DETERMINATION OF AN
EMPIRICAL FORMULA
1
Empirical Formulas
• Empirical formula (“formula unit”) -the
lowest whole number ratio of atoms in an
ionic compound.
• There are two ways to determine the
empirical formula for an ionic compound.
1. Using charges (use your ions and this is
easy). C
CRISS
O
S
S
2. Mathematically (yippee!!!)
2
Method #1 – Charges
EX: Write the empirical formula
for the compound formed by
Na & P
c. K & Ne
Na3P
Sr & Cl
SrCl2
No compound
d. Cu & Cl
CuCl or CuCl2
3
Method #2 –
Mathematically
• Step 1:Use the information given in the
problem and dimensional analysis with
the atomic mass of the element (from
periodic table- round to 3 SDs) to find
the number of moles you have
• Step 2:Take all the mole values and
divide them by the SMALLEST one to
figure out a ratio
• Step 3: Use the answers as subscripts 4
in the empirical formula
Reference-Empirical Formulas
7.06 g of silver combine with an excess of
fluorine to produce 8.30 g of a compound
Silver
7.06g
+
Fluorine
1.24g

Ag?F?
8.30g
Found by subtracting!
7.06 g Ag X
𝟏 𝒎𝒐𝒍𝒆 𝑨𝒈
𝟏𝟎𝟖 𝒈 𝑨𝒈
.𝟎𝟔𝟓𝟒
= .0654 moles Ag
.𝟎𝟔𝟓𝟑
= 1.00 or 1
Ag1F1
1.24 g F X
𝟏 𝒎𝒐𝒍𝒆 𝑭
𝟏𝟗.𝟎 𝒈 𝑭
= .0653 moles F
.𝟎𝟔𝟓𝟑
.𝟎𝟔𝟓𝟑
= 1.00 or 1
ANSWER = AgF
5
Reference-Empirical Formulas
A compound contains 24.58% K, 35.81% Mn,
and 40.50% O. Find the empirical formula
(assume working with 100 grams of the
compound and change percentages to grams)
24.58 g K X
35.81 g Mn X
40.50 g O
X
1 𝑚𝑜𝑙𝑒 𝐾
39.1 𝑔 𝐾
= .629 mole K
.𝟔𝟐𝟗
.𝟔𝟐𝟗
= 1.00 or 1
1 𝑚𝑜𝑙𝑒 𝑀𝑛
= .652 moles Mn
54.9 𝑔 𝑀𝑛
.𝟔𝟓𝟐
.𝟔𝟐𝟗
= 1.04 or 1
1 𝑚𝑜𝑙𝑒 𝑂
16.0 𝑔 𝑂
𝟐.𝟓𝟑
.𝟔𝟐𝟗
= 4.02 or 4
= 2.53 moles O
6
ANSWER = KMnO4
Uneven Empirical Formulas
• When figuring empirical formulas
mathematically, sometimes the resulting
numbers don’t come out so clean
• You can’t just assume and round how you
choose
7
Reference- .05 Rule
• Values used in these problems are obtained by
experimentation. The 0.05 rule allows for
experimental error:
– If the value is within .05 of a whole number (+0.05
or - 0.05), then the value may be rounded to that
whole number
– Examples: 1.96 can be rounded to 2
• 1.07 cannot be rounded to 1
• 3.02 could be rounded to 3
• 1.93 cannot be rounded to 2
• IF one of the values is not within .05 of a whole
number, all the values must be multiplied by an
integer so that all values fall within .05 of whole
8
numbers
Reference-Uneven Empirical
Formulas
4.35 g sample of zinc is combined with an excess
of the element phosphorus. 5.72 g of compound
are formed. Calculate the empirical formula.
Zinc + Phosphorous  Zn?P?
4.35g
1.37g Found by
5.72g
4.35 g Zn X
𝟏 𝒎𝒐𝒍𝒆 𝒁𝒏
𝟔𝟓.𝟒 𝒈 𝒁𝒏
= .0665 moles Zn
subtracting!
