Transcript Link to Notes - Coweta County Schools

Chp 6
The Mole
Counting by Weighing
 Every item has a certain average mass.
 We can use that mass to calculate how
much we would expect 50, 1000 or 1000000
of that item to weigh.
 Then we could weigh out that amount rather
than count out 1000000 of that item.
– This is how we can count atoms.
– The atomic mass is the amount that 1 mole of
that element weighs.
The Mole
 Just like the term “a dozen” represents the
number 12, a mole represents the number
6.023 x 1023.
– That’s 602,300,000,000,000,000,000,000!!!
– A number that big is hard to comprehend so a
few facts:
 If you distributed a mole of pennies between
everyone alive on Earth, we’d each get a trillion $.
 If you stacked a mole of paper, it would reach the
moon and back 80 billion times.
 A mole of marshmallows would completely cover the
Earth in a pile 12 miles deep.
The Practical Application
 So 12.01 g of carbon contains 6.023 x 1023
atoms of carbon.
 Therefore 24.02 g of carbon contains 1.2046
x 1024 atoms (2 times 6.023 x 1023).
 We can use this in dimensional analysis to
figure out fractions of this:
– Atomic mass = 1 mole
– 1 mole = 6.023 x 1023 atoms
An Example
 How many moles are in 0.5 g of hydrogen?
– 1 mole H = 1.007 g
0.5 g H
1 mole H
= 0.496 mol H
1.007 g H
 How many atoms is this?
– 1 mole = 6.023 x 1023 atoms
0.496 mol H 6.023 x 1023 atoms H = 2.99 x 1023
1 mol H
atoms H
What About Compounds?
 We can do the same thing for a compound
by first adding the individual masses to get
the molar mass of the compound.
– Ex. CH4
C = 12.01 g/mol
H = 1.007 g/mol x 4
So CH4 = 16.04 g/mol
– How many moles are in 0.5 g of CH4?
0.5 g CH4 1 mol CH4
= 0.03 mol CH4
16.04 g CH4
Mass Percent
 How much of a particular element makes up a
compound.
 Mass % =
mass element
x 100
molar mass of compound
 Ex. What is the mass % of C in CH4?
12.01 x 100 = 74.9%
16.04
What about H?
4.028 x 100 = 25.1%
16.04
** Should add to 100%
Empirical Formulas


The simplest, whole number ratio of atoms
in a compound.
Rules:
1. Get the mass of each element present.
2. Convert grams to moles.
3. Divide all answers by the smallest number
from step 2.
4. Multiply by an integer, if necessary, to get
whole numbers.
5. Those numbers are the subscripts of the
empirical formula.
An Example
What is the empirical formula of a compound
containing 4.151 g Al and 3.692 g O?
1. Masses are listed above for us.
2. 4.151 g Al 1 mol Al
= 0.1539 mol Al
26.98 g Al
3.692 g O 1 mol O
15.99 g O
= 0.2308 mol O
An Example
3. 0.1539 is smallest so:
0.1539 = 1 mol Al
0.1539
0.2308 = 1.5 mol O
0.1539
4. Since we don’t have whole #s, we need to
multiply by an integer. Let’s try 2.
1 mol Al x 2 = 2 mol Al
1.5 mol O x 2 = 3 mol O
5. Now we have whole numbers, so the empirical
formula is
Al2O3.
Molecular Formulas
 The actual ratio of atoms in a compound.
 May be the same as the empirical or a
multiple of it.
– Ex. CH4 is an empirical formula
CH4, C2H8, C3H12 could all be molecular
formulas based on that E.F.
 Found by dividing the actual molar mass by
the molar mass of the empirical compound.
If the answer isn’t 1, you multiply the E.F.
subscripts by that number.
An Example
 What is the molecular formula of a
compound with a molar mass of 283.33
g/mol if the empirical formula is P2O5?
The mass of P2O5 is:
2 x 30.97 = 61.94 g
5 x 15.99 = 79.95 g
141.89 g/ mol
An Example
So, now we divide the real M.M. by the M.M.
of the empirical formula.
283.33 g = 2
141.89 g
Now we multiple the subscripts of the
empirical formula by that 2.
2 (P2O5) = P4O10 is our answer!