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EIT Review
S2007
Dr. J.A. Mack
Atom: The smallest divisible unit of an element
Compound: A substance made of two or more atoms
Ion: A charged atom or molecule
Cation: Positive ion
Anion: Negative ion
www.csus.edu/indiv/m/mackj/
NOMENCLATURE
Format for naming chemical compounds using prefixes,
suffixes, and other modifications of the names of elements
which constitute compounds.
Part 1
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2
Sometimes things get confusing…
Elements:
inert or “noble” gasses:
H = hydrogen
O = oxygen
C = carbon
Compounds:
H2 = hydrogen
O2 = oxygen
H2O = water
Metals form Cations
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non-metals form anions
Metalloids can do both
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1
Ion Charges:
Compounds fall into one of two classes:
+1
+3
-3 -2
-1
Inorganic Salts
Molecules
metal cation
non-metal
+
+
non-metal or
polyatomic anion
non-metal
(no cations or anions)
+2
Variable
The two use different formalisms for naming…
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Binary Compounds: Metal & non-Metal
Metals of variable charge (transition) with a non-metal
Metal of fixed oxidation (charge) state combined with a
non-metal.
Examples:
Cation
non-metal takes on “ide” suffix
Examples:
Cation
Anion
Formula
K+
Cl−
KCl
Potassium chloride
Ca2+
O2-
CaO
Calcium Oxide
Na+
S2-
Na2S
Sodium sulfide
Al3+
S2-
Al2S3
Aluminum sulfide
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Pb2+
Name
modify transition metal name with roman numeral
Anion
Formula
Name
Cl−
PbCl2
lead (II) chloride
pronounced: lead - two - chloride
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Pb4+
Cl−
PbCl4
lead (IV) chloride
Fe3+
O2−
Fe2O3
Iron (III) oxide
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Some common polyatomic ions:
Ternary Compounds: Those with three different elements
Type A: metal of fixed charge with a complex ion
NH4+
ammonium
H3O+
hydronium
CO32–
carbonate
HCO32– hydrogen or bicarbonate
NO2–
nitrite
NO3–
nitrate
SO42–
sulfate
SO32–
sulfite
PO43–
phosphate
C2H3O2– acetate
Cation
Anion
Formula
Name
K+
OH−
KOH
Potassium hydroxide
Ca2+
OH−
Ca(OH)2
Calcium hydroxide
Na+
SO 24−
Na2SO4
Sodium sulfate
Al3+
SO 24−
Al2(SO4)2
Aluminum sulfate
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Metal of variable charge transition) with a complex ion
Cation
Anion
Formula
Fe3+
NO3−
Fe(NO3)3
Iron (III) nitrate
Fe2+
NO −2
Fe(NO2)2
Iron (II) nitrite
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NonNon-metal with a nonnon-metal
When non-metals combine, they form molecules.
They may do so in multiple forms:
Name
CO
CO2
Because of this we need to specify the number of each
atom by way of a prefix.
1 = mono
5 = penta
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2 = di
3 = tri
6 = hexa
4 = tetra
7 = hepta
12
3
Examples:
Formula
D) Writing formulas for acids and Bases
Name:
BCl3
boron trichloride
SO3
sulfur trioxide
NO
nitrogen monoxide
we don’t write:
N2O4
•An acid is a substance that produces H+ when dissolved
in water.
•Certain gaseous molecules become acids when dissolved
in water.
nitrogen monooxide
or mononitrogen monoxide
•A base produces OH− when dissolved in water.
•Bases often are Group I and Group II hydroxide salts.
dinitrogen tetraoxide
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H+ and S2-
Type I Acids: Acids derived from –ide anions.
it takes 2 H+ to
cancel one S2-
The names for these acids follows the formula:
“hydro”
+
the root of the ide anion
+
ic “acid”
Anion:
Acid:
chloride
HCl
hydrochloric acid
fluoride
HF
hydrofluoric acid
H2S
Name:
hydro sulfuric acid
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Examples:
Practice:
Anion:
(nitrate)
NO3−
(sulfate)
2−
4
(acetate)
SO
Acid:
Name:
HNO3
nitric acid
C2 H 3O 2−
H2SO4
N2SO4
sodium sulfate
barium carbonate
BaCO3
FeO
Iron (II) oxide
zinc phosphide
Zn3P2
NiBr2
nickel (II) bromide
sulfuric acid
HC2H3O2
acetic acid
vinegar
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Common Names:
CO2
carbon dioxide
carbon monoxide
CO
P2O5
diphosphorous pentaoxide
nitrogen trihydride
NH3
(ammonia)
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H2O
water
ammonia
NH3
CH4
methane
NO
nitric oxide
N2O
nitrous oxide
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Modern Atomic Theory:
•Atoms are made of protons, neutrons and electrons.
please go to my chem. 1A web site and download a
nomenclature practice worksheet if you need more review.
http://www.csus.edu/indiv/m/mackj/chem1a/chem1A_lab.html
•The nucleus of the atom carries most of the mass.
