Stoichiometry

The study of quantities of substances in chemical reactions

Interpreting Chemical Equations

N 2 + 3 H 2



2 NH 3 Particles:

1 molecule of Nitrogen reacts with 3 molecules of Hydrogen to produce 2 molecules of Ammonia (NH 3 ).



Moles:

1 mole of Nitrogen reacts with 3 moles of Hydrogen to produce 2 moles (NH 3 ).

of Ammonia

The important thing to notice is that the coefficients determine the ratio of each reactant to each product.

Interpreting Chemical Equations

N 2 + 3 H 2



2 NH 3



Mass

Balanced reactions must obey the law of conservation of mass . Using the mole idea you can determine the mass of species in this equation:

used

1 mole N 2 3 mole H 2

produced

= 28 g = 6 g 2 moles NH 3 = 34 g 

Volume

@STP conditions

22.4 L of N 2 reacts with 67.2 L of H 2 to produce

44.8L of NH 3

Notice that volume of gases is NOT conserved

22.4 + 67.2 ≠ 44.8

Which of the following are conserved in a chemical reaction?

C.

Mass A.

Volume of Gas B.

Moles C.

Mass D.

# of Particles The rearrangement of atoms during the reaction can change the # of everything else. But mass is conserved.

Mole-Mole Calculations

N 2 + 3 H 2



2 NH 3



Relating moles of reactants to moles of products.

 If this reaction started with 2 moles of N 2 , many moles of NH 3 could be produced?

how  We do not always use full moles of reactants, or want full moles of products.

 We more often deal with partial moles.

1:2 --- 2:4 It’s a simple ratio For more complicated ratios use this guide

Mole-Mole Calculations

N 2 + 3 H 2



2 NH 3

 So if we start with .75 moles of N 2 , how much ammonia is produced?

 How many moles of H 2 must be used?

Known moles x Coefficient of what you want to find out Coefficient of what you know  If we want to produce 3.18 moles of ammonia, how many moles of hydrogen do we need to use?

4Al + 3 O

2 

2 Al

2

O

3 Write three mole ratios from this equation How many moles of Aluminum metal are needed to produce 3.7 moles of Aluminum Oxide?

How many moles of O 2 are required to react completely with 14.8 moles of Al?

Calculate the moles of Al 2 O 3 formed when 0.78n of O 2 react with Al.

Mass-Mass Calculations

N 2 + 3 H 2



2 NH 3

Our tools do not directly measure moles.

The balances measure… Mass (g)  1.

Solving mass-mass problems: So we often must convert to moles from mass, and mass from moles.

2.

3.

Convert the given quantity to MOLES.

(Divide mass by molar mass) Calculate unknown moles by multiplying by a mole ratio.

(unknown/known) Convert the answer in # 2 to mass.

(Multiply by molar mass)

Mass-Mass Calculations

N 2 + 3 H 2



2 NH 3

If 8.5 grams of Nitrogen are reacted, what mass of ammonia can be produced?

 If 16.4 grams of Hydrogen are reacted, how many grams of Nitrogen are needed to fully react?

 Solving mass-mass problems: 1.

2.

3.

Convert the given quantity to MOLES.

(Divide mass by molar mass) Calculate unknown moles by multiplying by a mole ratio.

(unknown/known) Convert the answer in # 2 to mass.

(Multiply by molar mass)

K

2

O + H

2

O



2 KOH

1.

Calculate the mass of KOH produced when 4.80 g of K 2 O is reacted.

 Solving mass-mass problems: 2.

3.

How many grams of water are needed for this reaction?

1.

2.

If 50.0 g of KOH is produced, what is the mass of Water used?

3.

Convert the given quantity to MOLES.

(Divide mass by gfm) Calculate unknown moles by multiplying by a mole ratio.

(unknown/known) Convert the answer in # 2 to mass.

(Multiply by gfm)

2 Al + 3 Cl

2 

2 AlCl

3

If 70.9 g of Chlorine Gas are used, how many moles of Aluminum Chloride can be produced?

What is the mass of the AlCl 3 ?



Solving mass-mass problems:

1.

Convert the given quantity to MOLES.

(Divide mass by molar mass) 2.

Calculate unknown moles by multiplying by a mole ratio.

(unknown/known) 3. Convert the answer in # 2 to mass.

(Multiply by molar mass)

Other legs of the mole wheel

 If given a mass of a reactant you can find # of particles produced  If given the volume (@STP) of a reactant you can find the mass produced  If given the # of molecules of a product you can find # molecules of reactant

2

H

2

O

2

MnO 2

 2

H

2

O + O

2 Starting with 29.2 g of hydrogen peroxide How many molecules of Oxygen are produced?

What is the volume of the Oxygen gas produced? (assume STP conditions) What mass of

water

is produced?

2 SO

2

+ O

2 

2 SO

3 How many liters of Oxygen are needed to produce 24g of Sulfur Trioxide?

(All species are in the gas phase @ STP)

Limiting Reagent analogy

IF you have: 40 slices of turkey 28 slices of cheese 6 heads of lettuce 1 mole of tomatoes Jars of mayo and mustard 2 slices of bread  How many sandwiches can you make?

 What limits your production of sandwiches?

Bread is your “limiting reagent”

Limiting Reagent analogy

Your reaction is your recipe  Seldom do you have perfect amounts of each ingredient.

