STOICHIOMETRY STOICHIOMETRY BASICSBASICSUnit 12Unit 12

ChemistryChemistry

LangleyLangley

**Corresponds to Chapter 12 in the Prentice Hall Chemistry Textbook

INTRODUCTION INTRODUCTION

StoichiometryStoichiometry Dervived from two Greek words meaning measured Dervived from two Greek words meaning measured

and elementand element 4 types4 types

Mole-MoleMole-Mole Mass-MoleMass-Mole Mass-MassMass-Mass Volume-VolumeVolume-Volume

Mathematically basedMathematically based Balanced equations very important; first place to Balanced equations very important; first place to

startstart

Interpreting Chemical Interpreting Chemical ReactionReaction

From a balanced chemical equation, you From a balanced chemical equation, you can describe a chemical reaction in terms can describe a chemical reaction in terms of quantities of products and reactants. of quantities of products and reactants. These quantities include the number of These quantities include the number of molecules, atoms, moles, mass, and molecules, atoms, moles, mass, and volume of the compounds used as well volume of the compounds used as well as each element used to make the as each element used to make the respective compoundsrespective compounds

Interpreting a Chemical Interpreting a Chemical ReactionReaction

NN22 + 3H + 3H22 2NH 2NH33 (Figure 12.3, pg. 357) (Figure 12.3, pg. 357) Number of atoms: 2 atoms of nitrogen, 6 atoms of Number of atoms: 2 atoms of nitrogen, 6 atoms of

hydrogen, 8 atoms of ammoniahydrogen, 8 atoms of ammonia Number of molecules: 1 molecule of nitrogen, 3 Number of molecules: 1 molecule of nitrogen, 3

molecules, 2 molecules of ammonia (1:3:2 ratio)molecules, 2 molecules of ammonia (1:3:2 ratio) Number of relative moles: 1 mole of nitrogen, 3 Number of relative moles: 1 mole of nitrogen, 3

relative moles of hydrogen, 2 relative moles of relative moles of hydrogen, 2 relative moles of ammoniaammonia Molecules and moles are not the same thingMolecules and moles are not the same thing

Mass: 28.0 g of nitrogen, 6.0 g of hydrogen, 34 g of Mass: 28.0 g of nitrogen, 6.0 g of hydrogen, 34 g of ammoniumammonium

Volume: 22.4L of nitrogen, 67.2L of hydrogen, Volume: 22.4L of nitrogen, 67.2L of hydrogen, 44.8L of ammonia44.8L of ammonia Because 1 mol of any gas at STP occupies a volume of Because 1 mol of any gas at STP occupies a volume of

22.4L22.4L

Molar Mass of Molar Mass of CompoundsCompounds

The molar mass (MM) of a compound is The molar mass (MM) of a compound is determined the same way, except now you add determined the same way, except now you add up all the atomic masses for the molecule (or up all the atomic masses for the molecule (or compound)compound) Ex. Molar mass of CaClEx. Molar mass of CaCl22 Avg. Atomic mass of Calcium = 40.08gAvg. Atomic mass of Calcium = 40.08g Avg. Atomic mass of Chlorine = 35.45gAvg. Atomic mass of Chlorine = 35.45g Molar Mass of calcium chloride = Molar Mass of calcium chloride =

40.08 g/mol Ca + (2 X 35.45) g/mol Cl40.08 g/mol Ca + (2 X 35.45) g/mol Cl 110.98 g/mol CaCl 110.98 g/mol CaCl22

20

Ca  40.08 17

Cl

35.45

Molar Mass CalculationMolar Mass Calculation

Calculate the Molar Mass of calcium Calculate the Molar Mass of calcium phosphatephosphate Formula = (NH4)Formula = (NH4)33(PO(PO44))

Masses elements:Masses elements:

Molar Mass = Molar Mass =

Atoms or Molecules

Moles

Mass (grams)

Divide by 6.02 X 1023

Multiply by 6.02 X 1023

Multiply by atomic/molar mass from periodic table

Divide by atomic/molar mass from periodic table

molar mass Avogadro’s numbermolar mass Avogadro’s number Grams Grams MolesMoles particles particles

Everything must go Everything must go through Moles!!!through Moles!!!

