Chapter 2
description
Transcript of Chapter 2
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CHAPTER 2
Atoms, Molecules & Stoichiometry
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Learning Outcomes
define the terms relative atomic, isotopic, molecular and formula
masses, based on the 12C scale
define the term mole in terms of the Avogadro constant
analyse mass spectra in terms of isotopic abundances
calculate the relative atomic mass of an element given the relativeabundances of its isotopes, or its mass spectrum
define the terms empirical and molecular formulae
calculate empirical and molecular formulae, using combustion data
or composition by mass
write and/or construct balanced equations
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Learning Outcomes
perform calculations, including use of the mole concept, involving:
(i) reacting masses (from formulae and equations)
(ii) volumes of gases (e.g. in the burning of hydrocarbons)
(iii) volumes and concentrations of solutions
deduce stoichiometric relationships from calculations
describe and explain redox processes in terms of electron transfer
and/or of changes in oxidation number (oxidation state)
(calculation of oxidation number is required)
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Relative Mass
Relative Mass
Relative Atomic
Mass
-Ratio of the average massof one atom of the element
to 1/12 the mass of an atom
of 12C isotope,expressed on
the12C scale
Relative IsotopicMass
- Ratio of the mass of one
atom of the isotope to 1/12
the mass of an atom of 12C
isotope,expressed on the12C scale
Relative Molecular
Mass- Ratio of the average mass of
one molecule of the substance
to 1/12 the mass of an atom of12C isotope,expressed on the12C scale
Relative FormulaMass
- Ratio of the average
mass of one formula unit of
the compound to 1/12 the
mass of an atom of 12C
isotope,expressed on the
12C scale
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A Mole of Atoms
A moleis a collection that contains
6.02 x 1023atoms of an element (Avogadros
number).
1 mole element Number of Atoms
1 mole C = 6.02 x 1023C atoms
1 mole Na = 6.02 x 1023Na atoms
1 mole Au = 6.02 x 1023Au atoms
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A Mole of a Compound
A mole
Of a covalent compound has Avogadros number of
molecules.
1 mole CO2= 6.02 x 1023CO2molecules
1 mole H2O = 6.02 x 1023H2O molecules
Of an ionic compound contains Avogadros number
of formula units.
1 mole NaCl = 6.02 x 1023NaCl formula units
1 mole K2SO4 = 6.02 x 1023K2SO4formula units
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One Mole of Four
Elements
One mole each of helium,
sulfur, copper, and mercury.
How many atoms of helium
are present? Of sulfur? Of
copper? Of mercury?
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Using Avogadros Number
Avogadros number is used toconvert molesof a
substance toparticles.
How many Cu atoms are in 0.50 mole Cu?
0.50 mole Cu x 6.02 x 1023Cu atoms
1 mole Cu= 3.0 x 1023Cu atoms
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Using Avogadros Number
Avogadros number is used to convertparticlesof a
substance tomoles.
How many moles of CO2are in 2.50 x 1024molecules CO2?
2.50 x 1024
molecules CO2x 1 mole CO2
6.02 x 1023molecules CO2
= 4.15 moles CO2
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Molar Mass
The molar massis
The mass of one mole of a substance.
The atomic mass of an element expressed in grams.
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The mass spectrometer is an instrument used to measure
the masses and relative (natural) abundances of the
isotopes present in a sample of an element
AVERAGE ATOMIC MASS
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Mass Spectrometer
In a mass spectrometer, charged particles are passed through a magnetic
field.
The path of the stream of particles will be bent by the magnetic field
depending on the mass and charge of the particles.
When naturally occurring samples of most elementsare charged andpassed through a mass spectrometer, the spectrum indicates more than
one form of the element.
These are isotopes of the element:the number of protons is the same but
the number of neutrons/atom differs for each isotope giving each isotope aslightly different mass.
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boron-10 23
boron-11 100
There are 2 isotopes for boron:
Suppose you had 123 typical atoms of boron.
23 of these would be 10B and 100 would be 11B.
