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nomenclature
Empirical formula
Molecular formula
Types of Chemical Formulas
A chemical formula is comprised of element symbols and numerical
subscripts that show the type and number of each atom present in the
smallest unit of the substance.
An empirical formula indicates the relative number of atoms of
each element in the compound. It is the simplest type of formula.
The empirical formula for hydrogen peroxide is HO.
A molecular formula shows the actual number of atoms of
each element in a molecule of the compound.
The molecular formula for hydrogen peroxide is H2O2.
A structural formula shows the number of atoms and the
bonds between them, that is, the relative placement and
connections of atoms in the molecule.
The structural formula for hydrogen peroxide is H-O-O-H.
Empirical and Molecular Formulas
Empirical Formula The simplest formula for a compound that agrees with
the elemental analysis and gives rise to the smallest set
of whole numbers of atoms.
Molecular Formula The formula of the compound as it exists, it may be a
multiple of the empirical formula.
Determining the Empirical Formula from Masses
of Elements
PROBLEM: Elemental analysis of a sample of an ionic compound showed
2.82 g of Na, 4.35 g of Cl, and 7.83 g of O. What are the
empirical formula and name of the compound?
PLAN:
Once we find the relative number of moles of each element,
we can divide by the lowest mol amount to find the relative
mol ratios (empirical formula).
SOLUTION: 2.82 g Na
mass(g) of each element
divide by M(g/mol)
4.35 g Cl
amount(mol) of each element
use # of moles as subscripts
preliminary formula
change to integer subscripts
empirical formula
7.83 g O
mol Na
22.99 g Na
mol Cl
35.45 g Cl
mol O
16.00 g O
Na1 Cl1 O3.98
= 0.123 mol Na
= 0.123 mol Cl
= 0.489 mol O
NaClO4
NaClO4 is sodium perchlorate.
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
PROBLEM: During physical activity. lactic acid (M=90.08 g/mol) forms in
muscle tissue and is responsible for muscle soreness.
Elemental anaylsis shows that this compound contains 40.0
mass% C, 6.71 mass% H, and 53.3 mass% O.
(a) Determine the empirical formula of lactic acid.
(b) Determine the molecular formula.
PLAN:
assume 100g lactic acid and find the
mass of each element
divide each mass by mol mass(M)
amount(mol) of each element
molecular formula
use # mols as subscripts
preliminary formula
convert to integer subscripts
empirical formula
divide mol mass by
mass of empirical
formula to get a
multiplier
Determining a Molecular Formula from Elemental
Analysis and Molar Mass
continued
SOLUTION:
Assuming there are 100. g of lactic acid, the constituents are
40.0 g C mol C
12.01g C
3.33 mol C
C3.33
3.33
6.71 g H mol H
53.3 g O mol O
1.008 g H
16.00 g O
6.66 mol H
3.33 mol O
H6.66 O3.33
3.33 3.33
molar mass of lactate
CH2O
empirical formula
90.08 g
3
mass of CH2O
30.03 g
C3H6O3 is the
molecular formula
Combustion train for the determination of the
chemical composition of organic compounds.
m
m
CnHm + (n+ ) O2 = n CO(g) +
H O(g)
2
2 2
Determining a Molecular Formula from Combustion
Analysis
PROBLEM:
Vitamin C (M=176.12g/mol) is a compound of C,H, and O
found in many natural sources especially citrus fruits. When a
1.000-g sample of vitamin C is placed in a combustion chamber
and burned, the following data are obtained:
mass of CO2 absorber after combustion
=85.35g
mass of CO2 absorber before combustion
=83.85g
mass of H2O absorber after combustion
=37.96g
mass of H2O absorber before combustion
=37.55g
What is the molecular formula of vitamin C?
