~ 1 ~ Synthesis of the elements. When two pure elements react, they combine to form a binary compound: E + E → BC Ex.1) Zinc metal is heated in pure oxygen: Ex.2) Hydrogen gas reacts with solid sulfur (S 8 ): Ex.3) Sodium metal is heated with chlorine: Ex.4) Synthesis of magnesium nitride 15.1 Synthesis/Decomposition Review
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15.1 Synthesis /Decomposition Revie15+Notes,+18-19.pdf~ 9 ~ Acid rain is rain with a particularly low (acidic) pH. “Clean” rain is naturally acidic with a pH of about 5.6, since
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~ 1 ~
Synthesis of the elements. When two pure
elements react, they combine to form a binary
compound: E + E → BC
Ex.1) Zinc metal is heated in pure oxygen:
Ex.2) Hydrogen gas reacts with solid sulfur (S8):
Ex.3) Sodium metal is heated with chlorine:
Ex.4) Synthesis of magnesium nitride
15.1 Synthesis/Decomposition Review
~ 2 ~
Metal chlorides and oxygen. Metal chlorides react
with oxygen to form metal chlorates:
MCl + O2 → MClO3
Ex.5) sodium chloride is heated in oxygen gas:
Ex.6) formation of barium chlorate:
Ex.7) iron (III) chloride and oxygen gas:
Ex.8) synthesis of cobalt (VI) chlorate
~ 3 ~
Metal oxides and carbon dioxide. Metal oxides
react with carbon dioxide to produce metal
carbonates:
MO + CO2 → MCO3
Ex.9) calcium oxide and carbon dioxide:
Ex.10) formation of copper (II) carbonate:
Ex.11) synthesis of potassium carbonate:
Ex.12) aluminum oxide and carbon dioxide:
~ 4 ~
Metal oxides and water. Metal oxides react with
water to form metal hydroxides. As a result, they
are often called basic oxides.
MO + H2O → MOH
Ex.13) Magnesium oxide is added to water:
Ex.14) Sodium oxide is added to water:
Ex.15) Formation of silver hydroxide:
Ex.16) Synthesis of lead (IV) hydroxide:
~ 5 ~
Decomposition patterns are the exact reverse of
their synthesis counterparts and are often brought
about by exposure to intense heat. Carbonates also
decompose readily with acids.
Ex.17) magnesium carbonate is heated...
Ex.18) decomposition of potassium chlorate
Ex.19) copper (II) hydroxide is heated in a crucible
Ex.20) nitrogen triiodide explosively decomposes
~ 6 ~
Non-metal oxides and water. Many (but not all)
non-metal oxides react with water to form oxoacids.
We call them these substances acidic oxides.
NMO + H2O → Oxoacid
(All of the polyatomic acids w/ oxygen = oxoacids)
Only certain non-metal oxides are acidic. To find
them, dehydrate an existing oxoacid by removing all
hydrogens and exactly ONE oxygen atom:
Ex.1) Sulfuric acid: H2SO4 - H2O =
Ex.2) Carbonic acid: H2CO3
Ex.3) Chlorous acid:
15.2 Acidic Oxides and Acid Rain
~ 7 ~
If a non-metal oxide can NOT be created by this
dehydration method, it is unlikely to be acidic in
water:
Ex.4) Working backwards from sulfuric and sulfurous
acids, which of the following sulfur oxides are
acidic and which one(s) is/are not?
SO SO2 SO3
As a result, not all NMO's react with water:
Ex.5) water and carbon dioxide:
Ex.6) water and carbon monoxide:
~ 8 ~
Ex.7) synthesis of sulfuric acid:
Ex.8) nitrogen dioxide and water:
There are many "non-traditional" oxides which form
oxoacids in water. Typically, the substance in
question will contain a larger number of oxygens
and will form the oxoacid with the greatest number
of oxygen atoms.
Ex.9) Phosphorus pentoxide (P4O10) + water
Ex.10) Cl2O7 + water
~ 9 ~
Acid rain is rain with a particularly low (acidic) pH.
“Clean” rain is naturally acidic with a pH of about
5.6, since the CO2 from respiration combines with
water to make carbonic acid, H2CO3. However, the
addition of nitrogen oxides (NOx) and sulfur dioxide
(SO2) to the atmosphere causes rain to become
even more acidic, as water combines with those
two to become STRONG acids.
Complete and balance the following:
CO2 + H2O → (pH ≈ 5.6)
NO2 + H2O → (pH ≈ 4.0)
SO2 + H2O → (pH ≈ 4.0)
H2SO3 + O2 → (pH ≈ 4.0)
SO2 is the most important gas in acid rain formation.
Sulfur dioxide can be formed by volcanic eruptions,
and nitrogen oxides by lightning strikes. These
naturally occuring processes provide key nutrients
for life (ex. nitrogen in plant life). However, both are
being formed in excessive ammounts due to hu
industrial processes. This includes electricity
generation (main), factories, and motor vehicles.
