Chapter 11 Study Guide – The Mole
Chemistry - Dr. Hazlett
1
I. Key Equations
A. KNOW THIS!
÷ Molar Mass (M)
-------------------------
x NA
----------------------
Mass
Moles
Number
(Grams)
n
Particles
--------------------------------------------x Molar Mass (M)
÷ NA
* NA = 6.022314 x 1023
B. NA is Avogardro’s Number and equals 6.022 x 1023
1. This became equated with the “mole” (abbreviated mol)
a. 1 mole of anything = 6.022 x 1023 of that item
C. Molar Mass (M) is the number of grams per mole of a substance (g/mol)
1. For elements on Periodic Chart – look for atomic mass (amu)
under symbol and change this to xx.xxx g/mol
2. For compounds – add up all masses of all elements in the compound
a. Careful to watch for subscripts and distributive property
b. For compounds – this is called Molecular Molar Mass
3. Example (some of these are rounded off):
Element/Compound
C
He
K
H2O
C6H12O6
Atomic/Molecular Mass
12.011 amu (u)
4.00 u
39.098 u
(2 x 1.007) + 15.999 = 18 u
(6 x 12) + (12 x 1) + (6 x 16) = 180 u
Molar Mass
12.011 g
4.00 g
39.098 g
18 g
180 g
D. Moles (mols or n)
1. An encompassing term for 6.022 x 1023 units of whatever is being
counted, regardless of their mass
a. Therefore, 1 mol of whatever units/particles = 6.022 x 1023
b. Derived from C-12 atom and its atomic mass
2. # mols = # particles ÷ NA
3. # mols = mass ÷ molar mass
4. # mols = concentration (Molar) ÷ volume (Liter)
Chapter 11 Study Guide – The Mole
Chemistry - Dr. Hazlett
2
5. Mass = # mols x molar mass
6. # particles = # mols x NA
7. Atomic mass = mass of element ÷ # mols
E. Numerous examples were demonstrated in class during the lectures!
II. Percent Composition of Compounds
A. Percentage by mass is the percent by mass of each element in a compound
B. % Composition of element = n (molar mass element) x 100
Molar mass of compound
1. n is the number of moles or simply the subscript number of element
C. Examples:
1. What is the percentage of C and O in the compound CO2?
Carbon has 1 mole (subscript 1) so it is 1 x molar mass of 12.011 divided by
molar mass of CO2 which is 44.01 g/mol multiplied by 100
g/mol C x
g/mol CO2
100
2 x 16 g Oxygen =
=
12.01 g/mol C x
44.01 g/mol CO2
32 g/mol
44.01 g/mol
100
= 27.29%
x 100 = 72.71%
2. What is the formula for a compound that has 92.2% C and 7.8% H.
M = 52.1 g/mol


Assume the percentages are in grams, so total mass of compound becomes
100 g.
Find # of mols of each element by dividing mass by molar mass.
92.2 g C ÷ 12.01 g/mol C = 7.68 mol C
7.8 g H ÷ 1.01 g/mol H = 7.72 mol H


In this case, the mole ratio is 1:1
Divide all moles calculated by the lowest one found
7.68 mol C = 1.00 mol C
7.68
7.72 mol H = 1.01 mol H
7.68
Chapter 11 Study Guide – The Mole
Chemistry - Dr. Hazlett


3
Therefore the empirical formula (reduced form) is C1H1, but this does not
mean it is the true molecular formula
To discover molecular formula, take the M compound / M empirical
formula, and multiply the empirical formula by the result
M compound given 52.1 g/mol = 4, so (CH)4 = C4H4
M empirical unit
13.02 g/mol
3. Given 38.7% C, 9.7% H and 51.6% O, with M = 62.0 g/mol. What is
molecular formula?





First, find empirical formula following previous example. CH3O
Find formula unit’s molar mass.
31.0 g/mol
Divide molecular molar mass given by empirical formula molar mass
62 / 31 = 2
Multiply each subscript (which represents n or moles of each element) by
the result of 2
n(CH3O) = (CH3O)2 = C2H6O2
Result of this is molecular formula!
III. Other Examples
A. How many mols of He atoms are there in 6.46 g He?
 Find molar mass of He = 4.0003 g/mol He
 Using flowchart, 6.46 g He divided by molar mass He = # mols
 So, 6.46 / 4.003 = 1.61 mol He
B. How many grams of Zn are in 0.356 mol?
 1 mol Zn has a mass of 65.39 g as determined by Periodic Table
 0.356 mol Zn x 65.39 g Zn = 23.3 g Zn
C. How many atoms are in 16.3 g S?
 1 mol S = 32.07 g (from Periodic Table)
 1 mol contains 6.022 x 1023 particles (atoms, ions, molecules, etc.)
 So: 16.3 g S divided by 32.07 g/mol S multiplied by 6.022 x 1023
 Result is 3.06 x 1023 atoms S
D. How many mols of methane (CH4) are in 6.07 g of CH4?
 M of CH4 = 12.01 g/mol C + (4 x 1.008) g/mol H = 16.04 g/mol
 6.07 g CH4 divided by 16.04 g/mol CH4 = 0.378 mols CH4
E. How many H atoms are in 25.6 g (NH2)2CO (urea). M = 60.6 g/mol.
 Process will be: grams urea  mols urea  molecules urea  H atoms
 25.6 g (given mass of urea) ÷ 60.6 g/mol (molar mass of urea) x (6.022
x 1023) to find molecules of urea x 4 H atoms per molecule
 Result is 1.03 x 1024 atoms H
Chapter 11 Study Guide – The Mole
Chemistry - Dr. Hazlett
4
F. Mass of a single atom
 Take the molar mass of the element and divide it by NA to get g/atom
 What is the mass of one atom Si?
M of Si = 28.09 g/mol
28.09 g/mol Si
6.022 x 1023

=
4.665 x 10-23 g/atom Si
Note the exponent becomes a negative since dividing!