determination of an empirical formula
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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 – ChargesEX: Write the empirical formula
for the compound formed byNa & P c. K &
Ne
Sr & Cl d. Cu & Cl
No compoundNa3P
SrCl2 CuCl or CuCl23
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 in the empirical formula
4
Reference-Empirical Formulas 7.06 g of silver combine with an excess of fluorine to produce 8.30 g of a compound
Silver + Fluorine Ag?F?
7.06g 1.24g 8.30g7.06 g Ag
1.24 g F
X
X
= .0654 moles Ag
= .0653 moles F
= 1.00 or 1
= 1.00 or 1
Found by subtracting!
Ag1F1
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
35.81 g Mn
40.50 g O
X
X
X
= .629 mole K
= .652 moles Mn
= 2.53 moles O
= 1.00 or 1
= 1.04 or 1
= 4.02 or 4
ANSWER = KMnO46
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 numbers 8
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 5.72gFound by subtracting!
4.35 g Zn
1.37 g P
X
X
= .0665 moles Zn
= .0442 moles P
= 1.50
= 1.00
Not within .05 of a whole number
X 2 = 3.00 or 3
X 2 = 2.00 or 2
ANSWER = Zn3P2
9
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 FeX —————— 55.8 g Fe
= 1.30 mole Fe
27.7g O 1 mole O X —————— 16.0 g O
= 1.73 mole O
1.30 mole 1.30 mole
1.73 mole 1.30 mole
= 1.00
=1.33
X 3 = 3.00 = 3
X 3 = 3.99 = 4
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 Cl 35
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
can’t be rounded to 2 14
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 potassium4 carbon5 hydrogen6 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 x 6.02 x1023 formula unit KC4H5O6 1 formula unit 1 mole KC4H5O6 KC4H5O63.61 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 wheels1 dozen bicycles = 2 dozen wheels
1 mole bicycles = 2 mole wheels23
How many dozen hydrogen atoms are in 1 dozen water molecules?
1 water = 2 hydrogen1 dozen water = 2 dozen hydrogen1 mole water = 2 mole hydrogen
H2O24
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 16.0 g O = 96.0 g O1 mole KC4H5O6 1 mole O mole
KC4H5O6 26
Let’s Do It!!!• How many atoms in one formula
unit?Magnesium Acetate: Mg(C2H3O2)2
1-Mg4-C6-H4-O
15 total atoms 27
Let’s Do It!!!• How many formula units in one
mole ofMagnesium 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 Oxygen31
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 mole32
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 (for SD use place
value!)
34
Reference- Molar MassCalculate the molar mass (grams in
one mole) of ammonium sulfite, (NH4)2SO3
In 1 mole of the compound there are: 2 moles of N X 14.0 g N/mole= 28.0g N 8 moles of H X 1.01 g H/mole = 8.08gH 1 mole of S X 32.1 g S/mole =32.1 g S 3 moles of O X 16.0 g
O/mole=48.0g O SD by place value 116.2gunits always g/mole 1mole
(NH4)2SO3
35
• 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
Molar Mass
36
• 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
Reference- Molar Mass as a Conversion Factor
37
• 75.2 g (NH4)2SO3 is how many moles of (NH4)2SO3?
? moles (NH4)2SO3 =
1 mole (NH4)2SO3
75.2 g (NH4)2SO3 X ---------------------- 116.2 g (NH4)2SO3
= .647 moles (NH4)2SO3
Reference- Molar Mass as a Conversion Factor
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)32 moles Al
3 moles H3 moles P12 moles O
(27.0g Al/mole) =(1.01g H/mole) =(31.0 g P/mole) =(16.0 g O/mole)=
54.0 g Al3.03 g H93.0 g P192 g O_
342 gmole Al2(HPO4)3MASS OF ONE MOLE!!!
So 342 g Al2(HPO4)3 = 1 mole Al2(HPO4)3 40
Let’s Do It!!!!• Then find the number of moles in
87.4 g of aluminum monohydrogen phosphate• 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
Obviously, this is less than one mole!
