Micro World atoms & molecules
Macro World grams
By definition:
1 atom 12 C “weighs” 12 amu
On this scale
1 H = 1.008 amu
16 O = 16.00 amu
3.1
The mole (mol) is the amount of a substance that contains as many elementary entities as there are atoms in exactly 12.00 grams of 12 C
1 mol = N
A
= 6.0221367 x 10 23
Avogadro’s number ( N
A
)
3.2
1

Molar mass
Na atoms
Kr atoms
Li atoms
1 mole 12 C atoms = 6.022 x 10 23 atoms = 12.00 g
1 mole 12 C atoms = 12.00 g 12 C
1 mole lithium atoms = 6.941 g of Li
3.2
I.
Write Unit of Answer
II. Write Starting Quantity from Problem
III. Add appropriate unit factor(s) to cancel out starting quantity and put in unit of answer.
How many atoms are in 0.551 g of potassium (K) ?
1 mol K = 39.10 g K
1 mol K = 6.022 x 10 23 atoms K
0.551 g K x
1 mol K
39.10 g K x 6.022 x 10
23 atoms K
1 mol K
=
8.49 x 10 21 atoms K
3.2
2

Avogadro’s
Number
Molar Mass
Molecular mass (or molecular weight) is the sum of the atomic masses (in amu) in a molecule.
SO
2
1S 32.07 amu
2O + 2 x 16.00 amu
SO
2
64.07 amu
1 molecule SO
2
1 mole SO
2
= 64.07 amu
= 64.07 g SO
2
3.3
3

Percent composition of an element in a compound = n x molar mass of element molar mass of compound x 100% n is the number of moles of the element in 1 mole of the compound
C
2
H
6
O
%C =
%H =
%O =
2 x (12.01 g)
46.07 g
6 x (1.008 g)
46.07 g
1 x (16.00 g)
46.07 g x 100% = 52.14% x 100% = 13.13% x 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
3.5
The balanced chemical equation represents the ratio
Of one species to any others in the equation:
NaOH (aq) + HCl (aq) Æ NaCl (aq) + H
2
O (l)
1mole of NaOH produces 1mole of NaCl
1 mol NaCl
1mol NaOH
2H
2
O
2
(l) Æ 2 H
2
O (l) + O
2
(g)
2 moles of H
2
O
2
1 mol O
2
2 mol H
2
(l) produces 1 mole of O
2
O
2
(l) gas
Mole composition gives ratio of each element
In a compound
0.345 moles of Al
2
(CO
3
)
3
Al in the compound: has the following moles of
Mol ratio of Al in Al
2
(CO
3
)
3
:
0.345 mol Al
2
(CO
3
)
3
X 2 mol Al = 0.690 mol Al
1 mol Al
2
(CO
3
)
3
Moles of compound X mol ratio (element/molecule)
= mol of element
4

Mass of Aluminum is obtained from the molar mass:
0.690 mol Al X 26.98 g = 18.62 g Al mol Al
Mol X grams/mol = mass
The mass of any substance can be determined if
The molar mass is known. The molar mass comes
From the periodic table.
Al
2
(CO
3
)
3 continued:
C: 0.345 mol X 3 mol C = 1.04 mol C X 12.011g/mol
= 12.43 g C
1 mol Al
2
(CO
3
)
3
O: 0.345 mol X 9 mol O = 3.11 mol O X 15.999g/mol
= 49.677 g O
1 mol Al
2
(CO
3
)
3
Formula Weight converted to mass from moles:
2 mol X 26.98g/mol (Al) =53.96g
3mol X 12.011g/mol (C) =36.033g
9 mol X 15.999g/mol (O) = 143.991 g total molar mass 233.984 g in 1 mol Al
2
(CO
3
)
3
233.984 g/mol X 0.345 mol = 80.724 g Al
2
(CO
3
)
3
The real usefulness of balanced equations can be
Observed in the following series of predictions:
I Predict moles of any products of a reaction
II Predict the amount of any and all reactants needed for a given amount of product
III Determine precisely how much of all reactants are used, and which are present in excess.
The basis of chemical calculations for reactions
Is the balanced reaction equation.
5