The total mass of these would be (23 x 10) + (100 x 11) = 1330
The average mass of these 123 atoms would be 1330 / 123 = 10.8 (to 3 significant
figures).
10.8 is the relative atomic mass of boron.
Mass Spectrometer
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The data can be summarized as follows:
Isotope Mass` Abundance
90Zr 90.00 amu 51.5 %91Zr 91.00 amu 11.2 %92Zr 92.00 amu 17.1 %94Zr 94.00 amu 17.4 %96Zr 96.00 amu 2.80 %
atomic mass of isotope 100 %
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Calculate the weighted average mass of zirconium using the data below.
Change each percent to a decimal by dividing by 100.
Multiply by the mass.
Add it all together.
Isotope Mass` Abundance90Zr 90.00 amu 51.5 %91Zr 91.00 amu 11.2 %92Zr 92.00 amu 17.1 %94Zr 94.00 amu 17.4 %96Zr 96.00 amu 2.80 %
0.515(90.00) + 0.112(91.00) + 0.171(92.00) + 0.174(94.00) +0.0280(96.00)
= 91.3 amu
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Calculate the average atomic mass for Germanium
20.52 %
69.92428 amu
27.43 %
71.92174 amu
7.76 %
72.9234 amu
36.54 %
73.92115 amu
7.76 %
75.9214 amu
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Chlorine consists of molecules, not individual atoms. Whenchlorine is passed into the ionisation chamber, an electron is
knocked off the molecule to give a molecular ion , Cl2+.
These ions won't be particularly stable, and some will fall apart to
give a chlorine atom and a Cl+ion.
The mass spectrum of chlorine
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Empirical and Molecular Formula
The empir ical formu la of a compound is the simplest formula which shows
the ratio of the atoms of the different elements in the compound.
Question:
69.58% Ba, 6.090% C, 24.32% O.
What is the empirical formula?
Assume you have 100 g of sample,
1: 69.58 g Ba, 6.090 g C, 24.32 g O
2: Ba: 69.58 g 137.33 g/mol = 0.50666 mol Ba
C: 6.090 g 12.01 g/mol = 0.50708 mol C
O: 24.32 g 16.00 g/mol = 1.520 mol O
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3.
4: the simplest formula is BaCO3
mol (reduced)
mol
1.520/
0.50666
= 3.000
0.50708/
0.50666
= 1.001
0.50666/
0.50666
= 1
1.5200.507080.50666
OCBa
Empirical Formula
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Empirical Formula
CxHy+ (x+ )O2(g) x CO2(g) + H2O(g)y
2
y
2
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PROCEDURE FOR...
Obtaining an Empirical
Formula from Combustion
Analysis
EXAMPLE 3.19
Obtaining an Empirical
Formula from Combustion
Analysis
1. Write down asgiven the masses of
each combustion product and the
mass of the sample (if given).
2. Convert the masses of CO2and
H2O from step 1 to moles by usingthe appropriate molar mass for each
compound as a conversion factor.
GIVEN: 1.83 g CO2, 0.901 g H2O
FIND: empirical formula
3. Convert the moles of CO2
and
moles of H2Ofrom step 2 to moles
of C and moles of H using the
conversion factors inherent in the
chemical formulas of CO2and
H2O.
Upon combustion, a compound containing only carbon andhydrogen produces 1.83 g CO2 and 0.901 g H2O. Find the
empirical formula of the compound.
2011 Pearson Education, Inc.
continued
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4. If the compound contains an element other
than C and H, find the mass of the other
element by subtracting the sum of the
masses of C and H (obtained in step 3) from
the mass of the sample. Finally, convert themass of the other element to moles.
5. Write down a pseudoformula for the
compound using the number of moles of each
element (from steps 3 and 4) as subscripts.
6. Divide all the subscripts in the formula
by the smallest subscript. (Round allsubscripts that are within 0.1 of a whole
number.)
No other elements besides C and H, so
proceed to next step.
7. If the subscripts are not whole
numbers, multiply all the subscripts
by a small whole number to get
whole-number subscripts.
