Stoichiometry

Chapters 2, 3, 4

Significant Figures - A.K.A. Sig. Fig.’s

In almost every experiment, scientists measure data. These measured numbers help determine many things in life - as such, we need a way to recognize the error in each measurement.

• We do this by using significant figures.• Numerically 3.0, 3.00, 3.000 are of the same value,

but 3.000 shows that it was measured with the more precise instrument.

• The zeroes in all three numbers are considered "significant figures".

• They are shown to indicate the precision of the measurements.

• If we take away the zeroes, the value does not change.

Types of Numbers

Exact NumbersApproximate Numbers

• numbers obtained through counting.

• These stand as is.

• numbers obtained through measurement.

• With these numbers we must determine how precise they are when we use them in calculations.

Determining Sig. Figs

• All non-zero integers are significant.

• Captive zeros are always significant.

• Leading zeros are NEVER significant.

• Trailing zeros in a decimal are significant.

• Assumed decimals are not significant. Ex. 500

4 sig. figs

3 sig. figs

2 sig. figs

4 sig. figs

1 sig. figs

Ex. 256.5

Ex. 206

Ex. 0.0056

Ex. 25.00

Rules For Calculations:Addition and Subtraction Multiplication and Division

• Do the math. • The answer should

contain the same number of decimal places as the number with the least decimal places.

Ex. 25.46 + 12.5 + 82.91 =

only one dec. place

120.9

• Do the math. • The answer should

contain the same number of Sig. Figs as the number with the least number of Sig. Figs

Ex. 0.14 x 98.54 x 12.5 =

only 2 sig. fig

170

120.87 172.445

Chapter 2.1 Isotopes and Average Atomic

MassIsotopes

Recall that the mass number tells the total number of protons and neutrons.

• All neutral atoms have the same number of protons and electrons, however, the number of neutrons may vary.

• Isotopes - atoms of an element that have the same number of protons but different number of neutrons.

atoms of the same element do not necessarily have the same mass number.

Ex. Oxygen naturally occurs in 3 different forms:

Carbon – exists as 3 isotopes...carbon-12 carbon-13 carbon-14

6 p+, 6n

6p+, 7n 6p+, 8n

The Relative Mass of An Atom

• The mass of an atom is expressed in atomic mass units, u.

•Atomic masses are relative - all are based

on the mass of carbon-12, 12 u.

the mass of all other elements are defined in comparison to carbon-12.

Isotopic Abundance• An element may have several naturally occurring

isotopes, but not all of its isotopes will exist in the same amount.

Ex. Magnesium exists in 3 isotopes – Mg-24, Mg-25, and Mg-26

• All sources of magnesium occur in the following amounts:

Mg-24 79%, Mg-25 10%, Mg-26 11%

• Isotopic Abundance - the relative amount in which each isotope is present in an element.

• Determined with a sophisticated machine called a mass spectrometer. It uses a magnetic field to separate the different masses in a sample.

Complete #1-4 Page 45

Average Atomic Mass And the Periodic Table

• Average atomic mass - the average of all the masses of all the element’s isotopes – accounting for the abundance of all the isotopes.

• the average atomic mass is the mass given in the Periodic Table.

To Calculate the Average Atomic Mass

 • 1. Multiply the mass

of each isotope by the relative abundance (you must change the abundance to a decimal).

• 2. Add all these totals together.

Average Atomic Mass = 106.9 u (.518) + 108.9 u(.482) = 107.9 u  Note: This is the mass number given on the Periodic Table. No atom of silver actually has a mass of 107.9 u - it is just an average.

The Mole

• chemists need a unit of measure that works for many things – they use the mole.

Mole•A quantity, an amount.• symbol – mol

▫1 mol element= 6.02 x 1023 atoms▫ 1 mol compound = 6.02 x 1023 molecules▫ 1 mol gas = 22.4 L @STP▫ 1 mol = mass of substance

2.3 Molar Mass• the mass of one mole of a substance.• expressed in g/mol.Molar Mass of an Element• a mass expressed in grams that is

numerically equivalent to the element’s average atomic mass.

• 1 mol Na = 22.99 g• 1 mol Ar = 39.95 g

Molar Mass Of Compounds

• the mass of one mole of a compound is equal to the molar mass of all the elements it contains.