• It consists of the protons and neutrons surrounded by a
cloud of electrons.
The charge on the electron is –1
The charge on the proton is +1
The link to the worksheet is at the top of the page.
There is no charge on the neutron
The Atomic Number or number of protons in the
nucleus defines an element.
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75
The isotope 34
Se is used medically for diagnosis of
Isotopes, Atomic Numbers, and Mass Numbers
pancreatic disorders. How many protons, neutrons, and
electrons does an atom of
•Atomic number (Z) = number of protons in the nucleus.
75
34 Se
have?
•Mass number (A) = total number of nucleons in the
nucleus (i.e., protons and neutrons).
75 protons +
neutrons
•One nucleon has a mass of 1 amu
(Atomic
Atomic Mass Unit)
Dalton”
Unit a.k.a “Dalton
75
34 Se
34 protons
•Isotopes have the same Z but different A.
•The elements are arranged by Z on the periodic table.
By convention, for element X, we write
protons = 34
A
ZX
electrons = 34
neutrons = 75-34 = 41
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Molar Masses (Molecular Weights) of Compounds:
Avagadro’
Avagadro’s Number
The molar mass of a molecular compound is the sum of the
molar masses of its atoms.
Since one mole of 12C has a mass of 12g (exactly), 12g of 12C
contains 6.022142 x 1023 12C-atoms.
Example:
But carbon exists as 3 isotopes: C-12, C-13 &C-14
The molar mass of CO2 is:
The average atomic mass of carbon is 12.011 u.
From this we conclude that 12.011g of carbon contains
1 x (12.01 g/mol)
6.022142 x 1023 C-atoms
+ 2 x (16.00 g/mol)
Yes, since NA is so large,
the statistics hold.
Is this a valid
assumption?
= 44.01 g/mol
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How many oxygen atoms are there in 25.1g of chromium (III) acetate?
Cr(C2H3O2)3
step 2: calculate the molar mass…
Question: What is the weight % of each element in C2H6?
229.13 g/mol
First determine the molar mass of C2H6:
1 mol of C 2 H 6 has a mass of 30.07 g
step 3: use dimensional analysis to solve the problem…
25.1g Cr(C2 H 3O 2 )3 × 1mol Cr(C2 H 3O 2 )3
229.13g Cr(C2 H 3O 2 )3
×
6.022 × 1023 O − atoms
=
1mol O
3.96 ×
1023
×
Percent Composition:
The relative amounts of each atom in a molecule or
compound can be represented fraction of the whole.
step 1: write the correct chemical formula…
Cr3+ & C2H3O2−
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(2×12.01 + 6 ×1.008) g/mol
Next determine the mass of hydrogen in 1 mol of the
compound:
6mol O
1mol Cr(C2 H3 O 2 )3
1 mol C2 H 6 ×
O-atoms
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6 mol H
1.0079 g H
×
= 6.047 g H
1 mol C2 H 6
1 mol H
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Determining a Formula from Percent Composition:
Now relate the mass of H in one mol of the
compound to the molar mass of the compound
Given the relative percentages of each element in a
compound,
1
× 100 = 20.11% H
6.047 g H ×
30.07g C2 H 6
10 % X,
Since there is only C as the remaining element:
% C = 100 − % H
20 % Y,
30 % Z
etc…
one can find the empirical formula of the compound.
= 79.89 % C
The empirical formula of a compound or molecule
represents the simplest ratio of each element in 1 mol
of the compound or molecule.
The compound C2H6 is 20.11% H & 79.89% C
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Example: A compound is found to be 64.82 % carbon,
21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
Solution:
Example: A compound is found to be 64.82 % carbon,
21.59 % oxygen and 13.59 % hydrogen.
What is the empirical formula for this compound?
2. Convert the grams of each element to moles.
determine X, Y & Z in (CXHYOZ)
(g element X → mole X etc…)
1. Since the percentages for each element sum to 100%,
if one equates % to grams (g), the sum of the masses must
be 100g.