So far, we have seen mole ratio’s from an equation
That tells us how much to expect in the reaction:
N
2
(g) + H
2
(g) Æ NH
3
(g)
N
2
(g) + 3H
2
(g) Æ 2NH
3
(g) (Balanced equation)
Think: mols reactant x moles product = mols product mols reactant
What we are asked for goes on top of the ratio. What we
Are given goes on bottom.
How many moles of NH
3 can be produced from 5.00 mol H
2
5.00 mol H
2
(given) x 2 mol NH
3
3 mol H
2
= 3.33 mol NH
3 formed
In the laboratory, we must measure the mols by
Mass. This requires an initial conversion to mols
From the given mass of substance:
How many moles of NH
Of N
2
:
3 are formed from 33.6 g
33.6 g x 1 mol N
2
28.0 g N x 2 mol NH
3
2
1 mol N
2
= 2.40 mol NH
3
Mass reactant x 1 x mol product = mol product
F.W react. mol reactant
Finally, we must determine the mass of a product
In the laboratory, since we cannot evaluate moles:
What mass of H
2 is needed to produce 119 g of NH
3
What is asked for: mass of H
2
What is given: mass of NH
3
Mass reactant x 1 x mol product x FW prod =mass
F.W react. mol reactant product
119 g NH
3 x1 mol NH
3
17.0 g NH
3 x 3 mol H
2
2 mol NH x 2.02 g H
2
3 mol H
2
= 21.2 g H
2
6

Memorize these steps:
Mass react x 1 mol react x mol product x prod. g = mass grams react mol react mol prod.
Mass react x 1 mol react x mol product x prod. g = mass grams react mol react mol prod.
SiO
2
SiO
2 is etched with HF (aq):
(s) + 4 HF (aq) Æ SiF
4
(g) + 2 H
2
O (l)
If there is 4.86 moles HF and 60 g SiO
2
, will there be
Any glass left over?
4.86 mol HF x 1 mol SiF
4 x 104 g SiF
4 mol HF mol SiF
4
4
= 126 g SiF
4
;
126 g SiF
4 x 1 mol SiF
4
104 g SiF
4 x SiO
2
SiF
4 x 60g SiO
2 mol SiO
2
= 72.7 g SiO
2
Since I have only 60g (1 mol) of SiO
2
Glass before I use up the acid!
Æ SiO
, I will run out of
2 is LIMITING
4 Ag (s) + 2 H
2
S (g) + O
2
(g) Æ 2 Ag
2
S (s) + 2 H
2
O (l)
IF there is 0.145 mol Ag, 0.0872 mol H
O
2
, how much Ag
2
S, and excess
2
S is produced and what mass of
Reactant is in excess?
Ag: 0.145 mol x 2 mol Ag
4 mol Ag
2
S = 0.0725 mol Ag
2
S
H
2
S: 0.0872 mol H
2
S x 2 mol Ag
2
2 mol H
2
S
S = 0.0872 mol Ag
2
S
Ag is limiting and H
2
S is present in excess. We must use the limiting reagent quantity to predict actual products :
0.0725 mol Ag
2
S x 248 g Ag
2 mol Ag
2
S
S = 18.0 g Ag
2
S formed
7

A reaction in the laboratory results in a product mass
That is not quite what the balanced equation calculation
Predicts.
The difference between the theoretical amount and the
Actual one is called Percent Yield :
Actual Yield (mass) x 100% = percent yield
Theoretical yield (mass)
Thermo-chemical equations
A balanced chemical equation that indicates the heat
Energy transformation in the reaction.
Heat of reaction: ∆ H
Exothermic: a reaction in which heat energy is liberated
∆ H = - value
Endothermic: a reaction in which heat energy is absorbed
∆ H= + value
Heat energy is measured in Kilojoules
2 C
2
H
2
(g) + 5O
2
(g) Æ 4CO
2
(g) + 2H
2
O (l) ∆ H=-2602 KJ
This reaction gives off heat. (Acetylene Torch)
If a sample of acetylene is burned, and releases 550 KJ of
Energy, how many grams of CO
2 can we expect to form?
2 C
2
H
2
(g) + 5O
2
(g) Æ 4CO
2
(g) + 2H
2
O (l) H=-2602 KJ
We can relate the moles of CO
2 to the amount of heat, since
It is part of the balanced equation:
550 KJ x 4 mol CO
2
2602 KJ x 44.0 g CO
2 mol CO
2
= 37.2 g CO
2
8

