summer chemistry ap exam review copyright © 1999 by harcourt brace & company all rights...

SUMMER SUMMER CHEMISTRYCHEMISTRY

AP EXAM REVIEWAP EXAM REVIEW

Copyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to:Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida 32887-6777

Copyright (c) 1999 by Harcourt Brace & CompanyAll rights reserved

The Language of The Language of ChemistryChemistry

• CHEMICAL ELEMENTSCHEMICAL ELEMENTS - pure substances that cannot - pure substances that cannot be decomposed by ordinary means to other substances.be decomposed by ordinary means to other substances.

CHEMICAL COMPOUNDS CHEMICAL COMPOUNDS are composed of are composed of atoms and so can be decomposed to those atoms.atoms and so can be decomposed to those atoms.


The Language of The Language of ChemistryChemistry• The elements, their names, The elements, their names,

and symbols are given on and symbols are given on the the PERIODIC TABLEPERIODIC TABLE

• How many elements are How many elements are there?there?


Copper Copper atoms on atoms on silica silica surface.surface.

See See Screen Screen 1.41.4

• An An atomatom is the smallest particle of an is the smallest particle of an element that has the chemical element that has the chemical properties of the element.properties of the element.


ATOM ATOM COMPOSITIONCOMPOSITION

•protons and neutrons in protons and neutrons in the nucleus.the nucleus.

•the number of electrons is equal to the the number of electrons is equal to the number of protons.number of protons.

•electrons in space around the nucleus.electrons in space around the nucleus.

•extremely small. One teaspoon of water has extremely small. One teaspoon of water has 3 times as many atoms as the Atlantic Ocean 3 times as many atoms as the Atlantic Ocean has teaspoons of water.has teaspoons of water.

The atom is mostlyThe atom is mostlyempty spaceempty space


ATOMIC COMPOSITIONATOMIC COMPOSITION• ProtonsProtons

– + electrical charge+ electrical charge

– mass = 1.672623 x 10mass = 1.672623 x 10-24-24 g g

– relative mass = 1.007 atomic relative mass = 1.007 atomic mass units (amu)mass units (amu)

• ElectronsElectrons– negative electrical chargenegative electrical charge

– relative mass = 0.0005 amurelative mass = 0.0005 amu

• NeutronsNeutrons– no electrical chargeno electrical charge

– mass = 1.009 amumass = 1.009 amu


A A MOLECULEMOLECULE is the smallest unit of is the smallest unit of a compound that retains the chemical a compound that retains the chemical characteristics of the compound.characteristics of the compound.Composition of molecules is given by a Composition of molecules is given by a

MOLECULAR FORMULAMOLECULAR FORMULA

H2O C8H10N4O2 - caffeine


KINETIC NATURE OF MATTERKINETIC NATURE OF MATTERMatter consists of atoms and molecules Matter consists of atoms and molecules

in motion.in motion.


PROBLEM: Mercury (Hg) has a density PROBLEM: Mercury (Hg) has a density of 13.6 g/cmof 13.6 g/cm33. What is the mass of 95 mL . What is the mass of 95 mL of Hg? In grams? In pounds?of Hg? In grams? In pounds?

PROBLEM: Mercury (Hg) has a density PROBLEM: Mercury (Hg) has a density of 13.6 g/cmof 13.6 g/cm33. What is the mass of 95 mL . What is the mass of 95 mL of Hg? In grams? In pounds?of Hg? In grams? In pounds?

Solve the problem using Solve the problem using DIMENSIONAL DIMENSIONAL ANALYSIS.ANALYSIS.


PROBLEM: Mercury (Hg) has a density PROBLEM: Mercury (Hg) has a density of 13.6 g/cmof 13.6 g/cm33. What is the mass of 95 mL . What is the mass of 95 mL of Hg?of Hg?


First, note that First, note that 1 cm1 cm33 = 1 mL = 1 mL


Then, use dimensional analysis to Then, use dimensional analysis to calculate mass.calculate mass.



First, note that First, note that 1 cm1 cm33 = 1 mL = 1 mL

95 cm3 • 13.6 g

cm3 = 1.3 x 103 g


Use dimensional analysis.Use dimensional analysis.

What is the mass in pounds? (1 lb = 454 g)What is the mass in pounds? (1 lb = 454 g)



95 cm3 • 13.6 g

cm3 = 1.3 x 103 g


Use dimensional analysis.Use dimensional analysis.

What is the mass in pounds? (1 lb = 454 g)What is the mass in pounds? (1 lb = 454 g)



95 cm3 • 13.6 g

cm3 = 1.3 x 103 g

1.3 x 103 g • 1 lb

454 g = 2.8 lb

Counting AtomsCounting Atoms• Mg burns in air (OMg burns in air (O22) to ) to

produce white produce white magnesium oxide, MgO. magnesium oxide, MgO.

• How can we figure out How can we figure out how much oxide is how much oxide is produced from a given produced from a given mass of Mg?mass of Mg?



Counting AtomsCounting Atoms

• Chemistry is a quantitative science—Chemistry is a quantitative science—we need a “counting unit.”we need a “counting unit.”

• The The MOLEMOLE• 1 mole is the amount of substance that 1 mole is the amount of substance that

contains as many particles (atoms, contains as many particles (atoms, molecules) as there are in 12.0 g of molecules) as there are in 12.0 g of 1212C.C.


Particles in a MoleParticles in a Mole

6.0221367 x 106.0221367 x 102323

Avogadro’s NumberAvogadro’s Number

Amedeo AvogadroAmedeo Avogadro1776-18561776-1856

There is Avogadro’s number of particles in a mole of any substance.There is Avogadro’s number of particles in a mole of any substance.


Molar MassMolar Mass

1 mol of 1 mol of 1212C = 12.00 g of CC = 12.00 g of C = 6.022 x 10 = 6.022 x 102323 atoms of C atoms of C

12.00 g of 12.00 g of 1212C is its MOLAR MASSC is its MOLAR MASS

Taking into account all of the isotopes of Taking into account all of the isotopes of

C, the molar mass of C is 12.011 g/molC, the molar mass of C is 12.011 g/mol

13

Al

26.9815

atomic number

symbol

atomic weight

13

Al

26.9815

atomic number

symbol

atomic weight

Find molar Find molar mass from mass from periodic periodic tabletable


PROBLEM: How many moles PROBLEM: How many moles are represented by 0.200 g of are represented by 0.200 g of Mg?Mg?





Mg has a molar mass of 24.3050 g/mol.





0.200 g • 1 mol

24.31 g = 8.23 x 10-3 mol





How many atoms in this piece of Mg?