.𝟎𝟔𝟔𝟓
.𝟎𝟒𝟒𝟐
= 1.50 X 2 = 3.00 or 3
Not within .05 of a
whole number
𝟏 𝒎𝒐𝒍𝒆 𝑷
𝒈𝑷
1.37 g P X 𝟑𝟏.𝟎
= .0442 moles P .𝟎𝟒𝟒𝟐 = 1.00 X 2 = 2.00 or 2
.𝟎𝟒𝟒𝟐
9
ANSWER = Zn3P2
Let’s Do It!!!
A compound is found to contain 72.3%
Fe and 27.7% O by weight. Calculate
the empirical formula.
Assume in 100 g of compound there
would be 72.3 g Fe and 27.7 g O
10
72.3g Fe
1 mole Fe
X ——————
55.8 g Fe
1.30 mole
= 1.00
1.30 mole
= 1.30 mole Fe
X 3 = 3.00 = 3
1 mole O
27.7g O X —————— = 1.73 mole O
16.0 g O
1.73 mole =1.33 X 3 = 3.99 = 4
1.30 mole
Fe3O4
11
Review
• Number your paper from 1-5 and answer the
following questions. Two will be cumulative
review!
• 1. Which of these is the correct symbol for
Chlorine-35?
• a. 35 Cl
17
• b. 17
Cl
37
• c.
17
35
Cl
12
Review
• A
• 2. Which is the correct answer with the
right number of SD’s if we add 75g and
25.0g?
– a. 105 g
– b. 10Ō g
– c. 100.0 g
– d. 105.0 g
13
Review
• B
• 3. Which is correct if we consider our
rounding rule?
– a. 4.02 couldn’t be rounded to 4 but
1.93 can be rounded to 2
– b. 4.02 couldn’t be rounded to 4 and
1.93 can’t be rounded to 2
– c. 4.02 can be rounded to 4 and 1.93 can
be rounded to 2
– d. 4.02 can be rounded to 4 and 1.93
14
can’t be rounded to 2
Review
• D
• 4. Zinc + Phosphorous  Zn?P?
4.35g
?g
5.72g
– a. 4.35 g
– b. 2.37 g
– c. 1.37 g
– d. 1.47 g
15
Review
• C
• 5. If I have 32.0g of O, then how many
moles is this?
– a. 2.0 moles
– b. 2.5 moles
– c. 2 moles
– d. 20 moles
16
Review
• A
17
Molar Mass
• KC4H5O6 is the empirical formula for
cream of tartar
• How many atoms of oxygen are in 1
formula unit of cream of tartar?
6 atoms O
18
Molar Mass
• KC4H5O6
• How many atoms are in 1 formula
unit of cream of tartar?
16 atoms
1 potassium
4 carbon
5 hydrogen
6 oxygen
19
Molar Mass
• KC4H5O6
• How many formula units are in 1 mole
of cream of tartar?
6.02 x 1023 formula units
20
Molar Mass
• KC4H5O6
• How many atoms of oxygen are in
(PER) 1 mole of cream of tartar?
• ? atoms O/mole KC4H5O6 =
6 atoms O
1 formula unit
KC4H5O6
x 6.02 x1023 formula unit KC4H5O6
1 mole KC4H5O6
3.61 x 1024 atoms O/mole KC4H5O6
21
Molar Mass
• KC4H5O6
• How many moles of oxygen atoms are
in 1 mole of cream of tartar?
• Let’s put this in perspective……….
22
How many wheels are in a
dozen bicycles?
1 bicycle = 2 wheels
1 dozen bicycles = 2 dozen wheels
23
1 mole bicycles = 2 mole wheels
How many dozen hydrogen
atoms are in 1 dozen water
molecules?
H2O
1 water = 2 hydrogen
1 dozen water = 2 dozen hydrogen
1 mole water = 2 mole hydrogen 24
Molar Mass
• So………
• KC4H5O6
• How many moles of oxygen atoms are
in 1 mole of cream of tartar?
