Transcript Chemistry 30 Review of Basic Chemistry 20

Chemistry 30 – Review
of Basic Chemistry 20
Polyatomic Molecular Elements:
H2(g)
hydrogen
N2(g)
nitrogen
O2(g)
oxygen
F2(g)
fluorine
Cl2(g)
chlorine
Br2(l)
bromine
I2(g)
iodine
S8(s)
sulfur
P4(s)
phosphorus
H2O2(l)
O3(g)
hydrogen peroxide CH3OH(l)
ozone
C2H5OH(l)
NH3(g)
C6H12O6(s)
ammonia
glucose
C12H22O11(s
sucrose
)
H2S(g)
CCl4(l)
methanol
ethanol
hydrogen sulfide
carbon tetrachloride
Naming Molecular Compounds:
Combining elements of the periodic table that come only from the nonmetals (right side
of the “staircase” only) forms a molecular compound.
Common molecular compounds include:
CH4(g)
methane C6H14(l)
hexane
C2H6(g)
ethane
C7H16(l)
heptane
C3H8(g)
propane
C8H18(l)
octane
C4H10(g)
butane
C9H20(l)
nonane
C5H12(l)
pentane
C10H22(l)
decane
When naming molecular compounds:
write the first name as given on the periodic
table of elements.
2.
write the last name using an “ide” ending.
3.
place the appropriate prefix in front the first
and last name to describe the number of
atoms there are of each element.
4.
where the first element has only one atom,
“mono” is not necessary.
Example: P4O3(g) = tetraphosphorus trioxide
1.
4 atoms
3 atoms
Naming Ionic Compounds:
Combining elements of the periodic table that
come from the metals and nonmetals (left and
right side of the “staircase” only) forms an
ionic compound.
2.
When naming ionic compounds:
3.
write the first name as given on the periodic
table of elements.
4.
write the last name using an “ide” ending.
5.
use no prefixes.
Example: CaCl2 - calcium chloride
1.
When writing simple ionic formulas:
1.
2.
3.
4.
put down the metallic element first.
put down the nonmetallic element last.
cross the elements’ ionic charge to become the subscript for each
other element.
numerically simplify the subscripts.
Example:
magnesium phosphide - Mg2+ and P3– join to
produce MgP
Use the charge of one element to be the subscript for the other
element – Mg3P2
Example:
calcium oxide - Ca2+ and O2– join to produce CaO
Use the charge of one element to be the subscript for the other
element – Ca2O2
Now simplify – CaO
all ionic compounds are solids at room temperature.
When writing ionic formulas
involving complex ions:

use the same format as above but whenever a complex ion is
named, use brackets to keep that complex ion as a group.
Example:
sodium sulfate Na+ and SO42–
Put the two together grouping the complex ion: Na+ (SO42–)
Now cross the charges:
Na2(SO4)1
Since 1’s are not necessary:
Na2SO4
Example:
Example:
calcium nitrate - Ca2+ and NO3–
Ca2+(NO3–)
Ca(NO3)2
sodium hydroxide - Na+ and OH–
Na+(OH–)
NaOH
When writing ionic formulas involving elements
with more than one charge:

use the first ion listed as the most common. For example, Cu2+ is
more common than Cu+, so Cu2+ would be used if no choice is
given.
When naming these compounds containing
elements with more than one charge:
• use Roman numerals to indicate the charge of the ion used.
Example:
CuCl is copper (I) chloride
Example:
CuCl2 is copper (II) chloride
Hydrated Compounds
When writing hydrated compounds, follow all
ionic rules described above. Then use a dot
along with the number of water molecules
required.
 When naming hydrated compounds, follow all
ionic rules described above. Then use a prefix in
front of the word “hydrate”.
Example: CuSO4  6H2O is:
copper (II) sulfate hexahydrate
Example: aluminum chloride trihydrate is:
AlCl3  3H2O