0.200 g • 1 mol

24.31 g = 8.23 x 10-3 mol






0.200 g • 1 mol

24.31 g = 8.23 x 10-3 mol

8.23 x 10-3 mol • 6.022 x 1023 atoms

1 mol






= 4.95 x 1021 atoms Mg

0.200 g • 1 mol

24.31 g = 8.23 x 10-3 mol

8.23 x 10-3 mol • 6.022 x 1023 atoms

1 mol


Device for Remembering Device for Remembering How to Do Mass/Mole How to Do Mass/Mole

CalculationsCalculations

mass

# moles MW

massMW# moles # moles=

mass # moles MW=

MW=mass


RadioactivityRadioactivity• One of the pieces of evidence for the One of the pieces of evidence for the

fact that atoms are made of smaller fact that atoms are made of smaller particles came from the work of particles came from the work of

Marie CurieMarie Curie (1876-1934). (1876-1934).

• She discovered She discovered radioactivityradioactivity, the , the spontaneous disintegration of some spontaneous disintegration of some elements into smaller pieces.elements into smaller pieces.


The modern view of the atom was developed by The modern view of the atom was developed by

Ernest RutherfordErnest Rutherford (1871-1937).(1871-1937).


Atomic Number, ZAtomic Number, Z

All atoms of the same element All atoms of the same element have the same number of have the same number of protons in the nucleus, protons in the nucleus, ZZ

13

Al

26.9815

atomic number

symbol

atomic weight


Mass Number, AMass Number, A• C atom with 6 protons and 6 neutrons C atom with 6 protons and 6 neutrons

is the mass standard is the mass standard

• = 12 atomic mass units= 12 atomic mass units

• Mass NumberMass Number = # protons + # neutrons= # protons + # neutrons

• A boron atom can have A boron atom can have A = 5 p + 5 n = 10 amuA = 5 p + 5 n = 10 amu


Mass Number, AMass Number, A• C atom with 6 protons and 6 neutrons C atom with 6 protons and 6 neutrons

is the mass standard is the mass standard

• = 12 atomic mass units= 12 atomic mass units

• Mass NumberMass Number = # protons + # neutrons= # protons + # neutrons

• A boron atom can have A boron atom can have A = 5 p + 5 n = 10 amuA = 5 p + 5 n = 10 amu

A

Z

10

5B

A

Z

10

5B


IsotopesIsotopes• Atoms of the same element (same Z) Atoms of the same element (same Z)

but different mass number (A).but different mass number (A).

• Boron-10 (Boron-10 (1010B) has 5 p and 5 nB) has 5 p and 5 n

• Boron-11 (Boron-11 (1111B) has 5 p and 6 nB) has 5 p and 6 n

10B

11B


IsotopesIsotopes

• Because of isotopes, the mass of a an element Because of isotopes, the mass of a an element on the Periodic Table has an average value.on the Periodic Table has an average value.

• Average mass = Average mass = ATOMIC WEIGHTATOMIC WEIGHT• Boron is 20% Boron is 20% 1010B and 80% B and 80% 1111B. That is, B. That is, 1111B is B is

80 percent abundant on earth. 80 percent abundant on earth.

• For boron atomic weightFor boron atomic weight

= 0.20 (10 amu) + 0.80 (11 amu) = 10.8 amu= 0.20 (10 amu) + 0.80 (11 amu) = 10.8 amu

You should be able to duplicate this calculationYou should be able to duplicate this calculation..

10B

11B


Radioactivity Changes Elements

• In radioactive decay, one element changes to another by emitting (giving off) a particle from the nucleus

• Alpha particles ) = Helium nucleus 42He

• Beta particles () = electron 0-1e-

• In a nuclear reaction, mass (A) is conserved, atomic number (Z) is conserved

• Alpha decay: 23892U ---> 234

90Th + 42He

• Beta decay: 23490Th ---> 234

91Pa + 0-1e-


Periodic TablePeriodic Table• Dmitri Mendeleev developed the Dmitri Mendeleev developed the

modern periodic table. Argued modern periodic table. Argued that element properties are that element properties are periodic functions of their atomic periodic functions of their atomic weights.weights.

• We now know that element We now know that element properties are periodic functions properties are periodic functions of their of their ATOMIC NUMBERSATOMIC NUMBERS..

• See CD-ROM, Screen 2.16.See CD-ROM, Screen 2.16.


GroupsGroups in the Periodic Table in the Periodic Table


Compounds Compounds & Molecules& Molecules

Compounds Compounds & Molecules& Molecules

• COMPOUNDSCOMPOUNDS are a combination of 2 or are a combination of 2 or

more elements in definite ratios by mass. more elements in definite ratios by mass.

• The character of each element is lost when The character of each element is lost when

forming a compound.forming a compound.

• MOLECULESMOLECULES are the smallest unit of a are the smallest unit of a

compound that retains the characteristics of compound that retains the characteristics of

the compound. the compound.


MOLECULAR FORMULASMOLECULAR FORMULAS

• Formula for glycine is Formula for glycine is CC22HH55NONO22• In one molecule there areIn one molecule there are

– 2 C atoms2 C atoms

– 5 H atoms5 H atoms

– 1 N atom1 N atom

– 2 O atoms2 O atoms


WRITING FORMULASWRITING FORMULAS

• Can also write glycine formula as

–H2NCH2COOHto show atom ordering

• or in the form of a structural formulastructural formula

C

H

H C

H

H

O

O HN


IONS AND IONIC COMPOUNDSIONS AND IONIC COMPOUNDSsee Screen 3.7see Screen 3.7

IONS AND IONIC COMPOUNDSIONS AND IONIC COMPOUNDSsee Screen 3.7see Screen 3.7

• IONSIONS are atoms or groups of atoms with a are atoms or groups of atoms with a

positive or negative charge. positive or negative charge.

• Taking away an electron from an atom gives a Taking away an electron from an atom gives a

CATIONCATION with a positive charge with a positive charge

• Adding an electron to an atom gives an Adding an electron to an atom gives an

ANIONANION with a negative charge. with a negative charge.


Formation of Cations Formation of Cations & Anions& Anions

Formation of Cations Formation of Cations & Anions& Anions

A cation forms A cation forms when an atom when an atom loses one or loses one or more electrons.more electrons.

An anion forms An anion forms when an atom when an atom gains one or gains one or more electronsmore electrons

Mg --> Mg2+ + 2 e- F + e- --> F-


PREDICTING ION CHARGESPREDICTING ION CHARGESPREDICTING ION CHARGESPREDICTING ION CHARGES

In generalIn general

• metalsmetals (Mg) (Mg) lose electrons lose electrons ---> ---> cationscations

• nonmetalsnonmetals (F) (F) gain electronsgain electrons ---> ---> anionsanions

•See CD-ROM Screen 3.7 and book Figure 3.6See CD-ROM Screen 3.7 and book Figure 3.6


Charges on Common IonsCharges on Common Ions

+1

+2 +3

-4 -3 -2 -1


POLYATOMIC IONSPOLYATOMIC IONSScreen 3.8Screen 3.8

Groups of atoms with a charge.Groups of atoms with a charge.

