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