Download B) Examples of Avagadro`s Number

AP CHEMISTRY II
Chapter 3 Notes
Stoichiometry
I.
Writing Chemical Equations
A) All chemical reactions involve chemical changes in substances:
1) Reactants—the substances you start with
2) Products—the substances you end up with
Reactants
Products
Yields
B) Chemical bonds are broken and new bonds are formed as reactants are
converted into products
C) Chemical equations are the "recipes" that tell chemists what amounts of
reactants to mix and what amounts of products to expect.
D) Balanced equation: each side of the chemical equation has the same
number of atoms of each element- this follows the Law of Conservation
of Mass (matter is neither created nor destroyed, it is always conserved)
Massreactants = Massproducts
4Fe (s) + 3O2 (g) → 2Fe2O3 (s)
E) Some reactions require a catalyst: a substance that speeds up a reaction
without being used in the reaction itself
H2O2 (aq)
H2O (l) + O2 (g)
MnO2
Without the catalyst MnO2, this reaction moves very slowly. Heat is one
of the most common catalysts
F) The calculation of quantities in chemical equations is called
stoichiometry.
II. Symbols used in equations
A) The arrow
separates the reactants from the products—read
“reacts to form” or “yields”.
B) The plus sign (+) = “and”
C) (s) after the formula indicates a solid
D) (g) after the formula indicates a gas
E) (l) after the formula indicates a liquid
F) (aq) after the formula indicates aqueous solution
G) (ppt) after the formula indicates a solid precipitate has formed
H) ↑ indicates a gas has evolved
I) ↓ indicates a precipitate has formed
J)
indicates a reaction is reversible
K) 

shows heat has bee added to the reaction
L)
indicates a catalyst is used (in this case MgO2)
MgO2
M)
III.
indicates that electricity has been enduced
Rules for balancing chemical equations
A) CHECK FOR DIATOMIC MOLECULES!!!! H2, N2, O2, F2, Cl2, Br2, I2—as
free elements these are ALWAYS found in pairs!—Memorize them—test
tomorrow)
B) Make sure all formulas for reactants and products are correct, (use
subscripts to balance them) and make sure all reactants are to the left of
the arrow while all products are to the right of the arrow
C) Count the number of atoms each element on each side to see what
needs balanced
Remember, unless it decomposes, count a polyatomic ion as a single
unit instead of separate atoms
D) Balance each element using the smallest whole number coefficients
(numbers in front of the substances) possible:
1) If you have an odd/even situation with a particular atom, try doubling
the odd one
2) Normally, don’t balance the hydrogen and/or oxygen until the very
last
E) Check the equation to make sure all atoms are balanced
F) Make sure all elements are in the lowest whole-number ratio possible
Example 1- balance:
AgNO3 (aq) + Cu (s) → Cu(NO3)2 (aq) + Ag (s)
The number of silver is the same on both sides, but the number of
nitrates is not. We need to place a 2 in front of the Silver nitrate in order
to get the 2 nitrates in our product.
2AgNO3 (aq) + Cu (s) → Cu(NO3)2 (aq) + Ag (s)
However this unbalances the number of silver, so we need to put the
coefficient 2 in front of the silver in our product:
2AgNO3 (aq) + Cu (s) → Cu(NO3)2 (aq) + 2Ag (s)
Now recount. We have 2 silver on both sides, 2 nitrates on both sides
and one copper on both sides. The equation is balanced!
Example 2- balance:
Al (s) + O2 (g) → Al2O3 (s)
IV.