6 moles O
25
Molar Mass
• KC4H5O6
• What is the mass of the oxygen
atoms in (PER) 1 mole of cream of
tartar?
? g O/ mole KC4H5O6
6 moles O x
1 mole KC4H5O6
16.0 g O =
96.0 g O
1 mole O
mole KC4H5O6
26
Let’s Do It!!!
• How many atoms in one formula unit?
Magnesium Acetate: Mg(C2H3O2)2
1-Mg
4-C
6-H
4-O
15 total atoms
27
Let’s Do It!!!
• How many formula units in one
mole of
Magnesium Acetate: Mg(C2H3O2)2
6.02 x 1023
28
Let’s Do It!!!!
• How many oxygen atoms are in
one formula unit?
Magnesium Acetate: Mg(C2H3O2)2
4 oxygen atoms
29
Let’s Do It!!!
• How many oxygen atoms are in one
mole of this substance?
Magnesium Acetate: Mg(C2H3O2)2
(4) (6.02 x 1023)
= 2.41 x 1024 atoms O
30
Let’s Do It!!!
• How many moles of oxygen
atoms are in one mole of this
substance?
Magnesium Acetate: Mg(C2H3O2)2
4 moles of Oxygen
31
Let’s Do It!!!
• What is the mass of oxygen atoms in
one mole of this substance?
Magnesium Acetate: Mg(C2H3O2)2
• ? g O=
1 mole Mg(C2H3O2)2 x 4 moles O x 16.0 g O
1 mole Mg(C2H3O2)2 1 mole O
= 64.0 g O = 64.0g/ 1 mole
32
Molar Mass
• We can use this information about what
makes up a compound to figure out a
compound’s total molar mass
• Molar mass- the mass in grams of 1 mole
of a compound
• It’s the mass of one mole:
grams KC4H5O6/mole KC4H5O6
• Also called formula weight, gram formula
weight, molecular weight
33
Reference- Molar Mass
• Calculate the molar mass of magnesium
iodide, MgI2 from its parts. How many
grams are in 1 mole?
Mg 1 mole (24.3g Mg/mole )= 24.3 g Mg
I 2 mole (127 g I/mole) = 254 g I
278 g/mole MgI2
34
(for SD use place value!)
Reference- Molar Mass
Calculate the molar mass (grams in one
mole) of ammonium sulfite, (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
SD by place
units always
X 14.0 g N/mole= 28.0g N
X 1.01 g H/mole = 8.08gH
X 32.1 g S/mole =32.1 g S
X 16.0 g O/mole=48.0g O
value
116.2g
g/mole
1mole (NH4)2SO353
Molar Mass
• Molar mass can be used to convert between
moles and grams
• For an element:
1 mole = 6.02x1023 atoms = atomic mass
• For a compound:
• 1 mole = 6.02x1023 formula units = molar
mass
36
Reference- Molar Mass as a
Conversion Factor
• What is the mass of 1.35 moles of
(NH4)2SO3?
• Earlier we found the molar mass:
116.2g (NH4)2SO3 = 1mole (NH4)2SO3
? g (NH4)2SO3 =
1.35 moles (NH4)2SO3 x
116.2 g (NH4)2SO3
1 mole (NH4)2SO3
= 157 g (NH4)2SO3
37
Reference- Molar Mass as a
Conversion Factor
• 75.2 g (NH4)2SO3 is how many moles of
(NH4)2SO3?
? moles (NH4)2SO3 =
75.2 g (NH4)2SO3
1 mole (NH4)2SO3
X ---------------------116.2 g (NH4)2SO3
= .647 moles (NH4)2SO3
38
Let’s Do It!!!!
• How many moles are in 87.4 g of aluminum
monohydrogen phosphate Al2(HPO4)3 ?
• This is a two part problem: remember you
first have to find the molar mass!
