1. Stoichiometry
Chapter 12

2. Introduction
Learn how to use the chemical equation to obtain data on quantities of reactants and products.
Learn what the coefficients of a chemical equation mean.
Learn how to convert between the different components of a reaction.
Learn what a limiting reagent is and how it is used to calculate the yield of a reaction.

3. The Arithmetic Equation (Section 12.1)
Using Everyday Equations
Using Balanced Equations
Interpreting Chemical Equations
Mass Conversion in Chemical Reactions

4. I. Using Everyday Equations
Everyday we are given instructions for doing something.
Instructions for cooking, putting together a toy, etc.
These instruction often describe the type of components and the number of the components.
Chemical equations are instructions for making particular chemical products.

5. II. Using Balanced Chemical Equations
Chemical equations can help us estimate the amounts of products obtained if we start with a certain amount of reactants.
Amounts in chemistry is often expressed in terms of grams or moles.
Stoichiometry: A subject in chemistry that deals with the calculation of quantities in chemical reactions.

6. Stoichiometry
Stoichiometric calculations are calculations using balanced equations.
Stoichiometry allows chemist to estimate the amounts of reactants and products using ratios of moles or representative particles.

7. III. Interpreting Chemical Equations
A balanced chemical equation gives us information about the types of reactants and products as well as the quantities of each.
These quantities can be interpreted in terms of numbers of atoms, molecules, or moles; mass; and volume.

8. Coefficients as Number of Atoms
A balanced chemical equation indicates the numbers and types of each atom in the reaction.
The number and type should remain the same in either side of the arrow.
N2(g) + 3H2(g) → 2NH3(g)

9. Coefficients as Number of Molecules
A balanced chemical equation also indicates the number and types of molecules or formula units that are involved in a reaction.
The ratios between the reactants and products will also remain the same for a given reaction
N2(g) + 3H2(g) → 2NH3(g)

10. Coefficients as Number of Moles
A balanced chemical equation gives the number of moles of reactants and products.
This use of the coefficients is the most useful for chemical purposes.
We can calculate the amounts of reactants and products in a reaction using the coefficient as moles.
N2(g) + 3H2(g) → 2NH3(g)

11. Coefficients as Mass Units
Coefficients do not relate the amount of reactants and products directly.
We must use the coefficients as moles and convert to mass units using the molar masses of the reactants and products.
Masses of reactants will equal the masses of products to preserve the law of conservation of mass.
N2(g) + 3H2(g) → 2NH3(g)

12. Coefficients as Volume Units
At STP, the coefficients can be used to indicate the volume of reactants and products directly if we recall that 1 mol = 22.4 L.
N2(g) + 3H2(g) → 2NH3(g)

13. IV. Mass Conservation in Chemical Reactions
Remember: Mass and atoms are always conserved in a chemical reaction.
Molecules, formula units, moles, and volume are not always going to be conserved.

14. Sample problem
Interpret the following equation in terms of: a.) representative particles and moles b.) masses of reactants and products 2H2S(g) + 3O2(g) →2SO2(g) + 2H2O(g)

15. Chemical Calculations (Section 12.2)
Writing and Using Mole Ratios
Other Stoichiometric Calculations
The Haber Process

16. I.) Writing and Using Mole Ratios
A balanced chemical equation is required for calculations involving amounts of reactants and products.
The coefficients in a balanced chemical equation allow us to compare the moles of the reactants and products to each other.
Such a comparison can yield a ratio that can be used for conversions.

17. The mole ratio. A conversion factor derived from the
coefficients of a balanced chemical equation
interpreted I terms of moles.
Mole ratios are used to convert between moles of reactants, moles of products, or moles of reactants and moles of products.
Use these conversion factors like any other conversion factor we have learned about.

18. Sample problem
What are the three mole ratios that can be derived from the following equation? N2(g) + 3O2(g) →2 NH3(g)

19. Sample problem
Using the mole ratios above, calculate how many moles of ammonia are produced when 0.60 mole of nitrogen reacts with hydrogen.

20. Mass-Mass Calculations
Moles are not efficient units to use in real world situations- no balances ca measure out moles.
Mass is the more useful unit for carrying out reactions in the real world.
Mole to mass conversion require the use of a compound’s molar mass.

21. Sample problem
Calculate the number of grams of NH3 produced by the reaction of 5.40 g. of hydrogen with an excess of nitrogen. The balanced equation was given earlier.

