changing matter matter can be changed two ways: –physically physical reaction physical change...
TRANSCRIPT
Changing Matter• Matter can be changed two ways:
–Physically • Physical reaction
• Physical change
–Chemically• Chemical reaction
• Chemical change
Physical Changes
• Do NOT CHANGE THE TYPE OF MATTER
– Nothing new or different is formed– Could be a change in:
• Mass• Volume• Density• Change in state • Color• Shape
Size
Examples of Physical Changes
• Boiling, vaporization…. Any state change • Dissolving • Breaking • Making a mixture
– 2 or more types of matter (substances) mixed together
• Not in specific amounts• Can be separated physically
Chemical Changes
• Atoms have electrons arranged in energy levels or energy shells
• Electrons in the last (outermost) shell are called valence electrons
• Valence electrons let atoms bond with other atoms– Ionic bonding
• Gaining or losing electrons
– Covalent bonding• Sharing electrons
Chemical Changes
• Atoms that bond form molecules– May be the same type (nonmetals) of atom or,– Different types (metal + nonmetal) of atoms
• Different types compounds
Chemical Changes
• Molecules can bond and “unbond” – Atoms can re-arranged in different
combinations– For example:
CaCO3 (1 atom Ca, 1 atom C, 3 atoms O)
Add heat to re-arranged the atoms:
CaO
CO2
Chemical Changes
• Evidence of a chemical reaction– Formation of gas– Formation of precipitate– Change in color– Change in energy
• Endothermic– Absorbs heat energy (gets cold)
• Exothermic– Releases heat energy (gets hot)
Chemical Changes
• Chemical reactions can be represented by equations
CaCO3 CaO + CO2
Reactants Products
Chemical Changes
• Atoms are re-arranged, NOT created or destroyed– Law of Conservation of Matter– Law of Conservation of Mass
Chemical Changes
• Matter is conserved type of atoms does not change– Nothing is created or destroyed
• Mass is conserved amount of atoms cannot change– Nothing is created or destroyed
Chemical Changes
• To show conservation of mass Balance equations– Make sure there are the same number of
each type of atom in the products and in the reactants
Balancing Equations
The equation for the burning of methane gas in oxygen is:
CH4 + 2 O2 → CO2 + 2 H2O
Subscript
Shows # of atoms
Coefficient
Shows # of molecules
Balancing Equations
• No subscript or coefficient is understood to be 1
CH4 + 2 O2 → CO2 + 2 H2O =
C1H4 + 2 O2 → C1O2 + 2 H2O1
1 C 1 C 4 H 4 H4 O 4 O
Periodic Properties Post Lab
• Periodic Trends affect properties because of Zeff– Ionization energy
• Energy required to remove an electron
– Electronegativity• Ability to attract electrons
– Atomic Radius• Distance of the valence electrons to the nucleus
Other Periodic Trends
• Groups have similar properties– Valence electrons
• Members of the same representative family have the same number of valence electrons
– Reactivity• Because they have the same number of electrons,
they react similarly.– CaCl2, MgCl2, SrCl2, etc.
– Density = mass/volume• % error = I actual – theoretical I
theoretical
Sn = 7.265Pb = 11.34Si = 2.33Ge = 5.323
Solubility Trends
• Common in Double Replacement Rxns
Double Replacement Rxns
• Two compounds react to form two new compounds– Metal replaces metal– Remember basic formula writing rules
Ex: Ca(NO3)2 + Na2CO3 CaCO3 + NaNO32
ppt
Type of Reaction
Definition Equation
Synthesis
Decomposition
Single Replacement
Double Replacement
Some Types of Chemical Reactions
A + B → AB
AB → A + B
AB + C → AC + B
AB + CD → AC + BD
Two or more elements or compounds combine to make a more complex
substance
Compounds break down into simpler substances
Occurs when one element replaces another one in a
compound
Occurs when different atoms in two different
compounds trade places
Identifying Chemical Reactions
____ P + O2 → P4O10 ____ Mg + O2 → MgO
____ HgO → Hg + O2 ____ Al2O3 → Al + O2
____ Cl2 + NaBr → NaCl + Br2 ____ H2 + N2 → NH3
S = Synthesis D = Decomposition SR = Single Replacement DR = Double Replacement
MOLAR MASS
Molar Mass is shown below the element symbol on the Periodic Table.
