/1) X (1 mole/ molar mass)

(6.84 g sucrose ) ( 1 mol sucrose ) 1 342.0g sucrose

= 2.0 X10-02 moles of sucrose

1. 16.0g sulfur dioxide, SO2 2. 68.0g ammonia, NH3 3. 17.5g copper(II) oxide, CuO

1. 16.0g sulfur dioxide, SO2 (16.0g/1) (1mole/64.1g ) = 0.250 mol SO2

2. 68.0g ammonia, NH3

( 68.0g/1) (1 mole/ 17.0g) = 4.00 mol NH3

3. 17.5g copper(II) oxide, CuO ( 17.5g/1) (1 mole/ 79.1g) = 0.22 mol CuO

Steps:1. Find the molar mass of the compound2. Use the molar mass to convert the given

number of moles to a mass (moles) X (g/mol)

3. Solve4. Check using dimensional analysis (make

sure units cancel and leaves only grams)

1. Find the molar mass of the compound (H2O) H - 2 atoms – 1.0 = 2.0 O - 1 atom - 16.0 = 16.0 18.0 g/mol 2. Use the molar mass to convert the given number of

moles to a mass (moles) X (g/mol) (7.5 mol H2O) ( 18.0 g H2O) ( 1 mol H2O)

2. Solve : 7.5 X 18.0g H2O = 135 g H2O3. Check using dimensional analysis (make sure units

cancel and leaves only grams) “mol H2O” cancel each other out, units are correct!

1. 3.52 mol Si2. 1.25 mol aspirin, C9H8O4 3. 0.550 mol F2

4. 2.35 mol Barium Iodide, BaI2

(moles) X (g/mol)1. What mass of Si = 3.52 mol Si (3.52 mol Si) (28.1g Si) = 98.9g Si 1 (1 mole Si)

2. What mass of C9H8O4 = 1.25 mol aspirin, C9H8O4 C -9 atoms – 12.0 – 108.0H- 8 atoms – 1.0 - 8.0O – 4 atoms – 16.0 - 64.0

180.0g/mol

(1.25 mol C9H8O4) (180.0g C9H8O4) = 225.0g C9H8O4

1 (1 mole C9H8O4)

3. What mass of F2 = 0.550 mol F2

F- 2 atoms – 19.0 = 38.0 g/mol (0.550 mol F2 ) (38.0 g F2) = 20.9g F2

1 (1 mole F2)

4. What mass of BaI2 = 2.35 mol Barium Iodide, BaI2Ba-1 atom – 137.3 - 137.3I – 2 atoms – 126.9 - 253.8

391.1g/mol

(2.35 mol BaI2) (391.1g BaI2) = 919.1g BaI2 1 (1 mole BaI2)

Know:1. What stoichiometry is2. What a mole is3. How to calculate molar mass of an element and of a compound1. How to determine the number of atoms or

formula units in a given mass of sample 2. How to determine the number of moles in a

given mass of a sample3. How to determine the mass of a given molar

quantity

Review of Calculation Rules To Find molar mass (g/mol)

(atomic mass of each atom) X (amount of each atom) Then add together mass of all atoms(g/mol)

To Find the # atoms in a given mass

(given mass) X (1mole) /(molar mass (g)) X (# atoms) /(1 mole)

To Find the # moles in a given mass

(given mass) X (1mole)/(molar mass (g)) X (#atom)/(1mole)

To Find the mass (g) of a given molar quantity

(#moles) X (grams/1 mole) (from molar mass)

Balanced chemical equations relate moles of reactants to moles of products Just like when baking, reactants have to be mixed in the

proper proportions to make a certain amount of the desired product

Specific amounts of reactants produce specific amounts of product

We can use balanced chemical equations and moles to PREDICT the masses of reactants or products

When one of the reactants in a reaction are used up, the reaction stops. The reactant that is used up is called a limiting reagent

You can not move directly from the mass of one substance to the mass of the second

Steps1. Write a balanced equation2. Convert the given mass to moles first!2. The coefficients of balanced reactions tell

you the NUMBER OF MOLES of each chemical in the reactant. (these are used as the conversion factor)

3. Once you know the number of moles of any reactant or product, use the coefficients in the equation to convert the moles of the other reactants and products to mass

Ammonia gas is synthesized from nitrogen gas and hydrogen gas according to the balanced equation : N2 + 3H2 2NH3

How many grams of hydrogen gas are required for 3.75g of nitrogen gas to react completely? What mass of ammonia is formed?

Reactants and products are related in terms of moles

The amount of H2 needed depends on the moles of N2 present in 3.75g and the ratio of moles of H2 to moles of N2 in the equation.

The amount of ammonia formed depends on the ratio of moles N2 to moles of ammonia

How many grams of Hydrogen are required for 3.75g of nitrogen to react completely? What mass of Ammonia is formed?N2 + 3H2 2NH3Convert the given mass to moles Find the # of moles of N2 using molar mass(3.75g N2) (1 mol N2) (28.0 g N2)To find the mass of H2 needed : The coefficients of the balanced equation shows 3 mol of H2 react with 1 mole of N2 . Multiply moles of N2 by this ratio.

2.(3.75g N2) (1 mol N2) ( 3 mol H2) (28.0 g N2) (1 mol N2)Once you know the number of moles of any reactant H2 use the coefficients in the equation to convert the moles of the other reactants and productsTo find the mass of hydrogen, multiply the moles of H2 by the mass of 1 mole of H2.(3.75g N2) (1 mol N2) ( 3 mol H2) ( 2.0g H2) =3.75 X1 X3 X2.0=0.084 g H2

(28.0 g N2) (1 mol N2) (1 mol H2) 28.0

To find the mass of ammonia produced:1. Use the mole ratio of ammonia molecules to

nitrogen molecules to find the moles of ammonia formed.