39
Let’s Do It!!!
• Find the molar mass of aluminum
monohydrogen phosphate, Al2(HPO4)3
2 moles Al (27.0g Al/mole) =
3 moles H (1.01g H/mole) =
3 moles P (31.0 g P/mole) =
12 moles O (16.0 g O/mole)=
54.0 g Al
3.03 g H
93.0 g P
192 g O_
342 g
mole Al2(HPO4)3
MASS OF ONE MOLE!!!
40
So 342 g Al2(HPO4)3 = 1 mole Al2(HPO4)3
Let’s Do It!!!!
• Then find the number of moles in 87.4 g
of aluminum monohydrogen
Obviously, this is
phosphate
less than one mole!
• Molar mass = 342 g/mole
• ? moles Al2(HPO4)3 =
87.4 g Al2(HPO4)3 x 1 mole Al2(HPO4)3
342 g Al2(HPO4)3
= .256 moles Al2(HPO4)3
41
Review
• Number your paper from 1-5 and answer
the following questions. Two will be
cumulative review!
• 1. What does Avogadro’s number, 6.02 x
1023, mean?
– a. There are that many grams in a mole
– b. There are that many moles in a gram
– c. There are that many moles in an atom
– d. There are that many atoms in a mole
42
Review
• D
• 2. Which of these is false?
• a. protons repel protons
• b. protons attract neutrons
• c. protons attract electrons
• d. neutrons don’t repel anything
because they have no charge
43
Review
• B
• 3. How many elements are in this
compound? KC4H5O6
– a. 4
– b. 15
– c. 16
– d. 10
44
Review
A
• 4. How many moles of oxygen would be in 3
moles of KC4H5O6 ?
– a. 3
– b. 6
– c. 18
– d. 48
45
Review
• C
• 5. If H2O has a molar mass of 18, what
does this mean?
– a. There are 18 moles of water in a
gram
– b. There are 18 moles of oxygen in a
mole of water
– c. There are 18 grams in a mole of
water
– d. There are 10 moles of oxygen and 8
moles of hydrogen in each water
molecule
46
Review
• C
47
% Composition
• As we learned before, when we have
multiple elements that make up a
compound, each one has a certain
ratio
• We call this the compound’s
empirical formula
• From an empirical formula we can
figure out each element’s percent
composition
48
Reference-% Composition
• Find percent composition of Al(C2H3O2)3
(1 moles Al)(27.0 g/mole) = 27.0
(6 moles C) (12.0 g/mole) = 72.0
(9 moles H) (1.01 g/mole) = 9.09
(6 moles O)(16.0 g/mole) = 96.0
27.0
72.0
9.09
96.0
g Al /204.1
g C /204.1
g H /204.1
g O /204.1
g Al
gC
gH
gO
204.1 g Al(C2H3O2)3
g Al(C2H3O2)3 x 100 =
g Al(C2H3O2)3 x 100 =
g Al(C2H3O2)3 x 100 =
g Al(C2H3O2)3 x 100 =
13.2 %
35.3 %
4.45 %
47.0 %49
~100%
% Composition
• We can use this information further
• How many grams of aluminum can be
obtained from 1.50 moles of aluminum
acetate?
• ? grams Al =
1.50 moles Al(C2H3O2)3 x 𝟐𝟎𝟒.𝟏 𝐠 𝐀𝐥 𝐂𝟐𝐇𝟑𝐎𝟐
𝟏 𝐦𝐨𝐥𝐞 𝐀𝐥 𝐂𝟐𝐇𝟑𝐎𝟐
𝟐𝟕. 𝟎 𝒈 𝑨𝒍
X
𝟐𝟎𝟒. 𝟏 𝒈 𝑨𝒍 𝑪𝟐𝑯𝟑𝑶𝟐
𝟑
𝟑
= 40.5 g Al
OR
1.50 moles Al(C2H3O2)3 x
𝟏 𝐦𝐨𝐥𝐞 𝐀𝐥
𝟏 𝐦𝐨𝐥𝐞 𝐀𝐥 𝐂𝟐𝐇𝟑𝐎𝟐
X
𝟑
27.0 g Al
1 mole Al
= 40.4 g Al
50
𝟑
Let’s Do It!!!