C0.0416H0.100
2011 Pearson Education, Inc.
The correct empirical formula is C5H12.
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Molecular Formula
The molecular formula of a compound is one which shows the actual
number of atoms of each element present in one molecule of a
compound.
The molecular formula can be obtained if the empirical formula and Mrare known.
Example: A compound has the empirical formula CH2Br. Its relative
molecular mass is 187.8. Deduce the molecular formula of this
compound.n(CH2Br) = 187.8
n(93.9) = 187.8
n = 2
Molecular formula = C2H
4Br
2
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The subscriptsin a formula give
The relationship of atoms in the formula.
The moles of each element in 1 mole of
compound.
Glucose
C6H12O6
In 1 molecule: 6 atoms C 12 atoms H 6 atoms O
In 1 mole: 6 moles C 12 moles H 6 moles O
Chemical Formula and Mole Concept
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Chemical equation
Uses chemical symbols and chemical formulas to
describe the changes occur in a chemical reaction
Reactant is a starting material
Written on left side
Product is a substance produced
Written on right sideCH4+ 2O2 CO2+ 2H2O
Writing and Balancing Chemical Equation
Reactant Product
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Law of Conservation of Mass
The Law of Conservation of Massindicates that in
an ordinary chemical reaction,
Matter cannot be created or destroyed.
No change in total mass occurs in a reaction.
Mass of products is equal to mass of reactants.
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Example:2Ag + S Ag2S
in this equation the number of reactants are equal to
that of the products.(2 Ag and 1 S)
Balance the following chemical equations.
Fe(s) + S(s) Fe2S3(s)
NH3+ O2 N2+ H2O
HCl + O2 Cl2+ H2O
Writing and Balancing Chemical Equation
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a chemical equation indicates the relative amounts
of each reactant and product in a given chemical
reaction.
Example:
CH4 + 2O2 CO2 + 2H2O
The equation shows that 1 mol of methane reactswith 2 mol of oxygen to produce 2 mol of water.
Chemical Equations and The Mole Concept
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A balanced chemical equation can be used to calculate the
relative amounts of substances involved in chemical reactions.
Example:
When 18.6 g ethane gas C2H6burns in oxygen, how
many grams of CO2are produced?
2C2H6(g) + 7O2(g) 4CO2(g) + 6H2O(g)
18.6 g ? g
Chemical Calculations Using Chemical Equations
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Chemical Calculations Using Chemical Equations
Solution:g C2H6 mole C2H6 mole CO2 g CO2
molar mole-mole molar
mass C2H6 factor mass CO2
= 18.6 g C2H6x 1 mole C2H6x 4 moles CO2x 44.0 g CO2
30.1 g C2H6 2 moles C2H6 1 mole CO2
molar mole-mole molarmass C2H6 factor mass CO2
= 54.4 g CO2
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Balancing Ionic Equations
When ionic compounds dissolve in water, the ions separate from each
other.
Example : NaSO4 (s) + aq Na+(aq) + SO4
-2 (aq)
The ions that play no part in the reaction are called spectator ions.
Molecular Equation:
K2CrO4+ Pb(NO3)2 PbCrO4 + 2 KNO3
Soluble Soluble Insoluble Soluble
Total Ionic Equation:(cancell ing spectator ions)
2 K++ CrO4-2+ Pb+2+ 2 NO3
-
PbCrO4(s) + 2 K++ 2 NO3
-
Net Ionic Equation:
CrO4-2+ Pb+2 PbCrO4(s)
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Limiting Reactants
You can make cookies
until you run out of one
of the ingredients Sugar is the limiting
ingredient in making
the cookies (present in
small amount). Sugar will limit the
amount of cookies you
can make
Cookies Analogy:
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Limiting Reactants
The limiting reactant is the reactant present in
the smallest stoichiometric amount
In other words, its the reactant youll run out of first (in
this case, the H2) O2would be the excess reagent
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Theoretical yield
Maximum amount of product that can be
obtained in a reaction from the given amounts
of reactant.
