Download Chapter 3 Molecules, Moles, and Chemical Equations

CHEM 101-GENERAL CHEMISTRY
CHAPTER 3
MOLECULES, MOLES & CHEMICAL EQUATIONS
INSTR : FİLİZ ALSHANABLEH
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CHAPTER 3
MOLECULES, MOLES & CHEMICAL EQUATIONS
• Chemical Formulas and Equations
. Writing Chemical Equations
. Balancing Chemical Equations
• Aqueous Solution and Net Ionic Equations
. Solutions, Solvents and Solutes
. Chemical Equations for Aqueous Solutions
• Interpreting Equations and The Mole
. Avogadro’s Number and the Mole
. Determinig Molar Mass
• Calculations Using Moles and Molar Masses
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Explosions
1. Explosions release a large
amount of energy when a
fairly complex molecule
decomposes into smaller,
simpler compounds.
2. Explosions occur very
quickly.
3. Modern explosives are
generally solids.
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Explosions
•
Dynamite is an explosive
made from liquid nitroglycerin
and an inert binder to form a
solid material.
Solids are easier to handle
than liquids
The destructive force of
explosions is due in part to
expansion of gases, which
produces a shockwave.
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Chemical Formulas and Equations
•
Chemical formulas provide a concise way to represent chemical
compounds.
• Nitroglycerin, shown earlier, becomes C3H5N3O9
•
A chemical equation builds upon chemical formulas to concisely represent
a chemical reaction.
•
Chemical equations represent the transformation of one or more chemical
species into new substances.
•
Chemical formulas represent reactants and products.
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Writing Chemical Equations
• Phase labels follow each formula.
• solid = (s)
• liquid = (l)
• gas = (g)
• aqueous (substance dissolved in water) = (aq)
• Some reactions require an additional symbol placed over
the reaction arrow to specify reaction conditions.
• Thermal reactions: heat (∆)
• Photochemical reactions: light (hν)
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Writing Chemical Equations
• Different representations for the reaction between hydrogen
and oxygen to produce water.
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Balancing Chemical Equations
• The law of conservation of matter: matter is neither created
nor destroyed.
• Chemical reactions must obey the law of conservation of
matter.
• The same number of atoms for each element must
occur on both sides of the chemical equation.
• A chemical reaction simply rearranges the atoms into
new compounds.
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Balancing Chemical Equations
• Balanced chemical equation for the combustion of methane.
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Balancing Chemical Equations
• Chemical equations may be balanced via inspection, which
really means by trial and error.
• Numbers used to balance chemical equations are called
stoichiometric coefficients.
• The stoichiometric coefficient multiplies the number of
atoms of each element in the formula unit of the
compound that it precedes.
• Stoichiometry is the various quantitative relationships
between reactants and products.
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Exercise
Which of the following correctly balances the chemical
equation given below? There may be more than one
correct balanced equation. If a balanced equation is
incorrect, explain what is incorrect about it.
CaO + C → CaC2 + CO2
I.
II.
III.
IV.
CaO2 + 3C → CaC2 + CO2
2CaO + 5C → 2CaC2 + CO2
CaO + (2.5)C → CaC2 + (0.5)CO2
4CaO + 10C → 4CaC2 + 2CO2
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Aqueous Solutions and Net Ionic Equations
• Reactions that occur in water are said to take place in
aqueous solution.
• Solution: homogeneous mixture of two or more
substances.
• Solvent: solution component present in greatest
amount.
• Solute: solution component present in lesser amount.
• The preparation of a solution is a common way to
enable two solids to make contact with one another.
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Solutions, Solvents, and Solutes
• For solutions, the concentration is a key piece of information.
• Concentration: relative amounts of solute and solvent.
• Concentrated: many solute particles are present.
• Dilute: few solute particles are present.
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Solutions, Solvents, and Solutes
• Compounds can be characterized by their solubility.
• Soluble compounds dissolve readily in water.
• Insoluble compounds do not readily dissolve in water.
• Solubility can be predicted using solubility guidelines.
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Solutions, Solvents, and Solutes
• Solubility guidelines for soluble salts
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Exercise
• Which of the following compounds would you predict are
soluble in water at room temperature?
a) Mg(NO3)2
b) CaCO3
c) BaSO4
d) KMnO4
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Solutions, Solvents, and Solutes
• Electrolytes are soluble compounds that conduct current
when dissolved in water.