CalculationsCalculations

Chocolate Chip Cookies!!Chocolate Chip Cookies!!1 cup butter

1/2 cup white sugar

1 cup packed brown sugar

1 teaspoon vanilla extract

2 eggs

2 1/2 cups all-purpose flour

1 teaspoon baking soda

1 teaspoon salt

2 cups semisweet chocolate chips

Makes 3 dozen

How many eggs are needed to make 3 dozen cookies?

How much butter is needed for the amount of chocolate chips used?

How many eggs would we need to make 9 dozen cookies?

How much brown sugar would I need if I had 1 ½ cups white sugar?

Recipes and EquationsRecipes and Equations

Just like chocolate chip cookies have recipes, Just like chocolate chip cookies have recipes, chemists have recipes as wellchemists have recipes as well

Instead of calling them recipes, we call them Instead of calling them recipes, we call them reaction equationsreaction equations

Furthermore, instead of using cups and Furthermore, instead of using cups and teaspoons, we use molesteaspoons, we use moles

Lastly, instead of eggs, butter, sugar, etc. we Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredientsuse chemical compounds as ingredients

Chemistry RecipesChemistry Recipes Looking at a reaction tells us how much of Looking at a reaction tells us how much of

something you need to react with something you need to react with something else to get a product (like the something else to get a product (like the cookie recipe)cookie recipe)

Be sure you have a balanced reaction Be sure you have a balanced reaction before you start!before you start!

Example: 2 Na + ClExample: 2 Na + Cl2 2 2 NaCl 2 NaCl This reaction tells us that by mixing 2 moles of This reaction tells us that by mixing 2 moles of

sodium with 1 mole of chlorine we will get 2 moles sodium with 1 mole of chlorine we will get 2 moles of sodium chlorideof sodium chloride

What if we wanted 4 moles of NaCl? 10 moles? What if we wanted 4 moles of NaCl? 10 moles? 50 moles?50 moles?

Starting Point-Balanced Starting Point-Balanced EquationEquation

Write the balanced reaction for hydrogen gas Write the balanced reaction for hydrogen gas reacting with oxygen gas.reacting with oxygen gas.

2 H2 H22 + O + O22 2 H 2 H22OO How many moles of reactants are needed?How many moles of reactants are needed? What if we wanted 4 moles of water?What if we wanted 4 moles of water? What if we had 3 moles of oxygen, how much What if we had 3 moles of oxygen, how much

hydrogen would we need to react and how much hydrogen would we need to react and how much water would we get?water would we get?

What if we had 50 moles of hydrogen, how much What if we had 50 moles of hydrogen, how much oxygen would we need and how much water oxygen would we need and how much water produced? produced?

Mole-Mole Calculations Mole-Mole Calculations These mole ratios can be used to These mole ratios can be used to

calculate the moles of one chemical from calculate the moles of one chemical from the given amount of a different chemical the given amount of a different chemical

Example: How many moles of chlorine is Example: How many moles of chlorine is needed to react with 5 moles of sodium needed to react with 5 moles of sodium (without any sodium left over)?(without any sodium left over)?

2 Na + Cl2 Na + Cl22 2 NaCl 2 NaCl

5 moles Na 1 mol Cl2

2 mol Na= 2.5 moles Cl2

Mole-Mole CalculationsMole-Mole Calculations

How many moles of ammonia are How many moles of ammonia are produced when 0.60 moles of nitrogen produced when 0.60 moles of nitrogen reacts with hydrogen?reacts with hydrogen? NN22(g) + 3H(g) + 3H22 (g) (g) 2NH 2NH33(g)(g)

0.6 mol N22.0 mol NH3

1.0 mol N2

= 1.2 mol NH3

Mass-Mole CalculationsMass-Mole Calculations Sometimes you are going to start with mass Sometimes you are going to start with mass

and will have to convert to moles of product or and will have to convert to moles of product or another reactantanother reactant

