AP Notes Chapter 4 AP Notes Chapter 4

StoichiometryStoichiometry

Atoms are “Letters”Compounds are “Words”Chemical Equations are the

“Sentences of Chemistry”

Reactants Products

Experimental Evidence Shows that in chemical reactions

Mass and atoms conserved

Moles not conserved

The MoleThe Mole The mole is a number.The mole is a number. A very large number, but still, just a A very large number, but still, just a

number.number. 6.022 x 106.022 x 102323 of anything is a mole of anything is a mole A large dozen.A large dozen. The number of atoms in exactly 12 The number of atoms in exactly 12

grams of carbon-12.grams of carbon-12. Makes the numbers on the table the Makes the numbers on the table the

mass of the average atom.mass of the average atom.

1.Balance to conserve atoms:

Mg(s) + O2(g) MgO(s)

2. Balance the equation in #1 for one mole of product.

Interpretations of Equation:2 Mg + O2 2 MgO

Microscopic

Macroscopic2 moles magnesium and one

mole oxygen produce 2 moles magnesium oxide

+ +

Molar massMolar mass

Mass of 1 mole of a substance.Mass of 1 mole of a substance. Often called molecular weight.Often called molecular weight. To determine the molar mass of an To determine the molar mass of an

element, look on the table.element, look on the table. To determine the molar mass of a To determine the molar mass of a

compound, add up the molar masses compound, add up the molar masses of the elements that make it up.of the elements that make it up.

Find the molar mass ofFind the molar mass of

CHCH44

MgMg33PP22

Ca(NOCa(NO33))22

AlAl22(Cr(Cr22OO77))33

CaSOCaSO44 · 2H · 2H22OO

Chemical EquationsChemical Equations

Are sentences.Are sentences. DescribeDescribe what happens in a chemical what happens in a chemical

reaction.reaction. Reactants Reactants Products Products Equations should be Equations should be balanced.balanced. Have the same number of each kind Have the same number of each kind

of atoms on both sides because ...of atoms on both sides because ...

Balancing equationsBalancing equations

CH4 + O2 CO2 + H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2

Balancing equationsBalancing equations

CH4 + O2 CO2 + 2 H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2 4

Balancing equationsBalancing equations

CH4 + O2 CO2 + 2 H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2 4

4

Balancing equationsBalancing equations

CH4 + 2O2 CO2 + 2 H2OReactantsReactants ProductsProducts

C1 1

O2 3

H4 2 4

44

AbbreviationsAbbreviations

((ss) ) ((gg) ) ((aqaq)) heatheat

catalystcatalyst

PracticePractice

Ca(OH)Ca(OH)22 + H + H33POPO44 H H22O + CaO + Ca33(PO(PO44))22 Cr + SCr + S88 Cr Cr22SS33 KClOKClO33((ss) ) ClCl22((gg) + O) + O22((gg))

Solid iron(III) sulfide reacts with gaseous Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas.chloride and hydrogen sulfide gas.

FeFe22OO33((ss) + Al() + Al(ss) ) Fe( Fe(ss) + Al) + Al22OO33((ss))

Write and balance the equation for the reaction in which ammonium nitrate decomposes to form dinitrogen oxide and water.

Write and balance the equation for the reaction of calcium phosphate and phosphoric acid (hydrogen phosphate) to form calcium dihydrogen phosphate.

MeaningMeaning A balanced equation can be used to A balanced equation can be used to

describe a reaction in molecules and describe a reaction in molecules and atoms.atoms.

Not grams.Not grams. Chemical reactions happen molecules Chemical reactions happen molecules

at a timeat a time or dozens of molecules at a timeor dozens of molecules at a time or moles of molecules. or moles of molecules.

Stoichiometry

Calculations based on chemical formulas or equations

StoichiometryStoichiometry

Given an amount of either starting Given an amount of either starting material or product, determining the material or product, determining the other quantities.other quantities.

use conversion factors fromuse conversion factors from molar mass (g - mole)molar mass (g - mole) balanced equation (mole - mole)balanced equation (mole - mole)

keep track.keep track.

Map of the World of Stoichiometry

grams grams

Equationmole 1 mole 2

atoms ormolecules

atoms ormolecules

1. How many grams of dinitrogen oxide are produced from the decomposition of 1.00 x 103 grams of ammonium nitrate?

2. Calculate the mass of chlorine required to react with 10.0 grams of sodium metal.

3. Calculate the mass of sodium nitrate that must decompose to produce 128 grams of sodium nitrite.

4. Calculate the mass of nitrogen needed to make 1000. grams of nitric acid by the process given.

5. What is the percent yield if 106 grams of sodium chlorite are isolated from the reaction of 202.3 grams of ClO2 with excess sodium hydroxide?

6. A volume of 3.42 mL of SiCl4 (density = 1.48 g/mL) reacts with excess hydrogen sulfide giving HSSiCl3. The hydrogen chloride produced reacts with 0.449 gram of NaOH. What is the percent yield?

ExamplesExamples

One way of producing OOne way of producing O22((gg) involves the ) involves the

decomposition of potassium chlorate decomposition of potassium chlorate into potassium chloride and oxygen gas. into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is A 25.5 g sample of Potassium chlorate is decomposed. How many moles of Odecomposed. How many moles of O22(g) (g)

are produced? are produced? How many grams of potassium chloride?How many grams of potassium chloride? How many grams of oxygen?How many grams of oxygen?

