Chapter 7 Part II: Chemical Formulas and

Chapter 7 Part II: Chemical Formulas and

Chapter 7 Part II:
Chemical Formulas and
Equations
Mr. Chumbley
Chemistry 1-2
SECTION 3: USING CHEMICAL FORMULAS
Molecules and Formula Unit
• We have not yet discussed the different ways in which chemical
compounds combine and form
• We have mentioned that there are two types of compounds:
• Molecular compounds
• Ionic compounds
• Since they are physically different, the identification of a simplest
amount of each quantity is different
• Molecular compounds have a simplest amount called a molecule
• A molecule is neutral group of atoms that are held together by covalent
bonds
• Ionic compounds have a simplest unit called a formula unit
• A formula unit is the simplest collection of atoms from which an ionic
compound’s formula can be written
Formula Mass
• Similar to elements, chemical
compounds also have
characteristic average masses
• This mass can be obtained by
adding the average atomic
mass of all of the atoms
present in the simplest unit
average atomic mass of H: 1.01 u
average atomic mass of O: 16.00 u
2 H atoms ×
1.01 u
= 2.02 u
H atom
16.00 u
• The formula mass of any
1 O atom ×
= 16.00 u
O atom
molecule, formula unit, or ion
is the sum of the average
atomic masses of all atoms
average mass of H2 O molecule = 18.02 u
represented in its formula
Sample Problem F (p. 226)
Find the formula mass of potassium chlorate, KClO3.
• The mass of the formula unit of KClO3 is found by summing
the masses of one K atom, one Cl atom, and three O atoms.
39.10 u
1K atom ×
= 39.10 u
K atom
35.45 u
1 Cl atom ×
= 35.45 u
Cl atom
16.00 u
3 O atoms ×
= 48.00 u
O atom
formula mass of KClO3 = 122.55 u
Molar Mass of Compounds
• The molar mass of a
compound can be
determined by adding
the molar mass of each
of the moles of the
elements in the
compound
• A compound’s molar
mass is numerically
equal to its formula
mass
molar mass of H: 1.01 g/mol
molar mass of O: 16.00 g/mol
2 mol H ×
1.01 g H
= 2.02 g/mol
1 mol H
16.00 g O
1 mol O ×
= 16.00 g/mol
1 mol O
molar mass of H2 O = 18.02 g/mol
Sample Problem G (p. 227)
Find the molar mass of barium nitrate, Ba(NO3)2.
• The mass of the formula unit of Ba(NO3)2 is found by summing the
masses of one mole of Ba and two moles of NO3-. The two moles of
NO3- result in two moles N and six moles O.
137.33 g Ba
1mol Ba ×
= 137.33 g Ba
1 mol Ba
2 mol N ×
14.01 g N
= 28.02 g N
1 mol N
16.00 g O
6 mol O ×
= 96.00 g O
1 mol O
molar mass of Ba(NO3)2 = 261.35 g/mol
Homework:
• Practice F (p. 226)
• Practice G (p. 227)
Whiteboarding!
• Get out your homework from last night!
• Check your playing card. Get together
with the other person(s) with matching
numbers and whiteboard your solution
to the corresponding problem.
• You have 5 minute to discuss your
answers and whiteboard them before
we begin presentations
• Be sure that both of you are capable of
explaining, because the presenter will
be selected at random!!!
Problem #
Card #
F 1a
Ace
F 1b
2
F 1c
3
F 1d
4
Ga
5
Gb
6
Gc
7
Conversions using Formula
Mass and Molar Mass
• Pure elements could be converted back and forth from the
quantities of mass, amount in moles, and number of atoms
• Compounds can be converted in a similar way
• The exception is that one mole of a compound contains
Avogadro’s number of molecules or formula units
Mass of a
compound
in grams
=
×
𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝
×
𝟏 𝐦𝐨𝐥
𝟏 𝐦𝐨𝐥
=
𝐦𝐨𝐥𝐚𝐫 𝐦𝐚𝐬𝐬 𝐨𝐟 𝐜𝐨𝐦𝐩𝐨𝐮𝐧𝐝
Amount of
a
compound
in moles
=
×
𝟏 𝐦𝐨𝐥
×
𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐮𝐧𝐢𝐭𝐬
𝟔. 𝟎𝟐𝟐 × 𝟏𝟎𝟐𝟑 𝐮𝐧𝐢𝐭𝐬
=
𝟏 𝐦𝐨𝐥
Number of
molecules
or formula
units of a
compound
Sample Problem H (p. 228)
What is the mass in grams of 2.50 mol of oxygen gas?
• Step 1: Analyze
• Given: 2.50 mol of O2
• Unknown: mass of O2 in grams
• Step 2: Plan
• amount of O2 in moles  mass of O2 in grams
• To convert amount of O2 in moles to mass od O2 in grams, multiply by the
molar mass of O2
amount of O2 mol × molar mass of O2 g mol = mass of O2 (g)
Sample Problem H (p. 228)
What is the mass in grams of 2.50 mol of oxygen gas?