• Weak electrolytes dissociate partially into ions in solution.
• Strong electrolytes dissociate completely into ions in
solution.
• Nonelectrolytes do not dissociate into ions in solution.
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Solutions, Solvents, and Solutes
a) Sugar, a nonelectrolyte, does not conduct current when dissolved in water.
b) Acetic acid, a weak electrolyte, weakly conducts current when dissolved in
water.
c) Potassium chromate, a strong electrolyte, strongly conducts current when
dissolved in water.
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Chemical Equations for Aqueous Reactions
• When a covalently bonded material dissolves in water and the
molecules remain intact, they do not conduct current. These
compounds are nonelectrolytes.
→ C6 H12O 6 (aq)
C6 H12O 6 (s) 
• When an ionic solid dissolves in water, it breaks into its
constituent ions. This is called a dissociation reaction. These
compounds conduct electric current and are electrolytes.
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Chemical Equations for Aqueous Reactions
•
Aqueous chemical reactions can be written as a molecular equation. The
complete formula for each compound is shown.
→ NH 4 NO 3 (aq)
HNO 3 (aq) + NH 3 (g) 
•
Dissociation of reactants and products is emphasized by writing a total
ionic equation.
→ NH +4 (aq) + NO −3 (aq)
H + (aq) + NO −3 (aq) + NH 3 (g) 
•
•
Spectator ions are ions uninvolved in the chemical reaction. When
spectator ions are removed, the result is the net ionic equation.
Net ionic equation:
H + (aq) + NH 3 (g) 
→ NH +4 (aq)
Spectator ion = NO −3
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Exercise
Write the molecular equation, total ionic equation, and net
ionic equation for the reaction between cobalt(II) chloride
and sodium hydroxide.
Molecular Equation:
CoCl2(aq) + 2NaOH(aq) → Co(OH)2(s) + 2NaCl(aq)
Total Ionic Equation:
Co2+(aq) + 2Cl−(aq) + 2Na+(aq) + 2OH−(aq) →
Co(OH)2(s) + 2Na+(aq) + 2Cl−(aq)
Net Ionic Equation:
Co2+(aq) + 2Cl−(aq) → Co(OH)2(s)
•
Na+ and Cl− are spectator ions.
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Acid-Base Reactions
• Acids are substances that dissolve in water to produce H+ (or
H3O+) ions.
• Examples: HCl, HNO3, H3PO4, HCN
• Bases are substances that dissolve in water to produce OH–
ions.
• Examples: NaOH, Ca(OH)2, NH3
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Acid-Base Reactions
• Strong acids and bases completely dissociate in water.
→ H 3O + (aq) + Cl− (aq)
HCl(g) + H 2O(l ) 
→ Na + (aq) + OH − (aq)
NaOH(s) 
• Weak acids and bases partially dissociate in water.


→ H 3O + (aq) + CH 3COO − (aq)
CH 3COOH(aq) + H 2O(l ) ←



→ NH +4 (aq) + OH − (aq)
NH 3 (aq) + H 2O(l ) ←

• Notice the two-way arrows, which emphasize that the reaction
does not proceed completely from left to right.
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Acid-Base Reactions
• Mixing an acid and a base leads to a reaction known as
neutralization, in which the resulting solution is neither acidic
nor basic.
• Net ionic equation for neutralization of strong acid and
strong base.
H 3O + (aq) + OH − (aq) 
→ 2H 2O(l )
Moleculer Equation:
NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l)
Ionic Equation:
Na+(aq) + OH−(aq) + H+(aq) + Cl−(aq) → Na+(aq) + Cl−(aq) + H2O(l)
Net Ionic Equation:
OH−(aq) + H+(aq) → H2O(l)
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Precipitation Reactions
• A precipitation reaction is an aqueous reaction that produces a
solid, called a precipitate.
• EXAMPLE: When aqueous silver nitrate and sodium chloride are
combined, the solution becomes cloudy white with solid silver chloride.