MEMORIZEMEMORIZE the names and formulas the names and formulas

in Table 3.1, page 110.in Table 3.1, page 110.


COMPOUNDS COMPOUNDS FORMED FORMED

FROM IONSFROM IONS

COMPOUNDS COMPOUNDS FORMED FORMED

FROM IONSFROM IONS

CATION + CATION + ANION ---> ANION --->

COMPOUNDCOMPOUND

NaNa++ + Cl + Cl- - --> NaCl--> NaCl

A neutral compd. A neutral compd. requires equal requires equal number of + number of +

and - and - charges.charges.


IONIC COMPOUNDSIONIC COMPOUNDS

NH4+

Cl-

ammonium chloride, NHammonium chloride, NH44ClCl


Some Ionic CompoundsSome Ionic CompoundsCaCa2+2+ + 2 F + 2 F-- ---> --->

CaFCaF22

Name = calcium fluorideName = calcium fluoride

MgMg2+2+ + NO + NO33-- ----> ---->

Mg(NOMg(NO33))22

magnesiummagnesium nitratenitrate

FeFe2+2+ + PO + PO443-3- ----> ---->

FeFe33(PO(PO44))22

iron(II) phosphateiron(II) phosphate

(See CD, Screen 3.13 for naming practice)(See CD, Screen 3.13 for naming practice)

calcium fluoridecalcium fluoride

Properties of Ionic CompoundsProperties of Ionic CompoundsForming NaCl from Na and ClForming NaCl from Na and Cl22

Properties of Ionic CompoundsProperties of Ionic CompoundsForming NaCl from Na and ClForming NaCl from Na and Cl22

• A metal atom can A metal atom can transfer an transfer an electron to a electron to a nonmetal.nonmetal.

• The resulting The resulting cation and anion cation and anion are attracted to are attracted to each other by each other by

electrostatic electrostatic forcesforces..Copyright © 1999 by Harcourt Brace & Company

All rights reserved.Requests for permission to make copies of any part of the work should be mailed to:Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida 32887-6777

MOLECULAR WEIGHT MOLECULAR WEIGHT AND MOLAR MASSAND MOLAR MASS

Molecular weightMolecular weight is the sum of the is the sum of the

atomic weights of all atoms in the atomic weights of all atoms in the

molecule.molecule.

Molar massMolar mass = molecular weight in = molecular weight in

gramsgrams



What is the molar mass of What is the molar mass of ethanol, Cethanol, C22HH66O?O?

1 mol contains1 mol contains

2 mol C (12.01 g C/1 mol) = 24.02 g C2 mol C (12.01 g C/1 mol) = 24.02 g C

6 mol H (1.01 g H/1 mol) = 6.06 g H6 mol H (1.01 g H/1 mol) = 6.06 g H

1 mol O (16.00 g O/1 mol) = 16.00 g O1 mol O (16.00 g O/1 mol) = 16.00 g O

TOTAL = molar mass = 46.08 g/molTOTAL = molar mass = 46.08 g/mol


How many moles of alcohol are How many moles of alcohol are there in a “standard” can of beer if there in a “standard” can of beer if there are 21.3 g of Cthere are 21.3 g of C22HH66O?O?

(a) Molar mass of C2H6O = 46.08 g/mol

(b) Calc. moles of alcohol


How many moles of alcohol are How many moles of alcohol are there in a “standard” can of beer if there in a “standard” can of beer if there are 21.3 g of Cthere are 21.3 g of C22HH66O?O?

(a) Molar mass of C2H6O = 46.08 g/mol

(b) Calc. moles of alcohol

21.3 g • 1 mol

46.08 g = 0.462 mol


How many How many moleculesmolecules of alcohol of alcohol are there in a “standard” can of are there in a “standard” can of beer if there are 21.3 g of Cbeer if there are 21.3 g of C22HH66O?O?

= 2.78 x 1023 molecules

We know there are 0.462 mol of C2H6O.

0.462 mol • 6.022 x 1023 molecules

1 mol


How many How many atoms of C atoms of C are there in are there in a “standard” can of beer if there are a “standard” can of beer if there are 21.3 g of C21.3 g of C22HH66O?O?

= 5.57 x 1023 C atoms

We know there are 2.78 x 1023 molecules.

Each molecule contains 2 C atoms.

Therefore, the number of C atoms is

2.78 x 1023 molecules • 2 C atoms1 molecule


Empirical and Molecular Empirical and Molecular FormulasFormulas

A pure compound always consists of the A pure compound always consists of the same elements combined in the same same elements combined in the same proportions by weight.proportions by weight.

Therefore, we can express molecular Therefore, we can express molecular composition as composition as PERCENT BY PERCENT BY WEIGHTWEIGHT

Ethanol, CEthanol, C22HH66OO

52.13% C, 13.15% H, 52.13% C, 13.15% H,

34.72% O34.72% O


Percent CompositionPercent Composition

Consider NOConsider NO22, Molar mass = ?, Molar mass = ?

What is the weight percent of N and of O?What is the weight percent of N and of O?

Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%Wt. % O 2 (16 .0 g O per mole )46 .0 g

x 100 % 69 .6%

Wt. % N = 14.0 g N

46.0 g NO2 • 100% = 30.4 %Wt. % N =

14.0 g N46.0 g NO2

• 100% = 30.4 %


Determining FormulasDetermining Formulas

In chemical analysis we determine the % In chemical analysis we determine the % by weight of each element in a given by weight of each element in a given amount of pure compound and derive amount of pure compound and derive thethe EMPIRICALEMPIRICAL or or SIMPLESTSIMPLEST formula.formula.

PROBLEM: A compound of B and PROBLEM: A compound of B and H is 81.10% B. What is its H is 81.10% B. What is its empirical formula?empirical formula?


A compound of B and H is A compound of B and H is 81.10% B. What is its empirical 81.10% B. What is its empirical formula?formula?

• Because it contains only B and H, Because it contains only B and H, it must contain 18.90% H.it must contain 18.90% H.

• In 100.0 g of the compound there In 100.0 g of the compound there are 81.10 g of B and 18.90 g of H.are 81.10 g of B and 18.90 g of H.

• Calculate the number of moles of Calculate the number of moles of each constitutent.each constitutent.


A compound of B and H is 81.10% B. What is its A compound of B and H is 81.10% B. What is its empirical formula?empirical formula?

