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Back to Moles • number of atoms in exactly 12 grams carbon-12

Chem Unit5

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Page 1: Chem Unit5

Back to Moles

• • number of atoms in exactly 12 grams carbon-12

Page 2: Chem Unit5

Avogadro’s Number• There are 6.02 x 1023 particles in a

mole of any substance.

Page 3: Chem Unit5

Molar Mass

• Mass of one MOLE of pure substance

• Helium: 4.00 grams -------> 4 grams per mole

• Lithium 6.94 grams ---------> 6.94 grams per mole

• Mercury 200.59 grams -------> 200.59 grams per mole

Page 4: Chem Unit5

0.1 Moles of Uranium

•(0.1 moles) x 238 grams/mole

= 23.8 grams

Page 5: Chem Unit5

0.1 moles Uranium

•How many atoms?

• 6.022 x 1023 atoms/mole x 0.1 moles

= 6.022 x 1022 atoms

Page 6: Chem Unit5

Molecules

•5 moles of CO2

= 5 moles x 6.022 x 1023 molecules/mole

= 3.1 x 1024 molecules of CO2

Page 7: Chem Unit5

Molar Mass

•Mass in grams of one moleof the substance

CO2: mass C + 2 (mass O)

= 12 + 2(16) = 44 grams

Page 8: Chem Unit5

Mass -> Moles

•Converting from g to moles:

For 112 grams of C2H4:

Grams of SubstanceGrams of

Substance Molar mass MOLES

Page 9: Chem Unit5

Mass -> Moles•Converting from g to moles:

For 112 grams of C2H4:

112 g x 1 mole/28 g = 4 moles

Grams of SubstanceGrams of

Substance Molar mass MOLES

Page 10: Chem Unit5

Moles -> Molecules• Converting from moles to Molecules

For 4 moles of C2H4:4 moles x 6.02 x 1023 molecules/mole = 2.41 x 1024

molecules

MolesMolesAvogadro’s Number molecules

Page 11: Chem Unit5

Law of Definite Proportions

• Joseph Proust

• Regardless of the source, a compound always has the same proportion of elements

Page 12: Chem Unit5

Law of Definite Proportions

•Copper (II) Carbonate

CuCO3 -> Copper + Carbon + Oxygen

51.36% 9.72% 38.85%

Page 13: Chem Unit5

Percent Composition

•CO CO2

28 g/mole 44 g/mole

C: 12/28 = 42.86% C: 12/44 = 27.27%O: 16/28 = 57.14% O: 32/44 = 72.73%

Two different compound with different percentcompositions and different properties

Page 14: Chem Unit5

Empirical Formula

• Smallest whole number ratio of atoms in a compound

• CaffeineMolecular formulaC8H10N4O2

Empirical formulaC4H5N2O

Page 15: Chem Unit5

Percent Composition -> Empirical Formula

•Vinegar

•C: 40%H: 6.67%

•O: 53.33%

Page 16: Chem Unit5

Percent Composition -> Empirical Formula

•Vinegarassume 100 g

C: 40 gramsH: 6.67 gramsO: 53.33 grams

Page 17: Chem Unit5

Percent Composition -> Empirical Formula

C: 40 g x 1mole/12g = 3.33 moles

H: 6.67 g x 1 mole/1g = 6.67 moles

O: 53.33 g x 1 mole/16 g = 3.33 moles

Page 18: Chem Unit5

Percent Composition -> Empirical Formula

C: 40 g x 1mole/12g = 3.33 moles

H: 6.67 g x 1 mole/1g = 6.67 moles

O: 53.33 g x 1 mole/16 g = 3.33 moles

Empirical Formula:

CH2O

Page 19: Chem Unit5

Percent Composition -> Empirical Formula

C: 40 g x 1mole/12g = 3.33 moles

H: 6.67 g x 1 mole/1g = 6.67 moles

O: 53.33 g x 1 mole/16 g = 3.33 moles

Actual Formula:

C2H4O2

Page 20: Chem Unit5

Percent Composition -> Empirical Formula

C: 40 g x 1mole/12g = 3.33 moles

H: 6.67 g x 1 mole/1g = 6.67 moles

O: 53.33 g x 1 mole/16 g = 3.33 moles

Actual Formula:

C2H4O2

Page 21: Chem Unit5

Empirical Formula-> Molecular Formula

Empirical Formula: CH2O

Molecular Formula: C2H4O2

Molecular Formula is a whole number multiple of empirical formula

Page 22: Chem Unit5

Empirical Formula-> Molecular Formula

Butyric Acid = butter flavor Empirical formula = C2H4O

Molar mass = 88 g/mole

Molecular formula =

Page 23: Chem Unit5

Empirical Formula-> Molecular Formula

Butyric Acid = butter flavor Empirical formula = C2H4O

Molar mass = 88 g/mole Mass of emp. formula = 44 Molecular formula =

Page 24: Chem Unit5

Empirical Formula-> Molecular Formula

Butyric Acid = butter flavor Empirical formula = C2H4O

Molar mass = 88 g/mole Mass of emp. formula = 44 Molecular formula = C4H8O2

Page 25: Chem Unit5

Formula -> Percent Composition

Empirical formula Butyric Acid: C2H4O

Molar mass C x 2 = 12 x 2 = 24 Molar mass H x 4 = 1 x 4 = 4

Molar mass O = 16

44 g/mole

Page 26: Chem Unit5

Formula -> Percent Composition

Empirical formula Butyric Acid: C2H4O

C = 24/44 = 0.545 = 54.5% H = 4/44 = 0.091 = 9.1%

O = 16/44 = 0.364 = 36.4%

Page 27: Chem Unit5

Chemical VS. Physical Change

Physical Chemical

Page 28: Chem Unit5

Chemical VS. Physical Change

Physical Chemical• change in phase • atomic rearrangement

• change in appearance, • energy change

shape (absorption or release)

Page 29: Chem Unit5

Chemical Equations

Shorthand symbols:(s), (l), (g) = solid, liquid, gas phases

(aq) = substance in aqueous solution

--> = reaction event AX -> BY

<=> reaction in equilibrium (reversible)

Page 30: Chem Unit5

Chemical Equations

Silicon tetrachloride and water react to form silicon dioxide and hydrochloric acid

Page 31: Chem Unit5

Chemical Equations

Silicon tetrachloride and water react to form silicon dioxide and hydrochloric acid

SiCl4 + H2O SiO2 + HCl

Reactants Products

Page 32: Chem Unit5

Chemical Equations

Silicon tetrachloride and water react to form silicon dioxide and hydrochloric acid

SiCl4 + 2 H2O SiO2 + 4 HCl

Balanced

Page 33: Chem Unit5

Balancing Chemical Equations

Conservation of Matter:

H2 + O2 ---------------> H2O

Page 34: Chem Unit5

Balancing Chemical Equations

Conservation of Matter:

2 H2 + O2 ---------------> 2 H2O

Page 35: Chem Unit5

Balancing Chemical Equations

propane gas + oxygen gas -> carbon dioxide gas + water vapor

Page 36: Chem Unit5

Balancing Chemical Equations

propane gas + oxygen gas -> carbon dioxide gas + water vapor

C3H8(g) O2(g) -> CO2(g) H2O(g)

Page 37: Chem Unit5

Balancing Chemical Equations

propane gas + oxygen gas -> carbon dioxide gas + water vapor

C3H8(g) O2(g) -> 3 CO2(g) H2O(g)

Page 38: Chem Unit5

Balancing Chemical Equations

propane gas + oxygen gas -> carbon dioxide gas + water vapor

C3H8(g) O2(g) -> 3 CO2(g) 4 H2O(g)

Page 39: Chem Unit5

Balancing Chemical Equations

RULES:

•Identify reactants and products

•Count atoms

•Insert coefficients

•Check your work

Page 40: Chem Unit5

Balancing Chemical Equations

aluminum metal and iron (II) nitrate react to form aluminum nitrate and iron metal

Page 41: Chem Unit5

Balancing Chemical Equations

aluminum metal and iron (II) nitrate react to form aluminum nitrate and iron metal

Al + Fe(NO3)2 Al(NO3)3 + Fe

Page 42: Chem Unit5

Balancing Chemical Equations

aluminum metal and iron (II) nitrate react to form aluminum nitrate and iron metal