(i.e. one can assume 100g of the compound)
64.82 g C
21.59 g O
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13.59 g H
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64.82g C ×
1 mol C
12.011 g C
= 5.397 mol C
21.59g O ×
1 mol O
16.00 g O
= 1.349 mol O
13.59g H ×
1 mol H
= 13.48 mol H
1.0079 g H
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3. Divide each of the individual moles by the smallest number of
Rounding to the nearest whole numbers:
moles to gain the molar ratios for each element in the compound.
These are the formula subscripts. (X2Y3 etc…)
Subscript for C
Subscript for H
X = 4.001 = 4
**If the ratios are fractional (0.5, 1.5 or 0.333) multiply each ratio by a
whole number to get even number formula subscripts.
0.5 × 2 = 1
Z = 1.000 = 1
Subscript for O
13.48
1.349
5.397
= 9.992 Z =
= 1.000
X=
= 4.001 Y =
1.349
1.349
1.349
Examples:
Y = 9.992 =10
0.25 × 4 = 1
C4H10O
The empirical formula is:
The results of this calculation tells us only about the
empirical formula of the compound.
To determine the molecular formula, we need more
information. This will be shown in a later example.
0.333 × 3 = 1
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Balancing Chemical Reactions: Example
Chemical Equations:
C2H6 +
Mass is conserved in a chemical reaction.
→
O2
2 C’s & 6 H’s
2 O’s
balance last
CO2
+
H2O
1 C & 2 O’s
2 H’s & 1 O
balance H first
Total mass of
reactants
=
Total mass of
products
2 2H6
___C
+
→
O2
CO2
3 H2O
6
___
+
balance C next
Chemical equations must therefore be balance for mass
2C2H6 +
balance O
The number and type of atoms on either side of the equation must
be equal!
2C2H6 +
O2
7 O2
____
4 C’s 12 H’s 14 O’s
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→
4 CO2
___
→
+
4CO2 +
6H2O
6H2O
4 C’s 12 H’s 14 O’s
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Reduction and Oxidation Reactions: RedOx
Oxidation Numbers
+1
+2
+3
-3 -2 -1
Oxidation involves an atom or compound losing electrons
Variable
Reduction involves an atom or substance gaining electrons
Niether process can occur alone… that is, there must be an exchange
of electrons in the process.
Metals take on their formal charge:
Non-metals do the same
The substance that is oxidized is the reducing agent
The substance that is reduced is the oxidizing agent
Chemists use oxidation numbers to account for the transfer of electrons
in a RedOx reaction.
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Balancing REDOX reactions:
Electrochemistry: Oxidation numbers
Fe + O2 → Fe2O3
In the compound potassium bromate (KBrO3), the
oxidation number of bromine (Br) is?
The compound is neutral so the sum of the oxidation numbers should
be zero.
+1
5
??
3×(-2) = -6
KBrO3
1+ ?? + (-6) = 0
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oxidation states:
0
0
+3
-2
oxidation half reaction:
{Fe → Fe+3
reduction half reaction:
{O2
+ 3e-} x 4
+ 4e- → 2O2-} x 3
Balance electrons transferred then sum the half RXN’s:
4Fe + 3O2 + 12e- → 2Fe2O3 + 12e-
?? = 5
4Fe + 3O2 → 2Fe2O3
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Quantitative calculations: Mass and moles
2HCl (aq) + Ba(OH)2 (aq) Æ 2H2O (l) + BaCl2(aq)
Consider the following reaction:
HCl (aq) + Ba(OH)2 (aq) Æ H2O (l) + BaCl2(aq)
g BaCl2 Æ mol BaCl2
Balacing:
Æ mol HCl
2 HCl (aq) + Ba(OH)2 (aq) Æ 2 H2O (l) + BaCl2(aq)
How many moles of HCl are consumed if 1.50 g of BaCl2 are
produced assuming that Ba(OH)2 is in excess?
Solution:
convert
to
g BaCl2
then convert
Æ
using the Molar mass of BaCl2
mol BaCl2
1.50 g BaCl 2 ×
1 mol BaCl 2
2 mol HCl
×
= 0.0144 mol HCl
208.24 g BaCl 2 1 mol BaCl 2
to
Æ
Molar mass BaCl2
mol HCl
Molar ratio for equation
using the Molar ratio for equation
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Stoichiometry: Conversions between masses & moles in
chemical reactions.
Limiting Reactant:
ÖWhen one reactant is present in an amount such that it
is completely consumed before all other reactants, we
say that it limits the reaction.
Ö The other reactants are said to be in excess.
excess
Ö The Theoretical Yield is determined by the stoichiometry
of the limiting reactant.
Ö The limiting reactant can only be determined through
molar ratios. It cannot be identified by mass.
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