Clues for a Chemical Reaction

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

Formation of a precipitate
Formation of a gas
Colour change
Energy change
Types of Chemical Reactions
Simple Composition
element + element  compound
2 Na(s) + Br2(g)  2 NaBr(s)
Simple Decomposition
compound  element + element + element
2CaCO3(s)  2Ca(s) + 2C(s) + 3O2(g)
Single Replacement
element + compound  element + compound
Mg(s) + 2 NaOH(aq)  2 Na(s) + Mg(OH)2(aq)
Double Replacement
compound + compound  compound + compound
3 HCl(aq) + Al(OH)3(aq)  AlCl3(s) + 3 HOH(l)
Hydrocarbon Combustion
hydrocarbon + oxygen  carbon dioxide + water vapour
C3H8(g) + O2(g)  3 CO2(g) + 4 H2O(g)
Writing Dissociation Equations
Compounds that dissolve in water may produce ions. These solutions
are called electrolytes.
Some compounds may dissolve in water but form no ions. These
solutions are called nonelectrolytes.
When electrolytes are formed, dissociation equations can be shown.
NaOH(aq)  Na+(aq) + OH–(aq)
Al2(SO4)3(aq)  2 Al3+(aq) + 3 SO42–(aq)
Use mole ratios to determine solution or ion concentrations.
Examples: NaOH(aq)  Na+(aq) + OH–(aq)
2.0 mol/L
?
?
2.0 mol/L
2.0 mol/L
Al2(SO4)3(aq)  2 Al3+(aq) + 3 SO42–(aq)
?
3.0 mol/L
?
1.5 mol/L
4.5 mol/L
Examples:
Writing Nonionic, Total Ionic and
Net Ionic Equations
Example: A silver nitrate solution reacts with a solution of barium chloride.
AgNO3(aq)
+
BaCl2(aq)

Ba(NO3)2(aq)
+
AgCl(s) (unbalanced)
Nonionic Equation: (regular balanced equation)
2 AgNO3(aq)
+
BaCl2(aq)

Ba(NO3)2(aq)
+
2 AgCl(s)
Total Ionic Equation: (list dissociations for electrolytes only)
2 Ag+(aq) + 2 NO3–(aq) + Ba2+(aq) + 2 Cl–(aq)  Ba2+(aq) + 2 NO3–(aq)
(do not write dissociations for solids, liquids or gases)
Net Ionic Equation: (list only what reacts or changes)
or, simplified:
2 Ag+(aq) + 2 Cl–(aq)  2 AgCl(s)
Ag+(aq) + Cl–(aq)  AgCl(s)
+
2 AgCl(s)
Significant Digits

All numbers listed are significant except
zeros before or after a decimal that must
be used as placeholders.
Example: 100.0010 - 7 significant digits
0.001010 - 4 significant digits
Significant Digits
Continued…

Multiplication or Division Rules:



Count the number of digits in each number
being multiplied or divided.
Perform the multiplication or division.
Round off to the least number of digits found
in each of the individual numbers being
multiplied.
Significant Digits
Continued…
Example: 2.34 x 3.342 x 0.012 = 0.09384336 = 0.094 (answer to 2 sig. dig)
3
4
2
Example: 3.54 x 120.4 x 0.10 = 42.6216 =
3
4
2
43
Example: 35.127 x 225.5 x 2.75 = 21783.13088 = 2.18 x 104
5
4
3
Example: 350.55  12 = 29.2125 = 29
2
Example:
12  350 = 0.034285714 = 0.034
2
3
Addition or Subtraction Rules:
Count the number of digits following the decimal in each number
being added or subtracted.
Perform the addition or subtraction.
Round off to the least number of digits found following the decimal.
Example:
2.34 (2)
+ 2.8 (1)
5.14
=5.1
Example:
3.54
(2)
– 1.134 (3)
2.406
=2.41
Note: Always save all number in the calculator and round off only for
your final answer.
Stoichiometry






Determine the balanced chemical equation.
Determine information given.
Determine what it is you are solving for.
Determine the number of moles of what is given.
Use a mole ratio to determine the number of moles of the unknown.
Solve for the answer.
Example: If 200 mL of 0.100 mol/L silver nitrate solution reacts with a piece of copper, determine
the mass of metal reacted.
2 AgNO3 (aq)
v = 0.200 L
C = 0.100 mol/L
+
Cu(s)

m=?
nAgNO3 = Cv
nAgNO3= (0.200 mol/L)(0.100 L)
nAgNO3 = 0.0200 mol
nCu = 0.0200 mol x ½ = 0.0100 mol
mCu = nM
mCu = (0.0100 mol)(63.55 g/mol)
mCu = 0.636 g
2 Ag(s)
+
Cu(NO3)2(aq)