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1.Molecules, Ions and
their Compounds
2. Chemical Equationsand Stoichiometry
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MOLECULAR FORMULAS
Formula for glycine is C2H5NO2
In one molecule there are
2 C atoms
5 H atoms
1 N atom
2 O atoms
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WRITING FORMULAS
Can also write glycine formula as
H2NCH2COOH
to show atom ordering
or in the form of a structuralformula
C
H
H C
H
H
O
O HN
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Empirical and Molecular Formulas
Empirical Formula -
Molecular Formula -
The simplest formula for a compound that agrees withthe elemental analysis and gives rise to the smallest set
of whole numbers of atoms.
The formula of the compound as it exists, it may be a
multiple of the empirical formula.
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MOLECULAR WEIGHT AND
MOLAR MASSMolecular weight= sum of the atomic
weights of all atoms in the molecule.
Molar mass= molecular weight in grams
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How many molesof alcohol are therein a standard can of beer if thereare 21.3 g of C2H6O?
(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
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How many moleculesof alcohol arethere in a standard can of beer ifthere are 21.3 g of C2H6O?
= 2.78 x 1023molecules
We know there are 0.462 mol of C2H6O.
0.462 mol 6.022 x 10
23molecules
1 mol
There are 2.78 x 1023molecules.
Each molecule contains 2 C atoms.
Therefore, the number of C atoms is
2.78 x 1023
molecules 2 C atoms
1 molecule
= 5.57 x 1023C atoms
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Table 3.1 Summary of Mass Terminology
Term Definition Unit
Isotopic mass Mass of an isotope of an element amu
Atomic mass
Molecular
(or formula) mass
(also calledmolecular weight)
Molar mass (M)
(also called
atomic weight)
(also called
gram-molecular weight)
amu
amu
g/mol
Averageof the masses of the naturallyoccurring isotopes of an element
weighted according to their abundance
Sum of the atomic masses of the atoms(or ions) in a molecule (or formula unit)
Mass of 1 mole of chemical entities(atoms, ions, molecules, formula units)
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Information Contained in the Chemical Formula of Glucose
C6H12O6( M = 180.16 g/mol)
Oxygen (O)
Mass/mole ofcompound
6 atoms
96.00 g
Table 3.2Carbon (C) Hydrogen (H)
Atoms/moleculeof compound
Moles of atoms/mole of compound
Atoms/mole ofcompound
Mass/moleculeof compound
6 atoms 12 atoms
6 moles
of atoms
12 moles
of atoms
6 moles
of atoms
6(6.022 x 1023)
atoms
12(6.022 x 1023)
atoms
6(6.022 x 1023)
atoms
6(12.01 amu)
=72.06 amu
12(1.008 amu)
=12.10 amu
6(16.00 amu)
=96.00 amu
72.06 g 12.10 g
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Interconverting Moles, Mass, and Number of Chemical Entities
Mass (g) = no. of moles xno. of grams
1 mol
No. of moles = mass (g) x
no. of grams
1 mol
No. of entities = no. of moles x6.022x1023entities
1 mol
No. of moles = no. of entities x6.022x1023entities
1 mol
g
M
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IONS AND IONIC COMPOUNDS
IONSare atoms or groups of atoms with a positive or negative charge.
Taking away an electron from an atom gives a CATIONwith a positive
charge
Adding an electron to an atom gives an ANIONwith a negative charge.
Mg --> Mg2++ 2 e- F + e- --> F-
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PREDICTING ION CHARGES
In general
metals(Mg) lose electrons ---> cations
nonmetals(F) gain electrons ---> anions
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METALS
M ---> n e- + Mn+
where n = periodic group
Na+ sodium ion
Mg2+ magnesium ion
Al3+ aluminum ion
Transition metals --> M2+or M3+
are common
Fe2+
iron(II) ionFe3+ iron(III) ion
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NONMETALS
NONMETAL + n e- ------> Xn-
where n = 8 - Group no.