Types of Chemical Reactions
A) Synthesis (Combination) Reaction—2 or more reactants combine to form
a single product: A + B → AB
Examples:
S (s) + O2 → SO2 (g)
2S (s) + 3O2 (g) → 2SO3 (g)
4Al (s) + 3O2 (g) → 2Al2O3 (s)
4Cu (s) + O2 (g) → 2Cu2O (s)
2Cu (s) + O2 (g) → 2CuO (s)
Two special types of Combination Reactions are:
1) No-metallic Oxide + Water → Acid
SO3 (g) + H2O (l) → H2SO4 (aq)
2) Metallic Oxide + Water → Base
CaO (s) + H2O (l) → Ca(OH)2 (aq)
B) Decomposition Reaction
1) A single reactant (compound) is broken down into 2 or more
products. AB → A + B
2) Normally, heat (Δ) or electricity (
occur
) is required for this reaction to
3) Examples:
a) CaCO3 (s)
Δ
b) 2H3O(l)
c) Ag2O(s)
d) NiCO3 (s)
CaO (s) + CO2 (g)
2H2 (g) +2O2 (g)
4Ag (s) + O2 (g)
Δ
Δ
NiO (s) + CO2 (g)
C) Single-Replacement Reaction
1) Atoms of a single element replace the atoms of a second element in
a compound
2) The general formula for this type of reaction is:
for metals: A + BC → AC + B
for 7A non-metals: D2 + BC → BD + C2
3) For metals:
a) What determines whether or not 1 metal will replace another
metal from a compound is determined by the relative reactivities
of the 2 metals
b) The Activity Series of Metals lists metals in order of decreasing
reactivity—p. 124
c) A reactive metal will replace any metal found below it in the
Activity Series- however, it will not replace a metal found above it
(no reaction, or NR, will occur)
d) Group 7A non-metals can also replace other Group 7A nonmetals from a compound based on their relative reactivities as
well
Examples:
(1)
Zn + H2SO4 → ZnSO4 + H2 ↑
(2)
2K + 2H2O → 2KOH + H2
(3)
Sn + NaNO3 → NR (No Reaction)
(4)
Cl2 + 2NaBr → 2NaCl + Br2
D) Double-Replacement Reaction
1) Involves the exchange of positive ions between 2 compounds
2) These reactions generally take place between 2 ionic compounds in
aqueous solution
3) For the reaction to occur, one of the following statements is usually
true concerning at least 1 of the products:
a) It is a precipitate ↓
b) It is a gas ↑
c) It is a molecular compound, such as water
4) The general formula for this type of reaction is AB + CD → AD + CB
5) Examples:
a) 2NaOH + H2SO4 → Na2SO4 + 2H2O
b) BaCl2 + K2CO3 → BaCO3 + 2KCl
c) FeS + 2HCl → FeCl2 + H2S
d) 3KOH + H3PO4 → K3PO4 + 3H2O
E) Combustion Reaction
1) This occurs when a substance is burned in oxygen, often producing
energy in the form of heat and light
2) It commonly involves the burning of hydrocarbons (fossil fuels)
3) Complete combustion always produces only 2 products: CO2 and H2O
4) Incomplete combustion (such as in your car engine) also produces C
and CO
Examples:
a) CH4 + 2O2 → CO2 + 2H2O
b) 2C6H6 + 15O2 → 12CO2 + 6H2O
F) Review of identifying Chemical Reactions:
1) Combination (Synthesis)
a) 2 reactants, 1 product
b) Example: Fe + S → FeS
2) Decomposition
a) 1 reactant, 2 (or more) products
b) Usually heat, a catalyst, or electricity is required for the reaction to
occur
c) Example: 2H2O2 → 2H2O + O2
3) Single Replacement
a) The 2 reactants are: a single element and a compound
b) Metals—the single element switches with the cation of the
compound
Example: Ca + H2SO4 → CaSO4 + H2
c) Group 7A Non-metals—the single element switches with the
anion of the compound
Example: Cl2 + 2NaBr → 2NaCl + Br2
d) Check the activity chart to see if these reactions will occur, if not
write NR (No Reaction)
4) Double Replacement
a) The reactants are 2 ionic compounds
b) The cation of the first switches with the cation of the second
Example: Pb(NO3)2 + 2KI → PbI2 + 2KNO3
5) Combustion
a) The 2 reactants are a hydrocarbon and Oxygen
b) The 2 products are Carbon dioxide and water
Example: 2C6H6 + 15O2 → 12CO2 + 6H2O
V.
How you measure how much?
 You can measure mass,
 or volume,
 or you can count pieces.
 We measure mass in grams.
 We measure volume in liters.
 We count pieces in MOLES.
A) What is a Mole?
1) There are three types of moles that live in North America: the Eastern
Mole, the Hairy-Tailed Mole and the Star-Nosed Mole.
2) The “Mexican” mole is a chocolate sauce or turkey stew, it comes
from the Aztec word “molli”.
3) Mole—(abbreviated: mol)
a) Defined as the number of carbon atoms in exactly 12 grams of
carbon-12.
b) 1 mole is 6.02 x 10
23
particles.
c) Treat it like a very large dozen
d) 6.02 x 10
23
is called Avogadro’s number
B) Examples of Avagadro’s Number
23
1) 6.02 X 10 watermelon seeds would be found inside a melon slightly
larger than the moon.