Percentage yield
A comparison of the amount actually obtainedto the amount it was possible to make.
Actual Yield
Theoretical YieldPercent Yield= x 100
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Silicon dioxide (quartz) is usually quite unreactive but reacts
readily with hydrogen fluoride according to the following
equation.
SiO2(s) + 4HF(g) SiF4(g) + 2H2O(l)
2.0 mol HF x 1 mol SiO24 mol HF
= 0.5 mol SiO2
2.0 mol ? mol
If 2.0 mol of HF are exposed to 4.5 mol of SiO2, which is the limiting reactant?
4.5 mol
4.5 mol SiO2 x 4 mol HF1 mol SiO2 = 18 mol HF
LR
LR
(smaller number)
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Calculations Involving Volumes of Gases
One mole of any gas occupies a volume of 22.4 dm3at
s.t.p. (or 24 dm3at r.t.p)
This is the molar volume of gases.
Example:
1. What volume is occupied by 4.21 moles of ammonia gas, NH3at STP?
4.21 mol 22.4 L = 94.3 L
1 mole
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Calculations Involving Volumes of Gases
2. What volume of hydrogen, measured at STP, can be released by 42.7
g of zinc as it reacts with hydrochloric acid?
Zn (s )+ 2 HCl (aq) H2(g )+ ZnCl2(aq)
42.7 g Zn 1 mol Zn 1 mol H2 22.4 L H2 = 14.6 L H265.4 g Zn 1 mol Zn 1 mol H2
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PrenticeHall 2005General Chemistry4thedition, Hill, Petrucci, McCreary, Perry
Chapter Three
Solute: the substance being dissolved.
Solvent: the substance doing the dissolving.
Concentrationof a solution: the quantity of a
solute in a given quantity of solution (or solvent).
A concentratedsolution contains a relatively large
amount of solute vs. the solvent (or solution).
A dilutesolution contains a relatively smallconcentration of solute vs. the solvent (or solution).
Concentrated and dilute arent very quantitative
Solutions and
Solution Stoichiometry
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CONCENTRATION OF AQUEOUS
SOLUTION
The concentration of aqueous solution may be expressed either as:
a) mass of solute per dm3of solution (units: g dm-3) or
b) mol of solute per dm3of solution (units: moldm-3)
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PrenticeHall 2005General Chemistry4thedition, Hill, Petrucci, McCreary, Perry
Chapter Three
Molarity(M), or molar concentration, is the amount
of solute, in moles, per liter of solution:
A solution that is 0.35 M sucrose contains 0.35
moles of sucrose in each liter of solution.
Keep in mind that molarity signifies moles of
solute per liter of solution, not liters of solvent.
Molar Concentration
moles of soluteMolarity=
liters of solution
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PrenticeHall 2005General Chemistry4thedition, Hill, Petrucci, McCreary, Perry
Chapter Three
Preparing 0.01000 M KMnO4
Weigh 0.01000
mol (1.580 g)
KMnO4.
Dissolve in water. How much
water? Doesnt matter, as long as
we dont go over a liter.
Add more water
to reach the
1.000 liter mark.
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PrenticeHall 2005General Chemistry4thedition, Hill, Petrucci, McCreary, Perry
Chapter Three
Dilutionis the process of preparing a more dilute
solution by adding solvent to a more concentrated
one.
Addition of solvent does not change the amount ofsolutein a solution but does change the solution
concentration.
It is very common to prepare a concentrated stocksolution of a solute, then dilute it to other
concentrations as needed.
Dilution of Solutions
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Vi li i th Dil ti f S l ti
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PrenticeHall 2005General Chemistry4thedition, Hill, Petrucci, McCreary, Perry
Chapter Three
Visualizing the Dilution of a Solution
We start and
end with the
same amount of
solute.
Addition of
solvent has
decreased the
concentration.
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PrenticeHall 2005General Chemistry4thedition, Hill, Petrucci, McCreary, Perry
Chapter Three
Dilution Calculations
couldnt be easier.
Moles of solute does not change on dilution.