C4-
,carbide
N3-, nitride
O2-, oxide
S2-, sulfide
F-, fluoride
Cl-, chloride
Group 7AGroup 6AGroup 4A Group 5A
Br-, bromide
I-, iodide
Name derived
by adding -ideto stem
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POLYATOMIC IONS
Note: many Ocontaining anions
have names ending
in ate (or -ite).
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Naming binary ionic compounds
The name of the cation is the same as the name of the metal.
Many metal names end in -ium.
The name of the anion takes the root of the nonmetal name
and adds the suffix -ide.
Calcium and bromine form calcium bromide.
The name of the cation is written first, followed by that of theanion.
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Naming Acids
1) Binary acids solutions form when certain gaseous compoundsdissolve in water.
For example, when gaseous hydrogen chloride(HCl) dissolves in
water, it forms a solution called hydrochloric acid. Prefix hydro-+
anion nonmetal root+ suffix -ic+ the word acid - hydrochloric
acid
2) Oxoacid names are similar to those of the oxoanions, except for
two suffix changes:
Anion -atesuffix becomes an -icsuffix in the acid.
Anion -ite suffix becomes an -oussuffix in the acid.
The oxoanion prefixes hypo-and per-are retained. Thus,
BrO4
-isperbromate, and HBrO4isperbromicacid; IO
2
-is iodite,
and HIO2is iodousacid.
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Properties of Ionic CompoundsForming NaCl from Na and Cl2
A metal atom cantransfer an electron to
a nonmetal.
The resulting cation
and anion are attractedto each other by
electrostaticforces.
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Electrostatic Forces
The oppositely charged ions in ionic compounds are attracted to one
another by ELECTROSTATIC FORCES.
These forces are governed by COULOMBS LAW.
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Electrostatic Forces
COULOMBS LAW
As ion charge increases, the attractive force_______________.
As the distance between ions increases, the
attractive force ________________.This idea is important and will come upmany times in future discussions!
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Molecular CompoundsCompounds without Ions
CH4 methane
CO2 Carbon dioxide
BCl3boron trichloride
All are
formed from
two or morenonmetals.
Ionic
compounds
generally
involve a metal
and nonmetal(NaCl)
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Empirical & Molecular
FormulasA pure compound always consists of the same
elements combined in the same proportions by
weight.
Therefore, we can express molecular composition asPERCENT BY WEIGHT
Ethanol, C2H6O
52.13% C13.15% H
34.72% O
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Percent CompositionConsider some of the family of nitrogen-oxygen compounds:
NO2, nitrogen dioxide and closely related, NO, nitrogen
monoxide (or nitric oxide)
Structure of NO2
Chemistry of NO,nitrogen monoxide
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Percent CompositionConsider NO2, Molar mass = ?
What is the weight percent of N and of O?
Wt. % O = 2 (16 .0 g O per mole )
46 .0 gx 100 % =69 .6%
Wt. % N =
14.0 g N
46.0 g NO2 100% = 30.4 %
What are the weight percentages ofN and O in NO?
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How to Determine a Formula?
Mass spectrometer
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Mass Spectrum of EthanolMass Spectrum of Ethanol(from the NIST site)
46
45
CH3CH2OH+
CH3CH2O+
H2C=OH+31
CH2O+
30
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DETERMINE THE FORMULA OF A
COMPOUND OF Sn AND I
Sn(s) + some I2(s) ---> SnIx
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Data to Determine the formula of a Sn
I Compound
Reaction of Sn and I2is done using excess Sn.
Mass of Sn in the beginning = 1.056 g
Mass of iodine (I2) used = 1.947 g Mass of Sn remaining = 0.601 g
Mass of Sn used = 0.455 g
Find moles of Sn used
0.455 g Sn 1 mol
118.7 g = 3.83 x 10-3 mol Sn
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Tin and Iodine Compound
Now find the number of moles of I2that combinedwith 3.83 x 10-3mol Sn. Mass of I2used was
1.947 g.
