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Regrade Policy• ALWAYS look over your quiz or exam when it is returned.
If there is an error you may request a regrade.
• Regrade requests are ONLY accepted within 1 week after an assignment is returned.
• For a regrade to be considered, there must be NO MARKS added to the quiz/exam being re-submitted.
• Regrade request must be written on a separate piece of paper attached to the original assignment.
• Graphing calculators are not allowed either! Only scientific or below.
Calculator Announcement
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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
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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-
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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
== =
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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
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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+
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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
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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
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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
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Nomenclature; Binary Acids• First, determine it’s a binary acid!
– hydrogen + nonmetal– Hydroelementic + acid– e.g. compound name– HCl– HBr
hydrochloric acid
hydrobromic acid
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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
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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
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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!)
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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
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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?
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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
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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%
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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)
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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
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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.
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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
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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
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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.