Units: grams mole
Use molar mass to convert between:Number of Moles
Mass (grams)Number of Molecules or Atoms
Molar Mass Examples
• sodium bicarbonate
• sucrose
NaHCO3
22.99 + 1.01 + 12.01 + 3(16.00)
= 84.01 g/mol
C12H22O11
12(12.01) + 22(1.01) + 11(16.00)
= 342.34 g/mol
MOLECULESATOMS
Multiply the number of moles
by Avogadro’s Constant to get the number of
atoms or molecules.
Divide the number of molecules or atoms by Avogadro’s Constant to get the number of
moles.
Divide the mass (in grams) by Molar Mass to get the
number of moles.
MOLARMASSgrams/
mol
MOLES
MASS inGRAMS
Multiply the moles
by the Molar Mass to get the
mass (in grams).
AVOGADRO’S CONSTANT6.022x1023
particles/mol
Molar Conversion Examples• How many moles of carbon are
in 26 g of carbon?
26 g C 1 mol C
12.01 g C= 2.2 mol C
Molar Conversion Examples• How many molecules are in
2.50 moles of C12H22O11?
2.50 mol 6.02 1023
molecules
1 mol= 1.51 1024
molecules C12H22O11
Molar Conversion Examples• Find the mass of 2.1 1024
molecules of NaHCO3.
2.1 1024
molecules 1 mol
6.02 1023
molecules
= 290 g NaHCO3
84.01 g
1 mol
Sample Problems – Moles and Atoms
• Determine the number of atoms present in 2.50 moles of strontium.
• Convert 5.01 x 1024 atoms of strontium to moles of strontium.
Sample Problems – Moles and Mass
• Determine the mass in grams of 2.50 moles of strontium.
• Determine the number of moles represented by 943.5 grams of strontium.
Sample Problems – Moles, Atoms, Mass
• Determine the mass in grams of (exactly) 5 atoms of strontium.
• Determine the number of atoms represented by 43.5 grams of strontium.
Percentage Composition
• the percentage by mass of each element in a compound
100mass total
element of massncompositio %
100 =
Percentage Composition
%Cu =127.10 g Cu
159.17 g Cu2S 100 =
%S =32.07 g S
159.17 g Cu2S
79.852% Cu
20.15% S
• Find the % composition of Cu2S.
%Fe =28 g
36 g 100 =78% Fe
%O =8.0 g
36 g 100 =22% O
• Find the percentage composition of a sample that is 28 g Fe and 8.0 g O.
Percentage Composition
• How many grams of copper are in a 38.0-gram sample of Cu2S?
(38.0 g Cu2S)(0.79852) = 30.3 g Cu
Cu2S is 79.852% Cu
Percentage Composition
100 =%H2O = 36.04 g
147.02 g 24.51%
H2O
• Find the mass percentage of water in calcium chloride dihydrate, CaCl2•2H2O?
Percentage Composition
Empirical Formula
C2H6
CH3
reduce subscripts
• Smallest whole number ratio of atoms in a compound
Empirical Formula1. Find mass (or %) of each element.
2. Find moles of each element.
3. Divide moles by the smallest # to find subscripts.
4. When necessary, multiply subscripts by 2, 3, or 4 to get whole #’s.
Empirical Formula• Find the empirical formula for a
sample of 25.9% N and 74.1% O.
25.9 g 1 mol
14.01 g = 1.85 mol N
74.1 g 1 mol
16.00 g = 4.63 mol O
1.85 mol
1.85 mol
= 1 N
= 2.5 O
Empirical Formula
N1O2.5Need to make the subscripts whole
numbers multiply by 2
N2O5
Molecular Formula
• “True Formula” - the actual number of atoms in a compound
CH3
C2H6
empiricalformula
molecularformula
?