I.
Formulas show chemistry at a standstill. Equations show chemistry in action.
A. Equations show:
1. the reactants which enter into a reaction.
2. the products which are formed by the reaction.
3.
the amounts of each substance used and each substance produced.
B. Two important principles to remember:
1.
Every chemical compound has a formula which cannot be altered.
2.
A chemical reaction must account for every atom that is used. This is an application of the Law of Conservation of Matter which states that in a chemical reaction atoms are neither created nor destroyed.
Some things to remember about writing equations:
1. The diatomic elements when they stand alone are always written H
2
N
2
, O
2
, F
2
, Cl
2
, Br
2
, I
2
2. The sign, ----->, means "yields" and shows the direction of the
, action.
3. A small delta, ( ∆ ), above the arrow shows that heat has been added.
4. A double arrow, <----->, shows that the reaction is reversible and can go in both directions.
5. Before beginning to balance an equation, check each formula to see that it is correct. NEVER change a formula during the balancing of an equation.
6. Balancing is done by placing coefficients in front of the formulas to insure the same number of atoms of each element on both sides of the arrow.
7. Always consult the Activity Series of metals and non-metals before attempting to write equations for replacement reactions.
8. If a reactant or product is solid, place (s) after the formula
9. If the reactant or product is a liquid, place (l) after the formula
10. If the reactant or product is a gas, place (g) after the formula
11. If the reactant or product is in water, place (aq) after the formula
12. A category of reaction will produce an unstable product which decomposes:
H
2
CO (aq) carbonic acid
Æ H
2
O (l) + CO
2
(g)
H
2
SO (aq) Æ sulfurous acid
H
2
O (l) + SO
2
(g)
NH
4
OH (aq) Æ NH ammonium hydroxide
(g) + H
2
O (l)
9

Rules for Writing Chemical Equations:
1. Write down the formula(s) for the reactants and add a “+” between them then put a “yields” arrow ( Æ ) at the end
2. Examine the formula(s) to determine which of four basic types of reactions will occur. On the basis of your decision, write down correct formula(s) for the products on the right side of the yield arrow.
3. Place a coefficient in front of each species in the reactants and the products to ensure that the conservation of matter is observed: “Balance” the reaction
II. Four Basic Types of Chemical reactions
A. Synthesis two or more elements or compounds combine to give a more complex product
Examples of Synthesis reactions:
1.
Metal + oxygen -----> metal oxide
+ O
2(g)
----> 2MgO
(s)
2Mg
2.
(s)
Nonmetal + oxygen -----> nonmetallic oxide
C
(s)
3.
+ O
2(g)
----> CO
2(g)
Metal oxide + water -----> metallic hydroxide
MgO
(s)
4.
+ H
2
O
(l)
----> Mg(OH)
2(s)
Nonmetallic oxide + water -----> acid
CO
2(g)
+ H
2
O
(l)
----> ; H
2
CO
3(aq)
5.
Metal + nonmetal -----> salt
2 Na
(s)
+ Cl
2(g)
----> 2NaCl
(s)
6.
A few nonmetals combine with each other.
2P
(s)
+ 3Cl
2(g)
----> 2PCl
3(g)
B. Decomposition:
•
•
A single compound breaks down into simpler compounds.
Basic form: AX -----> A + X
Examples of decomposition reactions (when heated):
1.
Metallic carbonates form metallic oxides and CO
2(g)
.
CaCO
2.
3(s)
----> CaO
(s)
+ CO
2(g)
Most metallic hydroxides decompose into metal oxides and water .
Ca(OH)
2(s)
----> CaO
(s)
+ H
2
O
(g)
3.
Metallic chlorates decompose into metallic chlorides and oxygen.
2KClO
3(s)
4.
----> 2KCl
(s)
+ 3O
2(g)
H
2
5.
Some acids decompose into nonmetallic oxides and water.
SO
4
----> H
2
O
(l)
+ SO
3(g)
Some oxides decompose.
2HgO
(s)
----> 2Hg
(l)
+ O
2(g)
6.
2H
Some decomposition reactions are produced by electricity.
2
O
(l)
2NaCl
(l)
----> 2H
2(g)
----> 2Na
(s)
+ O
2(g)
+ Cl
2(g)
10

