StoichiometryThis word sounds scary, but it’s really just a big word for the relationships between reactants and products in a chemical reaction.

-based on law of conservation of mass

mass reactants= mass products

Proportional Relationships

I have 5 eggs. Assuming I have plenty of the other ingredients, how many cookies can I make?

3/4 c. brown sugar1 tsp vanilla extract2 eggs2 c. chocolate chipsMakes 5 dozen cookies.

2 1/4 c. flour1 tsp. baking soda1 tsp. salt1 c. butter3/4 c. sugar

5 eggs5 dozen

cookies

2 eggs= 12.5 dozen cookies

Ratio of eggs to cookies

Proportional Relationships

• StoichiometryStoichiometry– mass relationships between substances in a chemical

reaction– based on the mole ratio

• Mole RatioMole Ratio– indicated by coefficients in a balanced equation

2 Mg + O2 Mg + O22 2 MgO 2 MgO

Mole Ratios

2 Mg + O2 Mg + O22 2 MgO 2 MgO• Relationships between coeffecients are

used as conversion factors in stoichiometry!

• A mole ratio is a ratio between the numbers of moles of any 2 substances.

Mole Ratios

2 Mg + O2 Mg + O22 2 MgO 2 MgO• What are the mole ratios in this equation?

2 molMg1 mol O2

2 molMg2 mol MgO

1 mol O2

2 mol Mg1 mol O2

2 mol MgO

Total of 6 mole ratios!

2 mol MgO2 mol Mg

2 mol MgO1 mol O2

Mole Ratios

2KClO2KClO33→2KCl + 3O→2KCl + 3O22

• What are the mole ratios in this equation?2 molKClO3

2 mol KCl2 molKClO33 mol O2

2 mol KCl2 mol KClO3

2 mol KCl3 mol O2

3 mol O2

2 mol KClO3

3 mol O2

2 mol KCl

Total of 6 mole ratios!

Stoichiometry Steps1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.

– Mole ratio - moles moles– Molar mass - moles grams– Molarity - moles liters soln– Molar volume - moles liters gas

Core step in all stoichiometry problems!!

– Mole ratio - moles moles

4. Check answer.

Mole ratios

• CO2(g) + 2LiOH(s) → Li2CO3(s) + H2O

• What is the ratio of• CO2 molecules to LiOH molecules?

• CO2 molecules to Li2CO3 molecules

• Li2CO3 molecules to LiOH molecules

Molar Conversions

Molar Mass(g/mol)

6.02 1023

particles/mol

MASSIN

GRAMSMOLES

NUMBEROF

PARTICLES

Chemical Equations and the Mole

P4O10 + 6 H2O 4 H3PO4

1 molecule of P4O10 reacts with 6 molecules of H2O to give 4 molecules of H3PO4

1 mole of P4O10 reacts with 6 moles of H2O to give 4 moles of H3PO4

Mole to Mole conversions

-use mole ratios from last notes

Formula used:

Moles of knownMoles of knownMoles of unknown

= moles of unknown

Mole to Mole conversions

Steps in Solving

1.Determine moles given in problem

2.Find mole ratio showing unknown on top and known on bottom

3.Fill into formula above and calculate moles of unknown

Stoichiometry and Chemical EquationsThe coefficients in a chemical equation not only describe the number of atoms/molecules being reacted/produced, they also describe the number of moles of atoms/molecules being reacted/produced.

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

Suppose 5.46 mol N2 completely react in this fashion:

5.46 mol N23 mol H2

1 mol N2

= 16.4 mol H2 are needed to react

5.46 mol N22 mol NH3

1 mol N2

= 10.9 mol NH3 must be produced

Stoichiometry PracticeBismuth(III) chloride will react with hydrogen sulfide to form bismuth(III) sulfide and hydrochloric acid. Write the balanced equation for this reaction, then calculate how many moles of acid would be formed if 15.0 mol of hydrogen sulfide react.

2 BiCl3 + 3 H2S Bi2S3 + 6 HCl

15.0 mol H2S6 mol HCl

3 mol H2S= 30.0 mol HCl

Mole to Mass conversions

-use if you know # moles of one thing and you want to know mass of another thing

Steps in solving

1.Convert moles of known to moles of unknown.

2.Convert moles of unknown to grams of unknown

Stoichiometry PracticeSodium nitride can be formed by reacting sodium metal with nitrogen gas. Write the balanced equation for this reaction, then calculate how many moles of sodium nitride can be produced from 25.0 g of sodium.

6 Na + N2 2 Na3N

25.0 g Na 1 mol Na 2 mol Na3N22.99 g Na 6 mol Na

= 0.362 mol Na3N

Stoichiometry Practice When magnesium burns in air, it combines with oxygen to form magnesium oxide according to the following equation:

• What mass (in grams) of magnesium oxide is produced from 2.00 mol of magnesium?

2Mg(s) + O2(g) 2MgO(s)

2.0 mol Mg 2 mol MgO 40.3g MgO2 mol Mg 1 mol MgO

= 80.6 g MgO

Mass to Mass conversions

Use if you know the mass of 1 thing and want to find mass of another

Steps in solving

1.Convert mass of known to moles of known

2.Do mole-mole conversion

3.Convert moles of unknown to mass of unknown

2KClO3 → 2KCl + 3O2Calculate the mass of O2 produced if

2.5 g of KClO3 is completely decomposed when burned

2.5 g KClO3 3 mol O2 32 g O2

2 mol KClO3 1 mol O2

= .98g O2

122.6g KClO3

1 mol KClO3

Mole ratio

NH4NO3 → N2O + 2H2O

Determine the mass of water produced from the decomposition of 25.0g of

solid ammonium nitrate

25.0 g NH4NO3 2 mol H2O 18.02 g H2O

1 mol NH4NO3 1 mol H2O

= 11.26g H2O

80.04 g NH4NO3

1 mol NH4NO3

Mole ratio

Summary of Conversions

Mole to Mole: 1 step

Mass to Mole: 2 steps

Mass to Mass: 3 steps

What setup do all conversions require?