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
ReviewA• 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)(6 moles C)(9 moles H)(6 moles O)
(27.0 g/mole) =(12.0 g/mole) =(1.01 g/mole) =(16.0 g/mole) =
27.0 g Al72.0 g C9.09 g H96.0 g O
204.1 g Al(C2H3O2)3/204.1 g Al(C2H3O2)3 x 100 =
/204.1 g Al(C2H3O2)3 x 100 =/204.1 g Al(C2H3O2)3 x 100 =/204.1 g Al(C2H3O2)3 x 100 =
13.2 %35.3 %4.45 %47.0 %~100%
27.0 g Al72.0 g C9.09 g H96.0 g O 49
% 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 X 𝟐𝟕 .𝟎𝒈 𝑨𝒍
𝟐𝟎𝟒 .𝟏𝒈 𝑨𝒍 (𝑪𝟐𝑯𝟑𝑶𝟐 )𝟑
= 40.5 g Al
1.50 moles Al(C2H3O2)3 x X 27.0 g Al 1 mole Al
= 40.4 g Al
OR
50
Let’s Do It!!!• Find the percent composition of iron
(III) dichromate, Fe2(Cr2O7)3
(2 moles Fe)(6 moles Cr)(21 moles O)
(55.8 g/mole) =(52.0 g/mole) =(16.0 g/mole) =
112 g Fe312 g Cr336 g O
76Ō g Fe2(Cr2O7)3
/76Ōg Fe2(Cr2O7)3 x 100 =/76Ōg Fe2(Cr2O7)3 x 100 =/76Ōg Fe2(Cr2O7)3 x 100 =
14.7 %41.1 %44.2 %~100%
112 g Fe312 g Cr336 g O
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 X
X = 1.38 moles Fe
525 g Fe2(Cr2O7)3 X 112 g Fe 76Ōg
X = 1.39 moles Fe
OR
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)(2 mole N)
(24.3 g Mg/mole) =(14.0 g N/mole) =
72.9 g Mg28.0 g N
100.9 g Mg3N2
2.00 g N x 72.9 g Mg 28.0 g N
= 5.21g Mg
72.9 g Mg + 28.0 g N = 100.9 Mg3N2
53
Reference- Advanced Molar Mass
• OR THE EASY WAY!!! Magnesium + Nitrogen Magnesium nitride (Mg3N2) ?g 2.00 g
? 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
72.9 g Mg + 28.0 g N = 100.9 Mg3N2
54
• 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
? g N =5.00 g Mg x 28.0 g N 72.9 g Mg
= 1.92 g N
Reference- Advanced Molar Mass
72.9 g Mg + 28.0 g N = 100.9 Mg3N2
55
• OR THE EASY WAY!!! Magnesium + Nitrogen Magnesium nitride Mg3N2
5.00 g ?g ? g N =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
Reference- Advanced Molar Mass
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 25g72.9 g Mg = 28.0 g N = 100.9 Mg3N2
? g Mg=
25 g Mg3N2 x 72.9 Mg = 18 g Mg 100.9 g Mg3N 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?
Magnesium + Nitrogen Magnesium nitride – Mg3N2
12.5 g ?g
72.9 g Mg = 28.0 g N = 100.9 Mg3N2
59
Let’s Do It!!!• How many grams of magnesium
nitride can be made from 12.5 g of magnesium?
Magnesium + Nitrogen Magnesium nitride – Mg3N2
12.5 g ?g
72.9 g Mg = 28.0 g N = 100.9 Mg3N2
12.5 g Mg x 100.9 g Mg3N 2
72.9 g Mg
= 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
? g Mg3N2 = 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
• 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
Review
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 = molar mass 10 g H 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 (non-water) part and a hydrous (water based) part
68
Hydrates
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.
Anhydrous compound Wate
r
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 heptahydrate
70
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
28.5 g H2O x 1 mole H2O = 1.58 mole H2O
18.0 g H2O
..525 = 1.00 = 1 .525
1.58 = 3.01 = 3 .525CaSO4 3 H2O
•When you are finding formulas of hydrates they ALWAYS come out even!
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 40.1 g/mole =
40.1 g1 mole S x 32.1 g/mole =
32.1 g4 mole O x 16.0 g/mole =
64.0 g7 mole H2O x 18.0 g/mole = 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% = 12.3 % S 262 g compound
% O = 64.0 g O X 100% = 24.4 % O 262 g compound
% H2O = 126 g H2O X 100% = 48.1 % H2O 262 g compound
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 CaSO47 H2O x 126 g H2O = 0.962 g H2O
262 g CaSO47 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
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