The Mole
Chapter 6
► Remember
The Mole
it is a
measurement for anything
► 1 mole means 6.022 x 1023 of
anything
► This is called Avogadro’s
Number
► A sample of an element with a
mass equal to that element’s
average atomic mass
expressed in grams contains 1
mol of atoms
► 26.98
1023
g of Aluminum has 6.022 x
atoms = 1mol
How can this help us?
► If
you know how many grams make 1 mole,
you can figure out how many moles you
have in a given mass
 If you have 0.00568 g of Silicon how many
atoms do you have?
 Know that 1 mol Si is 28.09 g and 1 mole is
6.022 x 1023 atoms, these are your conversion
factors
 So 0.00568 g x 1 mol/28.09 g = 0.000202 mol
Si
 0.000202 mol Si x 6.022 x 1023 atoms/ 1 mol =
1.22 x 1020 Si atoms
Molar Mass
► What
is the mass of 1 mole of a compound?
Simply add the mass of the elements moles to get
the compounds molar mass
► CH4 is made up of 1 mol of C and 4 mol of H
 1 mol C = 1 x 12.01 g = 12.01 g
 4 mol H = 4 x 1.008 g = 4.032 g
 So 1 mol of CH4 = 12.01 g + 4.032 g = 16.04 g
► Try to find molar mass of CaCO3
► Answer =
► Now figure out how many grams
4.86 mol
► Answer =
of CaCO3 are in
Practice Problem
► How
many moles and how many molecules
of C7H14O2 is in a bee sting if 1 gram is
released?






First find Molar Mass
Answer
Second find how many moles in 1 gram
Answer
Finally change moles to molecules
Answer
Enough for 1 day, Watch the Video
“The Mole”
Percent composition of Compounds
► Take
mass of 1 element and divide it by the
mass of entire compound and multiply by
100
► Example: C2H5OH (ethanol) What is the
mass percent of Carbon in this compound?






What is mass of 1 mole of ethanol?
Answer
What is the mass percent of carbon in ethanol?
Answer
Now do Hydrogen and Oxygen
Answers
6.6 Formulas of Compounds
► Empirical
Formula – simplest formula and
expresses the smallest whole-number ratio
of the atoms present
 C6H12O6 and C4H8O4 both have the same
empirical formula CH2O
► Molecular
Formula – the actual formula of a
compound that tells you the composition of
the molecules that are present
 C6H12O6 is the molecular formula for glucose
6.7 Calculation of Empirical Formulas
► Important
to learn the chemical formula of a
new compound




First figure out relative masses
Second convert masses to number of moles
Divide by the smallest number of moles found
Multiply the numbers from 3rd step by smallest
number that will make them all whole numbers,
this is the empirical formula
Example Problem
► An
oxide of aluminum is formed by the reaction of
4.151 g of aluminum with 3.692 g of oxygen.
What is the empirical formula of the compound
formed?
 Relative Masses must be converted to moles
► 4.151
g Al x 1 mol Al/ 26.98 g Al = 0.1539 mol of Al atoms
► 3.692 g O x 1 mol O/ 16.oo g O= 0.2308 mol of O atoms
 Divide by smallest number of moles
► 0.1539
mol Al / 0.1539 = 1.000 mol Al
► 0.2308 mol O / 0.1539 = 1.500 mol O
 Multiply by smallest number to make them all whole
► 1.500
O x 2 = 3 O atoms
► 1.000 Al x 2 = 2 Al atoms
 So formula is Al2O3
Practice Problem
► In
a lab experiment it was observed that 0.6884 g
of lead combines with 0.2356 g of chlorine to form
a binary compound. What is the empirical formula
of this compound?




Step 1 Relative Masses
Step 2 Convert to moles
Step 3 Divide by smallest number
Step 4 Multiply by smallest number that makes them all
whole numbers
► Rules
apply no matter how many elements are in
the compound
Calculation of Molecular Formulas
► Must
know percent composition and the
molar mass
► It’s always a multiple of the empirical
formula (Empirical Formula)n = molecular
formula or Empirical Formula x n =
molecular formula
► Compare the empirical formula to molar
mass
Example
► If
Empirical formula is P2O5 and molar mass
is 283.88 g, what is compound’s molecular
formula?
 Know we have 2 moles P and 5 moles O
►2
mol P = 2 x 30.97 g = 61.94 g
►5 mol O = 5 x 16.00 g = 80.00 g
►1 mol P2O5 = 141.94 g
 Know empirical formula x n = molecular formula
and that the molecular formula = molar mass
 This tells us n = molar mass/ empirical formula
 So 283.88 g / 141.94 = n = 2
 So (P2O5)2 means P4O10
Practice Problem
► What
are the empirical and molecular
formulas of a compound with these percent
compositions 71.65% Cl, 24.27% C, 4.07%
H and a molar mass of 98.96 g?