Transcript Chapter 1: Fundamental Concepts

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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 equation
Subscripts 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– e.g. LiF
lithium fluoride
Li
– MgCl2
magnesium chloride
Mg
+ F
Li+ + F-
LiF
e- Cl
e- Cl
Mg2+ + 2 ClMgCl2
Example Compounds
Molecular
CH4 =
Ionic
=
=
e-
NaCl
Na
+
Cl
Na+ Cl-
Formula Unit:
The smallest unit of a
compound. Shows the
smallest whole-number
ratio. (NaCl)
Ionic compounds must
be electrically neutral.
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
O
H
Ionic Compounds
•
Usually involve metals and/or polyatomic ions
1- anions
2- anions
1+ cations
2+ cations
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
Mg2+ and ClNa+ and O2Fe3+ and SO42-
compound
MgCl2
Na2O
Fe2(SO4)3
• Example: What compound should form between sulfur (S)
and potassium (K)? K2S
• Example: What compound will form between ammonium and
phosphate? (NH ) PO
4 3
4
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 ClMgCl2
magnesium chloride
sodium oxide
–
Na+ and O2Na2O
iron(II) nitride
–
Fe2+ and N3Fe3N2
– 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
calcium dichromate
–
Ca2+ and Cr2O72CaCr2O7
–
–
Co2+ and ClHg22+ and CN-
CoCl2•6H2O
Hg2CN2•H2O
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 diphosphorus tetrafluoride
– N2O5 dinitrogen pentoxide
• Exception: hydrogen plus one atom of a nonmetal, see next
section!
Nomenclature; Binary Acids
• First, determine it’s a binary acid!
– hydrogen + nonmetal
– Hydroelementic + acid
– e.g. compound
name
hydrochloric acid
–
HCl
–
HBr
hydrobromic acid
Nomenclature: Oxoacids and Their Salts
• oxoacid HxEOy (E = nonmetal)
• Removal of H+ yields polyatomic anions
oxoacid
polyatomic ions
salt example
H2SO4
sulfuric acid
SO42–
sulfate
Na2SO4
sodium sulfate
HSO4–
Hydrogen sulfate
NaHSO4
Sodium hydrogen
sulfate
SO32–
sulfite
CaSO3
Calcium sulfite
HSO3–
Hydrogen sulfite
Ca(HSO3)2
Calcium hydrogen
sulfite
polyprotic
acids
H2SO3
Sulfurous acid
acid
salts
Series of chlorine oxoacids and their salts: HClOx (x = 1,2,3,4)
Hypochlorous acid, chlorous acid, chloric acid, perchloric acid
Stoichiometric Equivalence
A 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 O
ratio 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}
Problem
What is the molecular mass of N2O5? (add the atomic masses!)
N2O5 = 2 N + 5 O
= 2(14.0) + 5(16.0) = 108.0
What 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
compound
Theoretical % composition -- from given formula
Example Problem
What 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=
= 19.36%
(12.01/62.00) x 100%
% O = [3 (16.00)/62.00] x 100%
Total
= 77.38%
100.00%
Empirical Formula Determination
Example Problem
A 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)
(6.08 g H) x (1 mol H/1.008 g H)
(57.38 g F) x (1 mol F/19.00 g F)
So, the mole ratio is:
C3.041H6.02F3.020
= 3.041 mol C
= 6.02 mol H
= 3.020 mol F
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
Problem
The 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
molecular formula is C2H4F2
so, n = 2
Sample Problem
Carboranes 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 Problem
Carboranes 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.