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Chapter 3: The “Language of Chemistry”

“alphabet” symbols for the elements, e.g. C, N, F, Mg, Fe, etc. (Know names/symbols for #1-88 except lanthanides, including spelling!)

“words” chemical formulas, e.g. H2O, N2, Fe2(CO3)3, etc.

counting atoms in formulas:

--1 molecule of H2O contains 2 hydrogen atoms and 1 oxygen

--the formula Fe2(CO3)3 represents:

2 iron atoms, 3 carbon atoms, and 9 oxygen atoms

“sentences” chemical equations (reactants and products)

 

Mg(OH)2(aq) + 2 HCl(aq) MgCl2(aq) + 2 H2O(l)

Coefficients are used to “balance” the equationSubscripts indicate states of matter (not always included)Balanced equation: same number of atoms of each element on both sides of arrow

Molecular and Ionic Compounds• Molecular Compounds

– Atoms linked together by “covalent chemical bonds” in discrete electrically neutral particles called molecules

– e.g. H2O CO2 PCl3 C12H22O11

• Ionic Compounds– Result from transfer of one or more electrons from

one atom to another to yield oppositely-charged particles called ions

– No discrete molecules; ions held together by electrostatic forces (“ionic bonds”) in a regular, 3-D pattern called a crystalline lattice

– e.g. LiF lithium fluoride

– MgCl2 magnesium chloride

Li + F Li+ + F-

LiF

Mg Cl

Cl

Mg2+ + 2 Cl-

MgCl2

e-

e-

e-

Example Compounds

Na + Cl Na+ Cl-

e-

Formula Unit: The smallest unit of a compound. Shows the smallest whole-number ratio. (NaCl)

Ionic compounds must be electrically neutral.

NaCl

Molecular

Ionic

CH4C

H

HH

H

== =

Types of Chemical Formulas• empirical formula shows the simplest ratio of the

elements present• molecular formula shows the actual number of atoms in

one molecule• structural formula shows how the atoms are connected

e.g. for “hydrogen peroxide” the three formulas are:

empirical: HO

molecular: H2O2

structural: H

O OH

Ionic Compounds• Usually involve metals and/or polyatomic ions

1+ cations 2+ cations

2- anions

1- anions

Other metals may form more than one cation, e.g. Fe2+, Fe3+, Sn2+, Sn4+

Writing Formulas for Ionic Compounds• Polyatomic ions--Table 3.5--KNOW formulas and names!!!

– Two or more atoms combined in a single charged unit

– e.g. NH4+ (ammonium),

– H3O+ (hydronium),

– NO3- (nitrate),

– PO43- (phosphate),

– HCO3- (hydrogen carbonate, or bicarbonate)

• Look for the simplest combination of cations (+) and anions (-) to yield an electrically neutral formula– e.g. ion combination compound

– Mg2+ and Cl- MgCl2– Na+ and O2- Na2O

– Fe3+ and SO42- Fe2(SO4)3

• Example: What compound should form between sulfur (S) and potassium (K)?

• Example: What compound will form between ammonium and phosphate?

K2S

(NH4)3PO4

Nomenclature for Ionic Compounds• First, determine if it’s ionic!

– metal(s) + nonmetal(s)

• Binary ionic compounds (2 different elements)– cation(charge if needed) + anionide– Know Tables 3.3 and 3.4 (not older names)– e.g. ion combination compound name

– Mg2+ and Cl- MgCl2– Na+ and O2- Na2O

– Fe2+ and N3- Fe3N2

– Other ionic compounds– With polyatomic ions; cation(charge if needed) + polyatomic ion

name– Hydrates; compound name (as above) + prefixhydrate (Know

prefixes, p92)– e.g. ion combination compound name

– Ca2+ and Cr2O72- CaCr2O7

– Co2+ and Cl- CoCl2•6H2O

– Hg22+ and CN- Hg2CN2•H2O

magnesium chloridesodium oxideiron(II) nitride

calcium dichromate

cobalt chloride hexahydrate

mercury(I) cyanide hydrate

Nomenclature: Molecular Compounds• First, determine it’s molecular!

– between nonmetals and/or metalloids• Binary molecular compounds (between 2 elements)

– prefixelement + prefixelementide– First element is most metallic (bottom left of per.

table)• Use prefixes to indicate numbers of each atom, e.g.

– PF3 phosphorus trifluoride

– P2F4

– N2O5

• Exception: hydrogen plus one atom of a nonmetal, see next section!

diphosphorus tetrafluoride

dinitrogen pentoxide

Nomenclature; Binary Acids• First, determine it’s a binary acid!

– hydrogen + nonmetal– Hydroelementic + acid– e.g. compound name– HCl– HBr

hydrochloric acid

hydrobromic acid

Nomenclature: Oxoacids and Their Salts

• oxoacid HxEOy (E = nonmetal)

• Removal of H+ yields polyatomic anions

Series of chlorine oxoacids and their salts: HClOx (x = 1,2,3,4)

oxoacid polyatomic ions salt example

H2SO4

sulfuric acid

SO42–

sulfate

Na2SO4

sodium sulfate

HSO4–

Hydrogen sulfate

NaHSO4

Sodium hydrogen sulfate

H2SO3

Sulfurous acid

SO32–

sulfite

CaSO3

Calcium sulfite

HSO3–

Hydrogen sulfite

Ca(HSO3)2

Calcium hydrogen sulfite

polyprotic acids

acid salts

Hypochlorous acid, chlorous acid, chloric acid, perchloric acid

Stoichiometric EquivalenceA chemical formula shows the ratio by atoms and by moles

of the elements in the formula.

e.g. in the compound N2O5:

ratio by atoms: 2 atoms N : 5 atoms Oratio by moles: 2 moles N : 5 moles O

in N2O5,

2 moles N 5 moles O (a chemical equivalence)

Problem:How many moles of N atoms are combined with 15 moles of

O in N2O5?

