I. Defining Stoichiometry

I. Defining Stoichiometry

A. Stoichiometry

• Stoichiometry - the calculation of quantities of substances involved in chem rxns

• Based on conservation of mass

• Helps predict amounts of reactants and products

• Must use balanced equation so you know mole ratios (the coefficients tell you # of moles)

• Quantities often expressed as mol, L, g, etc.

A. Stoichiometry

• Ex: For the following balanced equation, find the masses of the reactants and products.

4Fe + 3O₂ 2Fe₂O₃

• Convert each reactant from moles grams 4 mol Fe 55.85 g Fe 223.4 g Fe 1 mol Fe

3 mol O₂ 32.00 g O₂ 96.00 g O₂ 1 mol O₂

• Mass of reactants = 223.4 g + 96.00 g = 319.4 g

A. Stoichiometry

• Ex: 4Fe + 3O₂ 2Fe₂O₃

• Convert product from moles mass

2 mol Fe₂O₃ 159.7 g Fe₂O₃ 319.4 g Fe₂O₃ 1 mol Fe₂O₃

• Mass of reactants = mass of product

• Should always be equal because of conservation of mass.

B. Mole Ratio

• Mole Ratio – ratio between the amounts of any 2 substances involved in a chem rxn

• Usually written as a fraction

• The coefficients in the balanced equation are used to write mole ratios

2(H₂) = 2 mol H₂ O₂ = 1 mol O₂ 2(H₂) = 2 mol H₂ or 4 mol H or 4 mol H or 2 mol O 2(O) = 1 mol O₂ or 2 mol O

B. Mole Ratio

• Ex: What are the mole ratios for the elements and compounds in the following balanced equation?

N₂ + 3H₂ 2NH₃ 1 mol N2 or 3 mol H₂

3 mol H₂ 1 mol N₂

1 mol N₂ or 2 mol NH₃ 2 mol NH3 1 mol N₂ 3 mol H₂ or 2 mol NH₃

2 mol NH₃ 3 mol H₂

• Not all ratios include the #1

II. Stoichiometric Calculations

A. Mole-to-Mole Conversions

• Always begin with a balanced equation

• Ex: How many moles of ammonia are produced from 5.25 moles of hydrogen gas that reacts with an excess (plenty) of nitrogen gas?

• N₂ + 3H₂ 2NH₃

• Use dimensional analysis and mole ratio to find moles of NH₃

• Mole ratio H₂:NH₃ = 3:2 rearrange as fraction

• Convert moles of H₂ moles of NH₃

5.25 mol H₂ 2 mol NH₃ 3.50 mol NH₃ 3 mol H₂

A. Mole-to-Mole Conversions

• Begin with a balanced equation

• Ex: How many moles of hydrogen gas are needed to completely react with (use up all) 7.25 moles of nitrogen gas to form ammonia?

• N₂ + 3H₂ 2NH₃

• Use dimensional analysis & mole ratio to find moles of H₂

• Convert moles N₂ moles H₂

7.25 mol N₂ 3 mol H₂ 21.8 mol H₂ 1 mol N₂

A. Mole-to-Mole Conversions

• Begin with a balanced equation

• Ex: How many moles of nitrogen gas are needed to produce 4.30 moles of ammonia if there is

excess hydrogen?

• N₂ + 3H₂ 2NH₃

• Find moles of N₂

4.30 mol NH₃ 1 mol N₂ 2.15 mol N₂ 2 mol NH₃

B. Mole-to-Mass Conversions

• Ex: How many grams of ammonia are produced from 7.50 moles of nitrogen that reacts with an excess of hydrogen?

• N₂ + 3H₂ 2NH₃

• Convert: mol N₂ mol NH₃ g NH₃

• Mole ratio N₂:NH₃ = 1:2 write as a fraction

• M.M. NH₃ = 17.0 g/mol

7.50 mol N₂ 2 mol NH₃ 17.0 g NH₃ 255 g NH₃

1 mol N₂ 1 mol NH₃

B. Mole-to-Mass Conversions

• Ex: How many grams of nitrogen are needed to completely react with 11.7 moles of hydrogen to produce ammonia?

• N₂ + 3H₂ 2NH₃

• Convert: mol H₂ mol N₂ g N₂

• Mole ratio N₂:H₂ = 1:3

• M.M. N₂ = 28.0 g/mol

11.7 mol H₂ 1 mol N₂ 28.0 g N₂ 109 g N₂

3 mol H₂ 1 mol N₂

C. Mass-to-Mass Conversions

• Ex: How many grams of hydrogen are needed to produce 45 grams of ammonia if N₂ is in excess?

• N₂ + 3H₂ 2NH₃

• Convert: g NH₃ mol NH₃ mol H₂ g H₂

• Mole ratio NH₃:H₂ = 2:3

• M.M. NH₃ = 17.0 g/mol

• M.M. H₂ = 2.0 g/mol

45 g NH₃ 1 mol NH₃ 3 mol H₂ 2.0 g H₂ 7.9 g H₂ 17.0 g NH₃ 2 mol NH₃ 1 mol H₂

C. Mass-to-Mass Conversions

• Ex: How many grams of nitrogen are needed to completely react with 11.0 grams of hydrogen?