Molecular Formula1. Find the empirical formula.2. Find the empirical formula mass.3. Divide the molecular mass by the empirical mass.4. Multiply each subscript by the answer from step 3.
nmass EF
mass MF nEF
Molecular Formula• The empirical formula for ethylene is CH2. Find the molecular
formula if the molecular mass is 28.1 g/mol?
28.1 g/mol
14.03 g/mol = 2.00
empirical mass = 14.03 g/mol
(CH2)2 C2H4
Using the Basics…
The empirical formula of a
compound is found to be P2O5.
The molar mass of the compound
is 284 grams/mole.
What is the molecular formula for the compound?
Proportional Relationships
• StoichiometryStoichiometry–mass relationships between
substances in a chemical reaction–based on the mole ratio
• Mole RatioMole Ratio–indicated by coefficients in a balanced
equation
2 Mg + O2 Mg + O22 2 MgO 2 MgO
43
2
23 m
1 mol
ol H
N
N2 + 3H2 2NH31 mol 2 mol3 mol
44
1 mol 2 mol3 molN2 + 3H2 2NH3
2
32 mo
3 mol H
l NH
Stoichiometry Steps1. Write a balanced equation.2. Identify known & unknown.3. Line up conversion factors.
– Mole ratio - moles moles– Molar mass - moles grams– Molarity - moles liters soln– Molar volume - moles liters gas
Core step in all stoichiometry problems!!
• Mole ratio - moles moles
4. Check answer.
1 mol of a gas=22.4 Lat STP
Molar Volume at STP
Standard Temperature & Pressure0°C and 1 atm
Molar Volume at STP
Molar Mass(g/mol)
6.02 1023
particles/mol
MASSIN
GRAMSMOLES
NUMBEROF
PARTICLES
LITERSOF
SOLUTION
Molar Volume (22.4 L/mol)
LITERSOF GASAT STP
Molarity (mol/L)
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Mole-Mole Mole-Mole CalculationsCalculations
49
Phosphoric Acid• Phosphoric acid (H3PO4) is one of the
most widely produced industrial chemicals in the world.
• Most of the world’s phosphoric acid is produced by the wet process which involves the reaction of phosphate rock, Ca5(PO4)3F, with sulfuric acid (H2SO4).
Ca5(PO4)3F(s) + 5H2SO4 3H3PO4 + HF + 5CaSO4
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Mole Ratio
Calculate the number of moles of phosphoric acid (H3PO4) formed by the reaction of 10 moles of sulfuric acid (H2SO4).
Ca5(PO4)3F + 5H2SO4 3H3PO4 + HF + 5CaSO4
Step 1 Moles starting substance: 10.0 mol H2SO4
Step 2 The conversion needed is moles H2SO4 moles H3PO4
1 mol 5 mol 3 mol 1 mol 5 mol
3 42 4
2 4
3 mol H PO10 mol H SO x =
5 mol H SO3 46 mol H PO
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Step 2 The conversion needed is
moles Ca5(PO4)3F moles H2SO4
Calculate the number of moles of sulfuric acid (H2SO4) that react when 10 moles of Ca5(PO4)3 react.
Ca5(PO4)3F + 5H2SO4 3H3PO4 + HF + 5CaSO4
Mole Ratio
Step 1 The starting substance is 10.0 mol Ca5(PO4)3F
1 mol 5 mol 3 mol 1 mol 5 mol
2 45 4 3
5 4 3
5 mol H SO10 mol Ca (PO ) F x =
1 mol Ca (PO ) F 2 450 mol H SO
Stoichiometry Problems
• How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
9 mol O2 2 mol KClO3
3 mol O2
= 6 mol KClO3
2KClO3 2KCl + 3O2 ? mol 9 mol
53
Mole-Mass CalculationsMole-Mass Calculations
54Mole Ratio
2 4
3 4
5 mol H SO =
3 mol H PO
Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4.
Ca5(PO4)3F+ 5H2SO4 3H3PO4 + HF + 5CaSO4
Method 1 Step by Step
Step 1 The starting substance is 784 grams of H3PO4.
Step 2 Convert grams of H3PO4 to moles of H3PO4.
Step 3 Convert moles of H3PO4 to moles of H2SO4 by the mole-ratio method.
