Writing Formulas Oxidation Numbers & Chemical Formulas Mrs. Wolfe.
Chapter 3 Calculations with Chemical Formulas and...
Transcript of Chapter 3 Calculations with Chemical Formulas and...
Chapter 3
Calculations with
Chemical
Formulas and
Equations
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Contents and Concepts
Mass and Moles of Substances
Here we will establish a critical relationship
between the mass of a chemical substance and
the quantity of that substance (in moles).
1. Molecular Mass and Formula Mass
2. The Mole Concept
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Determining Chemical Formulas
Explore how the percentage composition and
mass percentage of the elements in a chemical
substance can be used to determine the
chemical formula.
3. Mass Percentages from the Formula
4. Elemental Analysis: Percentages of C, H, and O
5. Determining Formulas
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Stoichiometry: Quantitative Relations in
Chemical Reactions
Develop a molar interpretation of chemical
equations, which then allows for calculation
of the quantities of reactants and products.
6. Molar Interpretation of a Chemical Equation
7. Amounts of Substances in a Chemical Equation
8. Limiting Reactant: Theoretical and Percentage
Yield
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Learning Objectives
Mass and Moles of Substances
1. Molecular Mass and Formula Mass
a. Define the terms molecular and formula mass of a substance.
b. Calculate the formula mass from a formula.
c. Calculate the formula mass from molecular models.
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2. The Mole Concept
a. Define the quantity called the mole.
b. Learn Avogadro’s number.
c. Understand how the molar mass is related
to the formula mass of a substance.
d. Calculate the mass of atoms and molecules.
e. Perform calculations using the mole.
f. Convert from moles of substance to grams
of substance.
g. Convert from grams of substance to moles
of substance.
h. Calculate the number of molecules in a
given mass of a substance.
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Determining Chemical Formulas
3. Mass Percentages from the Formula
a. Define mass percentage.
b. Calculate the percentage composition of
the elements in a compound.
c. Calculate the mass of an element in a
given mass of compound.
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4. Elemental Analysis: Percentages of C, H, and O
a. Describe how C, H, and O combustion
analysis is performed.
b. Calculate the percentage of C, H, and O
from combustion data.
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5. Determining Formulas
a. Define empirical formula.
b. Determine the empirical formula of a binary
compound from the masses of its elements.
c. Determine the empirical formula from the
percentage composition.
d. Understand the relationship between the
molecular mass of a substance and its
empirical formula mass.
e. Determine the molecular formula from the
percentage composition and molecular
mass.
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Stoichiometry: Quantitative Relations in
Chemical Reactions
6. Molar Interpretation of a Chemical Equation
a. Relate the coefficients in a balanced
chemical equation to the number of
molecules or moles (molar interpretation).
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7. Amounts of Substances in a Chemical Equation
a. Use the coefficients in a balanced chemical
equation to perform calculations.
b. Relate the quantities of reactant to the
quantity of product.
c. Relate the quantities of two reactants or two
products.
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8. Limiting Reactant: Theoretical and Percentage Yield
a. Understand how a limiting reactant determines how many moles of product are formed during a chemical reaction and how much excess reactant remains.
b. Calculate with a limiting reactant involving moles.
c. Calculate with a limiting reactant involving masses.
d. Define and calculate the theoretical yield of chemical reactions.
e. Determine the percentage yield of a chemical reaction.
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Molecular Mass
The sum of the atomic masses of all the atoms in a
molecule of the substance.
Formula Mass
The sum of the atomic masses of all atoms in a
formula unit of the compound, whether molecular
or not.
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Calculate the formula weight of the
following compounds from their
formulas. Report your answers to three
significant figures.
calcium hydroxide, Ca(OH)2
methylamine, CH3NH2
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3 significant figures
74.1 amu
3 significant figures
31.1 amu
Total 74.095
2 O 2(16.00) = 32.00 amu
Ca(OH)21 Ca 1(40.08) = 40.08 amu
2 H 2(1.008) = 2.016 amu
CH3NH21 C 1(12.01) = 12.01 amu
5 H 5(1.008) = 5.040 amu
1 N 1(14.01) = 14.01 amu
Total 31.060
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What is the mass in grams of the nitric
acid molecule, HNO3?
First, find the molar mass of HNO3:
1 H 1(1.008) = 1.008
1 N 1(14.01) = 14.01
3 O 3(16.00) = 48.00
63.018 (2 decimal places)
63.02 g/mol
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Mole, mol
The quantity of a given amount of substance that
contains as many molecules or formula units as
the number of atoms in exactly 12 g of carbon-12.
