chapter 9 stoichiometry. uee using everyday equations stoichiometry- the calculations of quantities...

Chapter 9

StoIcHIomEtRYStoIcHIomEtRY

UUsing EEveryday EEquations

• Stoichiometry- the calculations of quantities of chemical reactions

Its a form of bookkeeping for chemist

Mole to Mole Calculations

• If you know the number of moles of one substance, the balanced chemical equation allows you to determine the number of moles of all other substances in the reaction.

• How to solve Mole to Mole problems:– write a balanced equation– use a Mole-to-Mole proportion to find moles of unknown (put

unknown on top of proportion)

– Note: The coefficients in the balanced equation are equal to the number of moles for that substance

Mole to Mole Proportion

• Coefficients from equation Problem

• Desired = Desired

• Given Given

Mole toMole Examples

A. How many moles of ammonia are produced when .60 mol of nitrogen reacts with hydrogen?

B. How many moles of Oxygen are needed to produce 2.5 moles of Water?

C. How many moles of Hydrogen are produced when 4.5 moles of Mg reacts with HCl?

Mass to Mass Calculations

• When given the number of grams of a substance in a chemical reaction you can calculate the specific number of grams of any other substance in that reaction.

• How to Solve Mass to Mass Problems:– Write a balanced equation– Convert grams given to moles– Use a mole-mole proportion to find moles for unknown– Convert moles of unknown to grams

Mass to Mass Examples

1. Calculate the number of grams of NH3 produced when 15.5 grams of nitrogen reacts with hydrogen

2. How many grams of Oxygen are produced when 40.5 grams of water decomposes?

3. How many grams of CO2 are produced during the combustion of 200 grams of methane?

• Theoretical yield- the maximum amount of product that will formed from given amounts of reactants.

• Actual yield- the amount of product that actually forms when the reaction is carried out. This is often less then the theoretical yield.

• Percent Yield- the ratio of the actual yield to the theoretical yield as a percent. This measures the efficiency of the reaction.

percent yield = actual yield theoretical yield

Calculating the Percent Yield

X 100

N2 + 3H2 2NH3

2 atoms N + 6 atoms H 2 atoms N & 6 atoms H

1 molecule N2 + 3 molecule H2 2 molecule NH3

10molecules N2 + 30molecules H2 20 molecules NH3

6.02X1023

molecules N2

+6.02X1023

molecules H2

6.02X1023

molecules NH3

1X 3X 2X

MOLES SUMMARY• We have recently learned that

Avogadro’s number of particles is the same as a mole of a substance.

• On the basis of the particle interpretation we just discussed, the equation also tells you the number of moles of reactants and products.

• 1 mole of N2 molecules reacts with 3 moles of H2 molecules to make 2 moles of NH3 molecules.

• The coefficients of the balanced chemical equation indicate the numbers of moles of reactants and products in a chemical reaction.

MOLES SUMMARY

• This is the most important information that a reaction equation provides.

• Using this information, you can calculate the amounts of reactants and products.

MOLES SUMMARY

• A balanced chemical equation must also obey the law of conservation of mass.–Mass can be neither created nor

destroyed in ordinary chemical or physical processes.

• Remember that mass is related to the number of atoms in a compound through the mole.

MASS SUMMARY

• The mass of 1 mol of N2 molecules is 28 g; the mass of 3 mols of H2 molecules is 6 g for a total mass of reactants of 34 g.

• The mass of 2 moles of NH3 molecules is 2 * 17g or 34 g.

• As you can see the reactants mass is equal to the mass of the products.

MASS SUMMARY

1 mol N2 + 3 mol N2 2 mol NH3

28 g N2 + 3 (2 g H2) 2 (17 g NH3)

34 g reactants 34 g products

+

22.4 L N2 67.2 L H2 44.8 L NH3

22.22.4 L4 L

22.22.4 L4 L

22.22.4 L4 L

22.22.4 L4 L

22.22.4 L4 L

22.22.4 L4 L

N2 + 3H2 2NH3

USING EQUATIONS• You can see how much

information is stored in a simple balanced reaction eqn

• We can combine this information with our knowledge of mole conversions to perform important common stoichiometric calculations.

MOLE – MOLE CALCULATIONS

• A balanced rxn eqn is essential for all calculations involving amounts of reactants and products.

• If you know the number of moles of 1 substance, the balanced eqn allows you to calc. the number of moles of all other substances in a rxn eqn.

• Let’s go back to our synthesis of ammonia rxn.

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

•The MOST important interpretation of this rxn is that 1 mole of N2 reacts with 3 moles of H2 to produce 2 moles of NH3.

MOLE – MOLE CALCULATIONS

• These connections of the coefficients allows us to set up conversion factors called mole ratios.

• The mole ratios are used to calculate the connections in moles of compounds in our reaction equation.

• We can start calculating…

MOLE – MOLE CALCULATIONS

• Sample Mole – Mole problem:How many moles of

ammonia are produced when .60 moles of N2 are

reacted with H2?

Given: .60 moles of N2

Uknown: ____ moles of NH3

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

MOLE – MOLE CALCULATIONS

• Sample Mole – Mole problem:How many moles of

ammonia are produced when .60 moles of N2 are

reacted with H2?

Given: .60 moles of N2

Uknnown: ____ moles of NH3

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

MOLE – MOLE CALCULATIONS

• Sample Mole – Mole problem:How many moles of

ammonia are produced when .60 moles of N2 are

reacted with H2?

Given: .60 moles of N2

Uknown: ____ moles of NH3

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

MOLE – MOLE CALCULATIONS

• According to the reaction equation, for every 1 mole of N2 reacted we form 2 mols of NH3.