3 4
3 4
1 mol H PO =
98.0 g H PO
3 48.00 mol H PO
2 413.3 mol H SO
3 4784 g H PO
3 48.00 mol H PO
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Mole Ratio
2 4
3 4
5 mol H SO =
3mol H PO
Calculate the number of moles of H2SO4 necessary to yield 784 g of H3PO4
Ca5(PO4)3F+ 5H2SO4 3H3PO4 + HF + 5CaSO4
Method 2 Continuous
grams H3PO4 moles H3PO4 moles H2SO4
The conversion needed is
3 4784 g H PO 3 4
3 4
1 mol H PO
98.0 g H PO
2 413.3 mol H SO
• How many grams of KClO3 are req’d to
produce 9.00 L of O2 at STP?
9.00 LO2
1 molO2
22.4 L O2
= 32.8 g KClO3
2 molKClO3
3 molO2
122.55g KClO3
1 molKClO3
? g 9.00 L
Stoichiometry Problems
2KClO3 2KCl + 3O2
57
Mass-Mass CalculationsMass-Mass Calculations
58
Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.N2 + 3H2 2NH3Method 1 Step by Step
Step 1 The starting substance is 112 grams of H2. Convert 112 g of H2 to moles.
grams moles2
2
1 mol H
2.02 g H
2112 g H 255.4 moles H
Step 2 Calculate the moles of NH3 by the mole ratio method.
3
2
2 mol NH=
3 mol H
255.4 moles H 336.9 moles NH
59
Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.N2 + 3H2 2NH3Method 1 Step by Step
Step 3 Convert moles NH3 to grams NH3.
moles grams
336.9 moles NH 3
3
17.0 g NH=
1 mol NH
3629 g NH
60
Calculate the number of grams of NH3 formed by the reaction of 112 grams of H2.N2 + 3H2 2NH3
grams H2 moles H2 moles NH3 grams NH3
2
2
1 mol H
2.02 g H
2112 g H 3
2
2 mol NH
3 mol H
Method 2 Continuous
3
3
17.0 g NH=
1 mol NH
3629 g NH
Stoichiometry Problems• How many grams of silver will be
formed from 12.0 g copper?
12.0g Cu
1 molCu
63.55g Cu
= 40.7 g Ag
Cu + 2AgNO3 2Ag + Cu(NO3)2
2 molAg
1 molCu
107.87g Ag
1 molAg
12.0 g ? g
63.55g Cu
1 molCu
Stoichiometry Problems
• How many grams of Cu are required to react with 1.5 L of 0.10M AgNO3?
1.5L
.10 molAgNO3
1 L= 4.8 g
Cu
Cu + 2AgNO3 2Ag + Cu(NO3)2
1 molCu
2 molAgNO3
? g 1.5L0.10M
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Limiting ReactantLimiting ReactantLimiting ReactantLimiting Reactant
64
• It is called the limiting reactant because the amount of it present is insufficient to react with the amounts of other reactants that are present.
• The limiting reactant limits the amount of product that can be formed.
• The limiting reactant is one of the reactants in a chemical reaction.
65
How many bicycles can be assembled from the parts shown?
From eight wheels four bikes can be constructed.
From four frames four bikes can be constructed.
From three pedal assemblies three bikes can be constructed.
The limiting part is the number of pedal assemblies.
9.2
66
H2 + Cl2 2HCl
+
7 molecules H2 can form 14 molecules HCl
4 molecules Cl2 can form 8 molecules HCl 3 molecules of H2 remain
H2 is in excess Cl2 is the limiting reactant
9.3
67
Steps Used to Determine Steps Used to Determine the Limiting Reactantthe Limiting Reactant
Steps Used to Determine Steps Used to Determine the Limiting Reactantthe Limiting Reactant
68
1. Calculate the amount of product (moles or grams, as needed) formed from each reactant.
2. Determine which reactant is limiting. (The reactant that gives the least amount of product is the limiting reactant; the other reactant is in excess.
3. Calculate the amount of the other reactant required to react with the limiting reactant, then subtract this amount from the starting quantity of the reactant. This gives the amount of the substance that remains unreacted.