1.947 g I2 1 mol
253.81 g
= 7.671 x 10-3 mol I2
How many mol of iodine atoms?
= 1.534 x 10-2mol I atoms
7.671 x 10-3
mol I22 mol I atoms
1 mol I2
$
%
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Tin and Iodine CompoundNow find the ratio of number of moles of moles of I and Sn
that combined.
1.534 x 10-2
mol I
3.83 x 10-3
mol Sn
=4.01 mol I
1.00 mol Sn
Empirical formula is SnI4
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Chemical Equations
Depict the kind of reactantsand productsand their relative amountsin a reaction.
4 Al(s) + 3 O2(g) ---> 2 Al2O3(s)
The numbers in the front are called stoichiometric coefficientsThe letters (s), (g), and (l) are the physical states of compounds.
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Chemical Equations
Because the same atoms are present in
a reaction at the beginning and at theend, the amount of matter in a system
does not change.
The Law of the Conservation of Matter
Demo of conservation of matter, See
Screen 4.3.
2HgO(s) ---> 2 Hg(liq) + O2(g)Lavoisier, 1788
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BalancingEquations
____C3H8(g) + _____ O2(g) ---->
_____CO2(g) + _____ H2O(g)
____B4H10(g) + _____ O2(g) ---->
___ B2O3(g) + _____ H2O(g)
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STOICHIOMETRY
- the study of thequantitative aspects ofchemical reactions.
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PROBLEM: If 454 g of NH4NO3decomposes, how much N2Oand H2O are formed? What is the theoretical yield of
products?
STEP 1
Write the balanced chemical
equation
NH4NO3 ---> N2O + 2 H2O
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454 g of NH4NO3--> N2O + 2 H2O
STEP 2 Convert mass reactant(454 g) --> moles
454 g
1 mol
80.04 g = 5.68 mol NH4NO3
STEP 3 Convert moles reactant
(5.68 mol) --> moles product
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454 g of NH4NO3--> N2O + 2 H2O
STEP 3 Convert moles reactant --> moles product
Relate moles NH4NO3to moles product expected.
1 mol NH4NO3--> 2 mol H2O
Express this relation as the STOICHIOMETRICFACTOR.
2 mol H2O produced
1 mol NH4NO3 used
5.68 mol NH4NO3 2 mol H2O produced
1 mol NH4NO3used= 11.4 mol H2O produced
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454 g of NH4NO3--> N2O + 2 H2O
11.4 mol H2O 18.02 g
1 mol= 204 g H2O
STEP 4 Convert moles product(11.4 mol) --> mass product
Called the
THEORETICAL YIELD
ALWAYS FOLLOW THESE STEPS INSOLVING STOICHIOMETRY PROBLEMS!
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Reactions Involving aLIMITING REACTANT
In a given reaction, there is not enough of one
reagent to use up the other reagent completely.
The reagent in short supply LIMITSthequantity of product that can be formed.
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An ice cream sundae analogy for limiting reactants.
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LIMITING REACTANTS
Demo of limiting reactantsMethanol combustion
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Rxn 1: Balloon inflates fully, some Zn left
* More than enough Zn to use up the 0.100 mol HCl
Rxn 2: Balloon inflates fully, no Zn left* Right amount of each (HCl and Zn)
Rxn 3: Balloon does not inflate fully, no Zn left.* Not enough Zn to use up 0.100 mol HCl
LIMITING REACTANTS
React solid Zn with 0.100
mol HCl (aq)
Zn + 2 HCl ---> ZnCl2+ H2
1 2 3
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Rxn 1 Rxn 2 Rxn 3
mass Zn (g) 7.00 3.27 1.31
mol Zn 0.107 0.050 0.020
mol HCl 0.100 0.100 0.100
mol HCl/mol Zn 0.93/1 2.00/1 5.00/1
Lim Reactant LR = HCl no LR LR = Zn
LIMITING REACTANTS
React solid Zn with 0.100 molHCl (aq)
Zn + 2 HCl ---> ZnCl2+ H2
0.10 mol HCl [1 mol Zn/2 mol HCl]= 0.050 mol Zn
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PROBLEM: Mix 5.40 g of Al with8.10 g of Cl
2
. What mass of Al2
Cl6
can form?