• 1 mole MgO = mass Mg + mass of O = 24.31 g + 16.00 g

= 40.31 g• 1 mole Ca3(PO4)2

= 3(Ca) + 2(P) + 8(O) = 3(40.08) + 2(30.97) + 8(16.00) = 310.18 g• 1 mole Ca3(PO4)2 = 310.18 g or molar mass of Ca3(PO4)2 = 310.18 g/mol

Conversion Flashback

• When 2 things are equal – we can set up conversion factors...

1 hour = 60 minutes

or

How many minutes are in 8.49 hours?8.49 hours x

1

60min

hour 60min

1hour

1

60min

hour 60min

1hour= 509 min

Converting Mass to Moles• When given the mass of a compound, we can

convert to moles using the molar mass.Steps:1. Determine molar mass.2. Start with what is given. 3. Set up factors so they cancel.

How many moles are in 43.38 grams of NaOH?Molar Mass NaOH = 40.00 g/mol

43.38 g NaOH x 1

40.00

molNaOH

g

= 1.085 mol

Converting Moles to Mass• When given the amount of moles of a compound,

we can convert to mass using the molar mass.

Steps:1. Determine molar mass.2. Start with what is given. 3. Set up factors so they cancel.

How many grams are in 5.48 moles of NaOH?Molar Mass NaOH = 40.00 g/mol

5.48 mol NaOH x 40.00

1

g

molNaOH = 219 g

Converting Moles To Particles• 1 mole contains 6.02 x 1023 particles

When talking about particles we must look at the substance to see what the particles are.

An Element – smallest particle is the atom. 1 mole of an element = 6.02 x 1023 atoms

A Compound • smallest particle for a covalent compound is a molecule.

H2O - one molecule of H2O - also 2 atoms of H and 1 atom of O

- or 3 atoms• smallest particle for an ionic compound is the formula unit.

NH4Cl – one form. unit of NH4Cl

- also one ion of NH4+ and one ion of Cl-

- also 1 N atom, 4 H atoms, and 1 Cl atom - or 6 atoms

Conversion Factors for Elements

• if 1 mole = 6.02 x 1023 atoms then

OR

How many moles are present in a sample of carbon that is made up of 5.82 x 1024 atoms of C.

5.82 x 1024 atoms x

23

1

6.02 10

mole

x atoms

236.02 10

1

x atoms

mole

23

1

6.02 10

moleC

x atoms= 9.67 mol of C

Conversions for Compounds

Covalent Compounds

• 1 mole = 6.02x1023 molecules

OR

Then look at one molecule and count:

• how many atoms are in one molecule

• how many atoms of a certain type.

Example: C3H8

1 mol C3H8 = 6.02x1023 molecules

1 molecule C3H8 = 11 atoms1 molecule C3H8 = 3 carbon atoms1 molecule C3H8 = 8 hydrogen

atoms

23

1

6.02 10

mole

x molecules

236.02 10

1

x molecules

mole

Conversions for Compounds

Ionic Compounds

• 1 mole = 6.02x1023 formula units

OR

Then look at one formula unit and count:

• how many ions are in one molecule• how many ions of a certain type• how many atoms are in one molecule • how many atoms of a certain type.

23

1

6.02 10 . .

mole

x FU

236.02 10 . .

1

x F U

mole

Example: Na2SO4

1 mol Na2SO4 = 6.02x1023 F.U.

1 F.U. Na2SO4 = 3 ions1 F.U. Na2SO4 = 2 Na+ ions1 F.U. Na2SO4 = 1 SO4

2- ions

1 F.U. Na2SO4 = 2 Na atoms1 F.U. Na2SO4 = 1 S atom1 F.U. Na2SO4 = 4 O atoms

The Mole Road Map

grams moles

Molecules or Formula Units

Atoms or Ions

Volume

1 mole = molar mass

1 mole = 6.02 x 1023 molecules

1 mole = what you

count

For Elements

For Compounds

1 mol=6.02x1023 atomsSTP 1

mol=22.4LSATP 1

mol=24.4L

3.1 Percent Composition• The Law of Definite Proportions:

Elements in a chemical compound are always present in the same proportions by mass.

• the mass of an element in a compound, expressed as a percent of the total mass of a compound, is the element’s mass percent.

Whether water, H2O, comes from a lake, a bottle, or a fountain, it will contain 11.2 % hydrogen and 88.8% oxygen.