Avogadro’s Number, NA
The number of atoms in exactly 12 g of carbon-12
NA = 6.02 × 1023 (to three significant figures).
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molecules10x6.02
mol1 x
mol
g63.0223
g10x41.04684385 22��������
figures)tsignifican(3
g10x1.05ismoleculeHNOoneofmassThe 223 .����
Next, convert this mass of one mole to one
molecule using Avogadro’s number:
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Molar Mass
The mass of one mole of substance.
For example:
Carbon-12 has a molar mass of 12 g or 12 g/mol
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A sample of nitric acid, HNO3, contains
0.253 mol HNO3. How many grams is
this?
First, find the molar mass of HNO3:
1 H 1(1.008) = 1.008
1 N 1(14.01) = 14.01
3 O 3(16.00) = 48.00
63.018 (2 decimal places)
63.02 g/mol
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mole1
g63.02xmole0.253
figures)tsignifican(3
g15.9����
Next, using the molar mass, find the mass of 0.253
mole:
= 15.94406 g
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Calcite is a mineral composed of
calcium carbonate, CaCO3. A sample
of calcite composed of pure calcium
carbonate weighs 23.6 g. How many
moles of calcium carbonate is this?
First, find the molar mass of CaCO3:
1 Ca 1(40.08) = 40.08
1 C 1(12.01) = 12.01
3 O 3(16.00) = 48.00
100.09 2 decimal places
100.09 g/mol
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g100.09
mole1xg23.6
g10x2.35787791 1��������
figures)tsignifican(3
g0.236org10x2.36 1��������
Next, find the number of moles in 23.6 g:
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The average daily requirement of the
essential amino acid leucine, C6H14O2N,
is 2.2 g for an adult. What is the average
daily requirement of leucine in moles?
First, find the molar mass of leucine:
6 C 6(12.01) = 72.06
2 O 2(16.00) = 32.00
1 N 1(14.01) = 14.01
14 H 14(1.008) = 14.112
132.182
2 decimal places
132.18 g/mol
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g132.18
mole1xg2.2
mol10x1.6643 2��������
figures)tsignifican(2
mol0.017ormol10x1.7 2��������
Next, find the number of moles in 2.2 g:
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The daily requirement of chromium in
the human diet is 1.0 × 10-6 g. How
many atoms of chromium does this
represent?
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mol1
atoms10x6.02x
g51.996
mol1xg10x1.0
236����
First, find the molar mass of Cr:
1 Cr 1(51.996) = 51.996
Now, convert 1.0 x 10-6 grams to moles:
=1.157781368 x 1016 atoms
1.2 x 1016 atoms
(2 significant figures)
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Lead(II) chromate, PbCrO4, is used as
a paint pigment (chrome yellow). What
is the percentage composition of
lead(II) chromate?
First, find the molar mass of PbCrO4:
1 Pb 1(207.2) = 207.2
1 Cr 1(51.996) = 51.996
4 O 4(16.00) = 64.00
323.196 (1 decimal place)
323.2 g/mol
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64.11%100%xg323.20
g207.2:Pb ����
16.09%100%xg323.20
g51.996:Cr ����
19.80%100%xg323.20
g64.00:O ����
Now, convert each to percent composition:
Check:
64.11 + 16.09 + 19.80 = 100.00
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The chemical name of table sugar is
sucrose, C12H22O11. How many grams
of carbon are in 68.1 g of sucrose.
First, find the molar mass of C12H22O11:
12 C 12(12.01) = 144.12
11 O 11(16.00) = 176.00
22 H 22(1.008) = 22.176
342.296 (2 decimal places)
342.30 g/mol
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figures)tsignifican(3
carbong26.0����
sucroseg342.30
carbong144.12xsucroseg61.8
Now, find the mass of carbon in 61.8 g sucrose:
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Percentage Composition
The mass percentage of each element in the
compound.
The composition is determined by experiment,
often by combustion. When a compound is burned,
its component elements form oxides—for example,
CO2 and H2O. The CO2 and H2O are captured and
weighed to determine the amount of C and H in
the original compound.
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Benzene is a liquid compound
composed of carbon and hydrogen; it is
used in the preparation of polystyrene
plastic. A sample of benzene weighing
342 mg is burned in oxygen and forms
1156 mg of carbon dioxide. What is the
percentage composition of benzene?
