### Present Remotely

Send the link below via email or IM

• Invited audience members will follow you as you navigate and present
• People invited to a presentation do not need a Prezi account
• This link expires 10 minutes after you close the presentation

Do you really want to delete this prezi?

Neither you, nor the coeditors you shared it with will be able to recover it again.

# The Mole

moles, empirical formula, dimensional analysis
by

## Nancy Cope

on 9 December 2014

Report abuse

#### Transcript of The Mole

The Mole
the mole
defined as the amount of matter in exactly 12 grams of carbon-12
quantified (counted) the mole to be 6.022x10 particles
why carbon-12? it is the most abundant non-gaseous isotope
23
602,200,000,000,000,000,000,000 particles
this is a very big number that is very difficult to understand
this tells us just how small atoms are
How big is a mole?
mole of paper stacked would wrap around the moon and back 100,000,000 times.
one mole of marshmallows would cover the earth 4 miles deep
How is the mole used?
one mole is the same as Avogadro's number
moles and molecules
moles and atoms
moles and particles
it is used as a conversion factor between
conversion factors
1 mole = 6.022x10 particles
23
23
23
1 mole = 6.022x10 molecules
1 mole = 6.022x10 atoms
try some
how many molecules are in 35 moles of CO ?
how many moles are 5.46 x10 atoms of Fe?
how many molecules are in 35 moles of CO ?
2
how many moles are 5.46 x10 atoms of Fe?
try some
35 moles
5.46x10 atoms
23
1 mole
6.022x10 molecules
33
1 mole
23
6.022x10 atoms
33
x
x
=
=
try some
2.1077 x 10 molecules of CO
9.066 x 10 moles of Fe
9
25
2
molar mass
the mass of one mole of a substance
measured in grams per mole
for elements, it can be found on the periodic table
for compounds, it is the mass of the individual elements added together
percent composition
the percent mass of each element in a compound
the mass of each element is divided by the mass of the whole compound
there is a percent for each element in the compound
NaCl
Na = 23 g/mol
Cl = 35 g/mol
NaCl = 58 g/mol
0.60344 x100 = 60.34% Cl
0.39655 x100 = 39.66% Na
58 g/mol
35 g/mol
23 g/mol
58 g/mol
=
=
Ca (PO )
3 4 2
3 Ca = 3 x 40 g/mol = 120 g/mol Ca
2 P = 2 x 31 g/mol = 62 g/mol P
8 O = 8 x 16 g/mol = 128 g/mol O
310 g/mol
128 g/mol O
62 g/mol P
120 g/mol Ca
Ca (PO )
3 4 2
310 g/mol
Ca (PO )
3 4 2
310 g/mol
Ca (PO )
3 4 2
=
=
=
x 100
x 100
x 100
20.00% P
41.29% O
38.71% Ca
NaCl
Ca (PO )
3 4 2
NaCl = 58 g/mol
Cl = 35 g/mol
Na = 23 g/mol
3 Ca = 3 x 40 g/mol = 120 g/mol Ca
2 P = 2 x 31 g/mol = 62 g/mol P
8 O = 8 x 16 g/mol = 128 g/mol O
Ca (PO )
3 4 2
= 310 g/mol
average atomic mass
the mass of each element as found on the periodic table
the average mass of all of the isotopes of that element weighted by their percent abundance
empirical formula
molecular formula
molar volume
can be used as a conversion unit between moles and liters
the amount of volume one mole of gas occupies at standard temperature and pressure
45 liters of CO
2
1 mole
2
22.4 L CO
=
x
1 mole of any gas occupies 22.4 L at STP
2 mol of CO
2
98.90 % carbon-12
1.10 % carbon-13
= 0.011 x 13 amu = 0.143 amu
= 0.989 x 12 amu = 11.868 amu
the formula with the lowest possible ratio of elements
can be determined from % composition in 3 steps
Step 1: Assume 100 grams
Step 2: Convert grams to moles
Step 3: Divide by lowest # of moles to create the ratio (and hence, subscripts)
by assuming a total of 100 grams, the percent of each elements is the same as the number of grams of each element present
convert grams to moles by dividing by the average atomic mass of the element
divide the moles of each element by the lowest number of moles. This creates a ratio between the elements. The ratio numbers become the subscripts in the formula.
commonly used for covalent molecules
shows exactly how many elements are in the compound
the formula with the ratio of elements that is not lowest terms
can be determined from % composition and molecular mass
Step 4: Multiply the subscripts of the empirical formula by the multiplier
Step 3: Divide to get the multiplier
Step 2: Calculate molar mass of empirical formula
Step 1: Determine empirical formula
Using the three step approach, determine the empirical formula from the % composition
Add together the masses of each of the atoms in the empirical formula
divide the mass of the molecular formula by the mass of the empirical formula. the answer is how many times bigger the molecular formula is than the empirical formula.
increases the subscripts of the empirical formula by the same factor creates a molecular formula
33
2
when the percent abundance is multiplied by the mass of the isotope, for all isotopes are added together, it is the average atomic mass.
11.868 + 0.143 = 12.011 amu
one amu or atomic mass unit is the mass of one proton or one neutron (they have the same mass) This is the unit for isotopes.
one amu is equal to one gram per mole
percent composition
Full transcript