Let’s write molecular, total ionic and net ionic equations:
•
•
Molecular Equation:
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
•
Total Ionic Equation:
Ag+(aq) + NO3−(aq) + Na+(aq) + Cl−(aq) → AgCl(s) + Na+(aq) + NO3−(aq)
•
•
Net İonic Equation:
Ag+(aq) + Cl−(aq) → AgCl(s)
Na+ and NO3− are spectator ions.
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Avogadro’s Number and the Mole
Avogadro’s Number and the Mole
• A mole is a means of counting the large number of particles in
samples.
• One mole is the number of atoms in exactly 12 grams of
12C (carbon-12).
• 1 mole contains Avogadro’s number (6.022 x 1023
particles/mole) of particles.
• The mass of 6.022 x 1023 atoms of any element is the
molar mass of that element.
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Avogadro’s Number and the Mole
•
One mole samples of various elements. All have the same number of
particles.
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Avogadro’s Number and the Mole
•
•
•
•
1 mole C = 6.022 x 1023 C atoms = 12.01 g C
1 mole H = 6.022 x 1023 H atoms = 1.008 g H
1 mole O = 6.022 x 1023 O atoms = 16.00 g O
, etc.
• Balanced chemical reactions also provide mole ratios
between reactants and products.
→ 2H 2 O(g)
2H 2 (g) + O 2 (g) 
• 2 moles H2 : 1 mole O2 : 2 moles H2O
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Determining Molar Mass
• The molar mass of a compound is the sum of the molar
masses of all the atoms in a compound.
•
Mass in grams of one mole of the substance:
Molar Mass of N = 14.01 g/mol
Molar Mass of H2O = 18.02 g/mol
(2 × 1.008 g) + 16.00 g
Molar Mass of Ba(NO3)2 = 261.35 g/mol
137.33 g + (2 × 14.01 g) + (6 × 16.00 g)
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Calculations Using Moles and Molar Mass
• Molar mass allows conversion from mass to number of
moles, much like a unit conversion.
•
1 mol C7H5N3O6 = 227.133 g C7H5N3O6
1 mol C 7 H 5 N 3O 6
300.0 g C 7 H 5 N 3O 6 ×
227.133 g C 7 H 5 N 3O 6
= 1.320 mol C 7 H 5 N 3O 6
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Calculations Using Moles and Molar Mass
• Avogadro’s number functions much like a unit conversion
between moles to number of particles.
• 1 mol C7H5N3O6 = 6.022 × 1023 C7H5N3O6 molecules
• How many molecules are in 1.320 moles of
nitroglycerin?
6.022 × 10 23 molecules C 7 H 5 N 3O 6
1.320 mol C 7 H 5 N 3O 6 ×
1 mol C 7 H 5 N 3O 6
= 7.949 × 10 23 molecules C 7 H 5 N 3O 6
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Exercise
Calculate the number of iron atoms in a 4.48 mole sample of
iron. (Fe: 55.85)
1 mole Fe = 6.022 x 1023 Fe atoms = 55.85 g Fe
1 mole Fe
6.022 x 1023 Fe atoms
4.48 mole Fe
X= 2.70×1024 Fe atoms
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Exercise
Which of the following is closest to the average mass of one atom of
copper?
(Atomic Mass of Copper= 63.55 amu)
a)
b)
c)
d)
e)
63.55 g
52.00 g
58.93 g
65.38 g
1.055 x 10-22 g
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Exercises
Prob1. Calculate the number of copper atoms in a 63.55 g
sample of copper.
6.022×1023 Cu atoms
Prob1. Calculate the number of oxygen atoms in a 44.01 g
sample of CO2 ( C:12.01, O:16.00)
1 mole of CO2 = 44.01 g = 6.022 × 1023 CO2 molecules = 2x 6.022 × 1023 O atoms
= 12.04 × 1023 O atoms
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Exercises
Prob3.
Which of the following 100.0 g samples contains the
greatest number of atoms? (Mg: 24.31 , Zn: 65.38 , Ag: 107.9)
a) Magnesium
b) Zinc
c) Silver
Prob4.
Rank the following 100.0 g samples (each) from greatest to least
number of oxygen atoms. (C:12.0 , H:1.01, O:16.0 )
a) C3H6O2
b) H2O,
c) CO2,
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Elemental Analysis: Determining Empirical
and Molecular Formulas
• Empirical formulas can be determined from an elemental
analysis.