Calculate the number of moles of each Calculate the number of moles of each element in 100.0 g of sample.element in 100.0 g of sample.

81.10 g B • 1 mol

10.81 g = 7.502 mol B81.10 g B •

1 mol10.81 g

= 7.502 mol B

18.90 g H • 1 mol

1.008 g = 18.75 mol H18.90 g H •

1 mol1.008 g

= 18.75 mol H


A compound of B and H is 81.10% B. What is its A compound of B and H is 81.10% B. What is its empirical formula?empirical formula?

But we need a whole number ratio. But we need a whole number ratio.

2.5 mol H/1.0 mol B = 5 mol H to 2 mol B2.5 mol H/1.0 mol B = 5 mol H to 2 mol B

EMPIRICAL FORMULA = BEMPIRICAL FORMULA = B22HH55

Take the ratio of moles of B and H. Take the ratio of moles of B and H. AlwaysAlways

divide by the smaller number. divide by the smaller number.

18.75 mol H7.502 mol B

= 2.499 mol H1.000 mol B

= 2.5 mol H1.0 mol B

18.75 mol H7.502 mol B

= 2.499 mol H1.000 mol B

= 2.5 mol H1.0 mol B


A compound of B and H is 81.10% B. A compound of B and H is 81.10% B. Its empirical formula is BIts empirical formula is B22HH55. What is . What is

its its molecular formulamolecular formula??Is the molecular formula BIs the molecular formula B22HH55, B, B44HH1010, ,

BB66HH1515, B, B88HH2020, etc.? , etc.?

B2H6

BB22HH66 is one example of this class of compounds. is one example of this class of compounds.



its molecular formula?its molecular formula?We need to do an EXPERIMENT to find We need to do an EXPERIMENT to find

the MOLAR MASS.the MOLAR MASS.

Here experiment gives 53.3 g/molHere experiment gives 53.3 g/mol

The mass of BThe mass of B22HH55 = 26.66 g/unit = 26.66 g/unit

Find the ratio of these masses.Find the ratio of these masses.

53.3 g/mol26.66 g/unit of B2H5

= 2 units of B2H5

1 mol


= 2 units of B2H5

1 mol



its molecular formula?its molecular formula?

Molecular formula = BMolecular formula = B44HH1010

We need to do an EXPERIMENT to find We need to do an EXPERIMENT to find

the MOLAR MASS.the MOLAR MASS.

Here experiment gives 53.3 g/molHere experiment gives 53.3 g/mol

The mass of BThe mass of B22HH55 = 26.66 g/unit = 26.66 g/unit

Find the ratio of these masses.Find the ratio of these masses.53.3 g/mol

26.66 g/unit of B2H5 =

2 units of B2H51 mol


= 2 units of B2H5

1 mol


Determine the formula of a Determine the formula of a compound of Sn and I using the compound of Sn and I using the following data.following data.

Determine the formula of a Determine the formula of a compound of Sn and I using the compound of Sn and I using the following data.following data.

• Reaction of Sn and IReaction of Sn and I22 is done using excess is done using excess Sn.Sn.

• Mass of Sn in the beginning = 1.056 gMass of Sn in the beginning = 1.056 g

• Mass of iodine (IMass of iodine (I22) used ) used = 1.947 g = 1.947 g

• Mass of Sn remaining Mass of Sn remaining = 0.601 g= 0.601 g

• See p. 139See p. 139


Find the mass of Sn that combined with Find the mass of Sn that combined with 1.947 g I1.947 g I22..

Mass of Sn initially = 1.056 gMass of Sn initially = 1.056 g

Mass of Sn recovered = 0.601 gMass of Sn recovered = 0.601 g

Mass of Sn used = 0.455 gMass of Sn used = 0.455 g

Find moles of Sn used:Find moles of Sn used:

Tin and Iodine CompoundTin and Iodine Compound


Find the mass of Sn that combined with Find the mass of Sn that combined with 1.947 g I1.947 g I22..

Mass of Sn initially = 1.056 gMass of Sn initially = 1.056 g

Mass of Sn recovered = 0.601 gMass of Sn recovered = 0.601 g

Mass of Sn used = 0.455 gMass of Sn used = 0.455 g

Find moles of Sn used:Find moles of Sn used:

0.455 g Sn • 1 mol

118.7 g = 3.83 x 10-3 mol Sn0.455 g Sn •

1 mol118.7 g

= 3.83 x 10-3 mol Sn




Now find the number of moles of INow find the number of moles of I22 that that combined with 3.83 x 10combined with 3.83 x 10-3-3 mol Sn. Mass mol Sn. Mass of Iof I22 used was 1.947 g. used was 1.947 g.




1.947 g I2 • 1 mol

253.81 g = 7.671 x 10-3 mol I21.947 g I2 •

1 mol253.81 g

= 7.671 x 10-3 mol I2




This is equivalent to 2 x 7.671 x 10This is equivalent to 2 x 7.671 x 10-3-3 or or1.534 x 101.534 x 10-2-2 mol iodine atoms mol iodine atoms

1.947 g I2 • 1 mol

253.81 g = 7.671 x 10-3 mol I21.947 g I2 •

1 mol253.81 g

= 7.671 x 10-3 mol I2


Tin and Iodine CompoundTin and Iodine CompoundNow find the ratio of number of moles of Now find the ratio of number of moles of

moles of I and Sn that combined.moles of I and Sn that combined.




1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =

4.01 mol I1.00 mol Sn

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =





Empirical formula is Empirical formula is SnISnI44

1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =


1.534 x 10-2 mol I

3.83 x 10-3 mol Sn =



Chemical EquationsChemical Equations

Depict the kind of Depict the kind of reactantsreactants and and productsproducts and their relative amounts in a reaction.and their relative amounts in a reaction.

4 Al(s) + 3 O4 Al(s) + 3 O22(g) ---> 2 Al(g) ---> 2 Al22OO33(s)(s)

The numbers in the front are calledThe numbers in the front are called

stoichiometric coefficientsstoichiometric coefficientsThe letters (s), (g), and (l) are the physical states of The letters (s), (g), and (l) are the physical states of

compounds.compounds.



4 Al(s) + 3 O4 Al(s) + 3 O22(g) (g) ---> 2 Al---> 2 Al22OO33(s)(s)

This equation meansThis equation means

4 Al atoms + 3 O4 Al atoms + 3 O22 molecules molecules ------give--->give--->

2 molecules of Al2 molecules of Al22OO33

4 moles of Al + 3 moles of 4 moles of Al + 3 moles of OO22 ---give--->---give--->

2 moles of Al2 moles of Al22OO33


Because of the principle of theBecause of the principle of the

conservation of matterconservation of matter, ,

an an equation must be equation must be balanced.balanced.