ExamplesExamples

A piece of aluminum foil 5.11 in x 3.23 A piece of aluminum foil 5.11 in x 3.23 in x 0.0381 in is dissolved in excess in x 0.0381 in is dissolved in excess HCl(aq). How many grams of HHCl(aq). How many grams of H22((gg) are ) are produced?produced?

How many grams of each reactant are How many grams of each reactant are needed to produce 15 grams of iron needed to produce 15 grams of iron form the following reaction? form the following reaction?

Fe Fe22OO33((ss) + Al() + Al(ss) ) Fe( Fe(ss) + Al) + Al22OO33((ss))

ExamplesExamples

KK22PtClPtCl44((aqaq)) + NH + NH33((aqaq)) Pt(NH Pt(NH33))22ClCl22((ss)) + KCl + KCl((aqaq))

What mass of Pt(NHWhat mass of Pt(NH33))22ClCl2 2 can be produced can be produced

from 65 g of Kfrom 65 g of K22PtClPtCl44 ? ?

How much KCl will be produced?How much KCl will be produced?

How much from 65 grams of NHHow much from 65 grams of NH33??

Limiting ReagentLimiting Reagent Reactant that determines the Reactant that determines the

amount of product formed. amount of product formed. The one you run out of first.The one you run out of first. Makes the least product.Makes the least product. Book shows you a ratio method.Book shows you a ratio method. It works. It works. So does mineSo does mine

YieldYieldHow much you get from an chemical How much you get from an chemical

reactionreaction

Limiting ReagentLimiting Reagent

To determine the limiting reagent To determine the limiting reagent requires that you do two requires that you do two stoichiometry problems.stoichiometry problems.

Figure out how much product each Figure out how much product each reactant makes.reactant makes.

The one that makes the least is the The one that makes the least is the limiting reagent.limiting reagent.

Limiting Reagent

Controls the amount of product generated in a reaction

1. What mass of S2Cl2 gas can be prepared from 32.0 grams of sulfur and 71.0 grams of chlorine gas?

2. What is the percent yield for a reaction in which 878 grams of ammonia gas are made from 1.00 x 103 g of nitrogen gas and 5.00 x 102 grams of hydrogen gas?

3. What remains in the reaction vessel after 150. grams of carbon tetrachloride liquid reacts with 100. grams of antimony(III) fluoride to form difluoro-dichloromethane (CCl2F2) and antimony(III) chloride solid?

ExampleExample

Ammonia is produced by the following Ammonia is produced by the following reactionreaction

N N22 + H + H22 NH NH3 3 What What

mass of ammonia can be produced mass of ammonia can be produced from a mixture of 100. g Nfrom a mixture of 100. g N22 and 500. g and 500. g

HH22 ? ?

How much unreacted material How much unreacted material remains?remains?

Excess ReagentExcess Reagent

The reactant you don’t run out of.The reactant you don’t run out of. The amount of stuff you make is the The amount of stuff you make is the

yield.yield. The The theoretical yield theoretical yield is the amount is the amount

you would make if everything went you would make if everything went perfect.perfect.

The The actual yield actual yield is what you make in is what you make in the lab.the lab.

Percent YieldPercent Yield

% yield = Actual x 100%% yield = Actual x 100% Theoretical Theoretical

% yield = what you got x % yield = what you got x 100%100% what you could have what you could have gotgot

ExamplesExamples

Aluminum burns in bromine Aluminum burns in bromine producing aluminum bromide. In a producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of with excess bromine. 50.3 g of aluminum bromide are produced. aluminum bromide are produced. What are the three types of yield.What are the three types of yield.

ExamplesExamples Years of experience have proven that Years of experience have proven that

the percent yield for the following the percent yield for the following reaction is 74.3%reaction is 74.3% Hg Hg + Br+ Br22 HgBr HgBr22

If 10.0 g of Hg and 9.00 g of BrIf 10.0 g of Hg and 9.00 g of Br22 are are

reacted, how much HgBrreacted, how much HgBr22 will be will be

produced?produced? If the reaction did go to completion, If the reaction did go to completion,

how much excess reagent would be how much excess reagent would be left?left?

ExamplesExamples Commercial brass is an alloy of Cu and Commercial brass is an alloy of Cu and

Zn. It reacts with HCl by the following Zn. It reacts with HCl by the following reaction reaction

Zn(s) + 2HCl(aq) Zn(s) + 2HCl(aq) ZnCl ZnCl22 (aq) + H (aq) + H22(g) (g)

Cu does not react. When 0.5065 g of Cu does not react. When 0.5065 g of brass is reacted with excess HCl, brass is reacted with excess HCl, 0.0985 g of ZnCl0.0985 g of ZnCl22 are eventually are eventually

isolated. isolated.

Pure O2 in xCO2 is absorbed

y H2O is absorbed

2

Sample is burned completely to form

CO2 and H2O

Stoichiometry & Combustion Analysis

C x H y

What is the balanced equation for the combustion of octane (C8H18)?

Problem 35 p. 142The aluminum in a 0.764 g sample of

an unknown material was precipitated as aluminum hydroxide Al(OH)3, which was then converted to Al2O3 by heating strongly. If 0.127 g of Al2O3 is obtained from the 0.764 g sample, what is the mass percent of aluminum in the sample?

Problem 30 p. 142Analysis of Mixtures

A mixture of CuSO4 and CuSO4 • 5H2O has a mass of 1.245 g. but, after heating to drive off all the water, the mass is only 0.832 g. What is the mass percent of CuSO4 • 5H2O in the mixture?