C. Replacement:
•
•
X a more active element takes the place of another element and frees the less active one.
Basic form: A + BX -----> AX + B or AX + Y -----> AY +
Examples of replacement reactions:
1.
Replacement of a metal in a compound by a more active metal.
Fe
(s)
+ CuSO
4(aq)
----> FeSO
4(aq)
+ Cu
(s)
2.
Replacement of hydrogen in water by an active metal.
2Na
(s)
Mg
(s)
+ 2H
+ H
2
O
2
O
(g)
(l)
----> 2NaOH
----> MgO
(s)
(aq)
+ H
+ H
2(g)
2(g)
3.
Replacement of hydrogen in acids by active metals.
Zn
(s)
+ 2HCl
(aq)
----> ZnCl
2(aq)
+ H
2(g)
4.
Replacement of nonmetals by more active nonmetals.
Cl
2(g)
+ 2NaBr
(aq)
----> 2NaCl
(aq)
+ Br
2(l)
NOTE: Refer to the Activity Series for metals and nonmetals to predict products of replacement reactions. If the free element is above the element to be replaced in the compound, then the reaction will occur.
If it is below, then no reaction occurs.
If the free element is above the element to be replaced in the compound,
Then the reaction will occur. If the free element is below it, no reaction occurs.
D. Ionic or Double Displacement:
• occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following:
1.
a precipitate
2.
a gas
3.
water or some other non-ionized substance.
• Basic form: AX + BY -----> AY + BX
Examples of ionic reactions:
1.
Formation of precipitate .
NaCl
(aq)
----> 2NaCl
(aq)
+ AgCl
(s)
+ BaSO
4(s)
BaCl
2(aq)
2.
HCl
3.
(aq)
+ AgNO
+ Na
2
Formation of a gas.
+ FeS
(s)
3(aq)
SO
----> NaNO
3(aq)
4(aq)
----> FeCl
2(aq)
+ H
2
S
(g)
HCl
(aq)
4.
Formation of water.
(If the reaction is between an acid and a base it is called a neutralization reaction.)
+ NaOH
(aq)
----> NaCl
(aq)
+ H
2
O
(l)
Formation of a product which decomposes.
CaCO
3(s)
+ HCl
(aq)
----> CaCl
2(aq)
+ CO
2(g)
+ H
2
O
(l)
11

Use the Solubility Rules to decide whether a product of an ionic reaction is insoluble in water and will thus form a precipitate. If a compound is soluble in water then it should be shown as being in aqueous solution, or left as separate ions. It is, in fact, often more desirable to show only those ions that are actually taking part in the actual reaction.
Equations of this type are called Net Ionic equations .
Combustion of Hydrocarbons:
Another important type of reaction, in addition to the four types above, is that of the combustion of a hydrocarbon. When a hydrocarbon is burned with sufficient oxygen supply, the products are always carbon dioxide and water vapor. If the supply of oxygen is low or restricted, then carbon monoxide will be produced. This is why it is so dangerous to have an automobile engine running inside a closed garage or to use a charcoal grill indoors.
•
) + O
2(g)
-----> CO
2(g)
•
Hydrocarbon (C x
H y
+ H
2
O
(g)
CH
4(g)
+ 2O
2(g)
----> CO
2(g)
+ 2H
2
O
(g)
•
2C
4
H
10(g)
+ 13O
2(g)
----> 8CO
2(g)
+ 10H
2
O
(g)
NOTE:
•
• attained.
•
Complete combustion means the higher oxidation number is attained.
Incomplete combustion means the lower oxidation number is
The phrase "To burn" means to add oxygen unless told otherwise.
Conservation of matter requires identical numbers of atoms on both sides
Begin by inspecting the number of each type of atom on each side.
Nitrogen must separate before the reaction
Hydrogen must separate before the reaction.
Re-combine in a different chemical species
Same atoms, different ratio!
12