23
2) 6.02 X 10 donut holes would cover the earth and be 5 miles deep.
23
3) 6.02 X 10 pennies would make at least 7 stacks that would reach
the moon.
23
4) 6.02 X 10 Grains of sand would be more than all of the sand on
Miami Beach.
23
5) 6.02 X 10 Blood cells: would be more than the total number of blood
cells found in every human on earth
2
6) A 1 liter bottle of water contains 55.5 moles of H2O. (6.02 X 10 3 X
55.5)
7) 3.5 lb bag of sugar contains 6.6 moles of C12H22O11. (6.02 X 1023
X 6.6)
C) Why is it such a big number?
1) Remember when calculating atomic mass, we use amu (atomic mass
units) 1amu = 1.66 x 10-24g.
2) The opposite of that equality is 1g = 6.02214 x 1023 amu.
3) That means that if you take 6.02214 x 1023 protons (or neutrons),
they would equal 1g.
4) So, 6.02214 x 1023 carbon atoms would weigh (6.02214 x 1023) x 12
amu, which is equal to 12g AND 6.02214 x 1023 iron atoms would
weigh, (6.02214 x 1023) X 55.8 amu = 55.8g
5) The atomic mass number on the periodic table is also the mass in
grams of 1 mole of those atoms.
D) Representative particles—the smallest pieces of a substance.
1) For a molecular compound: it is the molecule.
2) For an ionic compound: it is the formula unit (ions).
3) For an element: it is the atom.
4) Exception—there are 7 diatomic elements (made of molecules)
Hydrogen, Oxygen, Nitrogen, Fluorine, Chlorine, Bromine, Iodine
(Astatine is radioactive)
E) Molar Mass is the generic term for the mass of one mole of any
substance (in grams)
1) Gram Atomic Mass (gam) equals the mass of 1 mole of an element in
grams
2) The Gram Molecular Mass (gmm) is the mass of one mole of a
molecular compound.
3) Gram Formula Mass (gfm) is the mass of one mole of an ionic
compound.
F) How do you determine the molecular mass of molecules and formula
units?
1) First, answer how many atoms of each element are in the compound,
For example:
 NaCl
 H2O
 CaCO3
 Al2(SO4)3
2) Second, determine the molar mass of each of the elements and
multiply by the number of atoms
3) Add the total molar mass for each element




NaCl
H2O
CaCO3
Al2(SO4)3
Other Examples:
1) How many molecules of CO2 are there in 4.56 moles of CO2 ?
2) How many moles of water is 5.87 x 10
22
molecules?
3) How many atoms of carbon are there in 1.23 moles of C6H12O6 ?
24
4) How many moles is 7.78 x 10 formula units of MgCl2?
5) How much would 2.34 moles of carbon weigh?
6) How many moles of magnesium is 24.31 g of Mg?
7) How many atoms of lithium is 1.00 g of Li?
22
8) How much would 3.45 x 10 atoms of U weigh?
Calculate the molar mass of the following and tell what type it is:
9) Na2S
10) N2O4
11) C
12) Ca(NO3)2
13) C6H12O6
14) (NH4)3PO4
VI. Gases and the Mole
A) Many of the chemicals we deal with are gases, so they are difficult to
weigh. But we need to know how many moles of gas we have. Two
things effect the volume of a gas–Temperature and pressure. So we
need to compare them at the same temperature and pressure.
B) Standard Temperature and Pressure
1) Abbreviated STP
2) 0ºC (273K) and 1atm (760 mmHg, 760 torr, 101.3 kPa) in pressure
3) At STP 1 mole of gas occupies 22.4 L
4) This is called the molar volume
5) 1 mole = 22.4 L of any gas at STP
C) Examples
1) What is the volume of 4.59 mole of CO2 gas at STP?
2) How many moles is 5.67 L of O2 at STP?
3) What is the volume of 8.8 g of CH4 gas at STP?
D) Density of a gas
1) D = m / V
2) for a gas the units will be g/L
3) We can determine the density of any gas at STP if we know its
formula.
4) To find the density we need the mass and the volume.
5) If you assume you have 1 mole, then the mass is the molar mass
(from PT)
6) At STP the volume is 22.4 L.
E) Examples
1) Find the density of CO2 at STP.
2) Find the density of CH4 at STP.
F) Given the density, we can find the molar mass of the gas.
1) Again, pretend you have 1 mole at STP, so V = 22.4 L.