Mole:mole ratio

Limiting Reactants

How many bicycles can be assembled from the parts shown?

From eight wheels four bikes can be constructed.

From four frames four bikes can be constructed.

From three pedal assemblies three bikes can be constructed.

The limiting part is the number of pedal assemblies.

Likewise, in a chemical reaction, it is possible for one reactant to be used up before the other(s).A limiting reactant is a substance that is completely used up in a chemical reaction and causes the reaction to stop.Therefore, it is the limiting reactant that also determines how much product can actually be formed.

An excess reactant are the left-over reactants that are in abundance in a reaction.

The bicycle wheels and frames were in excess in our example

To determine which reactant is limiting in a given reaction:

1. Convert grams of each reactant to moles

2. Divide both #’s by the smallest # to determine actual mole ratio

3. Determine balanced mole ratio using moles in balanced equation

4. Which reactant isn’t as large as it’s supposed to be? THAT is your limiting reactant.

Limiting Reactant ExampleS8 + 4Cl2 → 4S2Cl2

If we are given 100 grams of Cl2 and 200 grams of S8, what is the limiting

reactant?

100 g Cl2 1 mol Cl2

70 g Cl2

= 1.4 mol Cl2

200 g S8 1 mol S8

256 g S8

= 0.8 mol S8

Step 1: Convert grams into moles

Step 2: Divide both #’s by the smallest # to determine actual mole ratio

For Cl2:

1.4 mol Cl2

0.8 mol S8

For S8:

0.8 mol S8

0.8 mol S8

= 1.75 =1

Divide by smallest #

Step 3: Determine balanced mole ratio using moles in

balanced equation

S8 + 4Cl2 → 4S2Cl2

Balanced mole ratio

1 mol S8

4 mol Cl2

Actual mole ratio

1 mol S8

1.75 mol Cl2

Step 4: Determine limiting reactant based on which reactant isn’t as

large as it’s supposed to be

Balanced mole ratio

1 mol S8

4 mol Cl2

Actual mole ratio

1 mol S8

1.75 mol Cl2

Cl2 is limiting b/c it’s actual mole ratio is smaller that the balanced mole ratio!!!

Calculating the amount of product from a limiting reactant:

1. Determine the limiting reactant

2. Determine the amount of limiting reactant (given in problem)

3. Multiply amount of limiting reactant by the mole ratio of product/limiting rgnt.

4. Convert moles of product into grams (multiply by molar mass)

S8 + 4Cl2 → 4S2Cl2

Step 1: Determine the limiting reactant

We know it’s Cl2 already!

Step 2: Determine the amount of limiting reactant- given in problem

1.4 mol Cl2

S8 + 4Cl2 → 4S2Cl2

Step 3: Multiply amount of limiting reactant by the mole ratio of product/limiting reactant from balanced equation

Step 4: Multiply by molar mass

1.4 mol Cl2 4 mol S2Cl2 135 g S2Cl2

4mol Cl2 1 mol S2Cl2

=190.4g S2Cl2

Calculating amount of excess reactant:

1. Determine the limiting reactant

2. Determine the amount of limiting reactant (given in problem)

3. Multiply amount of limiting reactant by the mole ratio of excess reactant/limiting rgnt.

4. Convert moles of excess reactant into grams (multiply by molar mass)

5. Subtract mass in #4 from original mass

S8 + 4Cl2 → 4S2Cl2

Step 1: Determine limiting reactant

Step 2: Determine amount (in moles of limiting reactant)

1.4 mol Cl2

S8 + 4Cl2 → 4S2Cl2

Step 3: Multiply amount of limiting reactant by the mole ratio of

Excess reactantLimiting reactant

Step 4: Multiply by molar mass of excess reactant

1.4 mol Cl2 1 mol S8 256.5g S8

4mol Cl2 1 mol S8

= 90.42 g S8

Step 5: Subtract mass in #4 from original mass given at the beginning of the problem. That is the amount left over

200g S8 - 90.42g S8 =

109.6g S8 in excess

Limiting Reactant Practice

Sodium metal will react with gaseous ammonia to produce solid sodium amide, NaNH2. The unbalanced equation for this reaction is…

Na + NH3 NaNH2 + H2

If 60.0 g of sodium are mixed with 48.0 g of ammonia, how many grams of sodium amide can be produced? Which substance is the limiting reactant?

Limiting Reactant Practice

Na + NH3 NaNH2 + H2

60.0 g Na 1 mol Na 2 mol NaNH2

22.99 g Na 2 mol Na= 2.61 mol NaNH2

48.0 g NH3 1 mol NH3 2 mol NaNH2

17.04 g NH3 3 mol NH3

= 2.82 mol NaNH2

smallest

Na is limiting

2 2 2

Theoretical, Actual & Percent Yields

Theoretical yield: the amount of product you would get if the reaction occurs with complete efficiency.

Actual yield: what you “actually” get when you perform the reaction.

Percent yield: the actual yield divided by the theoretical yield multiplied by 100.

% Yield = (Actual/Theoretical) x 100

Percent Yield

100yield ltheoretica

yield actualyield %

calculated on paper

measured in lab

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

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:

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