Massreactant
StoichiometricfactorMoles
reactantMolesproduct
Massproduct
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Step 1 of LR problem:compare actualmole ratioof reactants totheoreticalmole ratio.
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2 Al + 3 Cl2 --->
Al2Cl6Reactants must be in the mole ratio
Step 1 of LR problem:compare actual mole ratio of
reactants to theoreticalmole ratio.
mol Cl2
mol Al =
3
2
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Deciding on the Limiting Reactant
If
There is not enough Al to use up all
the Cl2
2 Al + 3 Cl2 ---> Al2Cl6
mol Cl2
mol Al
>3
2
Lim reag = Al
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If
There is not enough Cl2to use
up all the Al
2 Al + 3 Cl2 ---> Al2Cl6
mol Cl2
mol Al Al2
Cl6
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CALCULATIONS: calculate mass ofAl2Cl6expected.
Step 1: Calculate moles of Al2Cl6expected based on LR.
0.114 mol Cl2 1 mol Al2Cl6
3 mol Cl2
= 0.0380 mol Al2Cl6
0.0380 mol Al2Cl6
266.4 g Al2Cl6
mol = 10.1 g Al2Cl6
Step 2: Calculate mass of Al2Cl6expectedbased on LR.
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Cl2was the limiting reactant.
Therefore, Al was present in
excess. But how much?
First find how much Al was required.
Then find how much Al is in excess.
How much of which reactant willremain when reaction is complete?
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2 Al + 3 Cl2 products
0.200 mol 0.114 mol = LR
Calculating Excess Al
Excess Al = Al available - Al required
0.114 mol Cl2 2 mol Al
3 mol Cl2 = 0.0760 mol Al req'd
= 0.200 mol - 0.0760 mol
= 0.124 mol Al in excess
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Determining the Formula of a Hydrocarbonby Combustion
Active Figure 4.9
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Using Stoichiometry toDetermine a Formula
Burn 0.115 g of a hydrocarbon, CxHy, and produce 0.379 g
of CO2and 0.1035 g of H2O.
CxHy + some oxygen --->
0.379 g CO2+ 0.1035 g H2O
What is the empirical formula of CxHy?
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Using Stoichiometry toDetermine a Formula
First, recognize that all C in CO2and all H in H2O is from
CxHy.
CxHy + some oxygen --->
0.379 g CO2+ 0.1035 g H2O
Puddle of CxHy
0.115 g
0.379 g CO2+O2
+O2 0.1035 g H2O
1 H2O molecule forms for
each 2 H atoms in CxHy
1 CO2molecule forms for
each C atom in CxHy
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Using Stoichiometry toDetermine a Formula
First, recognize that all C in CO2and all H in H2O is from CxHy.
1. Calculate amount of C in CO2
8.61 x 10-3 mol CO2--> 8.61 x 10-3 mol C2. Calculate amount of H in H2O
5.744 x 10-3mol H2O -- >1.149 x 10-2 mol H
CxHy + some oxygen --->
0.379 g CO2+ 0.1035 g H2O
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Using Stoichiometry toDetermine a Formula
Now find ratio of mol H/mol C to find values of x and y in CxHy.
1.149 x 10 -2 mol H/ 8.61 x 10-3 mol C
= 1.33 mol H/ 1.00 mol C= 4 mol H/ 3 mol C
Empirical formula = C3H4
CxHy + some oxygen --->
0.379 g CO2+ 0.1035 g H2O