• N₂ + 3H₂ 2NH₃

• Convert: g H₂ mol H₂ mol N₂ g N₂

• Mole ratio N₂:H₂ = 1:3

• M.M. N₂ = 28.0 g/mol

• M.M. H₂ = 2.0 g/mol

11.0 g H₂ 1 mol H₂ 1 mol N₂ 28.0 g N₂ 51.3g N₂ 2.0 g H₂ 3 mol H₂ 1 mol N₂

III. Limiting Reactants

Each plastic bottle has the same amount of vinegar. Different amounts of baking soda (the “limiting reagent/reactant”) have been added to

produce different amounts of carbon dioxide gas.

A. Limiting & Excess Reactants

• Limiting Reactant (L.R.) - the reactant that limits or determines the amount of product that can be

made

• A.K.A Limiting Reagent

• A substance that is totally consumed when the chem rxn is completed

A. Limiting & Excess Reactants

• Excess Reactant (E.R.) – a reactant that is not completely used up in the rxn

• Some will be leftover at the end of the rxn

B. L.R. Problems

• Steps for solving limiting reactant problems 1. Start with the balanced equation

2. Determine the L.R.

– Find the mole ratio for the reactants (A:B) – Calculate required amount:

(Given amount of A) x (mole ratio) = required amt B OR

(Given amount of B) x (mole ratio) = required amt A

– Whichever substance is used up completely is the L.R.

The other is the E.R.

3. Use stoichiometry to find the desired quantity

– Use the L.R. to find out how much product is produced

B. L.R. Problems

• Ex: What is the L.R. and how many moles of NaCl are produced when 6.70 mol of Na react with 3.20 mol Cl₂?

• This problem has two parts:

– First you need to find the L.R.

– Then calculate how much NaCl is produced.

• 1. Balanced equation: 2Na + Cl₂ 2NaCl

2. Determine the L.R.

• What is the mole ratio?

– 1 mol Cl₂:2 mol Na or 2 mol Na:1 mol Cl₂

• Convert moles of one reactant to moles of the other 6.70 mol Na 1 mol Cl₂ 3.35 mol Cl₂

2 mol Na

• This calculation indicates 3.35 mol Cl₂ are required to react with 6.70 mol Na

• In the problem, only 3.20 mol of Cl₂ are available

• Cl₂ is the L.R. and will determine amount of product

• Na is the excess reactant (E.R.)

• Note: could have converted from mol Cl₂ mol Na.

• Go back to the question:

• What is the L.R. and how many moles of NaCl are produced when 6.70 mol of Na react with 3.20 mol Cl₂?

• 2Na + Cl₂ 2NaCl

• 3. Calculate moles of product

• From last step, Cl₂ is the L.R.

• Convert mol L.R. mol product

3.20 mol Cl₂ 2 mol NaCl 6.40 mol NaCl 1 mol Cl₂

C. L.R. Problems Involving Mass

• Remember: When in doubt, convert to moles!

• Ex: How many grams of copper(I) sulfide can be produced when 80.0 g of Cu reacts with 25.0 g S?

• Though not asked for directly, need to find the L.R.

1. Balanced equation:

2Cu + S Cu₂S

Here we are given g instead of moles. Need moles 1st 2. Calculate moles of Cu and S (using M.M.) and then

determine L.R. (using mole ratio)

80.0 g Cu 1 mol Cu 1.26 mol Cu 63.5 g Cu

25.0 g S 1 mol S 0.779 mol S 32.1 g S

1.26 mol Cu 1 mol S 0.630 mol S

2 mol Cu

• 0.630 mol S are needed to react with 1.26 mol Cu

• 0.779 mol S are available, so there is excess S

• S is E.R.

• Cu must be L.R.

• Again, could have gone from mol S mol Cu

• How many grams of copper(I) sulfide can be

produced when 80.0 g of Cu reacts with 25.0 g S?

• 2Cu + S Cu₂S

3. Use L.R. to calculate moles of product and then determine grams of product

• Cu is the L.R

• Convert mol Cu mol Cu₂S g Cu₂S

1.26 mol Cu 1 mol Cu₂S 159.1 g Cu₂S 100. g Cu₂S 2 mol Cu 1 mole Cu₂S

IV. Percent Yield

A. Theoretical and Actual Yield

• Theoretical Yield (T.Y.) - the max amount of

product that could be formed from given amounts of reactants

• Can be expressed as moles or grams

A. Theoretical and Actual Yield

• Actual Yield (A.Y.) – the amount of product that will actually form when the rxn is carried out

• Always less than T.Y.

• Why? Several possible reasons:

– rxns do not always go to completion – loss during filtration

– 100% recovery is near impossible – may be significant 'side' reactions

B. Percent Yield (%Y)

• Percent Yield (%Y) – The ratio of A.Y. to T.Y., expressed as a percent

• %Y = A.Y. x 100%

T.Y.

B. Calculating T.Y. and %Y

Ex: What is the T.Y. (in grams) of CaO if 24.8 g of CaCO₃ is decomposed by heat?

•Balanced equation: CaCO₃ CaO + CO₂

•g CaCO₃ mol CaCO₃ mol CaO g CaO

24.8 g CaCO₃ 1 mol CaCO₃ 1 mol CaO 56.1 g CaO 100.1 g CaCO₃ 1 mol CaCO₃ 1 mol CaO

•T.Y. = 13.9 g CaO

Ex: What is the %Y if only 13.1 g of CaO is produced?

•Use T.Y. from last problem

•Percent Yield = 13.1 g CaO x 100% = 94.2%

13.9 g CaO