2) m = D x V
3) m is the mass of 1 mole, since you have 22.4 L of the stuff.
4) What is the molar mass of a gas with a density of 1.964 g/L?
5) 2.86 g/L?
G) In Summary these four items are all equal:
1) 1 mole
2) molar mass (in grams)
3) 6.02 x 1023 representative particles
4) 22.4 L at STP
VII. Calculating Percent Composition of a Compound
A) Like all percent problems: Part/whole x 100%
B) First find the mass of each component
C) Then divide by the total mass.
D) If we know the formula, assume you have 1 mole. Then you know the
mass of the pieces and the whole.
E) Example:
1) Calculate the percent composition of a compound that is 29.0 g of Ag
with 4.30 g of S.
2) Calculate the percent composition of C2H4?
3) Aluminum carbonate?
VIII. The Empirical Formula
A) The lowest whole number ratio of elements in a compound.
B) The molecular formula = the actual ratio of elements in a compound.
C) The two can be the same.
1) CH2 is an empirical formula
2) C2H4 is a molecular formula
3) C3H6 is a molecular formula
4) H2O is both empirical & molecular
D) Calculating Empirical
1) Just find the lowest whole number ratio
2) C6H12O6
3) CH4N
E) The Empirical formula is not just the ratio of atoms, it is also the ratio of
moles of atoms.
1) In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.
2) In one molecule of CO2 there is 1 atom of C and 2 atoms of O.
F) Calculating the Empirical Formula from the percent composition.
1) Assume you have a 100 g.
2) The percentages become grams.
3) Convert grams to moles.
4) Find lowest whole number ratio by dividing by the smallest.
G) Examples
1) Calculate the empirical formula of a compound composed of 38.67 %
C, 16.22 % H, and 45.11 %N.
Assume 100 g so:
38.67 g C x 1mol C
16.22 g H x 1mol H
45.11 g N x 1mol N
= 3.220 mole C
= 16.09 mole H
= 3.219 mole N
The ratio is
3.220 mol C = 1 mol C
3.219 mol N
The ratio is
3.220 mol C = 1 mol N
3.219 mol N
The ratio is
16.09 mol H = 5 mol H
3.219 mol N
12.01g C
1.01g H
14.01g N
n= C1H5N1 or CH5N
2) A compound is 43.64 % P and 56.36 % O. What is the empirical
formula?
3) Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its
empirical formula?
H) Empirical Formula to Molecular Formula
1) Since the empirical formula is the lowest ratio, the actual molecule
would weigh more.
2) By a whole number multiple.
3) Divide the actual mass by the Empirical Formula mass
4) Caffeine has a molar mass of 194 g. what is its molecular formula?
I) Example
A compound is known to be composed of 71.65 % Cl, 24.27% C and
4.07% H. Its molar mass is known (from gas density) to be 98.96 g. What
is its molecular formula?
IX. Chemical Equations and Molar Mass
A) The production of ammonia (fertilizer) gives the following equation:
N2(g) + 3H2(g) → 2NH3(g)
B) What kinds of information can be learned from this equation?
1) Nitrogen + hydrogen forms ammonia
2) 1 molecule of nitrogen + 3 molecules of hydrogen forms 2 molecules
of ammonia
3) Mole Ratio: 1 mole of N2 + 3 moles of H2 yields 2 moles of NH3
4) Mass: (using the number of moles in the balanced chemical equation
a) 28 g of N2 + 6 g H2 yields 34 g of NH3
b) Note: the law of Conservation of Mass is followed:
28 g + 6 g = 34 g
5) Volume: (gases only at STP, again using the number of moles
a) 22.4 L of N2 + 67.2 L of H2 yields 44.8 L of NH3
b) Note: There is NO Law of Conservation of Volume!
6) Mole Ratios:
a) What is the mole ratio of the following:
N2 to H2
H2 to N2
N2 to NH3
H2 to NH3
1:3
3:1
1:2
3:2
or
or
or
or
1/3X
3X
1/2X
3/2X
b) What are the mole ratios in the following equation?