PLAN:
difference (after-before) = mass of oxidized element
find the mass of each element in its combustion product
find the mols
preliminary
formula
empirical
formula
molecular
formula
Determining a Molecular Formula from Combustion
Analysis
continued
SOLUTION:
CO2
H2O
85.35 g-83.85 g = 1.50 g
There are 12.01 g C per mol CO2
1.50 g CO2
37.96 g-37.55 g = 0.41 g
12.01 g CO2
= 0.409 g C
44.01 g CO2
There are 2.016 g H per mol H2O. 0.41 g H2O 2.016 g H2O
18.02 g H2O
O must be the difference:
0.409 g C
= 0.0341 mol C
12.01 g C
C1H1.3O1
1.000 g - (0.409 + 0.046) = 0.545
0.046 g H
= 0.0456 mol H
1.008 g H
C3H4O3
= 0.046 g H
176.12 g/mol
88.06 g
0.545 g O
= 0.0341 mol O
16.00 g O
= 2.000
C6H8O6
Table 3.3 Some Compounds with Empirical Formula CH2O
(Composition by Mass: 40.0% C, 6.71% H, 53.3% O)
M
(g/mol)
Molecular
Formula
Whole-Number
Multiple
formaldehyde
CH2O
1
30.03
disinfectant; biological preservative
acetic acid
C2H4O2
2
60.05
acetate polymers; vinegar(5% soln)
lactic acid
C3H6O3
3
90.09
sour milk; forms in exercising muscle
erythrose
C4H8O4
4
120.10
part of sugar metabolism
ribose
C5H10O5
5
150.13
component of nucleic acids and B2
glucose
C6H12O6
6
180.16
major energy source of the cell
Name
CH2O
C2H4O2
C3H6O3
C4H8O4
Use or Function
C5H10O5
C6H12O6
Table 3.4 Two Pairs of Constitutional Isomers
Property
Butane
2-Methylpropane
Ethanol
Dimethyl Ether
M(g/mol)
58.12
58.12
46.07
46.07
Boiling Point
-0.50C
-11.060C
78.50C
-250C
Density at 200C 0.579 g/mL 0.549 g/mL
(gas)
(gas)
Structural
formulas
H
H
H
H
H
H
C
C
C
C
H
H
H
H
H
C
C
C
C
H
H
H
H
H
H
H
H
Space-filling
models
0.789 g/mL 0.00195 g/mL
(liquid)
(gas)
H
H
C
C
H
H
H
H
H
H
OH H
C
H
O
C
H
H
The formation of HF gas on the macroscopic and molecular levels.
A three-level view of the chemical reaction in a flashbulb.
translate the statement
balance the atoms
adjust the coefficients
check the atom balance
specify states of matter
Balancing Chemical Equations
PROBLEM:
Within the cylinders of a car’s engine, the hydrocarbon octane
(C8H18), one of many components of gasoline, mixes with oxygen
from the air and burns to form carbon dioxide and water vapor.
Write a balanced equation for this reaction.
PLAN:
SOLUTION:
translate the statement
C8H18 +
O2
balance the atoms
C8H18 + 25/2 O2
H2O
CO2 +
H2O
8 CO2 +
9
adjust the coefficients
2C8H18 + 25O2
16CO2 + 18H2O
check the atom balance
2C8H18 + 25O2
16CO2 + 18H2O
specify states of matter
2C8H18(l) + 25O2 (g)
16CO2 (g) + 18H2O (g)
Summary of the mass-mole-number relationships
in a chemical reaction.
MASS(g)
of compound A
MASS(g)
of compound B
M (g/mol) of
compound A
AMOUNT(mol)
of compound A
molar ratio from
balanced equation
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound A
M (g/mol) of
compound B
AMOUNT(mol)
of compound B
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound B
Calculating Amounts of Reactants and Products
PROBLEM:
In a lifetime, the average American uses 1750 lb(794 g) of
copper in coins, plumbing, and wiring. Copper is obtained from
sulfide ores, such as chalcocite, or copper(I) sulfide, by a
multistep process. After an initial grinding, the first step is to
“roast” the ore (heat it strongly with oxygen gas) to form
powdered copper(I) oxide and gaseous sulfur dioxide.
(a) How many moles of oxygen are required to roast 10.0 mol of
copper(I) sulfide?
(b) How many grams of sulfur dioxide are formed when 10.0 mol
of copper(I) sulfide is roasted?
(c) How many kilograms of oxygen are required to form 2.86 kg
of copper(I) oxide?
write and balance equation
PLAN:
find mols O2
find mols SO2
find g SO2
find mols Cu2O
find mols O2
find kg O2
Calculating Amounts of Reactants and Products
continued
SOLUTION:
2Cu2S(s) + 3O2(g)
2Cu2O(s) + 2SO2(g)
(a) How many moles of oxygen are required to roast 10.0 mol of copper(I) sulfide?
