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Chemists describe a molecule (or the smallest unit of an infinite material) using formulae that list what elements the molecule contains and how many atoms of each element.
Chapter 3Molecules, Ions and Their
Compounds
Formulae
Three main types of formulae are used, depending on how much structural information is available or necessary:
A molecular formula gives no further information
A structural formula shows all of the bonding in a molecule, and is essentially a Lewis dot structure without the lone pairs.
A condensed formula indicates which atoms are grouped together but does not show any bonds.
C2H6OEx)
Ex) CH3CH2OH Ethanol Ex) CH3OCH3 Dimethyl ether
C CO
H
H
H H H
H C C
O
H
H
H
H
H
H
Types of Formulae
Molecular formulae tend to be used to describe simple molecules and ions that cannot be broken into stable multi-atom pieces.
Condensed formulae are used to describe more complex molecules that can be broken into stable multi-atom pieces
Structural formula are used when we want to be very clear about which atoms are bonded to which or the overall shape of the molecule.
Ex) H2O, CO2, NaCl
Ex) Cu(NO3)2 can be broken into a Cu2+ and two NO3- units
Organic chemists also use condensed formulae to describe what is bonded to each carbon atom in a chain
Ex) CH3CHBrCH3 is different from CH3CH2CH2Br
HC
C
CH
H H
H Br
H H
HC
C
CBr
H H
H H
H H IsomersMolecules which have the same molecular formula but different structural formulae
A structural formula in some cases is the only way to distinguishe between two isomers
IsomersMolecules which have the same molecular formula but different structural formulae
HC
C C
H
H HBr
H
HC
C C
HH H
BrH
Consider: CH3CHCHBr
Consider: CH3CHClBr
HC
C
ClH H
BrH
HC
C
Cl
H HBr
H
Structural Formulas and Isomers
Naming CompoundsIn order to communicate conveniently compounds need to be named unambiguously.
Naming Ionic Compounds
To name an ionic compound, name the cation then the anion.
Monoatomic CationsMost cations are single elements and are given the same name as the element.
Ex). Na+ = sodium Mg2+ = magnesium
If an element can give rise to different cations use a Roman numeral to indicate the charge.
Ex) Fe2+ = iron (II) Fe3+ = iron (III)Au+ = gold (I) Au3+ = gold (III)Pb2+ = lead (II) Pb4+ = lead (IV)
This is necessary for most transition metals as well as tin and lead
Monatomic cations and the Periodic Table
Monoatomic AnionsTo indicate that an element has formed an anion, we change its suffix to -ide.
Ex) F = fluorine F- = FluorideN = nitrogen N3- = NitrideO = oxygen O2- = OxideSe = selenium Se2- = Selenide
Tend to form exclusively from the non-metals and one metalliod
Charge = group # -18
Free ions do not exist with charges higher than 3 units.
carbides have littletrue ionic character. But the valence of carbon is still often 4- in many of its compounds
Polyatomic IonsNot all ions consist of just one atom. Some groups of covalently bonded atoms have overall charges. Because they are charged, they are ions (not molecules).
Polyatomic cations:
NH4+ Ammonium Best known
H3O+ Hydronium
Polyatomic Anions:
ClO- Hypochlorite (Bleach)
HCO3- Bicarbonate(baking Soda)
ExamplesCN- Cyanide CrO4
2- Chromate OH- Hydroxide Cr2O7
2- DichromateCO3
2- Carbonate MnO41- Permanganate
When H+ is added to a polyatomic atom the prefix “bi” is placed before the ‘old’ name.
HCO3- Bicarbonate HS- Bisulfide HSO4
- Bisulfate
Cl N S P
O hypochlorite(ClO-)
O2 chlorite(ClO2
-)nitrite(NO2
-)
O3 chlorate(ClO3
-)nitrate(NO3
-)sulfite(SO3
2-)phosphite
(PO33-)
O4 perchlorate(ClO4
-)sulfate(SO4
2-)phosphate
(PO43-)
The polyatomic ions ending with ‘ate’ all consist of an atom surrounded by a number of oxygen atoms.
Those ending in “ite” have one fewer O
Ionic CompoundsTo name an ionic compound, name the cation then the anion. The charge of each ion indicates how many are needed to make a neutral compound so no prefixes are necessary.
MgCl2
CuBr2
NaNO3
(NH4)2SO4
Sn(CO3)2
Sn(HCO3)2
NaNO2
Magnesium Chloride
Copper(II) Bromide
Sodium Nitrate
Ammonium Sulfate
Tin (IV) Carbonate
Exercise
Tin (II) Bicarbonate
Sodium Nitrite
Naming Covalent CompoundsList the elements in order from least to most electronegative, using a prefix to indicate how many atoms there are of each element.
Change the suffix of the final (most electronegative) element to “ide” as if it was an anion.
