Definition:
One mole is the amount of any pure substance containing
the same number of chemical units (atoms, molecules, ions,
electrons, etc.) as there are atoms in exactly 12 grams of
carbon-12.
Exactly 12 grams of carbon-12 contains:
6.02 x 1023 atoms of carbon
(Note: For atomic weights, we will round periodic atomic masses to the
nearest whole number. For example, H = 1, Li = 7, O = 16.)
To find a mole of a substance (element or
compound) use this two-step process:
Step 1:
Step 2:
Find the atomic mass of the element or the
molecular weight of the compound (add all
the masses of all the atoms in the
compound).
Add “grams” as the unit to the number
you found in step 1.
Example 1: Find one mole of lithium
Step 1.
The atomic mass of lithium is 7
Step 2.
One mole of lithium is 7 grams
Example 2: Find one mole of carbon dioxide, CO2
Step 1. The molecular weight of CO2 is 44
1C
1 x 12.0 = 12
2O
2 x 16.0 = 32
44
Step 2. One mole of CO2 is 44 grams
Moles to Grams
Grams to Moles
How many moles in 15 grams of lithium?
1. Find grams in one mole of the substance:
7g
1 mole = X
2. Set up a ratio:
7g
15 g
3. Cross multiply and divide to find unknown:
1 mole x 15 g = X x 7 g
7g
7g
X = 2.14 mol
Moles to Grams
Grams to Moles
How many moles in 15 grams of lithium?
1. Find grams in one mole of the substance
to set up conversion factor:
2. Set up the dimension:
15 g x 1 mol =
7g
3. Do the math:
15 g x 1 mole =
7g
2.14 mol
1 mol = 7 g
Moles to Grams
Grams to Moles
How many grams in 2.4 moles of sulfur?
1. Find grams in one mole of the substance:
2. Set up a ratio:
32 g
1 mole = 2.4 mol
32 g
X
3. Cross multiply and divide to find unknown:
1 mole x X = 2.4 mol x 32 g
1 mole
1 mole
X = 76.8 g
Solving Chemical Equation Word Problems
A six step process that can be used to solve just about any
chemical equation word problem that involves finding out
how much of one substance is needed to react with another
substance, or how much of a product is the result of
a chemical reaction.
This process requires the following six steps (skills):
1. writing chemical formulas (we have done)
2. finding molar masses
(we have done)
3. balancing chemical equations (we have done)
4. using coefficients to find molar equivalents (new)
(you have done) 5. setting up and using simple math ratio & proportion
(you have done) 6. cross multiplying and dividing
Solving Chemical Equation Word Problems
Use the (six) step by step process to solve the following problem:
How much nitrogen gas (grams) is released when ten grams of
ammonia (nitrogen trihydride) is combined with oxygen gas
to produce nitrogen gas and water?
Problem: How much nitrogen gas (grams) is released when ten grams of ammonia
(nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water?
Skill: Writing chemical formulas.
Step 1: Identify the reactants and products and then write
the chemical equation using the appropriate
chemical symbols and formulas.
Identify the reactants:
Identify the products:
nitrogen trihydride
NH3
+
nitrogen gas
N2
+
oxygen
O2
water
H2O
Write the chemical equation using appropriate
chemical symbols and formulas.
NH3 +
O2
→
N2
+
H2O
→
Problem: How much nitrogen gas (grams) is released when ten grams of ammonia
(nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water?
Skill: Finding molar masses.
Step 2: Before balancing the equation, use your periodic table to
find the mass of one mole of each element or compound.
Write the molar masses ABOVE the element or compound.
One mole = 17 g
32g
NH3 + O2
→
28 g
N2 +
18 g
H2O
(molar masses)
Problem: How much nitrogen gas (grams) is released when ten grams of ammonia
(nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water?
Skill: Balancing chemical equations.
Step 3: Balance the equation using correct coefficients.
One mole =
17 g
32g
4 NH3 + 3 O2
28 g
→ 2 N2
18 g
+
(molar masses)
6 H2O
Note: Coefficients will represent the
number of MOLES of a substance!!!!
Problem: How much nitrogen gas (grams) is released when ten grams of ammonia
(nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water?
Skill: Using coefficients to find molar equivalents.
Step 4: Multiply the molar masses by the coefficients (moles) to get the
balanced molar mass equivalents. Write this number below
the symbol of the element or compound.
One mole =
17 g
32g
4 NH3 + 3 O2
68 g
96 g
28 g
18 g
→ 2 N2
+ 6 H2O
56 g
108 g
(molar masses)
Problem: How much nitrogen gas (grams) is released when ten grams of ammonia
(nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water?
Skill: Setting up and using simple math ratio & proportion (fractions).
Step 5: Using your problem statement and the balanced mass
equivalents, set up a fractional ratio and proportion to
solve for your unknown.
A. Identify the unknown (the quantity you are trying to find = How
much nitrogen gas in grams?) and the information that you do know (you
have 10 grams of ammonia) and line them up right under the balanced mass
equivalents for the chemicals that represent the same measurements.
4 NH3 + 3 O2
68 g
10 g
96 g
→ 2 N2
+ 6 H2O
56 g
X
108 g
→
B. Since we only need two fractions to solve for one unknown,
we can now set up a simple equation to solve. (NOTE: Since
oxygen and water are not part of this particular problem, we
need them to balance the equation, but do not need them for the
final solution.)
68 g
10 g
=
56 g
X
Problem: How much nitrogen gas (grams) is released when ten grams of ammonia
(nitrogen trihydride) is combined with oxygen gas to produce nitrogen gas and water?
Skill: Cross multiplying and dividing.
Step 6: Cross multiply and divide to solve for the unknown (X)
quantity.
68 g
10 g
a. 68 g x X
=
=
56 g
X
10 g x 56 g
b. 68 X g = 560 g g
c. 68 X g = 560 g g
68 g
68 g
X = 8.2 g N2
Solving Chemical Equation Word Problems
Practice problem # 1:
Hydrogen gas (diatomic) combines with oxygen gas (diatomic) to
make water. If you start with 100 grams of hydrogen gas and all
the oxygen that you need to completely combine with the hydrogen,
then how much water will be produced?
hydrogen gas
1 mol =
2g
4 H2
8g
100 g
+
oxygen gas
→
32 g
+
O2
+
=
water
18 g
→
2 H2O
36 g
X
X = 450 g
Solving Chemical Equation Word Problems
Practice problem # 2:
You have 100 grams of sodium that you are going to combine with
chlorine gas (diatomic) to make sodium chloride (salt). How much
chlorine gas will you need to just exactly combine with the 100 grams
of sodium?
sodium +
1 mol =
chlorine gas
23 g
2 Na
46 g
100 g
→
sodium chloride
70 g
+
Cl2
=
70 g
X
58 g
→ 2
NaCl
X = 152 g
Complete and turn in the
worksheet (Due next class).