It must have the same It must have the same

number of atoms of the number of atoms of the

same kind on both sides.same kind on both sides.


Lavoisier, 1788Lavoisier, 1788

STOICHIOMETRYSTOICHIOMETRY

- the study of the - the study of the quantitative quantitative aspects of aspects of chemical chemical reactions.reactions.



PROBLEM: If 454 g of NHPROBLEM: If 454 g of NH44NONO33 decomposes, how much Ndecomposes, how much N22O and HO and H22O O are formed? What is the theoretical are formed? What is the theoretical yield of products?yield of products?

PROBLEM: If 454 g of NHPROBLEM: If 454 g of NH44NONO33 decomposes, how much Ndecomposes, how much N22O and HO and H22O O are formed? What is the theoretical are formed? What is the theoretical yield of products?yield of products?


PROBLEM: If 454 g of NHPROBLEM: If 454 g of NH44NONO33 decomposes, how much Ndecomposes, how much N22O is formed? O is formed? What is the theoretical yield of products?What is the theoretical yield of products?

PROBLEM: If 454 g of NHPROBLEM: If 454 g of NH44NONO33 decomposes, how much Ndecomposes, how much N22O is formed? O is formed? What is the theoretical yield of products?What is the theoretical yield of products?

STEP 1STEP 1

Write the balanced Write the balanced chemical equationchemical equation

NHNH44NONO33 ---> ---> NN22O + 2 HO + 2 H22OO


454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22O O 454 g of NH454 g of NH44NONO33 --> N --> N22O + 2 HO + 2 H22O O

STEP 2 STEP 2 Convert mass reactant Convert mass reactant (454 g) --> moles(454 g) --> moles

454 g • 1 mol

80.04 g = 5.68 mol NH4NO3



STEP 3 STEP 3 Convert moles reactant --> Convert moles reactant --> moles productmoles product

Relate moles NHRelate moles NH44NONO33 to moles product. to moles product.

1 mol NH1 mol NH44NONO33 --> 2 mol H --> 2 mol H22OO

Express this relation as the Express this relation as the STOICHIOMETRIC FACTORSTOICHIOMETRIC FACTOR..

2 mol H2O produced1 mol NH4NO3 used

2 mol H2O produced1 mol NH4NO3 used



= 11.4 mol H= 11.4 mol H22O producedO produced

5.68 mol NH4NO3 • 2 mol H2O produced1 mol NH4NO3 used

Convert moles reactant (5.68 mol) --> Convert moles reactant (5.68 mol) --> moles moles STEP 3 STEP 3 productproduct



ALWAYS FOLLOW THESE STEPS IN ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY SOLVING STOICHIOMETRY PROBLEMS!PROBLEMS!

18.02 gO

STEP 4 STEP 4 Convert moles product Convert moles product (11.4 mol) --> mass product(11.4 mol) --> mass product

11.4 mol H2O • 1 mol

= 204 g H2



Finally: Finally: Calculate the percent yieldCalculate the percent yield

If you isolated only 131 g of NIf you isolated only 131 g of N22O, what is the O, what is the

percent yield?percent yield?

This compares the theoretical (250. g) and This compares the theoretical (250. g) and

actual (131 g) yields. actual (131 g) yields.



% yield = actual yield

theoretical yield • 100%

Calculate the percent yieldCalculate the percent yield

% yield = 131 g250. g

• 100% = 52.4%


GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS

GENERAL PLAN FOR GENERAL PLAN FOR STOICHIOMETRY STOICHIOMETRY CALCULATIONSCALCULATIONS

Mass reactant

StoichiometricfactorMoles

reactantMoles product

Mass product

Reactions Involving aReactions Involving aLIMITING REACTANTLIMITING REACTANT

• In a given reaction, there is not enough of In a given reaction, there is not enough of one reagent to use up the other reagent one reagent to use up the other reagent completely.completely.

• The reagent in short supply The reagent in short supply LIMITSLIMITS the the quantity of product that can be formed.quantity of product that can be formed.



LIMITING REACTANTSLIMITING REACTANTS

Reactantseactants ProductsProducts

2 NO(g) + O2 (g) 2 NO2(g)

Limiting reactant = ___________Limiting reactant = ___________Excess reactant = ____________Excess reactant = ____________


PROBLEM: PROBLEM: Mix 5.40 g of Al with 8.10 g of Mix 5.40 g of Al with 8.10 g of ClCl22. How many grams of Al. How many grams of Al22ClCl66 can form? can form?PROBLEM: PROBLEM: Mix 5.40 g of Al with 8.10 g of Mix 5.40 g of Al with 8.10 g of ClCl22. How many grams of Al. How many grams of Al22ClCl66 can form? can form?

Mass reactant

StoichiometricfactorMoles

reactantMoles product

Mass product


Balance the Chemical EquationBalance the Chemical EquationBalance the Chemical EquationBalance the Chemical Equation

2 Al + 3 Cl2 ---> Al2Cl6


Finding the Limiting ReactantFinding the Limiting ReactantFinding the Limiting ReactantFinding the Limiting Reactant

•Complete the first three steps

•Examine the moles of productproduct formed

•Complete problem with SMALLEST SMALLEST moles of product


We have 5.40 g of Al and 8.10 g of ClWe have 5.40 g of Al and 8.10 g of Cl22

Calculate moles of each reactantCalculate moles of each reactantCalculate moles of each reactantCalculate moles of each reactant

5.40 g Al • 1 mol27.0 g

= 0.200 mol Al

8.10 g Cl2 • 1 mol70.9 g

= 0.114 mol Cl2


Apply Stoichiometric Ratio to Each ReactantApply Stoichiometric Ratio to Each ReactantApply Stoichiometric Ratio to Each ReactantApply Stoichiometric Ratio to Each Reactant

0.200 mol Al X 1 Al2Cl6 = 0.100 mol Al2Cl6

1 2 Al

0.114 mol Cl2 X 1 Al2Cl6 = 0.036 mol Al2Cl6

1 3 Cl2


Mix 5.40 g of Al with 8.10 g of ClMix 5.40 g of Al with 8.10 g of Cl22. . What mass of AlWhat mass of Al22ClCl66 can form? can form?

ClCl22 results in the smallest amount of product. Therefore:Limiting reactant = ClLimiting reactant = Cl22

Base all calculations on ClBase all calculations on Cl22

Complete the problem with ClComplete the problem with Cl22

ClCl22 results in the smallest amount of product. Therefore:Limiting reactant = ClLimiting reactant = Cl22

Base all calculations on ClBase all calculations on Cl22

Complete the problem with ClComplete the problem with Cl22

Using Stoichiometry to Using Stoichiometry to Determine a FormulaDetermine a Formula

Burn 0.115 g of a hydrocarbon, CxHy, and

produce 0.379 g of CO2 and 0.1035 g of

H2O.