• To determine the number of moles of NH3, the given quantity of N2 is multiplied by the mole ratio from the rxn eqn in such a way that the units of “mol N2” cancel

MOLE – MOLE CALCULATIONS

»Solve for the unknown:

.6 mol N21 mol N2

2 mol NH3

= 1.2 mol NH3

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

MOLE – MOLE CALCULATIONS

MOLE – MOLE EXAMPLE 2

• This equation shows the formation of aluminum oxide.

4Al(s) + 3O2(g) 2Al2O3(s)•How many moles of aluminum are needed to form 3.7 mol Al2O3?

Given: 3.7 moles of Al2O3

Uknown: ____ moles of Al

MOLE – MOLE EXAMPLE 2

• This equation shows the formation of aluminum oxide.

4Al(s) + 3O2(g) 2Al2O3(s)•How many moles of aluminum are needed to form 3.7 mol Al2O3?

Given: 3.7 moles of Al2O3

Uknown: ____ moles of Al

• Solve for the unknown:

3.7 mol Al2O32 mol Al2O3

4 mol Al

= 7.4 mol Al

Coefficients Coefficients in the in the

balanced balanced equationequation

Coefficients Coefficients in the in the

balanced balanced equationequation

4Al(s) + 3O2(g) 2Al2O3(s)

MOLE – MOLE CALCULATIONS

MASS – MASS CALCULATIONS

• No lab balance measures moles directly, instead the mass of a substance is usually measured in grams.

• From the mass of a reactant or product, the mass of any other reactant or product in a given chemical equation can be calculated.

• The mole – mole connection is still vital to do these calcs.

1. Then the mole ratio from the balanced equation can be used to calculate the number of moles of the unknown.

2. If the given sample is measured in grams, the mass can be converted to moles by using the molar mass.

MASS – MASS CALCULATIONS

3.If it is mass of the unknown that needs to be determined, the number of moles of the unknown can be multiplied by the molar mass of the desired compound.

• As in mole-mole calcs, the unknow can be either a reactant or a product.

MASS – MASS CALCULATIONS

• Again back to our synthesis of ammonia rxn.

N2(g) + 3H2(g) 2NH3(g)•Calculate the number of grams of NH3 produced by the reaction of 5.4 g of H2 with an excess of N2.

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

MASS – MASS CALCULATIONS

• What do we know?–Mass of H2 = 5.4 g H2

–3 mol H2 = 2 mol NH3 (from balanced equation - AKA mole ratio)

–Molar mass of H2 = 2.0 g H2

–Molar mass of NH3=17.0g NH3

• What are we asked for?–Mass of ammonia produced

MASS – MASS CALCULATIONS

Step 1: convert mass of given to moles of given using MM of G

5.4 g H22.0 g H2

1 mol H2

= 2.7 mol H2

Coefficients Coefficients in the in the

balanced balanced equationequation

Molar Molar mass of mass of

HH22

MASS – MASS CALCULATIONS

Step 2: convert mols of G to mols of U using the mole ratio

2.7 mol H23 mol H2

2 mol NH3

= 1.8 mol NH3

Coefficients Coefficients in the in the

balanced balanced equationequation

Coefficients Coefficients from the from the balanced balanced equationequation

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

MASS – MASS CALCULATIONS

Step 3: convert moles of desired compound to mass using MM of U

1.8 mol NH31 mol NH3

17.0 g NH3

= 31.0 g NH3

Coefficients Coefficients in the in the

balanced balanced equationequation

Molar Molar mass of mass of

NHNH33

MASS – MASS CALCULATIONS

• Mass to mass calculations always follow those same three steps.

• It uses the mole math that we have had lots of practice with (mass to moles and moles to mass)

• The only difference is the new middle step where we use our newly acquired mole ratio

MASS – MASS CALCULATIONS

• Let’s do another one:Acetylene gas (C2H2) is

produced by adding water to calcium carbide (CaC2).

CaC2 + 2H2O C2H2 + Ca(OH)2

How many grams of acetylene are produced by

adding water to 5.00 g CaC2?

CaC2 + 2H2O C2H2 + Ca(OH)2

MASS – MASS CALCULATIONS

• What do we know?–Mass of CaC2 = 5.0 g CaC2

–1 mol CaC2 = 1 mol C2H2 (from balanced equation)

–MM of CaC2 = 64.0 g CaC2

–MM of C2H2 = 26.0g C2H2

• What are we asked for?–Mass of C2H2 produced

MASS – MASS CALCULATIONS

Step 1: convert mass of given to moles of given using MM of G

5.0 g CaC264.0 g CaC2

1 mol CaC2

= .078mol CaC2

MASS – MASS CALCULATIONS

Step 2: convert mols of G to mols of U using the mole ratio

.078mol CaC21 mol CaC2

1 mol C2H2

= .078mol C2H2

CaC2 + 2H2O C2H2 + Ca(OH)2

MASS – MASS CALCULATIONS

Step 3: convert moles of desired compound to mass using MM of U

.078 mol C2H2

1 mol C2H2

26.0 g C2H2

= 2.03 g C2H2

MASS – MASS CALCULATIONS

MORE MOLE CALCULATIONS

• A balanced reaction equation indicates the relative number of moles of reactants and products.

• We can expand our stoichiometric calculations to include any unit of measure-ment that is related to the mole.

• The given quantity can be expressed in numbers of particles, units of mass, or volumes of gases at STP.

• The problems can include mass-volume, volume-volume, and particle-mass calculations.

MORE MOLE CALCULATIONS

• In any of these problems, the given quantity is first converted to moles.

• Then the mole ratio from the balanced equationn is used to convert from the moles of given to the number of moles of the unknown

MORE MOLE CALCULATIONS

• Then the moles of the unknown are converted to the units that the problem requests.

• The next slide summarizes these steps for all typical stoichiometric problems

MORE MOLE CALCULATIONS