69
ExamplesExamplesExamplesExamples
70
How many moles of HCl can be produced by reacting 4.0 mol H2 and 3.5 mol Cl2? Which compound is the limiting reactant?
Step 1 Calculate the moles of HCl that can form from each reactant.
24.0 mol H2
2 mol HCl
1 mol H
8.0 mol HCl
23.5 mol Cl2
2 mol HCl
1 mol Cl
7.0 mol HCl
H2 + Cl2 → 2HCl
Step 2 Determine the limiting reactant.
The limiting reactant is Cl2 because it produces less HCl than H2.
71
How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and 100.0 g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted?
Step 1 Calculate the grams of AgBr that can form from each reactant.
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
The conversion needed isg reactant → mol reactant → mol AgBr → g AgBr
250.0 g MgBr 102 g AgBr2
2
1 mol MgBr
184.1 g MgBr
2
2 mol AgBr
1 mol MgBr
187.8 g AgBr
1 mol AgBr
3100.0 g AgNO 110.5 g AgBr3
3
1 mol AgNO
169.9 g AgNO
3
2 mol AgBr
2 mol AgNO
187.8 g AgBr
1 mol AgBr
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How many moles of silver bromide (AgBr) can be formed when solutions containing 50.0 g of MgBr2 and 100.0 g of AgNO3 are mixed together? How many grams of the excess reactant remain unreacted?
Step 2 Determine the limiting reactant.
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
250.0 g MgBr 102 g AgBr2
2
1 mol MgBr
184.1 g MgBr
2
2 mol AgBr
1 mol MgBr
187.8 g AgBr
1 mol AgBr
3100.0 g AgNO 110.5 g AgBr3
3
1 mol AgNO
169.9 g AgNO
3
2 mol AgBr
2 mol AgNO
187.8 g AgBr
1 mol AgBr
The limiting reactant is MgBr2 because itforms less Ag Br.
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How many grams of the excess reactant (AgNO3) remain unreacted?
Step 3 Calculate the grams of unreacted AgNO3. First calculate the number of grams of AgNO3 that will react with 50 g of MgBr2.
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
250.0 g MgBr 392.3 g AgNO2
2
1 mol MgBr
184.1 g MgBr
3
2
2 mol AgNO
1 mol MgBr
3
3
169.9 g AgNO
1 mol AgNO
The conversion needed isg MgBr2 → mol MgBr2 → mol AgNO3 → g AgNO3
The amount of MgBr2 that remains is
100.0 g AgNO3 -92.3 g AgNO3 = 7.7 g AgNO3
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Reaction YieldReaction YieldReaction YieldReaction Yield
75
The quantities of products calculated from equations represent the maximum yield (100%) of product according to the reaction represented by the equation.
76
Many reactions fail to give a 100% yield of product.
This occurs because of side reactions and the fact that many reactions are reversible.
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• The theoretical yield of a reaction is the calculated amount of product that can be obtained from a given amount of reactant.
• The actual yield is the amount of product finally obtained from a given amount of reactant.
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• The percent yield of a reaction is the ratio of the actual yield to the theoretical yield multiplied by 100.
actual yield x 100 = percent yield
theoretical yield
79
187.8 g AgBr
1 mol AgBr
Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 1 Determine the theoretical yield by calculating the grams of AgBr that can be formed.
The conversion needed isg MgBr2 → mol MgBr2 → mol AgBr → g AgBr
2200.0 g MgBr 408.0 g AgBr2
2
1 mol MgBr
184.1 g MgBr
2
2 mol AgBr
1 mol MgBr
80
Silver bromide was prepared by reacting 200.0 g of magnesium bromide and an adequate amount of silver nitrate. Calculate the percent yield if 375.0 g of silver bromide was obtained from the reaction:
MgBr2(aq) + 2AgNO3 (aq) → 2AgBr(s) + Mg(NO3)2(aq)
Step 2 Calculate the percent yield.
actual yieldpercent yield = x 100
theoretical yield
percent yield = 375.0 g AgBr
x 100 =408.0 g AgBr
91.9%
must have same units
must have same units