Mass Percent =

Mass % Oxygen =

massofelement

massofcompound

16.00

18.02

g

gx 100% 88.8%

x 100%

Percent Composition• a statement of all the mass percents of every element in a compound.

• Vanillin is C8H8O3

▫ 63.1% C, 5.3% H, 31.6% O

Practice Problems #1-4 Page 82

Page 81

Percent Composition From a Formula

• When you are not given masses in a problem, you can still determine the percent composition.

• Assume you have one mole of the sample• Use molar masses and masses from PT to get

percents. Page 84

Practice Problems #5-8 Page 85

Empirical Formula of a Compound• The empirical formula (also simplest formula) of a

compound shows the lowest whole number ratio of the elements in a compound.

• The molecular formula (also the actual formula) shows the number of atoms of each element in a molecule.

• Many compounds can have molecular formulas that are the same as their empirical formulas.

Ex. NH3, H2O, CO2 etc.

 • Note: Ionic compounds do not have molecular formulas

because they do not exist as molecules. Because of the ions involved in the compound, there is only one way that they can combine - therefore the chemical formula is the empirical formula.

Write the Empirical Formula for Each of the Following:a. P4O6 b. C6H9

c. CH2OHCH2OH d. BrCl2e. C6H8O6 f. C10H22

g. Cu2C2O4 h. Hg2F2

Determining the Empirical Formula “Percent to mass,

mass to mole, divide by small, multiply until whole”

Page 88

Page 90

Practice Problems #9-12 Page 89Practice Problems #13-16 Page 91

1. A compound composed of: 72% iron (Fe) and 27.6% oxygen (O) by mass.

Fe3O4

2. A compound composed of: 9.93% carbon (C), 58.6% chlorine (Cl), and 31.4% fluorine (F).

CCl2F2

3. A compound composed of: .556g carbon (C) and .0933g hydrogen (H).

CH2

  

3.3 Molecular Formula

Name Molecular Formula

Multiple

Information

Formaldehyde CH2O 1 disinfectant

Acetic Acid C2H4O2 2 Vinegar

Lactic Acid C3H6O3 3 in muscles during exercise

Erythrose C4H8O4 4 formed from metabolizing sugar

Ribose C5H10O5 5 in nucleic acid and Vit.B

Glucose C6H12O6 6 sugar

• the molecular formula is a whole number ratio of the empirical formula

Determining Molecular Formula

1. Determine the empirical formula.2. molar mass of molecular formula

molar mass of empirical formula3. Multiply the ratio by the empirical formula.

= ratio

Page 97

1. NutraSweet is 57.14% C, 6.16% H, 9.52% N, and 27.18% O. Calculate the empirical formula of NutraSweet and find the molecular formula. (The molar mass of NutraSweet is 294.30 g/mol)

C14H18N2O5

2. An oxide of nitrogen, NxOy, contains 30.43% N. Its molecular formula is determined to be 138 g/mol. What is the molecular formula?

N3O6

3. A butane is a common fuel for lighters. When analyzed the compound if found to contain 82.63% carbon and 17.37% hydrogen. The molar mass of butane is 58.14 g/mol. What is the molecular formula?

C4H10Practice Problems # 17-20 Page 97

Hydrated Ionic Compounds

Many ionic compounds crystallize from a water solutionwith water molecules incorporated into their crystal structure, forming a hydrate.

• Hydrate - a compound with a specific number of water molecules chemically bonded to each formula unit.

Ex. MgSO4 7H2O

• every formula unit of MgSO4 has 7 water molecules chemically bonded to it.

• a raised dot in a chemical formula denotes a hydrated compound. ▫ It represents a weak bond between the ionic

compound and the water molecules.

When a hydrate is heated – we destroy the weak bonds between the ionic compound and evaporate the water molecules. This will form an anhydrous compound.

Anhydrous Compound or Anhydrate• a compound that has no water molecules incorporated into their chemical formula.  CaSO4 - anhydrous calcium sulfateCaSO4 2H2O - calcium sulfate dihydrate

Calculations Involving Hydrates

“Mass to Moles, Divide by small.”

•Sample Problem - page 102• A hydrate of barium hydroxide, Ba(OH)2

∙xH2O, is used to make barium salts and to prepare certain organic compounds. Since it reacts with CO2 from the air to yield BaCO3, it must be tightly stored.