(a)
10.0mol Cu2S
3mol O2
= 15.0mol O2
2mol Cu2S
(b) How many grams of sulfur dioxide are formed when 10.0 mol of copper(I) sulfide is roasted?
(b) 10.0mol Cu2S
2mol SO2
64.07g SO2
2mol Cu2S
mol SO2
= 641g SO2
(c) How many kilograms of oxygen are required to form 2.86 kg of copper(I) oxide?
(c) 2.86kg Cu2O
103g Cu2O
mol Cu2O
kg Cu2O 143.10g Cu2O
20.0mol Cu2O
3mol O2
2mol Cu2O
32.00g O2
mol O2
= 20.0mol Cu2O
kg O2
103g
O2
= 0.959kg O2
Writing an Overall Equation for a Reaction Sequence
PROBLEM:
Roasting is the first step in extracting copper from chalcocite,
the ore used in the previous problem. In the next step, copper(I)
oxide reacts with powdered carbon to yield copper metal and
carbon monoxide gas. Write a balanced overall equation for the
two-step process.
PLAN:
SOLUTION:
write balanced equations for each step
2Cu2S(s) + 3O2(g)
cancel reactants and products common
2Cu2O(s) + 2C(s)
2Cu2O(s) + 2SO2(g)
4Cu(s) + 2CO(g)
to both sides of the equations
sum the equations
2Cu2S(s)+3O2(g)+2C(s)
4Cu(s)+2SO2(g)+2CO(g)
An ice cream sundae analogy for limiting reactions.
Table 3.5 Information Contained in a Balanced Equation
Viewed in
Terms of
molecules
amount (mol)
mass (amu)
mass (g)
total mass (g)
Reactants
C3H8(g) + 5O2(g)
Products
3CO2(g) + 4H2O(g)
1 molecule C3H8 + 5 molecules O2
3 molecules CO2 + 4 molecules H2O
1 mol C3H8 + 5 mol O2
3 mol CO2 + 4 mol H2O
44.09 amu C3H8 + 160.00 amu O2
132.03 amu CO2 + 72.06 amu H2O
44.09 g C3H8 + 160.00 g O2
132.03 g CO2 + 72.06 g H2O
204.09 g
204.09 g
Using Molecular Depictions to Solve a LimitingReactant Problem
PROBLEM:
Nuclear engineers use chlorine trifluoride in the processing of
uranium fuel for power plants. This extremely reactive
substance is formed as a gas in special metal containers by the
reaction of elemental chlorine and fluorine.
(a) Suppose the box shown at left represents a container of the
reactant mixture before the reaction occurs (with chlorine
colored green). Name the limiting reactant, and draw the
container contents after the reaction is complete.
(b) When the reaction is run again with 0.750 mol of Cl2 and 3.00 mol of
F2, what mass of chlorine trifluoride will be prepared?
PLAN: Write a balanced chemical equation. Compare the number of
molecules you have to the number needed for the products.
Determine the reactant that is in excess. The other reactant is the
limiting reactant.
Using Molecular Depictions to Solve a LimitingReactant Problem
continued
SOLUTION:
Cl2(g) + 3F2(g)
2ClF3(g)
(a) You need a ratio of 2 Cl and 6 F for the reaction.
You have 6 Cl and 12 F.
6 Cl would require 18 F.
12 F need only 4 Cl (2 Cl2 molecules).
There isn’t enough F, therefore it must be the limiting reactant.
F2
Cl2
You will make 4 ClF2 molecules (4 Cl, 12 F) and
have 2 Cl2 molecules left over.
(b) We know the molar ratio of F2/Cl2 should be 3/1.
3.00 mol F2
0.750 mol Cl2
0.750 mol Cl2
4
=
1
Since we find that the ratio is 4/1,
that means F2 is in excess and
Cl2 is the limiting reactant.
2 mol ClF3 92.5 g ClF3
1 mol Cl
1 mol ClF3
=
139 g ClF3
Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
PROBLEM:
A fuel mixture used in the early days of rocketry is composed of
two liquids, hydrazine(N2H4) and dinitrogen tetraoxide(N2O4),
which ignite on contact to form nitrogen gas and water vapor.