Use the mole ratio as a conversion factor!(15 moles O) x (2 moles N/5 moles O) = 6.0 moles N

Formula Mass and Molecular Mass• formula mass = sum of all atomic masses of elements in a

formula(remember that atomic mass = the mass of a single atom)

• molecular mass = formula mass of a molecular substance{“formula weight” and “molecular weight” are often used instead}

ProblemWhat is the molecular mass of N2O5? (add the atomic masses!)

N2O5 = 2 N + 5 O

= 2(14.0) + 5(16.0) = 108.0What are the units? For 1 molecule: amu For 1 mole: grams

1 mole of a substance = its formula mass in grams e.g. 1 mole of N2O5 = 108.0 g N2O5

(just another conversion factor!)

Example Problems• What is the mass of 0.65 moles of N2O5?

(0.65 moles N2O5) x (108 g N2O5/1 mol N2O5) = 70 g N2O5

• What mass of iron combines with 5.00 g of oxygen to make Fe2O3?

Method: grams A --> moles A --> moles B --> grams B

(5.00 g O) x(1 mol O)/16.0 g O) x (2 mole Fe/3 mol O) x (55.85 g Fe/mole Fe) = 11.6 g Fe

Sample Problems• The label on my water bottle says there are 5.0 mg of

sodium in it; if that were pure sodium, how many atoms of sodium would that be?

• The same water bottle contains 16 oz of water. If I drink it all, how many moles of water did I drink?

Sample Problems• As I write this lecture I’m reading the label on my water

bottle. It says there are 5.0 mg of sodium in it; if that were pure sodium, how many atoms of sodium would that be?

• Answer: 1.3 x 1020 atoms Na

• The same water bottle contains 16 oz of water. If I drink it all, how many moles of water did I drink?

• Answer: 25 mol water

Percentage Composition• percentage composition -- mass % of elements in a

compoundTheoretical % composition -- from given formula

Example ProblemWhat is the percentage composition of H2CO3?

mole ratio = 2 mol H : 1 mole C : 3 mol O

molecular mass = 2(1.01) + 1(12.01) + 3(15.99) = 62.00 g/mol

% composition:

% H = [mass H / mass H2CO3] x 100%

= [2(1.01)/62.00] x 100% = 3.36%

% C = (12.01/62.00) x 100% = 19.36%

% O = [3 (16.00)/62.00] x 100% = 77.38%

Total 100.00%

Empirical Formula DeterminationExample ProblemA certain fluorocarbon is found to be 36.52% C, 6.08% H, and

57.38% F. What is the empirical formula for this compound?

We’re looking for the mole ratio of the elements.

In 100 g of the compound, there are:(36.52 g C) x (1 mol C/12.01 g C) = 3.041 mol C(6.08 g H) x (1 mol H/1.008 g H) = 6.02 mol H(57.38 g F) x (1 mol F/19.00 g F) = 3.020 mol F

So, the mole ratio is: C3.041H6.02F3.020

Now reduce to the simplest ratio (divide by the smallest number):

C3.041/3.020H6.02/3.020F3.020/3.020

= C1.007H1.99F = CH2F (the empirical formula)

Molecular Formula

Empirical formula combined with molecular mass = molecular formula

ProblemThe above fluorocarbon is found to have a molecular mass of

66.08 g/mole. What is the molecular formula?

n x (mass of empirical formula) = molecular mass (n = ?)Empirical formula = CH2F

Formula mass = 1 C + 2 H + F = 33.03 g/mole

n x (33.03 g/mole) = 66.08 g/mol so, n = 2

molecular formula is C2H4F2

Sample ProblemCarboranes are an interesting class of compounds that

contain carbon, hydrogen, and boron. One such carborane is found to have the following percentage composition: 28.18% C, 63.45% B, and 8.26% H. Determine the empirical formula of this carborane.

Sample ProblemCarboranes are an interesting class of compounds that

contain carbon, hydrogen, and boron. One such carborane is found to have the following percentage composition: 28.18% C, 63.45% B, and 8.26% H. Determine the empirical formula of this carborane.

• Answer: C2B5H7

Balancing Chemical Equations I• Adjust coefficients to get equal numbers of each kind of

element of both sides of arrow.• Use smallest, whole number coefficients.

e.g. start with unbalanced equation (for the combustion of butane):

C4H10 + O2 CO2 + H2O

Hint -- first look for an element that appears only once on each side; e.g. C

C4H10 + 13/2 O2 4 CO2 + 5 H2O

Multiply through by 2 to remove fractional coefficient:

2 C4H10 + 13 O2 8 CO2 + 10 H2O

Nomenclature: Organic Compounds• Compounds of carbon--organic chemistry

– If they have only C and H, hydrocarbons

– e.g. alkanes: methane CH4, ethane C2H6, propane C3H8

• general formula: CnH2n+2

– Know Table 3.7

• Functional Groups– R = hydrocarbon group

– e.g. alcohols: methanol CH3OH, ethanol C2H5OH

– e.g. amines: propyl amine, butylamine– Recognize functional groups in Table 3.8

Methanol (wood alcohol), CH3OH, is related to methane, CH4, by replacing one H with OH.