 The limiting reagent (reactant) is what determines the amount that is produced.

The other ingredients are considered to be excess reagents  Cheese, turkey, tomatoes…

Limiting Reagent Definition

Limiting Reactant: (a.k.a. Limiting Reagent) The substance that controls the quantity that can form In a chemical reaction.

Excess Reactant: (a.k.a. excess reagent) The substance that is not used up completely in a reaction.

Determining Limiting Reagent

2

Na + Cl

2  2

NaCl

Starting with 6.7n of Na, and 3.2n of Cl 2 How many moles of NaCl can be produced?

1.

Start with a known amount of one reactant.

A.

Determine limiting reagent B . Use limiting reagent to determine amount of product formed.

2.

3.

4.

Use the mole ratio to determine the amount of product.

Determine how much the other reactant would produce.

The reactant producing the LEAST is limiting.

2

Na + Cl

2  2

NaCl

Starting with 6.70n of Na, and 3.20n of Cl 2 How many moles of NaCl can be produced?

If 6.7n Na are used 1.

Start with a known amount of one reactant.

6.7 x ½ = 3.35n

Cl 2 needed Compare the 3.35n

Cl 2 needed, to 3.20n Cl 2 you have.

We have less Cl 2 than we need so

Chlorine is the limiting reagent.

Use the limiting reagent to determine the moles of NaCl produced.

2.

3.

4.

Use the mole ratio to determine the amount of product.

Determine how much the other reactant would produce.

The reactant producing the LEAST is limiting.

2

Na + Cl

2  2

NaCl

Starting with 6.70n of Na, and 3.20n of Cl 2 How many moles of NaCl can be produced?

We have determined the 3.2n of Cl 2 to be limiting.

3.2n Cl 2 x 2/1 = 6.4n NaCl

Now it becomes a regular mole-mole problem.

You could then find the mass of NaCl produced.

What determines the amount of product?

 The limiting reagent!

Mass

Moles

16 Cu + S

8 

8 Cu

2

S

Starting with 1.25n Cu and .78n of S determine the moles of Copper (I) Sulfide made.

0.625 n of Cu 2 S What mass of Cu 2 S is made?

99.4 g of Copper sulfide

Mass

moles

Mg + 2 HCl



MgCl

2

+ H

2 How many grams of Hydrogen can be produced when 6.00 g of HCl is reacted with 5.00 g of Mg?

1.

2.

3.

4.

Convert each reactant to moles.

Determine limiting reagent by multiplying by the mole ratio.

Use limiting reagent to determine moles of product.

Multiply moles of product by GFM to find mass Find the volume of this Hydrogen @ STP

Mass

Moles

2 H

2

+ O

2 

2 H

2

O

Given 40grams of Hydrogen and 90 grams of Oxygen, determine the limiting reagent.

Determine the mass of water that can be produced.

How much excess reagent remains?

Percent Yield

 Why don’t you always get 100% on your chemistry tests?

 There are many possible reasons  You went skiing during the important lectures.

 You worked on your AP history homework instead of studying for chem.

 Rippet is not as good as he thinks he is.

% Yield

In Chemical Reactions we don’t always get 100% of the products our equations predict.

Remember the Magnesium Burning Lab?

 2 Mg + O 2  2 MgO Theoretically all the Mg should have converted to MgO.

In reality, some things went wrong.

% Yield

An equation predicts the THEORETICAL YIELD When a reaction is carried out in the lab we get ACTUAL YIELD 2 Mg + O 2  2 MgO We reacted 2.5g of Mg, what mass of MgO could be produced?

Assume there is unlimited O 2 from the air so Mg is the limiting reagent.

0.103 n of Mg x 2/2 = 0.103n MgO We only had 3.61 grams produced… So we do NOT have 100% yield!

% Yield

Actual (lab results) Theoretical (equation prediction)

X 100 %

Multiply by 100, then add the “%” sign

3.61 / 4.15 = .869

.869 x 100 +”%” = 86.9% Yield

CaCO

3 

CaO + CO

2 What is the % Yield if 24.8 g of CaCO 3 give 13.1g

of CaO?

is heated to 1.

2.

3.

Convert g CaCO 3 to moles of CaCO 3 Only 1 reactant, so it’s the limiting reagent.

.

Mole ratio is 1:1 Convert moles of CaO to grams CaO.

Use % Yield Formula: actual x 100% theoretical

2 Al + 3 CuSO

4 

Al

2

(SO

4

)

3

+ 3 Cu

What is the % yield when 4.65g of Copper is made when 1.87g of Al reacts with excess Copper Sulfate.

1.

2.

3.

4.

5.

Excess means there is plenty, the other reagent is limiting. Convert grams Al to moles Al .

Find moles of Cu expected (multiply by mole ratio) 1.

Convert moles Cu to grams Cu.

This is the Theoretical amount of Copper Use the % Yield Formula.

Which balloon will have the greatest volume of gas?

NaHCO

3

+ HC

2

H

3

O

2



NaC

2

H

3

O

2

+ CO

2

+ H

2

O

Reaction 4 5 1 2 3

Mass of NaHCO 3 0.50g

1.00g

1.50g

2.00g

3.00g

Moles of Acetic Acid 0.02n

0.02n

0.02n

0.02n

0.02n