2 Al + 3 Fe(NO3)2 2 Al(NO3)3 + 3 Fe

Balanced

Page 43: Chem Unit5

Types of Chemical Equations

1. Combustion

2. Synthesis

3. Decomposition

4. Single Displacement

5.Double Displacement

6.Acid – Base*

Page 44: Chem Unit5

Combustion

Reaction of a carbon-based compound with oxygen

2 C4H10 + 13 O2 8 CO2 + 10 H2O

butane

Page 45: Chem Unit5

Combustion

Alcohol lamp

C2H5OH + 3 O2 2 CO2 + 3 H2O

Page 46: Chem Unit5

Synthesis

Forming a single compound from two or more reactants

H2 + O2 H2O

Page 47: Chem Unit5

Synthesis

Forming a single compound from two or more reactants

2 Na + Cl2 2 NaCl

Page 48: Chem Unit5

Decomposition

Breakdown of a single compound(Opposite of synthesis)

2 NaN3 2 Na + 3N2

Page 49: Chem Unit5

Decomposition

Breakdown of a single compound(Opposite of synthesis)

NH4NO3 N2O + 2 H2O

Page 50: Chem Unit5

Single Displacement

A single element reacts with a Compound and displaces another element from that compound

Zn + 2 HCl ZnCl2 + H2

Page 51: Chem Unit5

Single Displacement

A single element reacts with a Compound and displaces another element from that compound

2 K + 2 H2O 2 KOH + H2

Page 52: Chem Unit5

Activity Series

Element ReactivityK

CaNa

React with cold water and acids to replace hydrogen; react with oxygen to form oxides

MgAlZnFe

React with steam (not cold water) and acids to replace hydrogen; react with oxygen to form oxides

NiPb

Do not react with water; react with acids to replace hydrogen; react with oxygen to form oxides

H2

Cu React with oxygen to form oxides

AgAu Unreactive; form oxides only indirectly

Page 53: Chem Unit5

Activity Series

What will happen if iron (steel wool) is added to a copper(II) sulfate solution?

Page 54: Chem Unit5

Activity Series

Fe(s) + CuSO4 (aq) -> Cu(s) + FeSO4(aq)

Page 55: Chem Unit5

Double Displacement

Two compounds exchange ions andForm new compounds

Pb(NO3)2 + KI PbI2(s) + 2 KNO3

Page 56: Chem Unit5

Net Ionic Equations

When aqueous solutions are mixed, new ionic combinationsare made.

Mass and Charge must beconserved.

Na2SO4(aq) + Sr(NO3)2(aq) SrSO4(s) + 2 NaNO3(aq)

precipitate

Page 57: Chem Unit5

Net Ionic Equations

Precipitate = insoluble solid thatSeparates from solution

Na2SO4(aq) + Sr(NO3)2(aq) SrSO4(s) + 2 NaNO3(aq)

precipitate

Page 58: Chem Unit5

Net Ionic Equations

Na2SO4 - > 2 Na+ + SO42-

Sr(NO3)2 - > Sr2+ + 2 NO3-

NaNO3 - > Na+ + NO3-

Na2SO4(aq) + Sr(NO3)2(aq) SrSO4(s) + 2 NaNO3(aq)

precipitate

Page 59: Chem Unit5

Net Ionic Equations

Na2SO4 - > 2 Na+ + SO42-

Sr(NO3)2 - > Sr2+ + 2 NO3-

NaNO3 - > Na+ + NO3-

2 Na+ + SO42- + Sr2+ + 2 NO3

- SrSO4(s) + 2 Na+ + 2 NO3-

spectator ions precipitate

Page 60: Chem Unit5

Solubility Rules

All alkali metal and NH4+ salts are

soluble;

All nitrates (NO3) are soluble;

All Halides (F, Cl, Br, I) are solubleexcept Ag+ and Pb2+

Sulfates(SO4), except for Ca2+, Ba2+, Sr2+, Pb2+ are soluble

Carbonates (CO3) are insoluble except for alkali metal and NH4.

Page 61: Chem Unit5

Net Ionic Equations

Li2CO3 + BaBr2 BaCO3 + 2 LiBr

Page 62: Chem Unit5

Net Ionic Equations

Li2CO3 -> 2 Li+ + CO32-

BaBr2 -> Ba2+ + 2 Br-

Li2CO3 + BaBr2 BaCO3 + 2 LiBr

precipitate

Page 63: Chem Unit5

Unit 5 Review

Molar Mass g <-> atoms/molecules <-> moles

Percent Composition Empirical and Molecular Formula

Chemical equations: Word Equations, Balancing