• A) A 50.0 g sample of the hydrate contains 27.2 g of Ba(OH)2. Calculate the percent, by mass, of water.

• B) Find the formula of the hydrate.

• Percent By Mass• Ba(OH)2 = 27.2 g• water = 50.0g - 27.2 g = 22.8 g • % water = 22.8 g water x 100%• 50.0 g compound• = 45.6 %

• Formula• 27.2 g Ba(OH)2 x 1 mole Ba(OH)2 = 0.159 mol Ba(OH)2 =

1• 171.3 g Ba(OH)2 0.159

• 22.8 g H2O x 1 mole H2O = 1.27 mol H2O = 8

• 18.02 g H2O 0.159

• Therefore: Ba(OH)2 ∙8H2O

Stoichiometry Continued...• Balanced chemical equations are EXTREMELY

important in chemistry.

• The coefficients in front of the formulas and symbols for the compounds and elements tell you the ratio of particles involved in a chemical reaction.

• The coefficients also tell us the ratio of the number of molecules in the equation.

• The coefficients tell us the number of moles involved in the chemical equation.

the mole ratio or the mole-mole relationship.

Page 118

Mole-Mole Conversions

• when starting with one compound and making calculations about another compound you need to use the mole-mole relationship in conversion factors...

Common Everyday Example:

2 eggs + 75 mL oil + 125 mL water → 24 muffins

If you had 14 eggs, how many muffins could you make?

14 eggs x = 168 muffins24 muffins 2 eggs

Consider the following chemical rxn:

How many moles of ammonia, NH3, is formed from 13.3 moles of nitrogen gas reacts with excess hydrogen?

13.3 mol N2 x

2 mol NH3

1 mol N2

= 26.6 mol NH3

How many moles of hydrogen are needed to react completely with 0.589 moles of nitrogen gas?

0.589 mol N2 x 3 mol H2

1 mol N2

= 1.77 mol H2

Practice Problems #4-6 Page 115

Page 116

Practice Problems # 8-10 Page 117

Mass To Mass Conversions• We can also convert from the mass of one compound

to the mass of another compound through mole-mole relationships...

Consider the following reaction:

1 Co(NO3)2 + 2 NaOH → 2 NaNO3 + 1 Co(OH)2

How many grams of Co(OH)2 will be produced when 25.47 g of NaOH reacts with excess Co(OH)2?

25.47 g NaOH x 1 mol NaOH 40.00 g NaOH

x 1 mol Co(OH)2

2 mol NaOHx 92.95 g Co(OH)2

1 mol Co(OH)2= 29.59 g Co(OH)2

Mass to mass conversion

•P. 121 #11-14

•** In #14 you are also required to find the volume of a gas at STP. 22.4 L/mol will be needed for the conversion.

Mixed Stoichiometry

•P. 122 #15-18

•P. 125 # 19-22

These problems require conversions between mass, moles, particles, and volume to achieve the correct answer. (mole road map)

**All your conversion skills are used here.

The Limiting Reagent

•Suppose you place 13.65 g of Iron in a beaker with 8.50 g of copper (II) chloride. Copper metal and iron (II) chloride are produced.

A)Determine the limiting reactant in this reaction.

B) Determine the mass of copper that should be produced (theoretical yield).

Theoretical Yield & Percent Yield• Stoichiometry can be used to predict the amount of

product that could be expected if an experiment went perfectly to completion.

theoretical yield

Reasons why we do not ever achieve the theoretical yield

competing reactions

faulty or poor equipment

impure reactants

poor technique

• When a chemists actually does an experiment, they can measure the amount of product that they actually produced.

This is the Actual Yield.

• The Actual Yield is usually less than the theoretical yield.

% Yield = Actual Yield

Theoretical Yield

x 100%

Ex. When 2.3 g of NaOH reacts with excess Co(NO3)2 a precipitate is formed. A chemists completes this reaction and produces 1.74 g of Co(OH)2. Calculate the percent yield. 2.3 g NaOH x 1 mole NaOH x 1 mol Co(OH)2 x 92.95 g Co(OH)2 = 2.67 g 40.0 g NaOH 2 mol NaOH 1 mol Co(OH)2

 

% Yield = actual yield = 1.74 g x 100% = 65.2 % theoretical yield 2.67 g