These can come back to the surface either through
wet deposition (precipitation) or
(particles simply stick to the
~ 10 ~
Sulfur dioxide can be formed by volcanic eruptions,
and nitrogen oxides by lightning strikes. These
naturally occuring processes provide key nutrients
for life (ex. nitrogen in plant life). However, both are
being formed in excessive ammounts due to hu
industrial processes. This includes electricity
generation (main), factories, and motor vehicles.
These can come back to the surface either through
(precipitation) or dry deposition
(particles simply stick to the ground and plants)
Sulfur dioxide can be formed by volcanic eruptions,
and nitrogen oxides by lightning strikes. These
naturally occuring processes provide key nutrients
for life (ex. nitrogen in plant life). However, both are
being formed in excessive ammounts due to human
generation (main), factories, and motor vehicles.
These can come back to the surface either through
dry deposition
ground and plants)
~ 11 ~
Because of this, rain becomes too acidic and starts
to contaminate water (lakes, oceans, rivers), soil,
and forests, harming both animals and humans. In
addition, acid rain damages buildings and
monuments, especially those made of marble or
limestone: CaCO3 + HX → CaX + CO2 + H2O.
The effect of acid deposition is not localized to
the area where the sulfur/nitrogen was released;
the acid rain spreads around the world.
Mountainous regions tend to receive the most acid
rain simply because of their altitude (air cools as it
rises, causing precipitation).
Governments have the power to work together and
fix this problem. The Clean Air Act Amendments of
1990 have reduced SO2 emissions by 10 million tons
per year (65%). Using lime (CaO, calcium oxide), SO2
can be captured as it exits a power plant, a process
called flue-gas desulfurization:
CaO(s) + SO2(g) → CaSO3(s) → CaSO4 (inert)
~ 12 ~
Catalytic converters in cars serve a similar purpose.
Incomplete combustion of fuel (hydrocarbons)
produces nitrogen compounds, NOx, CO, and
unburnt CH compounds. A fine mesh of precious
metals within the converter captures these
molecules, allowing them to react with oxygen as it
becomes available. Once fully oxidized, they are
released:
2 NOx(g) → O2(g) + N2(g)
CO(g) → CO2(g)
CxHy(g) → CO2(g) + H2O(g)
This is why we have emissions tests for vehicles.
~ 13 ~
A solution is any homogenous mixture. Generally
speaking, we will refer to them as substances
(solute) dissolved in water (solvent). The
concentration of these solutions is most commonly
measured as their molarity:
Molarity, M = mol (of solute)
total volume (liters)
Molarity is one of several different methods
(molality, ppm, ppt, ppb, % v/v, % m/v, % m/m) of
measuring concentration and is expressed
differently on the AP and IB exams:
(AP) (IB)
Molarity, M = mol L-1
= mol dm-3
mol/L = mol/dm3
Other units: 1 mL = 1 cm3
1 L = 1 dm3
15.3 Solution Concentration and Prep.
~ 14 ~
Use the molarity equation to solve each of the
following problems. M = mol/L
Ex.1) Determine the molarity of 1.43 mol glucose which
has been dissolved to form 6.23 L of solution
Ex.2) Calculate the molarity of 431.5 mg NaCl (Mr =
58.443) dissolved in 235 cm3 of solution:
Ex.3) What mass of potassium iodide, in grams, would be
necessary to create 3.50 L of a 0.250 mol dm-3
solution?
~ 15 ~
The concentration of specific ions within a solution is
fairly easy. Just calculate the overall molarity of the
substance in question and multiply that value by the
number of ions into which it dissociates:
1.00 M NaK2PO4 → 1 x Na+ 2 x K
+ 1 x PO4
3-
[Na+] = 1.00 M [K
+] = 2.00 M [PO4
3-] = 1.00 M
Ex.4) 0.300 M Sr(NO3)2 [Sr2+
] = [NO3-] =
Ex.5) 0.750 mol L-1
iron (III) sulfate solution:
Ex.6) 0.80 mol of sodium carbonate in 500.0 mL solution:
Ex.7) 0.900 M ammonium dichromate
~ 16 ~
Concentrated solutions may also be diluted by the
addition of water or another solvent via the
equation below:
C1V1 = C2V2
"C" is used to denote any unit of concentration,
while "V" is any unit of volume. Be consistent on
both sides of the equation! The difference between
V1 and V2 is the amount ofwater which has been (or
needs to be) added to produce the dilution.
Ex.8) 50.0 mL of a stock 16.0 M HNO3 is diluted to a final
volume of 750.0 mL. Calculate [HNO3] at this point.
Ex.9) Given 2.5 L of a 1.3 M KCl solution, what is the new
molarity of the solution if 1.5 L of water is added?
~ 17 ~
Ex.10) 350.0 mL of a 0.500 mol L-1
sodium hydroxide
solution is left out over Spring Break. Upon
returning, you find that 78.4 mL of the solution has
evaporated. Calculate the [NaOH] at this point:
When a small sample (or aliquot) is taken from a
larger solution, the concentration of the aliquot
matches the concentration of the parent solution.