22. Mass-Mass Calculations
Molar
Mass
Molar
Mass
Yes
Moles
Moles
Mole Ratio

23. II.) Other Stoichiometric Calculations
Since the mole is related to mass, volume, and number of representative particles, we can do a number of different stoichiometric calculations from the balanced chemical equation.
It all starts with using the coefficients as moles and then applying the correct conversion factors.

24. The Mole and Conversions
Four types of conversion problems:
mol particles (atoms,etc)
1 mol = 6.02 x 1023 particles
mol mass
molar mass (g/mol)
mol volume
molar volume (1 mol = 22.4 STP)
density molar mass
molar volume

25. Solving problems using balanced chemical equations.
Balance the equation.
Identify the starting “unit.”
Identify the end “unit.”
Pick the conversion factor(s) that will get you from the starting unit to the end unit.
Set up the conversion problem so that all units cancel except the end unit.
Multiply or divide.

26. Mass-Particles Calculations
Molar
Mass
Avogadro’s #
Yes
Moles
Moles
Mole Ratio

27. Sample problem
Write the balanced equation then calculate the number of molecules of oxygen produced when 29.2 g. of water is decomposed by electrolysis?

28. Volume-Volume Calculations
Molar
Volume
Molar
Volume
Yes
Moles
Moles
Mole Ratio

29. Sample problem
Nitrogen monoxide and oxygen gas combine to form the brown gas nitrogen dioxide. How many liters of nitrogen dioxide gas are produced when 34 liters of oxygen reacts with an excess of nitrogen monoxide at STP?

30. Limiting Reagent and Percent Yield (Section 12.3)
Limiting and Excess Reagents
Percent Yield

31. I.) Limiting and Excess Reagents
In a chemical reaction, an insufficient quantity of any of the reactants will limit the amount of product that forms.

32. Limiting Reagent Excess Reagent
The reagent (reactant) that is used up first in a reaction and thus determines the amount of product that can be formed in a reaction.
Excess Reagent
The reagent that is not completely used up
and is left over after the reaction is complete.

33. How to determine the limiting reagent.
Balance the chemical equation.
Convert all units into mole units if they are not given already.
Use the mole ratio of the reactants to determine how many mole of one reactant will be needed to consume the other reactant.
Compare the mole value obtained with the given mole values.

34. Sample problem
Copper reacts with sulfur to form copper(I) sulfide. What is the limiting reagent when 80.0 g. of Cu reacts with 25.0 g. of S?

35. A quick note about limiting reagents.
From the previous problem the reactant in the smaller amount was not the limiting reagent.
The reactant present in the smaller amount by mass or by volume is not necessarily the limiting reagent.

36. Using limiting reagents to find the quantity of a product.
Start with a balanced equation.
Determine the limiting reagent.
Determine the mole ratio between the limiting reagent and the product(s).
Use this mole ratio to determine the mole of product(s) that will form.
Convert to other units if necessary.

37. Sample problem
What is the maximum number of grams of Cu2S that can be formed when 80.0 g. of Cu reacts with 25.0 g. of S?
2Cu(s) + S(s) → Cu2S(s)

38. II.) Percent Yield The ratio of the actual yield to the
theoretical yield expressed as a
percentage.
The percent yield is a measure of the efficiency of a reaction.
Percent yields are never 100% in real life.
Percent Yield= actual yield/theoretical yield x 100

39. Theoretical yield The maximum amount of product that
could be formed from given amount s
of reactants.
This number represents the amount of product formed if a reaction is run perfectly.
This value is never actually obtained.
It is the upper limit to what is expected from a reaction.

40. Actual yield The amount of product that actually forms
when a reaction is carried out in the
laboratory.
This is a measured, experimentally derived value.
Actual yields will almost always be lower than the theoretical yield.

41. Reasons why actual yields are lower than theoretical yields.
Reactions do not always run to completion.
Reactants are impure.
Product is lost during collection, transfer or filtration.
Reactions don’t always following theory.
Human error

42. Calculating percent yields.
Obtain a balanced chemical equation of the reaction.
Determine the limiting reagent for the reaction.
Calculate the amount of product(s) expected from the limiting reagent. This is the theoretical yield.
Obtain the actual yield.
Divide the actual yield by the theoretical yield and multiply by 100.

43. Sample problem
Calcium carbonate is decomposed by heating. The balanced chemical equation for the reaction is: CaCO3(s) → CaO(s) + CO2(g) What is the theoretical yield of CaO if 24.8 g. of CaCO3 is heated?

44. Sample problem
What is the percent yield if 13.1 g. of CaO is actually produced from the reaction given above?

45. Stoichiometry
Chapter 12
The End