Chapter 9Chapter 9

Stoichiometry

9.1 Intro. To Stoichiometry

• What is Stoichiometry?– The study of the quantitative relationships that

exist in chemical formulas and reactions.• Stoicheion- element• Metron- measure

2 Types of Stoichiometry:• Composition stoichiometry- mass relationships of

elements in a single compound. (ch3)• Reaction stoichiometry- mass relationships

between reactants and products in a chemical rxn.• Solved by using ratios in Balanced chemical equations.

Reaction-Stoichiometry Problems

• There are 3 types of Stoichiometry problems:– 1. Mass-Mass problems– 2. Mass-Volume problems– 3. Volume-Volume problems

In general, every problem will be solved in 3 steps:Quantity of given →mols of given →mols of unknown → quantity of unknown

Mole Ratios• Mole Ratio- a conversion factor that relates the

amounts in moles of any 2 substances involved in a chemical reaction.

• The coefficient in balanced chem. Equns. represent MOLE ratios for the substances in the rxn.– Ex.- N2H4 +2H2O2 → N2 + 4H2O

Means: 1mol N2H4 +2mol H2O2 → 1mol N2 + 4mol H2O

Molar Mass (review)

• From Ch 7:• “The mass in grams of 1 mole of a substance..” (M).• Molar Mass of…– Ca 40 amu –> 40 g/mol– C 12 amu -> 12 g/ mol– H2O2 34 amu -> 34 g/mol– NaCl– CaCl2

– C6H12O6

9.2 Idea Stoichiometric Calculations

• Knowing that the coefficients show MOLE RATIOS… we can now solve problems relating moles of one substance to moles of another.

• Ex– N2H4 +2H2O2 → N2 + 4H2O

How many mols of N2H4 are needed to react with 2 mols of H2O2 ?

More mol- mol practice

• 2S + 3O2 → 2SO3

How many moles of S are needed to react with 2.5 mols of O2?

• Cu + 4HNO3 → Cu(NO3)2 + 2NO2 + 2H2O How many moles of Cu(NO3)2 are produced from 12.3 moles of HNO3?

Mass-Mass Problems

• In these problems, you will be given the mass of one substance and asked to solve for the mass of another substance.

• Your mantra is: – Mass to moles, moles to moles, moles to mass.

• You will need a balanced chem. equation, the molar mass of the known and the unknown substance to solve.

Mass-mass practice

• 2H2 + O2 → 2H2O– How many grams of water will be produced from

23.5 g of oxygen?– *Mass to moles, moles to moles, moles to mass*

• Convert g to moles of O2, then mol O2 to mol H2O, and finally, mols of H2O to g of H2O.

More mass-mass practice• Zn + H2SO4 → ZnSO4 + H2– How many g of H2SO4 will react with 9.5 g of Zn?

• HCl + NaOH → NaCl + H2O– How many g of NaCl will be produced from 51.2 g of NaOH?

• 2Mg + O2 → 2 MgO– How many g of Mg are needed to produce 12.3g MgO?

Mass-Volume ProblemsIn these problems you will be given the MASS of

one substance and asked to find the VOLUME of a gas. You will be going from grams to liters (g → L).

Your Mantra is: Mass to moles, moles to moles, moles to volume.

To convert from mols to volume you will use the molar volume (22.4L/ mol).

Mass-Volume practice

• 2H2 + O2 → 2H2O– How many liters of hydrogen will be required to

produce 23.5 g of water?– Mass to moles, moles to moles, moles to volume.

• Convert g to moles of H2O, then mol H2O to mol H2, and finally, mols of H2 to L of H2.– Remember, use the molar volume: 22.4 L/mol

More mass-volume practice

• Zn + H2SO4 → ZnSO4 + H2

– How many L of H2 will be produced from 9.5 g of Zn?

• Sn + 2HF → SnF2 + H2– How many L of HF will be required to react with 30.0g of

Sn?• NH4NO3 → N2O + 2H2O– How many L of N2O will be produced from 52.6 g NH4NO3?

• Mg + O2 → MgO2

– How many L of O2 are needed to produce 12.3g MgO2?

Volume-Volume Problems

In these problems you will be given the VOLUME of one gas and asked to find the VOLUME of a different gas.

Your Mantra is: VOLUME to moles, moles to moles, moles to

volume.

To convert from mols to volume you will use the molar volume (22.4L/ mol).

Volume - Volume practice

• 2H2 + O2 → 2H2O– How many liters of hydrogen will be required to

react with 23.5 L of O2?– volume to moles, moles to moles, moles to

volume.

• Convert L to moles of O2, then mol O2 to mol H2, and finally, mols of H2 to L of H2.– Remember, use the molar volume: 22.4 L/mol

9.3 Limiting Reactants & Percent Yields

• The amount of product made depends on how much reactants are available. Sometimes, one of the reactants limits the number of products made…– How many bikes can be made from 11 bike frames

but only 7 tires?– Did the # of bike frames or the # of tires limit how

many bikes were made?• The reactant that limits the amount of product

formed is called the limiting reactant.• The reactant that is not completely used up is

called the excess reactant.

Identifying Limiting Reactants

• There are 3 general steps to identifying the limiting reactant.– 1. Begin with a balanced chemical equation.– 2. Calculate the amount of product formed by

EACH of the reactants. (pick only one product)– 3. The reactant that produces the least amount of

product is the limiting reactant.

Limiting Reactant Problems

• Cu + 2AgNO3 → Cu(NO3)2 + 2Ag

• Which is the limiting reactant if you have 6.0g Cu and 12.5g AgNO3 ? (hint: solve for Ag) – You will perform 2 mass-mass problems. • 1. mass of Cu to mass of Ag• 2. mass of AgNO3 to mass of Ag.

Percent Yield

The amount we calculate and what we actually make as the product are not necessarily the same amount.

Theoretical yield – amount of product that should be produced based on calculation.

Actual yield – the amount ACTUALLY obtained from the reaction (in lab).

• We need to know how much of the expected was made during the reaction. …did we only make 5% or 55%?

• Percent Yield – what % of the predicted amount was actually made.

% yield = (actual yield ÷ expected yield) X 100%

Percent Yield practice problems

• You burn Mg in O2 and produce 6.5g MgO. You expected to make 8.2g of MgO. What is your percent yield?

• Determine the %yield for 3.74g Na and excess O2 if 5.34g of Na2O2 is recovered?

We don’t know the theoretical yield so you will have to calculate it first, then solve %yield.