How many grams of nitrogen gas form when 1.00x102 g of N2H4
and 2.00x102 g of N2O4 are mixed?
PLAN: We always start with a balanced chemical equation and find the number
of mols of reactants and products which have been given.
In this case one of the reactants is in molar excess and the other will
limit the extent of the reaction.
mass of N2H4
mass of N2O4
divide by M
mol of N2H4
multiply by M
mol of N2O4
molar ratio
mol of N2
limiting mol N2
mol of N2
g N2
Calculating Amounts of Reactant and Product in
Reactions Involving a Limiting Reactant
continued
SOLUTION:
1.00x102g N2H4
2N2H4(l) + N2O4(l)
mol N2H4
32.05g N2H4
3.12mol N2H4
3 mol N2
= 3.12mol N2H4
4.68mol N2
mol
N2O4 N O
2 4
92.02g N2O4
2.17mol N2O4
3 mol N2
mol N2O4
N2H4 is the limiting reactant
because it produces less
product, N2, than does N2O4.
= 4.68mol N2
2mol N2H4
2.00x102g
3N2(g) + 4H2O(l)
= 2.17mol N2O4
= 6.51mol N2
28.02g N2
mol N2
= 131g N2
The effect of side reactions on yield.
A +B
C
(reactants)
(main product)
D
(side products)
Calculating Percent Yield
Silicon carbide (SiC) is an important ceramic material that is
made by allowing sand(silicon dioxide, SiO2) to react with
powdered carbon at high temperature. Carbon monoxide is also
formed. When 100.0 kg of sand is processed, 51.4 kg of SiC is
recovered. What is the percent yield of SiC from this process?
PLAN:
write balanced equation
find mol reactant & product
SOLUTION:
SiO2(s) + 3C(s)
100.0 kg SiO2
SiC(s) + 2CO(g)
103 g SiO2
mol SiO2
kg SiO2
60.09 g SiO2
= 1664 mol SiO2
mol SiO2 = mol SiC = 1664
find g product predicted
1664 mol SiC
mol SiC
actual yield/theoretical yield x 100
51.4 kg
percent yield
40.10 g SiC kg
66.73 kg
103g
x100 =77.0%
= 66.73 kg
Calculating the Molarity of a Solution
PROBLEM:
PLAN:
Glycine (H2NCH2COOH) is the simplest amino acid. What is the
molarity of an aqueous solution that contains 0.715 mol of
glycine in 495 mL?
Molarity is the number of moles of solute per liter of solution.
mol of glycine
divide by volume
concentration(mol/mL) glycine
103mL = 1L
molarity(mol/L) glycine
SOLUTION:
0.715 mol glycine
1000mL
495 mL soln
1L
= 1.44 M glycine
Summary of
mass-mole-number-volume
relationships in solution.
MASS (g)
of compound
in solution
M (g/mol)
AMOUNT (mol)
of compound
in solution
Avogadro’s number
(molecules/mol)
MOLECULES
(or formula units)
of compound
in solution
M (g/mol)
VOLUME (L)
of solution
Calculating Mass of Solute in a Given Volume of
Solution
PROBLEM:
PLAN:
A “buffered” solution maintains acidity as a reaction occurs. In
living cells phosphate ions play a key buffering role, so
biochemistry often study reactions in such solutions. How many
grams of solute are in 1.75 L of 0.460 M sodium monohydrogen
phosphate?
Molarity is the number of moles of solute per liter of solution.
Knowing the molarity and volume leaves us to find the # moles
and then the # of grams of solute. The formula for the solute is
Na2HPO4.
volume of soln
multiply by M
moles of solute
multiply by M
grams of solute
SOLUTION:
1.75 L 0.460 moles
1L
= 0.805 mol Na2HPO4
0.805 mol Na2HPO4 141.96 g Na2HPO4
mol Na2HPO4
= 114 g Na2HPO4
Laboratory preparation of molar solutions.
A
•Weigh the solid needed.
•Transfer the solid to a
volumetric flask that contains
about half the final volume of
solvent.
C Add solvent until the solution
reaches its final volume.
B Dissolve the solid
thoroughly by swirling.