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Strategy
1. Use the mass of CO2 to find the mass of carbon
from the benzene.
2. Use the mass of benzene and the mass of
carbon to find the mass of hydrogen.
3. Use these two masses to find the percent
composition.
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g10
mg1Cg10x23.15463758
3
1
����
��������
222
22
3
COmol1
Cg12.01x
COmol1
Cmol1x
COg44.01
COmol1COg10x1156 ����
= 315.5 mg C
First, find the mass of C in 1156 mg of CO2:
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Now, we can find the percentage composition:
����x100%mg342
mg26.5
����x100%mg342
mg315.5C92.3%
H7.7%
342 mg benzene
-315.5 mg C
26.5 mg H
(the decimal is not significant)
Next, find the mass of H in the benzene sample:
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Empirical Formula (Simplest Formula)
The formula of a substance written with the
smallest integer subscripts.
For example:
The empirical formula for N2O4 is NO2.
The empirical formula for H2O2 is HO
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Determining the Empirical Formula
Beginning with percent composition:
1. Assume exactly 100 g so percentages convert
directly to grams.
2. Convert grams to moles for each element.
3. Manipulate the resulting mole ratios to obtain
whole numbers.
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Manipulating the ratios:
Divide each mole amount by the smallest mole
amount.
If the result is not a whole number:
Multiply each mole amount by a factor.
For example:
If the decimal portion is 0.5, multiply by 2.
If the decimal portion is 0.33 or 0.67, multiply by 3.
If the decimal portion is 0.25 or 0.75, multiply by 4.
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Benzene is composed of 92.3% carbon
and 7.7% hydrogen. What is the
empirical formula of benzene?
Hmol7.64Hg1.008
Hmol1Hg7.7
Cmol7.685Cg12.01
Cmol1Cg92.3
����
����
17.64
7.64
17.64
7.685
����
����
Empirical formula: CH
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Molecular Formula
A formula for a molecule in which the subscripts
are whole-number multiples of the subscripts in the
empirical formula.
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To determine the molecular formula:
1. Compute the empirical formula weight.
2. Find the ratio of the molecular weight to the
empirical formula weight.
3. Multiply each subscript of the empirical formula
by n.
weightformulaempirical
weightmolecular����n
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Benzene has the empirical formula CH.
Its molecular weight is 78.1 amu. What
is its molecular formula?
613.02
78.1
amu13.02weightformulaEmpirical
����
����
Molecular formula
C6H6
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Sodium pyrophosphate is used in
detergent preparations. It is composed
of 34.5% Na, 23.3% P, and 42.1% O.
What is its empirical formula?
Omol2.631Og16.00
Omol1Og42.1
Pmol0.7523Pg30.97
Pmol1Pg23.3
Namol1.501Nag22.99
Namol1Nag34.5
����
����
Empirical formula
Na4P2O7
72x
22x
42x
����
����
����
3.500.7523
2.631
1.000.7523
0.7523
2.000.7523
1.501
����
����
����
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Hexamethylene is one of the materials
used to produce a type of nylon. It is
composed of 62.1% C, 13.8% H, and
24.1% N. Its molecular weight is 116
amu. What is its molecular formula?
Hmol1.720Ng14.01
Nmol1Ng24.1
Hmol13.69Hg1.008
Hmol1Hg13.8
Cmol5.171Cg12.01
Cmol1Cg62.1
����
����
����
11.720
1.720
81.720
13.69
31.720
5.171
����
����
����
Empirical formula
C3H8N
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The empirical formula is C3H8N.
Find the empirical formula weight:
3(12.01) + 8(1.008) + 1(14.01) = 58.104 amu
Molecular formula: C6H16N2
258.10
116��������n
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Stoichiometry
The calculation of the quantities of reactants and
products involved in a chemical reaction.
Interpreting a Chemical Equation
The coefficients of the balanced chemical equation
may be interpreted in terms of either (1) numbers
of molecules (or ions or formula units) or (2)
numbers of moles, depending on your needs.
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1. Convert grams of A to moles of A
� Using the molar mass of A
2. Convert moles of A to moles of B
� Using the coefficients of the
balanced chemical equation
3. Convert moles of B to grams of B
� Using the molar mass of B
To find the amount of B (one reactant or product)
given the amount of A (another reactant or
product):
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Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:
C3H8(g) + 5O2(g) � 3CO2(g) + 4H2O(g)
How many grams of CO2 are produced when 20.0 g of propane is burned?