• Find the percent composition of iron
(III) dichromate, Fe2(Cr2O7)3
(2 moles Fe)(55.8 g/mole) = 112 g Fe
(6 moles Cr)(52.0 g/mole) = 312 g Cr
(21 moles O)(16.0 g/mole) = 336 g O
76Ō g Fe2(Cr2O7)3
112 g Fe/76Ōg Fe2(Cr2O7)3 x 100 = 14.7 %
312 g Cr/76Ōg Fe2(Cr2O7)3 x 100 = 41.1 %
336 g O/76Ōg Fe2(Cr2O7)3 x 100 = 44.2 %
~100%
51
Let’s Do It!!!
• How many moles of iron can be
obtained from 525 g of iron (III)
dichromate, Fe2(Cr2O7)3?
• ? moles of Fe =
525 g Fe2(Cr2O7)3
= 1.38
𝟐 𝐦𝐨𝐥𝐞𝐬 𝐅𝐞
𝟏
𝐦𝐨𝐥𝐞
Fe
(Cr
O
)
2
2
7
3
X
X
𝟏 𝐦𝐨𝐥𝐞 Fe2(Cr2O7)3 moles Fe
𝟕𝟔𝟎 𝐠 Fe2(Cr2O7)3
OR
525 g Fe2(Cr2O7)3 X
𝟏 𝐦𝐨𝐥𝐞 Fe
112 g Fe
X
76Ōg Fe2(Cr2O7)3 𝟓𝟓.𝟖 𝒈 𝑭𝒆
= 1.39
moles Fe
52
Reference- Advanced Molar
Mass
• How many grams of Mg are needed to combine
with 2.00 g of N to make magnesium nitride?
Magnesium + Nitrogen Magnesium nitride (Mg3N2)
?g
2.00 g
(3 moles Mg) (24.3 g Mg/mole) = 72.9 g Mg
(2 mole N) (14.0 g N/mole) = 28.0 g N
100.9 g Mg3N2
72.9 g Mg + 28.0 g N = 100.9 Mg3N2
2.00 g N x 72.9 g Mg = 5.21g Mg
28.0 g N
53
Reference- Advanced Molar
Mass
• OR THE EASY WAY!!!
Magnesium + Nitrogen Magnesium nitride (Mg3N2)
?g
2.00 g
72.9 g Mg + 28.0 g N = 100.9 Mg3N2
? g Mg =
2.00 g N x 1 mole N x 3 mole Mg x 24.3 g =
14.0 g N
2 mole N
1 mole Mg
= 5.21g Mg
54
Reference- Advanced Molar
Mass
• How many grams of nitrogen are needed
to combine with 5.00 g Mg to make
magnesium nitride?
Magnesium + Nitrogen  Magnesium nitride Mg3N2
5.00 g
?g
72.9 g Mg + 28.0 g N = 100.9 Mg3N2
?gN=
5.00 g Mg x 28.0 g N = 1.92 g N
72.9 g Mg
55
Reference- Advanced Molar
Mass
• OR THE EASY WAY!!!
Magnesium + Nitrogen  Magnesium nitride Mg3N2
5.00 g
?g
?gN=
5.00 g Mg x 1 mole Mg x 2 mole N x 24.3 g =
24.3 g Mg
3 mole Mg 1 mole Mg
= 1.92 g N
56
Reference- Advanced Molar
Mass
• How many grams of Mg are needed to
make 25 g of magnesium nitride?
Magnesium + Nitrogen  Magnesium nitride – Mg3N2
?g
25g
72.9 g Mg = 28.0 g N = 100.9 Mg3N2
? g Mg=
25 g Mg3N2 x
72.9 Mg
100.9 g Mg3N
= 18 g Mg
2
57
Reference- Advanced Molar
Mass
OR THE EASY WAY!!!