SO2
NO2
CO
Sulfur dioxide
Nitrogen dioxide
Carbon monoxide
Carbon dioxide
Carbon tetrachloride
Compounds with hydrogen are often referred to by their common name or as ionic compounds
CH4 Methane NH3 Ammonia BH3 Borane
H2O Water
If there is only one atom of the first (least electronegative) element, no prefix is used before it.
Dinitrogen tetroxide
HF HCl
H2S
Hydrogen Fluoride Hydrogen Chloride
CO2
CCl4
N2O4
Hydrogen Sulfide
Hydrated CompoundsMany ionic compounds incorporate water molecules in the ionic lattice.
The electronegative oxygen atoms of the water molecules are attracted to the positively charged metal cations.
The water molecules can be removed by heating to “dehydrate” it.
Thus, the water is listed at the end of the chemical formula.
Ex) Hydrated CuSO4 is CuSO4·5H2O
The name of a hydrated compound, is composed of the name of the unhydrated molecule followed by a prefix before “hydrate” indicating the number of water molecules
Ex) CuSO4·5H2O copper(II) sulfate pentahydrate
If an ionic lattice contains no water, the compound is said to be anhydrous.
Exercise
2.46 g of a magnesium sulfate hydrate is dehydrated to give 1.20 g magnesium sulfate. What is the chemical formula for the hydrate?
MgSO4 M.W. = 24.3050 + 32.066 + 4*15.9994 g/mol
H2O M.W. = 2*1.0079 + 15.9994 g/mol =120.3686 g/mol
=18.0152 g/mol
Mol. MgSO4 = (1.20 g)/(120.3686 g/mol)= 0.0100 mol
Mol. H2O = (2.46g - 1.20 g)/(18.0152 g/mol)= 0.0700 mol
0.0100 mol MgSO4 : 0.07 mol H2O
1 MgSO4 : 7 H2O
MgSO4·7H2O mangesium sulfate heptahydrate
Empirical Formula and Molecular Formula
Recall that molecular formula indicates the total number of atoms of each element in a molecule.
Empirical formula indicates the ratio of atoms of each element in a molecule. It can be obtained by dividing the subscripts in the molecular formula by the largest common factor.
Write empirical formulae for the following molecules:
C6H6
C6H12O6
C2H4O
CH
CH2O
C2H4O
C10H12O4 C5H6O2
We can use the empirical formula to calculate the percent mass of each element in a molecule.
Ex). What is the percent composition of C, H and O in glucose, C6H12O6?
Percent Composition
Empirical formula? CH2O
mass of emp. form.= 12.011 + 2*1.0079+15.9994 g/mol= 30.0262 g/mol
Molecular Mass = (30.0262 g/mol)*6 = 180.1572 g/mol
% mass C = 100*mass C/total mass =100*(12.011 g/mol)/(30.1572 g/mol)
= 39.8280 %% mass H = 100*mass H/total mass =100*(2*1.0079 g/mol)/(30.1572 g/mol)
= 6.6843 %
% mass O = 100*mass O/total mass =100*(15.9994 g/mol)/(30.1572 g/mol)
= 53.0533 %
Percent composition can be used to determine the empirical formula.
Ex). A compound was subjected to elemental analysis and found to contain 58.01% C, 16.23% H and 25.76% O. What is its empirical formula?
Molecular formula cannot be calculated, since only the relative proportions are known, not their common factor.
Percent Composition & Empirical Formula
Assume 100 g of material, or any mass of choice
Compute mass of each element using % composition
Mass C = (% C)*(total mass)/100 = (58.01 %)*(100 g)/(100 %)
= 58.01 gMass H = (% H)*(total mass)/100 = (16.23 %)*(100 g)/(100 %)
= 16.23 g
Mass O = (% O)*(total mass)/100 = (25.76 %)*(100 g)/(100 %)
= 25.76 g
Compute the number of moles of each element.
Mol. C = (mass C)/ (molar mass C) = (58.01 g)/ (12.011 g/mol)
= 4.830 mol
Mol. C = (mass H)/ (molar mass H) = (16.23 g)/ (1.0079 g/mol)
= 16.10 mol
Mol. O = (mass O)/ (molar mass O) = (25.76 g)/ (15.9994 g/mol)
=1.610 mol
Find common ratio by dividing through by the smallest number
4.830 C : 16.10 H: 1.610 O Divide by 1.610
3.000 C : 10.00 H: 1.000 O
Empirical Formula is C3H10O
You are told that the molecular mass is 186.33 g, determine the molecular formula.
Recall that the empirical Formula is C3H10O
Mass of emp. form. = 3*12.011 + 10*1.0079 + 15.9994 g/mol
= 62.1114 g/mol
Ratio of molecular mass to mass of empirical formula gives the common factor we need.