• Step 3: Solve
• First, find the molar mass of O2
16.00 g O
2 mol O ×
= 32.00 g mass of one mole of O2
1 mol O
• The molar mass of O2 is therefore 32.00 g/mol. Now calculate the mass of
2.50 mol of O2
32.00 g O2
2.50 mol O2 ×
= 80.0 g O2
1 mol O2
• 2.50 mol of oxygen gas has a mass of 80.0 g
Sample Problem I (p. 229-230)
Ibuprofen, C13H18O2, is the active ingredient in many nonprescription pain relievers. Its molar mass is 206.31 g/mol.
a. If the tablets in a bottle contain a total of 33g of ibuprofen,
how many moles of ibuprofen are in the bottle?
b. How many molecules of ibuprofen are in the bottle?
c. What is the total mass is grams of carbon in 33g of
ibuprofen?
Sample Problem I (p. 229-230)
a.
If the tablets in a bottle contain a total of 33g of ibuprofen, how
many moles of ibuprofen are in the bottle?
Given:
• 33 g of C13H18O2
• molar mass =
206.31 g/mol
The mass in grams can be converted to amount in moles
using the molar mass.
mass  moles
g C13H18O2
1 mol C13H18O2
mol C13H18O2
206.31 g C13H18O2
Unknown:
• moles C13H18O2
33 g C13H18O2
1 mol C13H18O2
206.31 g C13H18O2
0.16 mol C13H18O2
Sample Problem I (p. 229-230)
b. How many molecules of ibuprofen are in the bottle?
Given:
• 33 g of C13H18O2
• molar mass =
206.31 g/mol
• 0.16 mol C13H18O2
Unknown:
• molecules
C13H18O2
The number of molecules can be found using Avogadro’s
number.
moles  molecules
mol C13H18O2
6.022 × 1023
molecules
molecules C13H18O2
1 mol
0.16 mol C13H18O2
6.022 × 1023
molecules
1 mol
9.6 × 1022 molecules
C13H18O2
Sample Problem I (p. 229-230)
c.
What is the total mass is grams of carbon in 33g of ibuprofen?
Given:
• 33 g of C13H18O2
• molar mass =
206.31 g/mol
• 0.16 mol C13H18O2
• 9.6 × 1022
molecules
C13H18O2
Unknown:
• grams C
The mass in grams of carbon can be found by
determining the amount in moles of carbon in ibuprofen,
and using the molar mass of carbon.
moles C13H18O2  moles C  grams C
mol C13H18O2
0.16 mol
C13H18O2
13 mol C
12.01 g C
1 mol C13H18O2
1 mol C
13 mol C
12.01 g C
1 mol C13H18O2
1 mol C
gC
25 g C
Homework:
• Practice I (p. 230)
• Answers:
•
•
•
•
•
1a. 0.0499 mol
1b. 61 mol
2a. 1.53 × 1023 molecules
2b. 2.20 × 1023 molecules
3. 1170 g
Practice!
• Begin working on the worksheet on your desk: Molar Mass of
Compounds Practice. Feel free to use the book, any example
problems, notes, etc.
• You have until 1:00 to work individually to finish all 10
problems.
• Pair up with the person either in front of or behind you,
depending on your seat. Compare your answers.
Answers!
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Fe2O3 – 159.70 g/mol
HClO3 – 84.46 g/mol
MnO4- – 118.94 g/mol
(NH4)2SO3 – 116.16 g/mol
51.8 g NH3
9.04 × 103 g C3H8
42.2 g Ca(NO3)2
0.0596 mol SO2
0.1559 mol CaCl2
0.9398 mol CuSO4
On a half sheet of paper (share with a neighbor if you
are so inclined), solve the following problem as I pass
out papers.
• How many moles of calcium phosphate are in a 100.0 g
sample?
• Formula: Ca3(PO4)2
• Molar mass: 310.18 g/mol
100.0 g Ca3(PO4)2
1 mol Ca3(PO4)2
310.18 g Ca3(PO4)2
0.3224 mol Ca3(PO4)2
Percent Composition
• A useful way to evaluate the amount of an element is its percentage
by mass
• The percent composition of a compound is the percentage by mass
of each element in a compound
• To find the percent composition, the mass of the element in a
sample of a compound is divided by the mass of the whole sample
and multiplied by 100
• Since the mass percentage is independent of the size of the sample,
it is often calculated using a sample size of one mole
mass of element in 1 mol of a compound
× 100 = % element in compound
molar mass of compound
Sample Problem J (p. 231)
Find the percentage composition of copper (I) sulfide, Cu2S.