What is the empirical formula of CxHy?

CxHy + some oxygen --->

0.379 g CO2 + 0.1035 g H2OCopyright © 1999 by Harcourt Brace & CompanyAll rights reserved.Requests for permission to make copies of any part of the work should be mailed to:Permissions Department, Harcourt Brace & Company, 6277 Sea Harbor Drive, Orlando, Florida 32887-6777



First, recognize that all C in CO2 and all H in H2O is from

CxHy.

1. Calculate moles of C in CO2

8.61 x 10-3 mol C

2. Calculate moles of H in H2O

1.149 x 10 -2 mol H

CCxxHHy y + some oxygen ---> + some oxygen --->

0.379 g CO0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO



Now find ratio of mol H/mol C to find values of x and y in Now find ratio of mol H/mol C to find values of x and y in CCxxHHyy..

1.149 x 10 1.149 x 10 -2 -2 mol H/ 8.61 x 10mol H/ 8.61 x 10-3 -3 mol C mol C

= 1.33 mol H / 1.00 mol C= 1.33 mol H / 1.00 mol C

= 4 mol H / 3 mol C= 4 mol H / 3 mol C

Empirical formula = CEmpirical formula = C33HH44

CCxxHHy y + some oxygen ---> + some oxygen --->

0.379 g CO0.379 g CO22 + 0.1035 g H + 0.1035 g H22OO


Aqueous SolutionsAqueous SolutionsHow do we know ions are present in aqueous How do we know ions are present in aqueous

solutions?solutions?

The solutions conduct electricity!The solutions conduct electricity!

They are called They are called ELECTROLYTESELECTROLYTES

HCl, KMnOHCl, KMnO44, MgCl, MgCl22, and NaCl are , and NaCl are strong strong electrolytes. electrolytes. They dissociate completely They dissociate completely (or nearly so) into ions.(or nearly so) into ions.

KMnOKMnO44(aq) ---> K(aq) ---> K++(aq) + MnO(aq) + MnO44--(aq)(aq)


NaCl dissolving NaCl dissolving in waterin water

Cl-

Na+

Negative O atom

Positive H atom


Aqueous SolutionsAqueous Solutions

Acetic acid ionizes only to a small extent, so Acetic acid ionizes only to a small extent, so it is a it is a weak electrolyte.weak electrolyte.

CHCH33COCO22H(aq) H(aq) ---> CH ---> CH33COCO22

--(aq) + H(aq) + H++(aq)(aq)


Ionized acetic acid

H+

Acetic acid — Weak Electrolyte


Aqueous SolutionsAqueous SolutionsSome compounds dissolve in water but Some compounds dissolve in water but

do not conduct electricity. They are do not conduct electricity. They are called called nonelectrolytes.nonelectrolytes.

Examples include:Examples include:

sugarsugar

ethanolethanol

ethylene glycol ethylene glycol (in antifreeze)(in antifreeze)


WATER SOLUBILITY OF WATER SOLUBILITY OF IONIC COMPOUNDSIONIC COMPOUNDS

Not all ionic compounds dissolve in Not all ionic compounds dissolve in water. Some are water. Some are INSOLUBLEINSOLUBLE..

See See Figure 5.4Figure 5.4

As long as one ion from the list is As long as one ion from the list is present in a compound, the present in a compound, the compound is water soluble.compound is water soluble.


WATER SOLUBILITY OF IONIC WATER SOLUBILITY OF IONIC COMPOUNDSCOMPOUNDS

Common minerals are often formed with anions that Common minerals are often formed with anions that lead to insolubility:lead to insolubility:

sulfidesulfide fluoridefluoride

carbonatecarbonate oxideoxide

Azurite, a copper carbonate

Iron pyrite, a sulfideOrpiment, arsenic sulfide


Net Ionic Net Ionic EquationsEquationsNet Ionic Net Ionic

EquationsEquationsMg(s) + 2 HCl(aq) --> HMg(s) + 2 HCl(aq) --> H22(g) + MgCl(g) + MgCl22(aq)(aq)

We really should writeWe really should write

Mg(s) + 2 HMg(s) + 2 H++(aq) + 2 Cl(aq) + 2 Cl--(aq) ---> (aq) ---> HH22(g) + Mg(g) + Mg2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

since HCl ionizes in water.since HCl ionizes in water.


Net Ionic EquationsNet Ionic EquationsNet Ionic EquationsNet Ionic Equations

• The two ClThe two Cl-- ions are ions are SPECTATOR IONSSPECTATOR IONS — — they do not participate. Could have used they do not participate. Could have used NONO33

- - and gotten the same result.and gotten the same result.

Mg(s) + 2 HCl(aq) Mg(s) + 2 HCl(aq) --> H--> H22(g) + MgCl(g) + MgCl22(aq)(aq)

We really should writeWe really should write

Mg(s) + 2 HMg(s) + 2 H++(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)

---> H---> H22(g) + Mg(g) + Mg2+2+(aq) + 2 Cl(aq) + 2 Cl--(aq)(aq)


Net Ionic EquationsNet Ionic EquationsNet Ionic EquationsNet Ionic Equations

Mg(s) + 2 HCl(aq) --> HMg(s) + 2 HCl(aq) --> H22(g) + MgCl(g) + MgCl22(aq)(aq)

Mg(s) + 2 HMg(s) + 2 H++(aq) + 2 Cl(aq) + 2 Cl--(aq) ---> (aq) --->

HH22(g) + Mg(g) + Mg2+2+(aq) + 2 Cl(aq) + 2 Cl--

(aq)(aq)

We leave the spectator ions out in writing We leave the spectator ions out in writing the the NET IONIC EQUATIONNET IONIC EQUATION

Mg(s) + 2 HMg(s) + 2 H++(aq) ---> H(aq) ---> H22(g) + Mg(g) + Mg2+2+(aq) (aq)


An acid -------> HAn acid -------> H++ in water in waterAn acid -------> HAn acid -------> H++ in water in water

ACIDSACIDSACIDSACIDS

Some Some strongstrong acids are acids are

HClHCl hydrochlorichydrochloric

HNOHNO33 nitricnitric

HClOHClO44 perchloricperchloric

HH22SOSO44 sulfuricsulfuricHNOHNO33


An acid -------> HAn acid -------> H++ in water in waterAn acid -------> HAn acid -------> H++ in water in water

ACIDSACIDSACIDSACIDS

• HCl(aq) ---> HHCl(aq) ---> H++(aq) + Cl(aq) + Cl--(aq)(aq)


The Nature The Nature of Acidsof Acids

The Nature The Nature of Acidsof Acids

HCl

H2O H3O+

Cl-

hydronium ion


ACIDSACIDSACIDSACIDSNonmetal oxides can be acidsNonmetal oxides can be acids

COCO22(aq) + H(aq) + H22O(liq) O(liq) ---> H---> H22COCO33(aq)(aq)

SOSO33(aq) + H(aq) + H22O(liq) O(liq) ---> H---> H22SOSO44(aq)(aq)

and can come from burning coal and and can come from burning coal and oil.oil.