• How many grams of Mg are needed to
make 25 g of magnesium nitride?
Magnesium + Nitrogen  Magnesium nitride Mg3N2
?g
25g
? g Mg =
25 g Mg3N2x 1 mole Mg3N2 x 3 mole Mg x 24.3 g =
100.9 g Mg3N2 1 mole Mg3N2 1 mole Mg
molar mass
=
18 g Mg
58
Let’s Do It!!!
• How many grams of magnesium nitride
can be made from 12.5 g of magnesium?
72.9 g Mg = 28.0 g N = 100.9 Mg3N2
Magnesium + Nitrogen  Magnesium nitride – Mg3N2
12.5 g
?g
59
Let’s Do It!!!
• How many grams of magnesium nitride
can be made from 12.5 g of magnesium?
72.9 g Mg = 28.0 g N = 100.9 Mg3N2
Magnesium + Nitrogen  Magnesium nitride – Mg3N2
12.5 g
12.5 g Mg x 100.9 g Mg3N
72.9 g Mg
?g
2
= 17.3 g Mg3N2
60
Let’s Do It!!!
OR THE EASY WAY!!!
• How many grams of magnesium nitride
can be made from 12.5 g of magnesium?
Magnesium + Nitrogen  Magnesium nitride Mg3N2
12.5g
? g Mg3N2 =
=
?g
molar mass
12.5 g Mg x 1 mole Mg x 1 mole Mg3N2 x 100.9g Mg3N2
24.0 g Mg
3 mole Mg
1 mole Mg3N2
= 17.5 g Mg3N2
61
Review
• Number your paper from 1-5 and answer the
following questions. Two will be cumulative
review!
• 1. Atoms with the same atomic number but
different mass numbers are
• a. different elements
• b. ions
• c. isotopes of the same element
• d. isotopes of different elements
62
Review
• D
• 2.An alpha particle
• a. has a negative charge
• b. can penetrate a sheet of aluminum
• c. is identical to a helium nucleus
• d. all of the above
63
Review
• C
• 3. How many moles of Cr are in a mole of
Fe2(Cr2O7)3
– a. 6
– b. 3
– c. 2
– d. 1
64
Review
• A
• 4. What is the percent composition of
hydrogen and oxygen in water?
– a. 80% O and 20% H
– b. 88.9% O and 11.2% H
– c. 75% O and 35% H
– d. 66.6% H and 33.3% O
65
Review
• B
• 1 mole O x 16.0 g O = 16.0 g 16.0/18.0 x 100
1 mole O
• 2 moles H x 1.01 g H = 2.02 2.02/18.0 x 100
1 mole H
• 5. How many grams of water can be made
from 10 grams of hydrogen and excess
oxygen?
–
–
–
–
a. 90 g
b. 10 g
c. 20 g
d. 0 g
66
Review
• A
• 5. How many grams of water can be made
from 10 grams of hydrogen? EASY WAY!!
–
–
–
–
a. 90 g
b. 10 g
c. 20 g
d. 0 g
? g H2O =
10 g H
molar mass
x 1 mole H x 1 mole H2O x 18.0 g H2O
1.01 g H
2 mole H
1 mole H2O
67
Hydrates
• Hydrated compounds- compounds with
molecules of water held in their crystal
structure
• These compounds contain an anhydrous (nonwater) part and a hydrous (water based)
part
68
Hydrates
Anhydrous compound
Water
Ex.) CaSO4 . 7H2O
• Compounds with molecules of water held in
their crystal structure
• Very common in nature
• Water can be removed by heating, leaving
behind what is called the anhydrous
compound.