(3.75g N2) (1 mol N2) ( 2 mol NH3) (28.0 g N2) (1 mol N2)

2. Use the molar mass of ammonia, 17.0g to find the mass of ammonia formed.

(3.75g N2) (1 mol N2) ( 2 mol NH3) (17.0g NH3) = 3.75 X1 X2 X17.0=4.55gNH3

(28.0 g N2) (1 mol N2) (1 mol NH3) 28.0

When potassium chlorate (KClO3) is heated, it decomposes to form potassium chloride and oxygen gas 2KClO3 2KCl + 3O2

a) How many grams of KCl are formed when 28.0g of KClO3 decompose?

(28.0g KClO3) (1 mol KClO3) ( 1mol KCl) (74.6g KCl) = 17.0gKCl (122.6 g KClO3) (1 mol KClO3) (1 mol KCl)

b) Use the mass of KCl you determined in part a to calculate the mass of oxygen gas produced.

(17.0g KCl) (1 mol KCl) ( 3 mol O2) (32g O2) = 10.9g O2

(74.6 g KCl) (2 mol KCl) (1 mol O2)

1. The combustion of propane (C3H8), a fuel used in backyard grills, produces carbon dioxide and water vapor. C3H8 + 5O2 3CO2 + 4H2O

What mass of carbon dioxide forms when 95.6g of propane burns?

2. Solid xenon hexafluoride is prepared by allowing xenon gas and fluorine gas to react.

Xe + 3F2 XeF6 How many grams of fluorine are required to

produce 10.0 g of XeF6? 3. Using the previous reaction, how many grams of

xenon are required to produce 10.0g of XeF6 ?

1. 287 g CO2

2. 4.65 g F2

3. 5.35 g Xe

Avogadro’s principle states that equal volumes of gasses at the same temperature and pressure contain equal numbers of moles of gasses.

The molar volume of a gas is the volume that a gas occupies at 1 atmosphere( 101 kPa, or 760 mm Hg) of pressure and a temp of 0.0° C or 273°K (STP).

At STP, the volume of 1 mole of any gas is 22.4L, (there masses may be different ).

Molar volume is often used in calculations, BUT BE SURE YOU ARE AT STP!

In the space shuttle, exhaled CO2 is removed from the air by passing it through canisters of lithium hydroxide. The following reaction takes place;

CO2 + 2LiOH Li2CO3 + H2OHow many grams of lithium hydroxide are required to

remove 500.0L of carbon dioxide gas at 101 kPa pressure and 25.0°C?

Use the molar volume to find the number of moles.

FIRST you must convert the volume of the gas at 25.0°C to a volume at standard temperature!

V = (500.0L CO2) (273 K) = 458 L CO2 (298 K)

Now find the number of moles of CO2(458 L CO2) ( 1 mol CO2 ) (22.4 L CO2 ) the molar ratio of LiOH to CO2 is 2 to 1 (from the

equation) Determine the number of moles of LiOH(458 L CO2) ( 1 mol CO2 ) ( 2 mol LiOH) (22.4 L CO2 ) ( 1 mol CO2) To Convert the number of moles of LiOH to

mass, use the molar mass, 23.9g/mol.(458 L CO2) ( 1 mol CO2 ) ( 2 mol LiOH) (23.9g LiOH) = 977 g LiOH (22.4 L CO2 ) ( 1 mol CO2) (1 mol LiOH)

What mass of glucose (C6H12O6) must be broken down in your body to produce 2.50 L of CO2 at 273°C and 1 atmosphere?

C6H12O6 + 6O2 6H2O + 6 CO2

(2.50L CO2) (1 mol CO2) (1 mol C6H12O6) (180g C6H12O6) = 3.35 g C6H12O6

( 22.4 L CO2) ( 6 mol CO2) (1 mol C6H12O6 )

When steel wool burns in air, this reaction occurs:

4Fe + 3O2 2 Fe2O3 What volume of oxygen, measured at 725 mm

Hg (760 mm Hg is 1 atm) and 25.0°C, is required to react with 100.0g of iron?

(100.g Fe) (1 mol Fe) ( 3 mol O2) (22.4 L ) (298K) (760 mm Hg) = 34.5 L O2

(55.8 g Fe) (4 mol Fe) (1mol O2) (273 K) (725 mm Hg)

1. What mass of sulfur must burn to produce 3.42 L of SO2 at 273°C and 101 kPa?

The reaction is S + O2 SO2 Answer: 2.45g S2. What volume of hydrogen gas can be

produced by reacting 4.20 g of sodium in excess water at 50.0°C and 106 kPa?

The reaction is 2Na + 2H2O 2NaOH + H2Answer:2.31 L H2

Review of Calculation and Rules To Predict the mass of Reactants and products

(grams) X (moles/gram) X (mole ratio reactants to products) X (moles/ mass )

Molar Volume

The molar volume of a gas is the volume that a gas occupies at 1 atmosphere( 101 kPa, or 760 mm Hg) of pressure and a temp of 0.0° C or 273°K (STP).

To Determine the # of grams of a reactant to make a volume of gas

Convert gas volume to volume at STP(gas volume at STP)X(1mol/22.4L) X(molar ratio) X (molar mass (g/mol)) = grams