We use molar mass and the mole ratio to get We use molar mass and the mole ratio to get to moles of the compound of interestto moles of the compound of interest Calculate the number of moles of ethane (CCalculate the number of moles of ethane (C22HH66) )

needed to produce 10.0 g of waterneeded to produce 10.0 g of water 2 C2 C22HH66 + 7 O + 7 O22 4 CO 4 CO22 + 6 H + 6 H220 0

10.0 g H2O 1 mol H2O 2 mol C2H6

18.0 g H2O 6 mol H20= 0.185

mol C2H6

Mass-Mole CalculationsMass-Mole Calculations

Calculate how many moles of oxygen are Calculate how many moles of oxygen are required to make 10.0 g of aluminum required to make 10.0 g of aluminum oxideoxide

Mole-Mass CalculationsMole-Mass Calculations Most of the time in chemistry, the amounts are Most of the time in chemistry, the amounts are

given in grams instead of molesgiven in grams instead of moles We still go through moles and use the mole ratio, We still go through moles and use the mole ratio,

but now we also use molar mass to get to gramsbut now we also use molar mass to get to grams Example: How many grams of chlorine are required Example: How many grams of chlorine are required

to react completely with 5.00 moles of sodium to to react completely with 5.00 moles of sodium to produce sodium chloride?produce sodium chloride?

2 Na + Cl2 Na + Cl22 2 NaCl 2 NaCl

5.00 moles Na 1 mol Cl2 70.90g Cl2

2 mol Na 1 mol Cl2

= 177g Cl2

Mole-Mass CalculationsMole-Mass Calculations

Calculate the mass in grams of Iodine Calculate the mass in grams of Iodine required to react completely with 0.50 required to react completely with 0.50 moles of aluminum.moles of aluminum.

Mass-Mass CalculationsMass-Mass Calculations Most often we are given a starting mass Most often we are given a starting mass

and want to find out the mass of a product and want to find out the mass of a product we will get (called theoretical yield) or how we will get (called theoretical yield) or how much of another reactant we need to much of another reactant we need to completely react with it (no leftover completely react with it (no leftover ingredients!)ingredients!)

Now we must go from grams to moles, Now we must go from grams to moles, mole ratio, and back to grams of mole ratio, and back to grams of compound we are interested incompound we are interested in

Mass-Mass CalculationsMass-Mass Calculations

Ex. Calculate how many grams of Ex. Calculate how many grams of ammonia are produced when you react ammonia are produced when you react 2.00g of nitrogen with excess hydrogen.2.00g of nitrogen with excess hydrogen.

NN2 2 + 3 H+ 3 H2 2 2 NH 2 NH33

2.00g N2 1 mol N2 2 mol NH3 17.06g NH3

28.02g N2 1 mol N2 1 mol NH3

= 2.4 g NH3

Mass-Mass CalculationsMass-Mass Calculations

How many grams of calcium nitride are How many grams of calcium nitride are produced when 2.00 g of calcium reacts produced when 2.00 g of calcium reacts with an excess of nitrogen?with an excess of nitrogen?

Volume-Volume Volume-Volume CalculationsCalculations

1 mole of any substance = 22.4L1 mole of any substance = 22.4L Nitrogen monoxide and oxygen gas Nitrogen monoxide and oxygen gas

combine to form the brown gas nitrogen combine to form the brown gas nitrogen dioxide, which contributes to dioxide, which contributes to photochemical smog. How many liters of photochemical smog. How many liters of nitrogen dioxide are produced when 34 L nitrogen dioxide are produced when 34 L of oxygen reacts with an excess of of oxygen reacts with an excess of nitrogen monoxide? Assume conditions nitrogen monoxide? Assume conditions of STP.of STP.