• An elemental analysis measures the mass percentage of
each element in a compound.
• The formula describes the composition in terms of the
number of atoms of each element.
• The molar masses of the elements provide the connection
between the elemental analysis and the formula.
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Percent Composition by Mass
•
Mass percent of an element:
mass of element in compound
mass % =
× 100%
mass of compound
•
For iron in iron(III) oxide, (Fe2O3):
mass % Fe =
2( 55.85 g)
111.70 g
=
× 100% = 69.94%
2( 55.85 g)+ 3( 16.00 g) 159.70 g
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Exercise
Consider separate 100.0 gram samples of each
of the following:
H2O, N2O, C3H6O2, CO2
 Rank them from highest to lowest percent
oxygen by mass.
Answer:
H2O, CO2, C3H6O2, N2O
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Determining Empirical and Molecular
Formulas
•
Assume a 100 gram sample size
•
Percentage element × sample size = mass element in compound. (e.g.,
16% carbon = 16 g carbon)
•
Convert mass of each element to moles using the molar mass.
•
Divide by smallest number of moles to get mole to mole ratio for
empirical formula.
•
When division by smallest number of moles results in small rational
fractions, multiply all ratios by an appropriate integer to give whole
numbers.
• 2.5 × 2 = 5, 1.33 × 3 = 4, etc.
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Determining Empirical and Molecular
Formulas
• A molecular formula is a whole number multiple of the
empirical formula.
• Molar mass for the molecular formula is a whole number
multiple of the molar mass for the empirical formula.
• If the empirical formula of a compound is CH2 and its
molar mass is 42 g/mol, what is its molecular formula?
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Determining Empirical and Molecular
Formulas
Empirical formula = CH
 Simplest whole-number ratio
• Molecular formula = (empirical formula)n
[n = integer]
• Molecular formula = C6H6 = (CH)6
 Actual formula of the compound
•
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Exercise
The composition of adipic acid is 49.3% C, 6.90% H, and 43.8% O (by
mass). The molar mass of the compound is about 146 g/mol. Find
empirical and molecular Formula of apidic acid.
(C:12.0 , H:1.01, O:16.0 )
Consider 100 g of CxHyOz



49.3 g C = 4.11 mol of C
6.90 g H = 6.83 mol of H
43.8 g O = 2.73 mol of O
The empirical formula?
C4.11H6.83O2.73 = C1.5H2.5O =
C3H5O2 : 73.1 g/mol
C 3 H 5 O2
The molecular formula: 146 = n( 73.1)
n(C3H5O2 ) = 2(C3H5O2 )=
C6H10O4
n= 2
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Molarity
• Molarity, or molar concentration, M, is the number of moles of
solute per liter of solution.
• Provides relationship among molarity, moles solute, and
liters solution.
moles of solute
Molarity (M ) =
liter of solution
•
n=MxV
3 M HCl =
6 moles of HCl
2 liters of solution
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Exercise
A 500.0-g sample of potassium phosphate (K3PO4 ) is
dissolved in enough water to make 1.50 L of solution. What is
the molarity of the solution? (K3PO4 : 212.4 g/mol)
Moles of Solute = n = 500.0 g / (212.4 g/mol) = 2.36 mol
Volume of Solution = 1.50 L
Molarity (M ) =
M = 2.36 / 1.5 =
moles of solute
liter of solution
1.57 M
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Dilution
• Dilution is the process in which solvent is added to a solution
to decrease the concentration of the solution.
• The number of moles of solute is the same before and
after dilution.
• Since the number of moles of solute equals the product of
molarity and volume (M × V), we can write the following
equation, where the subscripts denote initial and final
values.
M i × Vi = M f × Vf
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Concept Check
A 0.50 M solution of sodium chloride in an open beaker sits on a
lab bench. Which of the following would decrease the
concentration of the salt solution?
a)
b)
c)
d)
Add water to the solution.
Pour some of the solution down the sink drain.
Add more sodium chloride to the solution.
Let the solution sit out in the open air for a couple of
days.
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Exercise
What is the minimum volume of a 2.00 M NaOH solution
needed to make 150.0 mL of a 0.800 M NaOH solution?
M1V1 = M2V2
(2.00 M)(V1) = (0.800 M)(150.0 mL)
V1 = 60.0 mL
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