http://www.skanschools.org/webpages/rallen/files/notes%20-%20unit%205%20-%20moles%20&%20stoich_2012_key.pdf
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Monoatomic one atom of an element
that’s stable enough to stand on its own (very
rare)- not bonded to anything

Diatomic (diatoms) elements whose atoms
always travel in pairs (N2, O2, F2 etc.) bonded
to another atom of the same element

What would be the mass of one molecule of
oxygen (O2)?

O2 subscript tells you the total # of atoms in
molecule/compound
▪ O2= 2 x 16 amu = 32 amu

Formula Mass the mass of an atom,
molecule or compound in atomic mass units
(amu)

Gram Formula Mass the mass of one mole
of an atom, molecule or compound in grams

Mole 6.02 x 1023 units of a substance

Step 1: calculate the GFM for the compound

Ex. CaCl2
 Ca = 1 x 40.08 = 40.08
 Cl= 2 x 35. 453 = 70.906
“Parts”
Formula % composition by mass = mass of part x 100
Mass of whole
* Find the percent composition to the nearest 0.1%

A hydrate is a crystalline compound in which
ions are attached to one or more water
molecules

Ex. Na2CO3 10H2O


Notice how water molecules are built into the formula
Substances without water built into the formula are called
anhydrates

How many moles are in 4.75 g of sodium
hydroxide (NaOH)?

Na 1 x 22.98 = 22.98
Total
GFM = 39.977 g/mol

O 1 x 15.99 =

H 1 x 1.007 = 1.007
15.99

Plug in the given value and the GFM into the
“mole calculations” formula and solve for
moles

# of moles = 4.75 g
= .119 mol
39.977 g/mol

What is the mass of 4.5 moles of KOH?
Mass KOH = 4.5 moles x 56.087 g/mol
mass KOH = 252.39 g

A chemical equation is a set of symbols that
state the products and reactants in a
chemical reaction

Reactants the starting substances in a
chemical reaction (left side of arrow)

Products a substance produced by a
chemical reaction (right side of arrow)

Ex.
2Na + 2H2O  2NaOH + H2

Chemical equations must be balanced

Law of conservation of mass: mass can
neither be created nor destroyed in a
chemical reaction


The number of moles of each element on the
reactants side must be the same as the
number of moles of each element on the
products side

Coefficients and subscripts tell us how many
moles of each element we have

Ex. Balanced Equation

C + O2  CO2

1 mol of Carbon
Each side of the arrow
2 mol of Oxygen
This means the equation is balanced



Ex. Unbalanced Equation
 H2 + O2  H2O
Coefficient= integer in front of an element or
compound which indicates the # of moles present
 Subscript = the integer to the lower right of an
element which indicates # of atoms present
 Species= the individual reactants and products in a
chemical reaction


Ex. Unbalanced Equation
 H2 + O2  H2O


What do we use to balance equations?
 Coefficients
* we never change the subscripts in a formula

Balanced equation: 2H2 + O2  2H2O

Ex.

When balancing chemical equations,
polyatomic ions may be balanced as a single
element rather than as separate elements as
long as they stay intact during the reaction

Al2(SO4)3 + Ca(OH)2  Al(OH)3 + CaSO4


Polyatomic ions= sulfate and hydroxide
Polyatomic ions remain intact during the
reaction, can be considered one unit

Balance the equation:
Al2(SO4)3 + 3Ca(OH)2  2Al(OH)3 + 3CaSO4

Type 1: Single Replacement

Reaction where one species replaces another
(one species alone on one side and combined
on the other)

Ex. 3Ag + AuCl3  3AgCl + Au
2Cr + 3H2SO4  Cr2(SO4)3 + 3H2


Type 2: Double Replacement

Reaction where compounds react, switch
partners and produce 2 new compounds

Ex.
Pb(NO3)2 + 2 NaCl  PbCl2 + 2 NaNO3
Na3PO4 + 3 Ag NO3  Ag3PO4 + 3NaNO3
K2CO3 + 2AgNO3  Ag2CO3 + 2KNO3




Type 3: Synthesis

Reaction where we take more than one
reactant and create one product

Ex.
4Al + 3O2  2Al2O3
2H2 + O2  2H2O



Type 4: Decomposition

Reaction where we take one reactant and
create 2 products

Ex.
BaCO3  BaO + CO2
2H2O2  2H2O + O2
2Bi(OH)3 Bi2O3 + 3H2O




A chemical equation = recipe for reaction

Coefficients= tell the amount of reactants
and products needed

Reactants in an equation react in specific
ratios to produce specific amounts of
products

Method for solving mole-mole problems

Set up a proportion using your known and
unknown values

Cross multiply and solve for your unknown

Always check to make the equation is
balanced

Empirical formula= the reduced formula; a
formula whose subscripts cannot be reduced
any further

Molecular formula= the actual formula for a
compound; subscripts represent actual
quantity of atoms present

Step 1: always assume you have 100 g sample
(the total % for the compound must = 100, so
we can just change the units from % to g)

Step 2: convert grams to moles

Step 3: divide all mole numbers by the
smallest mole number

ex. a compound is 46.2% mass carbon and
53.8% mass nitrogen. What is its empirical
formula?

Step 1: assume 100 g sample
46.2 % C = 46.2 g C
53.8 % N= 53.8 g N



Step 2: Convert grams to moles (have g need
moles)


46. 2 g C= 3.85 mol C
12 g/mol C

* must have whole numbers for subscripts
53.8 g N = 3.84
14 g/mol

Step 3: divide each mole number by the
smallest mole number (we will round in this
step to the nearest integer)


3.85 mol C = 1
3.84 mol N

Empirical formula: CN
3.84 mol N = 1
3.84 mol N

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We know how to:
1. find an empirical formula from % mass
2. find an empirical formula from molecular
formula
Next how do we find out the molecular
formula from an empirical formula?

Ex. a compound is 80.0% C and 20.0% H by
mass. If its molecular mass is 75.0 g, what is
its empirical formula? What is its molecular
formula?

First, determine the empirical formula using
the 3 step process
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




Step 1: assume a 100 g sample
80.0 % C= 80.0 g C
20.0 % H = 20.0 g H
Step 2: convert g to moles
80.0 g C = 6.66 mol C 20.0 g H= 20 mol H
12 g/mol C
1 g/mol

Step 3: divide each by the smallest number of
moles and round to the nearest whole


C= 6.66 = 1
6.66

Empirical formula: CH3
H= 20.0 = 3.00
6.66

Empirical mass (the mass of 1 mol of CH3)= 15 g

Molecular mass = 75 g

Molecular mass is 5 times larger than empirical
mass

Molecular formula must be 5 times larger than
empirical formula

Multiply all subscripts in our empirical
formula by 5

Molecular formula (CH3)= C5H15