Base ---> OHBase ---> OH-- in water in waterBase ---> OHBase ---> OH-- in water in water

BASESBASESsee Screen 4.8 and Table 4.1see Screen 4.8 and Table 4.1

BASESBASESsee Screen 4.8 and Table 4.1see Screen 4.8 and Table 4.1

NaOH(aq) ---> NaNaOH(aq) ---> Na++(aq) + OH(aq) + OH--(aq)(aq)


Ammonia, NHAmmonia, NH33

An Important BaseNHNH33(aq) + H(aq) + H22O(liq) O(liq)

---> NH---> NH44++(aq) + OH(aq) + OH--(aq)(aq)


BASESBASESBASESBASES

Metal oxides are basesMetal oxides are bases

CaO(s) + HCaO(s) + H22O(liq) O(liq)

--> Ca(OH)--> Ca(OH)22(aq)(aq)

CaO in water. Indicator shows solution is basic.


Precipitation ReactionsPrecipitation Reactions

The “driving force” is the formation of The “driving force” is the formation of an insoluble compound — a an insoluble compound — a precipitate.precipitate.

Pb(NOPb(NO33))22(aq) + 2 KI(aq) ----->(aq) + 2 KI(aq) ----->

2 KNO2 KNO33(aq) + PbI(aq) + PbI22(s)(s)

Net ionic equationNet ionic equation

PbPb2+2+(aq) + 2 I(aq) + 2 I--(aq) ---> PbI(aq) ---> PbI22(s)(s)


Acid-Base ReactionsAcid-Base Reactions• The “driving force” is the formation of The “driving force” is the formation of

water.water.NaOH(aq) + HCl(aq) --->NaOH(aq) + HCl(aq) --->

NaCl(aq) + HNaCl(aq) + H22O(liq)O(liq)• Net ionic equationNet ionic equation OHOH--(aq) + H(aq) + H++(aq) ---> (aq) --->

HH22O(liq)O(liq)• This applies to ALL reactions of This applies to ALL reactions of STRONGSTRONG

acids and bases.acids and bases.


Acid-Base ReactionsAcid-Base Reactions• A-B reactions are sometimes called A-B reactions are sometimes called

NEUTRALIZATIONSNEUTRALIZATIONS because the solution is because the solution is neither acidic nor basic at the end.neither acidic nor basic at the end.

• The other product of the A-B reaction is a The other product of the A-B reaction is a SALTSALT, MX., MX.

HHXX + + MMOH ---> OH ---> MMXX + H + H22OO

MMn+n+ comes from comes from base base && XXn-n- comes from comes from acidacid

This is one way to make compounds!This is one way to make compounds!


Oxidation-Reduction ReactionsOxidation-Reduction Reactions

FeFe22OO33(s) + 2 Al(s)(s) + 2 Al(s)

---->---->

2 Fe(s) + Al2 Fe(s) + Al22OO33(s)(s)


Redox reactions are characterized Redox reactions are characterized byby ELECTRON TRANSFER ELECTRON TRANSFER between an electron donor and between an electron donor and electron acceptor.electron acceptor.

Transfer leads to— Transfer leads to— 1. 1. increase in oxidation numberincrease in oxidation number

of some element = of some element = OXIDATIONOXIDATION2.2. decrease in oxidation numberdecrease in oxidation number

of some element = of some element = REDUCTIONREDUCTION

REDOX REACTIONSREDOX REACTIONSREDOX REACTIONSREDOX REACTIONS


OXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERS

The electric charge an element The electric charge an element APPEARS to have when electrons are APPEARS to have when electrons are counted by some arbitrary rules:counted by some arbitrary rules:

1. Each atom in free element has ox. no. 1. Each atom in free element has ox. no. = 0. Zn O= 0. Zn O22 I I22 S S88

2. In simple ions, ox. no. = charge on 2. In simple ions, ox. no. = charge on ion.ion.

-1 for Cl-1 for Cl-- +2 for Mg +2 for Mg2+2+


OXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERSOXIDATION NUMBERS

3.3. O has ox. no. = -2O has ox. no. = -2

(except in peroxides: in H(except in peroxides: in H22OO22, O = -1), O = -1)

4. Ox. no. of H = +14. Ox. no. of H = +1 (except when H is (except when H is associated with a metal as in NaH where it is -1)associated with a metal as in NaH where it is -1)

5. Algebraic sum of oxidation numbers 5. Algebraic sum of oxidation numbers

= 0 for a compound = 0 for a compound

= overall charge for an ion= overall charge for an ion


Recognizing a Redox ReactionRecognizing a Redox ReactionRecognizing a Redox ReactionRecognizing a Redox Reaction

Corrosion of aluminumCorrosion of aluminum

2 Al(s) + 3 Cu2 Al(s) + 3 Cu2+2+(aq) ---> 2 Al(aq) ---> 2 Al3+3+(aq) + 3 Cu(s)(aq) + 3 Cu(s)

Al(s) --> AlAl(s) --> Al3+3+(aq) + 3 e(aq) + 3 e--

• Ox. no. of Al increases as eOx. no. of Al increases as e-- are donated by the are donated by the metal. metal.

• Therefore, Therefore, Al is OXIDIZED Al is OXIDIZED and is the and is the REDUCING AGENTREDUCING AGENT in this balanced in this balanced half-half-reaction.reaction.



Corrosion of aluminumCorrosion of aluminum


CuCu2+2+(aq) + 2 e(aq) + 2 e- - --> Cu(s) --> Cu(s)

• Ox. no. of Cu decreases as eOx. no. of Cu decreases as e-- are accepted by are accepted by the ion. the ion.

• Therefore, Therefore, Cu is REDUCED Cu is REDUCED and is the and is the OXIDIZING AGENTOXIDIZING AGENT in this balanced in this balanced half-half-reaction.reaction.



Notice that the 2 half-reactions add to give the Notice that the 2 half-reactions add to give the overall reaction if we use 2 mol of Al and 3 overall reaction if we use 2 mol of Al and 3 mol of Cumol of Cu2+2+..