Moles of solute =MV Therefore
MconcVconc= MdilVdil
46Example 3 26
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PrenticeHall 2005General Chemistry4thedition, Hill, Petrucci, McCreary, Perry
Chapter Three
Example 3.26
How many milliliters of a 2.00 M CuSO4stocksolution are needed to prepare 0.250 L of 0.400 M
CuSO4?
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Acid-base titration
A neutralization reaction in which a measured volume of an
acid or a base of known concentration is completely reacted
with a measured volume of a base or an acid of unknownconcentration
Acid-base indicator
A compound that exhibits different colours depending on the
pH of its solution
Acid-Base Titrations
S t f tit ti id ith b
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Setup for titrating an acid with a base
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The unknown concentration of the base or acid can be
calculated by using this formula:
M1V1 = M2V2
Acid-Base Titrations
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SAMPLE PROBLEM
In an acid-base titration, 17.45 mL of 0.180 M
nitric acid, HNO3, were completely neutralized
by 14.76 mL of aluminium hydroxide, Al(OH)3.
Calculate the concentration of the aluminiumhydroxide.
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SAMPLE ANSWER
The balanced equation for the reaction is:
3HNO3(aq) + Al(OH)3(aq) Al(NO3)3(aq) + 3H2O(l)
The number of moles of nitric acid used is:
y mol = 0.180 mol/L x 0.01745 L = 3.14 x 10-3mol HNO3
From the stoichiometry of the reaction, the number of moles ofaluminium hydroxide reacted is:
3.14 x 10-3mol HNO3 x 1 mol Al(OH)3= 1.05 x 10-3mol
3 mol HNO3
Therefore, the concentration of the aluminium hydroxide is:
1.05 x 10-3mol Al(OH)3= 0.0711 M
0.01476 L
EXERCISE
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EXERCISE
If 30 mL of 0.5 M HCl neutralizes 20.40 mL of Ba(OH)2
solution, what is the molarity of the Ba(OH)2solution? You
have to write a balanced equation for the neutralization.
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Redox Reactions
Oxidation-reduction (redox) reaction:
A chemical reaction in which there is a transfer of
electrons from one reactant to another reactant.
Oxidation number
of an atom is the charge that atom would have if
the compound was composed of ions.
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Rules for determining oxidation numbers:
Zero for element in its elemental state
Equal to the charge for the monatomic ion
For Groups IA and IIA in compounds always +1, +2
+1 for hydrogen in most compounds
-2 for oxygen in most compounds
In binary compounds, the more electronegative element is
assigned a negative oxidation number (as in its binary ioniccompounds)
The sum of the individual oxidation numbers is equal to zero for
a neutral compound, equal to the charge on the polyatomic ion
Redox Reactions
Oxidation number
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Oxidation number
The charge the atom would have in a molecule (or an
ionic compound) if electrons were completely transferred.
1. Free elements (uncombined state) have an oxidation
number of zero.
Na, Be, K, Pb, H2, O2, P4 = 0
2. In monatomic ions, the oxidation number is equal to
the charge on the ion.
Li+, Li = +1; Fe3+, Fe = +3; O2-, O = -2
3. The oxidation number of oxygen isusually2. In H2O2and O2
2-it is 1.
4 The oxidation number of hydrogen is +1 except when
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4. The oxidation number of hydrogen is 1exceptwhen
it is bonded to metals in binary compounds. In these
cases, its oxidation number is1.
6. The sum of the oxidation numbers of all the atoms in a
molecule or ion is equal to the charge on the
molecule or ion.
5. Group IA metals are +1, IIA metals are +2and fluorine is
always1.
HCO3-
O = -2 H = +1
3x(-2)+ 1+ ?= -1
C = +4
Oxidation numbers of allthe elements in HCO3
-?
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NaIO3
Na = +1 O = -2
3x(-2) + 1+ ?= 0
I = +5
K2Cr2O7
O = -2 K = +1
7x(-2) + 2x(+1) + 2x(?)= 0
Cr = +6
Oxidation numbers of all
the elements in the
following ?