AP Equations
Question 4
Molar Mass and % Composition
Chapter 3, Part 1
Directions:
For each of the following three reactions, write a
balanced equation for the reaction in part (i) and
answer the question about the reaction in part (ii).
In part (i), coefficients should be in terms of lowest
whole numbers. Assume that solutions are
aqueous unless otherwise indicated. Represent
substances in solutions as ions if the substances are
extensively ionized. Omit formulas for any ions or
molecules that are unchanged by the reaction. You
may use the empty space at the bottom of the next
page for scratch work, but only equations that are
written in the answer boxes provide will be scored.
Good News!
All the reactions on the test
occur!!!
The example on the test
• I don’t expect you to know the answer to this
one….yet! Just note the format…
An Example you know!
(a) Solid mercury(II) oxide decomposes as it is heated in
an open test tube in a fume hood.
(i) Balanced equation
2 HgO  2 Hg + O2
(ii) After the reaction is complete, is the mass of the
material in the test tube greater than, less than, or equal
to the mass of the original sample? Explain.
_________________________________
less; gaseous oxygen escapes
from the test tube
_________________________________
_________________________________
Add to Equation Sheet
• Composition Reaction A & B
If there is a metal that can have multiple
charges (like Fe+2 and Fe+3):
If there is limited amount of the metal (or
an excess amount of the other element) the ion
will have the higher charge.
Add to equation sheet!
• Ammonium carbonate decomposes into
ammonia, water and carbon dioxide
Ex: (NH4)2CO3  2NH3 + H2O + CO2
• Ammonium hydroxide decomposes into
ammonia and water
Ex: NH4OH  NH3 + H2O
Note:
• We are going to skip the acid reactions for
now! We will come back to them in Chapter 4
don’t worry 
Molar Mass
The mass of 1 mole of a substance (i.e., g/mol)
– The molar mass of an element is the
number we find on the periodic table
Molar Mass
Numerically equal to the:
- Formula weight (for ionic compounds)
- Molecular weight (for molecular
compounds)
The formula weight or molecular weight (in
amu’s) will be the same number as the molar
mass (in g/mol)
Calculate the molar mass of ammonium sulfite. The
formula is (NH4)2SO3.
In 1 mole of the compound there are:
2 moles of N
8 moles of H
1 mole of S
3 moles of O
X
X
X
X
14.0 g/mole = 28.0 g N
1.01 g/mole = 8.08 g H
32.1 g/mole = 32.1 g S
16.0 g/mole = 48.0 g O
116.2 g/mole
(SD done by place value)
(units are always g/mole)
Percent Composition
One can find the percentage of the mass of a compound
that comes from each of the elements in the
compound by using this equation:
molar mass of element in the compound
% of element = --------------------------------------------------- x 100%
molar mass of compound
Find the percent composition of
lithium perchlorate (LiClO4)
• Li: 1(6.94) = 6.94g Li
• Cl: 1(35.5) = 35.5 g Cl
• O: 4(16.0) = 64.0 g O
106.4 g LiClO4
%Li =
6.94𝑔
106.4𝑔
∗ 100% = 6.52% 𝐿𝑖
%Cl =
35.5𝑔
106.4𝑔
∗ 100% = 33.4% 𝐶𝑙
%O =
64.0𝑔
106.4𝑔
∗ 100% = 60.2% 𝑂
Example 1
How many moles of water are in 15.4 g of water?
1 mol = 6.02x1023 particles = Molar Mass (g/mol)
Molar Mass of H2O = 2(1.01) + 1(16.0) =18.0 g/mol
1 mol H2O
15.4 g H2O x ------------------ = 0.856 mol H2O
18.0 g H2O
Example 2
1 mol = 6.02x1023 particles = Molar Mass (g/mol)
How many molecules are in 5.0 moles of water?
6.02 x 1023 molecules H2O
5.0 mol H2O x ------------------------------------- =
1 mole H2O
3.0 x 1024
molecules H2O
Example 3
How many moles of oxygen are in 116.4 g of carbon
dioxide?
Molar Mass of CO2 = 1(12.0) + 2(16.0) = 44.0g/mol
116.4𝑔𝐶𝑂2 ∗
1𝑚𝑜𝑙 𝐶𝑂2
44.0𝑔𝐶𝑂2
∗
2 𝑚𝑜𝑙 𝑂
=
1𝑚𝑜𝑙 𝐶𝑂2
5.29 mol O
Example 4
• A sample of N2O4 contains 3.49x1024 nitrogen
atoms. How many moles of N2O4 are in the
sample?
1 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒 𝑁2 𝑂4
1 𝑚𝑜𝑙 𝑁2 𝑂4
3.49 ∗ 10 𝑎𝑡𝑜𝑚𝑠 𝑁 ∗
∗
= 2.89 𝑚𝑜𝑙
23
2 𝑎𝑡𝑜𝑚 𝑁
6.02𝑥10 𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑒𝑠 𝑁2 𝑂4
24
Example 5
How many grams of lithium can be
obtained from .0025 g of lithium
perchlorate?
• Method 1 (%):
. 0025𝑔𝐿𝑖𝐶𝑙𝑂4 ∗ .0652 = .00016𝑔 𝐿𝑖
Method 2 (DA):
. 0025 𝑔𝐿𝑖𝐶𝑙𝑂4 ∗
6.52𝑔 𝐿𝑖
=.00016g
106.4𝑔 𝐿𝑖𝐶𝑙𝑂4
Li