2 Al(s) --> 2 Al2 Al(s) --> 2 Al3+3+(aq) + 6 e(aq) + 6 e--

3 Cu3 Cu2+2+(aq) + 6 e(aq) + 6 e- - --> 3 Cu(s) --> 3 Cu(s)

----------------------------------------------------------------------------------------------------------------


Final eqn. is balanced for mass and charge.Final eqn. is balanced for mass and charge.


Common Oxidizing and Reducing Common Oxidizing and Reducing AgentsAgents

See Table 4.2See Table 4.2

Common Oxidizing and Reducing Common Oxidizing and Reducing AgentsAgents

See Table 4.2See Table 4.2

Metals Metals (Cu) are (Cu) are reducing reducing agentsagents

HNOHNO33 is an is an

oxidizing oxidizing agentagent

2 K + 2 H2 K + 2 H22O --> O -->

2 KOH + H2 KOH + H22

Metals Metals (Na, K, (Na, K, Mg, Fe) Mg, Fe) are are reducing reducing agentsagents

Cu + HNOCu + HNO33 --> -->

CuCu2+2+ + NO + NO22


Learn to recognize common Learn to recognize common oxidizing and reducing oxidizing and reducing agents. See Table 4.2.agents. See Table 4.2.

Learn to recognize common Learn to recognize common oxidizing and reducing oxidizing and reducing agents. See Table 4.2.agents. See Table 4.2.


TerminologyTerminologyTerminologyTerminology

In solution we need to define the -In solution we need to define the -

• SOLVENTSOLVENT

the component whose physicalthe component whose physical

state is preserved when state is preserved when solution formssolution forms

• SOLUTESOLUTE

the other solution componentthe other solution component


Concentration of SoluteConcentration of SoluteConcentration of SoluteConcentration of Solute

The amount of solute in a solution The amount of solute in a solution is given by its is given by its concentrationconcentration.

Molarity (M) = moles solute

liters of solution


PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 H•6 H22O in enough O in enough water to make 250 mL of solution. Calculate molarity.water to make 250 mL of solution. Calculate molarity.PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 H•6 H22O in enough O in enough water to make 250 mL of solution. Calculate molarity.water to make 250 mL of solution. Calculate molarity.


PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 H•6 H22O in enough O in enough water to make 250 mL of solution. Calculate molarity.water to make 250 mL of solution. Calculate molarity.PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 H•6 H22O in enough O in enough water to make 250 mL of solution. Calculate molarity.water to make 250 mL of solution. Calculate molarity.

Step 1: Step 1: Calculate moles of NiClCalculate moles of NiCl22•6H•6H22OO

5.00 g • 1 mol

237.7 g = 0.0210 mol


0.0210 mol0.250 L

= 0.0841 M

PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 H•6 H22O in O in enough water to make 250 mL of solution. enough water to make 250 mL of solution. Calculate molarity.Calculate molarity.

PROBLEM: Dissolve 5.00 g of NiClPROBLEM: Dissolve 5.00 g of NiCl22•6 H•6 H22O in O in enough water to make 250 mL of solution. enough water to make 250 mL of solution. Calculate molarity.Calculate molarity.

Step 1: Step 1: Calculate moles of NiClCalculate moles of NiCl22•6H•6H22OO

Step 2: Step 2: Calculate molarityCalculate molarity

5.00 g • 1 mol

237.7 g = 0.0210 mol


The Nature of a Na2CO3 SolutionThe Nature of a Na2CO3 Solution

This water-soluble compound is ionicThis water-soluble compound is ionic

NaNa22COCO33(aq) --> 2 Na(aq) --> 2 Na++(aq) + CO(aq) + CO332-2-(aq)(aq)

If [NaIf [Na22COCO33] = 0.100 M, then] = 0.100 M, then

[Na[Na++] = 0.200 M] = 0.200 M

[CO[CO332-2-] = 0.100 M] = 0.100 M

NaNa22COCO33


USING MOLARITYUSING MOLARITYUSING MOLARITYUSING MOLARITY

What mass of oxalic acid, HWhat mass of oxalic acid, H22CC22OO44, is required to make , is required to make 250. mL of a 250. mL of a 0.0500 M solution?0.0500 M solution?

Because Conc (M) = Because Conc (M) = moles/volume = mol/Vmoles/volume = mol/V

this means thatthis means that

moles = M • Vmoles = M • V


PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?PROBLEM: You have 50.0 mL of 3.0 M NaOH PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do?and you want 0.50 M NaOH. What do you do?

Add water to the 3.0 M solution to lower Add water to the 3.0 M solution to lower its concentration to 0.50 M its concentration to 0.50 M

Dilute the solution! Dilute the solution!


A shortcutA shortcut

MMinitialinitial • V • Vinitialinitial = M = Mfinalfinal • V • Vfinalfinal

The same equation is useful in TITRATIONSThe same equation is useful in TITRATIONS

Preparing Solutions Preparing Solutions by Dilutionby Dilution

Preparing Solutions Preparing Solutions by Dilutionby Dilution


ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

ACID-BASE REACTIONSACID-BASE REACTIONSTitrationsTitrations

HH22CC22OO44(aq) + 2 NaOH(aq) --->(aq) + 2 NaOH(aq) --->

acidacid basebase

NaNa22CC22OO44(aq) + 2 H(aq) + 2 H22O(liq)O(liq)

Carry out this reaction using a Carry out this reaction using a TITRATIONTITRATION..

Oxalic acid,Oxalic acid,

HH22CC22OO44


Titration setupTitration setupTitration setupTitration setup

Buret contains asolution whoseconcentration isknown exactly.

Solution ofunknownconcentration

Buret contains asolution whoseconcentration isknown exactly.

Solution ofunknownconcentration


TitrationTitrationTitrationTitration1. Add solution from the buret.1. Add solution from the buret.

2. Reagent (base) reacts with 2. Reagent (base) reacts with compound (acid) in solution in the compound (acid) in solution in the flask.flask.

3. Indicator shows when exact 3. Indicator shows when exact stoichiometric reaction has stoichiometric reaction has occurred.occurred.

4. Net ionic equation4. Net ionic equation

HH++ + OH + OH-- --> H --> H22OO

5. At equivalence point 5. At equivalence point

moles Hmoles H++ = moles OH = moles OH--


The same equation as The same equation as used for solutions:used for solutions:

MMacidacid • V • Vacidacid = M = Mbasebase • V • Vbasebase

is used to solve titration problems.is used to solve titration problems.

TitrationTitrationTitrationTitration


End of Midterm ReviewEnd of Midterm Review

Chemistry 180Chemistry 180

General ChemistryGeneral Chemistry

summer chemistry ap exam review copyright © 1999 by harcourt brace & company all rights...

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