2C2H2 + 5O2 → 4CO2 + 2H2O
C2H2 to CO2
C2H2 to H2O
O2 to CO2
CO2 to O2
O2 to C2H2
O2 to H2O
H2O to O2
2:4 or 1:2
2:2 or 1:1
5:4
4:5
5:2
5:2
2:5
or
or
or
or
or
or
or
1/2X
1X
5/4X
4/5X
5/2X
5/2X
2/5X
c) From the mole ratios of reactants and products in a balanced
chemical equation, we can determine how much product(s) is
produced from a given amount of reactant(s)
X. Mass-Mass Calculations
A) Procedure
1) Determine the balanced chemical equation
2) Using dimensional analysis:
a) Convert the given mass of the reactants to moles (using molar
mass)
b) Convert the moles of reactant to moles of desired product (using
the mole ratio)
c) Convert the moles of product to the mass of product (using molar
mass)
Example 1: If 20.0 g of Zinc reacts with excess hydrochloric acid, how
many grams of Zinc Chloride are produced?
Zn + HCl → ZnCl2 + H2
Example 2: How many grams of Chlorine gas must be reacted with
excess Sodium Iodide if 10.0 g of Sodium Chloride are needed?
NaI + Cl2 → NaCl + I2
Example 3: How many grams of Oxygen are produced in the
decomposition of 5.00 g of Potassium Chlorate?
KClO3 → KCl + O2
XI. Mass-Volume and Volume-Volume
A) When dealing with gases, it is much easier to measure the volume of a
gas than it is to measure its mass
B) Remember that at standard conditions (STP), the volume of all gases is
22.4 L per mole (the molar volume)
C) The calculations procedure is the same as in Mass-Mass problems, only
when you are dealing with volumes, use 22.4 L per mole as conversion
factor
1) Step 1: balance the equation
2) Step 2: dimensional analysis
a) Convert given information to moles (mass: use molar mass,
volume: use 22.4 L)
b) Use mole ratio to convert moles of given information to moles of
desired product
c) Convert moles of product back to either mass (using molar mass)
or volume (using 22.4 L)
Example 1: If 15.0 g of Zinc reacts with excess hydrochloric acid, how
many liters of hydrogen gas are produced?
Zn + HCl → ZnCl2 + H2
Example 2: Calculate the volume of Oxygen (in liters) that can be
produced by the electrolysis of 40.0 g of water.
H2O → H2 + O2
XII. Percent Yield
A) Up to this point, you have used a balanced chemical equation to
determine the amount of product obtained from a given amount of
reactants
B) This amount of product is called a theoretical yield: it is the maximum
amount of product obtained from the amounts of reactants started with-assuming everything goes perfectly
C) The amount of product you get in the lab is the actual yield--the observed
value that you actually measure (it is usually less than the theoretical
yield)
D) There are 2 measures used to illustrate the success and/or efficiency of
your results in the lab:
1) Percent Yield
a) The ratio of the actual yield to the theoretical yield expressed as a
percent
b) The equation for percent yield is:
% yield =
actual yield
X 100%
theoretical yield
2) Percent Error
a) The ratio of the difference in the actual yield and theoretical yield,
compared to the theoretical yield expressed as a percent
b) The equation for percent error is:
% error =
actual yield - theoretical yield X 100%
theoretical yield
Example 1: What is the percent yield of 5.74 g of Copper are produced
in the lab when 1.87 g of Aluminum is reacted with an excess of
Copper(II) Sulfate?
Al + CuSO4 → Al2SO4 + Cu
What is the percent error in the above example?
Note: % yield + % error = 100%
XIII. The Limiting Reactant (Reagent)
A) Most chemical reactions will continue until one of the reactants is
completely used up--then, no more product(s) can be formed
B) The reactant that is used up first, and therefore controls how much
product is formed, is called the limiting reactant (or limiting reagent)
C) The other reactant, in which there is more than enough, is said to be in
excess and is called the excess reactant (or excess reagent)
D) Limiting reactant problems are solved by comparing the number of moles
of product that each reactant can possible produce if it is all used up
1) Whichever reactant that produces the smallest number of moles of
product is therefore the limiting reactant
2) The limiting reactant will then determine exactly how much product
can be produced
3) Steps:
a) Determine the balanced chemical equation
b) Follow the first 3 steps of dimensional analysis of stoichiometry
with each of the reactants--this will tell you which reactant
produces the smallest number of moles of product (this is the
limiting reactant)
c) Complete step 4 of dimensional analysis of stoichiometry using
the small number of moles of product--as this is all the product
that can possible be produced
Example 1: According to the following chemical equation:
H3PO4 + MgCO3 → Mg3(PO4)2 + CO2 + H2O
If 40.0 g or H3PO4 reacts with 60.0 g of MgCO3:
Which reactant is the limiting reacting?
Which reactant is in excess, and by how many moles?
What volume of CO2 is produced?