Given:
• Cu2S
Unknown:
• % comp. Cu2S
formula  molar mass  mass percentage of each element
2 mol Cu ×
63.55 g Cu
= 127.10 g Cu
1 mol Cu
32.07 g S
1 mol S ×
= 32.07 g S
1 mol S
molar mass Cu2S = 159.17 g/mol
% Cu =
%S=
127.10 g Cu
× 100 = 79.85% Cu
159.17 g Cu2S
32.07 g S
× 100 = 20.15% S
159.17 g Cu2S
You try it!
Find the mass percentage of water in sodium carbonate decahydrate,
Na2CO3·10H2O.
Given:
• Na2CO3·10H2O
Unknown:
• molar mass of
hydrate
• % water by mass
Molar mass of Hydrate
2 mol Na ×
22.99 g Na
= 45.98 g Cu
1 mol Na
12.01 g C
1 mol C ×
= 12.01 g C
1 mol C
13 mol O ×
16.00 g O
= 208.00 g O
1 mol O
20 mol 𝐻 ×
1.01 g H
= 20.20 g H
1 mol H
molar mass Na2CO3·10H2O
= 𝟐𝟖𝟔. 𝟏𝟗 𝐠/𝐦𝐨𝐥
formula  mass water per mole %
water
10 mol H2O ×
18.02 g H2O
= 180.2 g H2O
1 mol H2O
In each mole of Na2CO3·10H2O,
there is 180.2 g of H2O.
% H 2O =
180.2 g H2O
× 100
286.19 g Na2CO3·10H2O
= 𝟔𝟐. 𝟗𝟕% 𝐇𝟐𝐎
SECTION 4: DETERMINING CHEMICAL
FORMULAS
Empirical Formulas
• Substances can be analyzed quantitatively to determine the
chemical formula of the quantity
• An empirical formula consists of the symbols for the elements
combined in a compound with subscripts showing the smallest
whole-number mole ratio of the different atoms in the
compound
• This is often times done with percentage composition, but can
also be done using known masses of an element within the
compound
Sample Problem L (p. 234)
Quantitative analysis shows that a compound contains 32.38%
sodium, 22.65% sulfur, and 44.99% oxygen. Find the empirical
formula of this compound.
Given:
• 32.38% Na
• 22.65% S
• 44.99% O
Unknown:
• NaxSxOx
percent comp.  mass comp.  molar comp.  empirical formula
32.38 g Na ×
1 mol Na
= 1.408 mol Na
22.99 g Na
22.65 g S ×
1 mol S
= 0.7063 mol S
32.07 g S
44.99 g O ×
1 mol O
= 2.812 mol O
16.00 g O
divide each molar amount by the smallest number
1.408 mol N 0.7063 mol S 2.812 mol O
:
:
0.7063
0.7063
0.7063
1.993 mol Na ∶ 1 mol S ∶ 3.981 mol O
The empirical formula of the compound is Na2SO4
Sample Problem M (p. 235)
Analysis of a 10.150 g sample of a compound known to contain only
phosphorus and oxygen indicates a phosphorus content of 4.433 g.
What is the empirical formula of the compound?
Given:
• 10.150 g PxOx
• 4.433f P
Unknown:
• PxOx
Molecular Formula
• While the empirical formula frequently provides a compound’s
chemical formula, it is not always the case
• Some molecular compounds share an empirical formula, but
have a different chemical formula that makes up a molecule of
that compound
• A molecular formula consists of the symbols for the elements
combined in a compound with subscripts showing the actual
number of the different atoms in the compound
Empirical versus Molecular
Empirical Formula
CH2
(85.6% C, 14.4% H)
Compound
Molecular Formula
methylene
CH2
ethene
C2H4
cyclopropane
C3H6
butene
C4H8
cyclohexane
C6H12
formaldehyde
CH2O
acetic acid
CH2O
(40.0% C, 6.7% H, 53.3% O) glyceraldehyde
glucose
C2H4O
C3H6O3
C6H12O6
Molecular Formula
• Molecular formulas will always be whole number multiples of
empirical formulas
• Since the formulas are whole number multiples, the molecular
formula mass is the same whole number multiple of the
empirical formula mass
𝑥 empirical formula mass = molecular formula mass
• This shows that multiple compounds can have the same
empirical formula but can be distinguished based on the mass
of the different molecular formulas
Sample Problem N (p. 236-237)
The empirical formula of a compound of phosphorus and oxygen was found
to be P2O5. Experimentation shows that the molar mass of this compound is
283.89 g/mol. What is the compound’s molecular formula?
Given:
• Empirical formula
P2O5
• Molar mass =
289.83 g/mol
Unknown:
• Molecular
formula
𝑥 empirical formula mass = molecular formula mass
molecular molar mass = 283.89 g mol
molecular formula mass = 283.89 u
determine the empirical formula mass
2 P atoms ×
5 O atom ×
30.97 u
= 61.94 u
1 P atom
16.00 u
= 80.00 g u
1 O atom
formula mass = 141.94 u
283.89 u
= 2.0001
141.94 u
The compound’s molecular formula is 2 times its empirical
formula, P4O10