Volume-Volume Volume-Volume CalculationsCalculations

2NO + O2NO + O22(g) (g) 2NO 2NO22(g)(g) Know: volume of oxygen = 34L OKnow: volume of oxygen = 34L O22

2 mol NO2 mol NO22/1 mol O/1 mol O22

1 mol O1 mol O22 = 22.4 L O = 22.4 L O22 (at STP) (at STP)

1 mole NO1 mole NO22 = 22.4 L NO (at STP) = 22.4 L NO (at STP)

34 L O2 1 mol O2

22.4 L O2

2 mol NO2

1 mol O2 1 mol NO2

22.4 L NO2 = 64 L O2

Limiting ReactantsLimiting Reactants1 cup butter

1/2 cup white sugar

1 cup packed brown sugar

1 teaspoon vanilla extract

2 eggs

2 1/2 cups all-purpose flour

1 teaspoon baking soda

1 teaspoon salt

2 cups semisweet chocolate chips

Makes 3 dozen

If we had the specified amount of all ingredients listed, could we make 4 dozen cookies?

What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies?

What if we only had one egg, could we make 3 dozen cookies?

Limiting ReactantLimiting Reactant

Most of the time in chemistry we have more of Most of the time in chemistry we have more of one reactant than we need to completely use one reactant than we need to completely use up other reactant.up other reactant.

That reactant is said to be in That reactant is said to be in excessexcess (there is (there is too much).too much).

The other reactant limits how much product The other reactant limits how much product we get. Once it runs out, the reaction. we get. Once it runs out, the reaction. This is called the This is called the limiting reactantlimiting reactant..

Limiting ReactantLimiting Reactant To find the correct answer, we have to try all of To find the correct answer, we have to try all of

the reactants. We have to calculate how much the reactants. We have to calculate how much of of aa product we can get from each of the product we can get from each of the reactants to determine which reactant is the reactants to determine which reactant is the limiting one.limiting one.

The lower amount of The lower amount of aa product is the correct product is the correct answer. answer.

The reactant that makes the least amount of The reactant that makes the least amount of product is the limiting reactant. Once you product is the limiting reactant. Once you determine the limiting reactant, you should determine the limiting reactant, you should ALWAYS start with it!ALWAYS start with it!

Be sure to pick Be sure to pick aa product! You can’t compare to product! You can’t compare to see which is greater and which is lower unless see which is greater and which is lower unless the product is the same!the product is the same!

Limiting Reactant:Limiting Reactant: 10.0g of aluminum reacts with 35.0 grams of 10.0g of aluminum reacts with 35.0 grams of

chlorine gas to produce aluminum chloride. Which chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how reactant is limiting, which is in excess, and how much product is produced?much product is produced?

2 Al + 3 Cl2 Al + 3 Cl22 2 AlCl 2 AlCl33 Start with Al:Start with Al:

Now ClNow Cl22::

10.0 g Al 1 mol Al 2 mol AlCl3 133.5 g AlCl3

27.0 g Al 2 mol Al 1 mol AlCl3

= 49.4g AlCl3

35.0g Cl2 1 mol Cl2 2 mol AlCl3 133.5 g AlCl3

71.0 g Cl2 3 mol Cl2 1 mol AlCl3

= 43.9g AlCl3

LimitingLimitingReactantReactant

Limiting ReactantLimiting Reactant We get We get 49.4g49.4g of aluminum chloride from the given of aluminum chloride from the given

amount of aluminum, but only amount of aluminum, but only 43.9g43.9g of aluminum of aluminum chloride from the given amount of chlorine. chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once Therefore, chlorine is the limiting reactant. Once the 35.0g of chlorine is used up, the reaction the 35.0g of chlorine is used up, the reaction comes to a complete .comes to a complete .

Limiting Reactant Limiting Reactant

15.0 g of potassium reacts with 15.0 g of 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is iodine. Calculate which reactant is limiting and how much product is made.limiting and how much product is made.

Finding the Amount of Finding the Amount of ExcessExcess

By calculating the amount of the excess By calculating the amount of the excess reactant needed to completely react with reactant needed to completely react with the limiting reactant, we can subtract that the limiting reactant, we can subtract that amount from the given amount to find the amount from the given amount to find the amount of excess.amount of excess.