Converting a concentrated solution to a dilute solution.
Preparing a Dilute Solution from a Concentrated
Solution
PROBLEM:
“Isotonic saline” is a 0.15 M aqueous solution of NaCl that
simulates the total concentration of ions found in many cellular
fluids. Its uses range from a cleaning rinse for contact lenses to
a washing medium for red blood cells. How would you prepare
0.80 L of isotomic saline from a 6.0 M stock solution?
It is important to realize the number of moles of solute does not
change during the dilution but the volume does. The new
volume will be the sum of the two volumes, that is, the total final
volume.
MdilxVdil = #mol solute = MconcxVconc
volume of dilute soln
SOLUTION:
multiply by M of dilute solution
PLAN:
moles of NaCl in dilute soln = mol NaCl
in concentrated soln
divide by M of concentrated soln
L of concentrated soln
0.80 L soln 0.15 mol NaCl = 0.12 mol NaCl
L soln
L solnconc
0.12 mol NaCl
= 0.020 L soln
6 mol
Calculating Amounts of Reactants and Products for
a Reaction in Solution
PROBLEM:
PLAN:
Specialized cells in the stomach release HCl to aid digestion. If
they release too much, the excess can be neutralized with
antacids. A common antacid contains magnesium hydroxide,
which reacts with the acid to form water and magnesium
chloride solution. As a government chemist testing commercial
antacids, you use 0.10M HCl to simulate the acid concentration
in the stomach. How many liters of “stomach acid” react with a
tablet containing 0.10g of magnesium hydroxide?
Write a balanced equation for the reaction; find the grams of
Mg(OH)2; determine the mol ratio of reactants and products;
use mols to convert to molarity.
mass Mg(OH)2
L HCl
divide by M
mol Mg(OH)2
divide by M
mol HCl
mol ratio
Calculating Amounts of Reactants and Products for
a Reaction in Solution
continued
SOLUTION:
Mg(OH)2(s) + 2HCl(aq)
0.10g Mg(OH)2
1.7x10-3
mol Mg(OH)2
3.4x10-3
= 1.7x10-3 mol Mg(OH)2
58.33g Mg(OH)2
mol Mg(OH)2
2 mol HCl
1 mol Mg(OH)2
1L
mol HCl
0.10mol HCl
MgCl2(aq) + 2H2O(l)
= 3.4x10-3 mol HCl
= 3.4x10-2 L HCl
Solving Limiting-Reactant Problems for Reactions
in Solution
PROBLEM:
PLAN:
Mercury and its compounds have many uses, from fillings for
teeth (as an alloy with silver, copper, and tin) to the industrial
production of chlorine. Because of their toxicity, however,
soluble mercury compounds, such mercury(II) nitrate, must be
removed from industrial wastewater. One removal method
reacts the wastewater with sodium sulfide solution to produce
solid mercury(II) sulfide and sodium nitrate solution. In a
laboratory simulation, 0.050L of 0.010M mercury(II) nitrate
reacts with 0.020L of 0.10M sodium sulfide. How many grams
of mercury(II) sulfide form?
As usual, write a balanced chemical reaction. Since this is a problem
concerning a limiting reactant, we proceed as we would for a limiting
reactant problem. Find the amount of product which would be made
from each reactant. Then choose the reactant that gives the lesser
amount of product.
Solving Limiting-Reactant Problems for Reactions
in Solution
continued
SOLUTION:
Hg(NO3)2(aq) + Na2S(aq)
L of Hg(NO3)2
0.050L Hg(NO3)2
multiply by M
x 0.010 mol/L
mol Hg(NO3)2
mol ratio
mol HgS
HgS(s) + 2NaNO3(aq)
0.020L Hg(NO3)2
x 0. 10 mol/L multiply by M
x 1mol HgS
x 1mol HgS
1mol Hg(NO3)2
1mol Na2S
= 5.0x10-4 mol HgS
= 2.0x10-3 mol HgS
Hg(NO3)2 is the limiting reagent.
5.0x10-4
mol HgS
232.7g HgS
1 mol HgS
L of Na2S
= 0.12g HgS
mol Na2S
mol ratio
mol HgS
An overview of the key mass-mole-number
stoichiometry relationships.