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Molar masses
C3H8: 3(12.01) + 8(1.008) = 44.094 g
CO2: 1(12.01) + 2(16.00) = 44.01 g
59.9 g CO2
(3 significant figures)
2
2
83
2
83
8383
COmol1
COg44.01
HCmol1
COmol3
HCg44.094
HCmol1HCg20.0
2COg359.8856987����
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Propane, C3H8, is normally a gas, but it is sold as a fuel compressed as a liquid in steel cylinders. The gas burns according to the following equation:
C3H8(g) + 5O2(g) � 3CO2(g) + 4H2O(g)
How many grams of O2 are required to burn 20.0 g of propane?
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Molar masses:
O2 2(16.00) = 32.00 g
C3H8 3(12.01) + 8(1.008) = 44.094 g
2
2
83
2
83
8383
Omol1
Og32.00
HCmol1
Omol5
HCg44.094
HCmol1HCg20.0
2Og572.5722320����
72.6 g O2
(3 significant figures)
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Limiting Reactant
The reactant that is entirely consumed when a
reaction goes to completion.
Once one reactant has been completely
consumed, the reaction stops.
Any problem giving the starting amount for more
than one reactant is a limiting reactant problem.
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All amounts produced and reacted are
determined by the limiting reactant.
How can we determine the limiting reactant?
1. Use each given amount to calculate the
amount of product produced.
2. The limiting reactant will produce the
lesser or least amount of product.
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Magnesium metal is used to prepare
zirconium metal, which is used to make
the container for nuclear fuel (the nuclear
fuel rods):
ZrCl4(g) + 2Mg(s) � 2MgCl2(s) + Zr(s)
How many moles of zirconium metal can
be produced from a reaction mixture
containing 0.20 mol ZrCl4 and 0.50 mol
Mg?
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ZrCl4 is the limiting reactant.
0.20 mol Zr will be produced.
Zrmol0.25Mgmol2
Zrmol1Mgmol0.50 ����
Zrmol0.20ZrClmol1
Zrmol1ZrClmol0.20
4
4 ����
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Urea, CH4N2O, is used as a nitrogen
fertilizer. It is manufactured from
ammonia and carbon dioxide at high
pressure and high temperature:
2NH3 + CO2(g) � CH4N2O + H2O
In a laboratory experiment, 10.0 g NH3
and 10.0 g CO2 were added to a reaction
vessel. What is the maximum quantity (in
grams) of urea that can be obtained?
How many grams of the excess reactant
are left at the end of the reactions?
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Molar masses
NH3 1(14.01) + 3(1.008) = 17.02 g
CO2 1(12.01) + 2(16.00) = 44.01 g
CH4N2O 1(12.01) + 4(1.008) + 2(14.01) +
1(16.00) = 60.06 g
CO2 is the limiting reactant.
13.6 g CH4N2O will be produced.
ONCHg13.6
ONCHmol1
ONCHg60.06
COmol1
ONCHmol1
COg44.01
COmol1COg10.0
24
24
24
2
24
2
22
����
ONCHg17.6
ONCHmol1
ONCHg60.06
NHmol2
ONCHmol1
NHg17.024
NHmol1NHg10.0
24
24
24
3
24
3
33
����
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10.0 at start
-7.73 reacted
2.27 g remains
2.3 g NH3 is left unreacted.
(1 decimal place)
reactedNHg7.73 3����
3
3
2
3
2
22
NHmol1
NHg17.02
COmol1
NHmol2
COg44.01
COmol1COg10.0
3NHg17.73460577����
To find the excess NH3, we find how much NH3
reacted:
Now subtract the amount reacted from the
starting amount:
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Theoretical Yield
The maximum amount of product that can be
obtained by a reaction from given amounts of
reactants. This is a calculated amount.
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Actual Yield
The amount of product that is actually obtained.
This is a measured amount.
Percentage Yield
x100%yieldltheoretica
yieldactualyieldpercentage ����
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2NH3 + CO2(g) � CH4N2O + H2O
When 10.0 g NH3 and 10.0 g CO2 are
added to a reaction vessel, the limiting
reactant is CO2. The theoretical yield is
13.6 of urea. When this reaction was
carried out, 9.3 g of urea was obtained.
What is the percent yield?
Theoretical yield = 13.6 g
Actual yield = 9.3 g
= 68% yield
(2 significant figures)
100%x g13.6
g9.3