Molecular mass/Mass of emp. form = (186.33 g/mol)/ 62.1114 g/mol)
= 3.000
Therefore the molecular formula is C9H30O3
Molecular Formula
Ex) 2.472 g of Manganese metal was completely reacted with 2.564 g of fluorine gas to produce the metal fluoride, MnxFy.
Information from a chemical reaction can be used to determine the empirical formula
Chemical Reaction
a) Calculate % composition of Mn and F in the product.
Complete reaction means that all the F and Mn have combined.
Mass of products = mass of reactants = 2.472 g Mn + 2.564 g F
= 5.036 g of product
% Mn =100*(mass Mn)/(mass product) = 100*(2.472 g)/(5.036 g) = 49.09 %
% F =100*(mass F)/(mass product) = 100*(2.564 g)/(5.036 g) = 50.91 %
b) Determine the empirical formula
We need to find the ratio of Mn atoms to F
Compute number of moles of Mn and F.
Mol Mn = mass Mn/ molar mass Mn = (2.472 g)/(54.938 g/mol)
= 0.04500 mol Mn
Mol F = mass F/ molar mass F = (2.564 g)/(18.9984 g/mol)
= 0.1350 mol Mn Ratio of Mn to F:
0.04500 mol Mn : 0.135 mol F Divide by 0.04500
3.000 mol Mn : 1.000 mol F
Empirical formula is MnF3
Concepts from Chapter 3
molecular vs. condensed vs. structural formula
naming ionic compounds (including polyatomic ions)
naming covalent compounds
hydrated vs. anhydrous compounds
empirical vs. molecular formula
calculating empirical formula from percent composition
calculating percent composition from empirical formula
ionic and covalent bonding (see chapter 9 notes)
the mole (see chapter 2 notes)
Problem #114 Ch 3A sample of CaCl2• 2H2O was weighed at 0.832 g. After heating it was weighed again giving a mass of 0.739 g.
Have they dehydrated the sample?
Assuming it was pure determine how many moles of CaCl2• 2H2O there is in the sample before heating
Moles CaCl2• 2H2O = mass/molar mass
= (0.832 g)/(40.0780 + 2*35.453 +2*(2*1.0079+15.9994))
= (0.832 g)/(147.0144 g/mol) = 0.00566 mol
Moles of water lost would have to be twice as many as the hydrate
What would happen if the sample were heated again?
Moles water = 0.01132 mol
Mass water = (0.01132 mol)(2*1.0079+15.9994) = 0.204 g
Therefore final weight should have been 0.832 -0.204 = 0.628 g
Was the sample pure?
Moles CaCl2 = mass/molar mass
= (0.739 g)/(40.078 + 2*35.453 g/mol)
Assuming the heating removed all the water compute moles of CaCl2
= (0.739 g)/(110.984 g/mol)
If the sample was pure then twice as many moles of water would be lost
mass water lost = (0.0133 mol)(18.0152 g/mol) = 0.240 g
Moles water lost = 0.0133 mol
= 0.00666 mol
Therefore the initial weight should have been 0.739 g + 0.240 g = 0.979 g
Both calculations tell us the sample is not dehydrated.
Compute how many moles of water was removed
Moles H2O = weight change/ molar mass
= (0.832 -0.739 g)/(2*1.0079+15.9994)
= 0.093 g/18.0152 g/mol = 0.0052 mol
If we had pure hydrate in the beginning half as many moles as water lost would have been in the sample
Moles hydrate = (0.0052 mol)/2 = 0.0026 mol
Mass of hydrate = (0.0026 mol)((147.0144 g/mol) = 0.38 g
The number of moles of hydrate before heating would be the same as the number of moles of CaCl2 after heating
Moles CaCl2 = 0.0026 mol
Mass CaCl2 = (0.0026 mol) (110.984 g/mol) = 0.28 g
Both disagree with observed weigh therefore 1) sample is not dehydrated or 2) was impure to begin with
Only by reheating the sample repeatedly until it stops losing weight can one determine the total water lost.
If half the moles of water lost corresponds to total moles of CaCl2 • 2H2O
before heating or number of moles of CaCl2 after heating we know the sample was pure.
If this is not the case the composition of the sample can be determined from the amount of water lost.
Concepts from Chapter 9
DRAWING LEWIS ELECTRON DOT DIAGRAMS
Octet rule
Resonance structures
Bond polarity (ionic, polar covalent and covalent bonds)
Ionic vs. covalent compounds
Electronegativity
Dipole vectors
Calculating formal charges and partial charges
Bond order
Bond lengths
VSEPR and predicting shapes of molecules
Polarity of molecules
Concepts from Chapter 10
MOLECULAR ORBITALS
LCAO theory
Correlation diagrams
Bonding, nonbonding and antibonding interactions
Calculating bond order using MO theory
hybridization (sp, sp2 and sp3 orbitals)
Sigma (σ) vs. pi (π) bonds
Composition of single, double and triple bonds
Resonance according to MO theory