Can we find the amount of excess Can we find the amount of excess potassium in the previous problem?potassium in the previous problem?

Finding ExcessFinding Excess 15.0 g of potassium reacts with 15.0 g of iodine. 15.0 g of potassium reacts with 15.0 g of iodine.

2 K + I2 K + I22 2 KI 2 KI We found that Iodine is the limiting reactant, and We found that Iodine is the limiting reactant, and

19.6 g of potassium iodide are produced.19.6 g of potassium iodide are produced.

15.0 g I2 1 mol I2 2 mol K 39.1 g K

254 g I2 1 mol I2 1 mol K= 4.62 g K USED!

15.0 g K – 4.62 g K = 10.38 g K EXCESS

Given amount of excess reactant

Amount of excess reactant actually used

Note that we started with the limiting reactant! Once you determine the LR, you should only start with it!

Limiting Reactant: RecapLimiting Reactant: Recap

Convert ALL of the reactants to the SAME product (pick Convert ALL of the reactants to the SAME product (pick any product you choose.)any product you choose.)

The lowest answer is the correct answer.The lowest answer is the correct answer. The reactant that gave you the lowest answer is the The reactant that gave you the lowest answer is the

LIMITING REACTANT.LIMITING REACTANT. The other reactant(s) are in EXCESS.The other reactant(s) are in EXCESS. To find the amount of excess, subtract the amount used To find the amount of excess, subtract the amount used

from the given amount.from the given amount. You can recognize a limiting reactant problem because You can recognize a limiting reactant problem because

there is MORE THAN ONE GIVEN AMOUNT.there is MORE THAN ONE GIVEN AMOUNT. If you have to find more than one product, be sure to If you have to find more than one product, be sure to

start with the limiting reactant. You don’t have to start with the limiting reactant. You don’t have to determine which is the LR over and over again!determine which is the LR over and over again!

Empirical vs. Molecular Empirical vs. Molecular FormulasFormulas

Empirical Formula of a compound shows the Empirical Formula of a compound shows the smalles whole number ration of the atoms in smalles whole number ration of the atoms in the compoundthe compound A compound is analyzed and found to contain A compound is analyzed and found to contain

25.9% nitrogen and 74.1% oxygen. What is the 25.9% nitrogen and 74.1% oxygen. What is the emprical formula of the compound?emprical formula of the compound? 25.9 g N * (1 mol/14.0g) = 1.85 mol N25.9 g N * (1 mol/14.0g) = 1.85 mol N 74.1 g O * (1 mol O/16.0g) = 4.63 mol O74.1 g O * (1 mol O/16.0g) = 4.63 mol O Empirical Formula = NEmpirical Formula = N22OO55

Empirical vs. Molecular Empirical vs. Molecular FormulasFormulas

Molecular Formula tells you the actual number Molecular Formula tells you the actual number of each kind of atom present in a molecule of of each kind of atom present in a molecule of the compoundthe compound

Calculate the molecular formula of a compound Calculate the molecular formula of a compound whose molar mass is 60.0 g and empirical whose molar mass is 60.0 g and empirical formula is CHformula is CH44N.N. Find empirical formula mass: 12+4+14=30gFind empirical formula mass: 12+4+14=30g Molar mass/efm = 60/30 = 2Molar mass/efm = 60/30 = 2 Molecular Formula = CMolecular Formula = C22HH88NN22

Percent, Theoretical, and Percent, Theoretical, and Actual YieldActual Yield

Actual Yield – the amount produced in a Actual Yield – the amount produced in a reaction (amount of products)reaction (amount of products)

Theoretical Yield – the amount that should Theoretical Yield – the amount that should have been produced in a reactionhave been produced in a reaction Maximum amount of product that could be formed Maximum amount of product that could be formed

from given amount of reactantsfrom given amount of reactants

Percent Yield – (actual/theoretical) X 100%Percent Yield – (actual/theoretical) X 100% Often less than the theoreticalOften less than the theoretical