Ex.11) 67.5 g of sodium phosphate (Mr = 163.94) is
dissolved to make 1.60 L of solution. How many mL
of this solution are required to create 1.00 L of a
0.100 M solution by dilution?
~ 18 ~
Ex.12) 10.0 g NaOH (Mr = 40) is used to produce 0.500 L of
solution "A". A 100.0 mL aliquout of solution A is
diluted to 500.0 mL to form solution "B". How
many mg of NaOH are contained by 10.0 mL of
solution B?
~ 19 ~
Precipitation reactions occur when two or more
dissolved reactants form at least one insoluble
product. For example:
KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)
The solid produced by these reactions is known as a
precipitate and can be collected, cleaned, and
purified via filtration. The mass of the precipitate
can be used to study the contents of the original
sample in a process known as gravimetric analysis.
This process is very similar to combustion analysis
and requires you to use the mass of the products to
"track" backwards to the mass(es) of the
reactant(s). You were introduced to this concept a
few weeks ago, in combustion analysis.
You WILL need to understand solubility concepts.
Review 10.3 if you're a little "fuzzy" on this topic.
15.4 Gravimetric Analysis
~ 20 ~
Gravimetric analysis is commonly used to determine
the purity of unknown samples. The precipitate is
collected and then used to calculate the amount of a
specific reactant within the sample:
Ex.1) You are given a 50.00-g mixture of barium chloride
(Mr = 208.234) and inert solid. After being dissolved
in water, an excess of silver nitrate (Mr = 169.875) is
added. A white precipitate is collected and, after
drying, found to have a mass of 17.48 g. Determine
the purity of the sample in terms of the chloride:
Give the balanced molecular equation for this rxn
and identify the white precipitate being described:
Using DA, work backwards from the mass of the
precipite to the mass of BaCl2 which produced it:
~ 21 ~
Determine the purity of the sample. Percent purity
is calculated as (mass of analyte / total mass of
sample) x 100:
Ex.2) A sample to be tested in lab contains a mixture of
barium chloride and sodium chloride. 3.725 g of
this mixture is dissolved in water and titrated with
an excess of potassium sulfate. A precipitate with a
mass of 2.734 g is collected and dried.
Give the balanced molecular equation for this rxn
and mark the precipitate which will be formed:
Determine the mass % of both chlorides within the
mixture:
~ 22 ~
Mass percentages or DA can be used to "pull" the
mass of an analyte directly from a precipitate.
Ex.3) A 3.00 g sample of an alloy containing lead and tin is
dissolved in nitric acid. An excess of sulfuric acid is
then added to the mixture, resulting in the
formation of 2.37 g of a precipitate.
Identify the precipitate which is being formed:
"Pull" the mass of the metal from the precipitate.
Determine the mass percentages of tin and lead in
this alloy:
~ 23 ~
Sometimes, you don't know the identify of the
substance being tested. In these instances, a pure
sample is used and the molar mass of the unknown
can be calculated.
Ex.4) A metal nitrate is known to have a general formula
of M(NO3)4. A 2.000-g sample of this nitrate is
dissolved in water and slowly titrated with a 0.2100
mol dm-3
solution of sodium phosphate.
Precipitation is complete after the addition of 42.91
mL of the phosphate solution.
Give the balanced molecular equation for this rxn.
Substitute "M" for the unknown metal.
How many moles of sodium phosphate were used?
~ 24 ~
How many moles of the unknown nitrate were
present in the sample?
Using the mass and moles of the unknown nitrate,
determine its molar mass.
Remove the mass of the nitrates and determine the
molar mass (and ID) of the unknown metal.
~ 25 ~
With a knowledge of solubility rules, net-ionic
equations can be written to simplify the normal (full
or molecular equations) we normally write.
In the context of this chapter, a net-ionic equation
would only include ions which form a precipitate.
The ions which do not form a precipitate remain
dissolved throughout the reaction and are called
spectator ions. For example:
The balanced molecular equation for the
precipitation of lead (II) iodide. This is what we
would normally write when doing stoichiometry:
Pb(NO3)2 (aq) + 2KI (aq) → PbI2 (s) + 2KNO3 (aq)
By showing the dissociation of soluble particles, we
create a complete ionic equation:
Pb2+
+ 2NO3- + 2K
+ + 2I
- → PbI2 + 2K
+ + 2NO3
-
15.5 Writing Net Ionic Equations
~ 26 ~
Since K+ and NO3
- appear on both sides, they are
considered spectator ions. Remove them and
balance the remaining particles to create the net
ionic equation:
Pb2+
+ 2NO3- + 2K
+ + 2I
- → PbI2 + 2K
+ + 2NO3
-
Net Ionic: Pb2+
(aq) + 2I- (aq) → PbI2 (s)
For examples 1-3, create the complete ionic and net