D. Ionic or Double Displacement:
• occurs between ions in aqueous solution. A reaction will occur when a pair of ions come together to produce at least one of the following:
1.
a precipitate
2.
a gas
3.
water or some other non-ionized substance.
• Basic form: AX + BY -----> AY + BX
Examples of ionic reactions:
1.
Formation of precipitate .
NaCl
(aq)
----> 2NaCl
(aq)
+ AgCl
(s)
+ BaSO
4(s)
BaCl
2(aq)
2.
HCl
3.
(aq)
+ AgNO
+ Na
2
Formation of a gas.
+ FeS
(s)
3(aq)
SO
----> NaNO
3(aq)
4(aq)
----> FeCl
2(aq)
+ H
2
S
(g)
HCl
(aq)
4.
Formation of water.
(If the reaction is between an acid and a base it is called a neutralization reaction.)
+ NaOH
(aq)
----> NaCl
(aq)
+ H
2
O
(l)
Formation of a product which decomposes.
CaCO
3(s)
+ HCl
(aq)
----> CaCl
2(aq)
+ CO
2(g)
+ H
2
O
(l)
Use the Solubility Rules to decide whether a product of an ionic reaction is insoluble in water and will thus form a precipitate. If a compound is soluble in water then it should be shown as being in aqueous solution, or left as separate ions. It is, in fact, often more desirable to show only those ions that are actually taking part in the actual reaction.
Equations of this type are called Net Ionic equations .
13

Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C
2
H
6
+ O
2
CO
2
+ H
2
O
2. Change the numbers in front of the formulas
( coefficients ) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.
2C
2
H
6
NOT C
4
H
12
Balancing Chemical Equations
3. Start by balancing those elements that appear in only one reactant and one product.
C
2
H
6
+ O
2
CO
2
+ H
2
O start with C or H but not O
2 carbon on left
C
2
H
6
+ O
2
1 carbon on right
2 CO
2
+ H
2
O multiply CO
2 by 2
6 hydrogen on left
C
2
H
6
+ O
2
2 hydrogen on right
2 CO
2
+ 3 H
2
O multiply H
2
O by 3
14

Balancing Chemical Equations
4. Balance those elements that appear in two or more reactants or products.
C
2
H
6
+ O
2
2 CO
2
+ 3 H
2
O multiply O
2 by
7
2
2 oxygen on left
C
2
H
6
+ O
2
2 C
2
H
6
+ 7 O
2
4 oxygen
(2x2)
+ 3 oxygen
(3x1)
= 7 oxygen on right
2 CO
2
+ 3 H
2
O remove fraction multiply both sides by 2
4 CO
2
+ 6 H
2
O
Balancing Chemical Equations
5. Check to make sure that you have the same number of each type of atom on both sides of the equation.
2 C
2
H
6
+ 7 O
2
14 O ( 7 x 2)
4 CO
2
+ 6 H
2
O
14 O ( 4 x 2 + 6 )
Reactants
4 C
12 H
14 O
Products
4 C
12 H
14 O
Chemical Equations
2 Mg + O
2
2 MgO
2 atoms Mg + 1 molecule O
2 makes 2 formula units MgO
2 moles Mg + 1 mole O
2 makes 2 moles MgO
48.6 grams Mg + 32.0 grams O
2 makes 80.6 g MgO
2 grams Mg + 1 gram O
2 makes 2 g MgO
15

2S (s) + 3O
2
(g) ->2 SO
3
(g) S
O
Reactant Mixture
Product Mixture
16