69
Reference-Naming Hydrated
Compounds
• The following is tacked on the name obtained
from the ions
H2O
monohydrate 2 H2O dihydrate
3 H2O trihydrate
4 H2O tetrahydrate
5 H2O pentahydrate 6 H2O hexahydrate
7 H2O heptahydrate 8 H2O octahydrate
9 H2O nonahydrate 10 H2O dekahydrate
CaSO4  7 H2O -- named as
calcium sulfate heptahydrate70
Reference-Empirical
Formula of Hydrates
• Find the empirical formula of a
hydrate of CaSO4 hydrate that is
28.5% H2O
• To solve this problem, find the
simplest mole ratio between the
anhydrous part of the compound
(CaSO4) and the water (H2O)
• H2O =28.5 %
CaSO4 =71.5 %
(100% - 28.5%)
71
Reference-Empirical
Formula of Hydrates
71.5 g CaSO4 x 1 mole CaSO4 = .525 mole CaSO4
136.2 g CaSO4
molar mass of CaSO4
..525 = 1.00 = 1
.525
28.5 g H2O x 1 mole H2O = 1.58 mole H2O
18.0 g H2O
•When you are finding formulas
of hydrates they ALWAYS
come out even!
1.58 = 3.01 = 3
.525
CaSO4  3 H2O
72
Reference- Molar Mass of a
Hydrate
• Find the molar mass of CaSO4  7 H2O
• First we find the molar mass for H2O
and treat the water like it’s an
element!
2 mole H X 1.01 g/mole
=
2.02 g
1 mole O X 16.0 g/mole
=
16.0 g
18.0 g/mole
(MEMORIZE THIS)73
Reference- Molar Mass of a
Hydrate
CaSO4  7 H2O
1 mole Ca x
1 mole S
x
4 mole O x
7 mole H2O x
40.1 g/mole
32.1 g/mole
16.0 g/mole
18.0 g/mole
= 40.1 g
= 32.1 g
= 64.0 g
= 126 g
262 g/mole
(SD by place value)
74
Reference- % Composition of
a Hydrate
• Find the % composition of CaSO4  7 H2O
• The water is treated like an element!
% Ca =
40.1 g Ca
X 100% = 15.3 % Ca
262 g compound
%S=
32.1 g S
X 100%
262 g compound
%O=
64.0 g O
262 g compound
% H2O =
= 12.3 % S
X 100% = 24.4 % O
126 g H2O X 100% =
262 g compound
48.1 % H2O
75
Using Molar Mass or %
Composition of a Hydrate
• How much water do we get when we heat
2.00 g of CaSO4  7 H2O?
2.00 g x .481 = .962 g H2O
OR
2.00 g CaSO47 H2O x
126 g H2O
= 0.962 g H2O
262 g CaSO47 H2O
76
Review
• Number your paper from 1-5 and answer
the following questions. Two will be
cumulative review!
– 1. What is the actual mass of a
hydrogen-3 atom?
• a. 6.02 x 1023 g
• b. 1.67 x 1024 g
• c. 5.01 x 1024 g
• d. 5.01 x 10-24 g
77
Review
• D
• 2. Which of these is true about the
discovery of Millikan’s oil drop experiment
– a. He discovered the electron
– b. He discovered the mass of the
neutron
– c. He discovered the mass and the
charge of the electron
– d. He discovered the proton
78
Review
• C
• 3. What is CaSO4  5 H2O called?
– a. calcium sulfate pentahydrate
– b. calcium sulfate heptahydrate
– c. calcium sulfate hydrate
– d. calcium sulfate trihydrate
79
Review
• A
• 4. What is the mass of water in CaSO4  5
H2O?
– a. 5 g
– b. 90.0 g
– c. 5.0 g
– d. 18 g
80
Review
• B
• 5 mole H2O x 18.0 g/mole = 90.0 g
• 5. Which of these would have the most
water given off when heated?
– a. calcium sulfate pentahydrate
– b. calcium sulfate heptahydrate
– c. calcium sulfate hydrate
– d. calcium sulfate trihydrate
81
Review
• B
82