SCH3U: G11 Chemistry
UNIT 2: QUANTITIES IN CHEMICAL REACTIONS
LESSON 4: CALCULATING HYDRATES AND THEIR CHEMICAL FORMULAE
Learning Targets
In today’s class, students will be learning to…
u Describe hydrate and anhydrous ionic
compounds.
u Use our understanding of % composition,
mass, and ratios to calculate molecular
formulas for hydrates.
u Use our understanding of % composition,
mass, and ratios to calculate the amount of
a hydrated compound is needed to make x
amount of anhydrous.
Success Criteria
I know I am successful when I can…
•
Describe and define the difference between hydrous and
anhydrous compounds.
•
Use % composition by mass to calculate simple molecular
ratios and molecular formulas for hydrates.
•
Use simple molar ratios to calculate how much of a
hydrated material I need to make a set amount of
anhydrous.
Warmer
If you recall, we talked about hydrous and anhydrous
compounds in the nomenclature lesson.
The following compounds are hydrate compounds:
1. CaBr2•3H2O
2. Potassium oxide tetra-hydrate
Task:
a. In your own words, describe what a hydrated
compound is.
b. Convert (1) into its’ written name using
nomenclature rules.
c. Convert (2) into its’ chemical formula.
Warmer - solutions
The following compounds are hydrate compounds:
1. CaBr2•3H2O
2. Potassium oxide tetra-hydrate
Task:
a. In your own words, describe what a hydrated compound is.
A hydrated compounds is any ionic compound that also has water included in
the compound.
b. Convert (1.) into its’ written name using nomenclature rules.
Calcium bromide tri-hydrate
c. Convert (2.) into its’ chemical formula.
K2O•4H2O
Reviewing Hydrates
• Many ionic compounds form crystals that
contain loosely held water molecules
within the crystal structure.
• These compounds are called hydrates,
and when the water is removed the
compound by heat, we call it anhydrous.
For example: Sodium chloride dihydrate
NaCl·2H2O
Viewing hydrates as ratios
Let’s look at our example here again of
NaCl·2H2O
Interestingly, this is also represents a ratio!
e.g. 1 formula unit of sodium chloride:
2 molecules of water
We can follow the same steps to
determine the ratio, % composition, and
mass within a hydrate.
For example – calculating % composition
and mass:
In 100g of NaCl•2H2O (total mass = 93AU):
NaCl = 23+35 = 57
57/93*100 = 61.3%
2H2O = 2(2 + 16) = 36
36/93*100 = 38.7%
NaCl has 61.3g and H2O has 38.7g
How to calculate empirical formula for
a hydrate
•
Watch the video in the resources folder
titled: ‘Calculating a hydrate empirical
formula’
This video will:
•
Re-introduce the concept of % composition
and mole ratios (and how to calculate
them)
•
Take you through the steps needed to
calculate the empirical formula of a
compound with 1 ionic compound : x H2O.
Worked example: calculating
molecular formulas
A 50g sample of hydrated barium hydroxide contains 27.2g of Ba(OH)2
Determine the formula of the hydrated sample.
Worked example: solution
A 50g sample of hydrated barium hydroxide contains 27.2g of Ba(OH)2
Determine the formula of the hydrated sample.
Moles of barium hydroxide
27.2g
22.8g
Ba(OH)2 ·
xH2O
Moles of water
50 grams
Now determine the
whole number ratio
between these values
Ba(OH)2 · 8H2O
Student task: Calculating molecular
formulas
A 75g sample of a hydrate Mo2S5•xH2O contains 65 grams of Mo2S5. How many water
molecules are bound to each formula unit of molybdenum sulfide?
Student task 1: Solution
A 75g sample of a hydrate Mo2S5•xH2O contains 65 grams of Mo2S5. How many
water molecules are bound to each formula unit of molybdenum sulfide?
65g
Mo2S5 ·
10g
xH2O
Moles of water
Moles of molybdenum sulphide
65 grams
Molecular mass = 370.2
Now determine the
whole number ratio
between these values
Mo2S5 · 3H2O
Student task 2:
An experiment requires 47.6g of MgCl2. However, only magnesium
chloride hexahydrate is available, what amount of this should be
used to yield 47.6g of anhydrous MgCl2?
Hint: How many mol is 47.6g? Use the mol ratio to figure out how many grams
of water needed to be added to 47.6g (total mass needed = 47.6g+mass of
water at molar ratio).
Student task 2: solution
An experiment requires 47.6g of MgCl2. However, only magnesium chloride
hexahydrate is available, what amount of this should be used to yield 47.6g of
anhydrous MgCl2?
MgCl2 · 6H2O
47.6 grams ?? grams
Moles of magnesium chloride
Using 1:6 ratio
Moles of H2O
Grams of H2O
A total of 101.6 grams
of the hydrated salt
should be used.
Comprehension check - Quiz
1. An ionic compound that has water heated out of it is called…
a. Anhydrous b. Hydrated
c. Dry d. Ratioed
2. Which of the following allows us to calculate an unknown quantity of water
in a hydrated solution?
a. Molecular mass
b. Molar ratio
c. Reactivity
d. Ionic structure
3. What is the % of water by mass of zinc(II) nitrate hexahydrate?
a. 57%
b. 63.7%
c. 36.3%
d. 34.3%
Comprehension check – Quiz solutions
1. An ionic compound that has water heated out of it is called…
a. Anhydrous b. Hydrated
c. Dry d. Ratioed
2. Which of the following allows us to calculate an unknown quantity of water
in a hydrated solution?
a. Molecular mass
b. Molar ratio
c. Reactivity
d. Ionic structure
3. What is the % of water by mass of zinc(II) nitrate hexahydrate?
a. 57%
b. 63.7%
c. 36.3%
d. 34.3%