Unit 2 – Section C

Unit 2 – Section C

Conserving Matter

HW 1

Read & take notes on Section C.1

C.1 – Keeping Track of Atoms

The law of conservation of matter – in a chemical reaction matter is neither created nor destroyed.

+ C O2 CO2

1 Carbon 1 oxygen 1 carbon dioxide atom (C) molecule (O2) molecule (CO2)

Molecules can be converted and decomposed by chemical processes: but atoms are forever.

C.1 – Keeping Track of Atoms(continued)

Reactants are placed on the left of the arrow;Products are placed on the right.

+ C O2 CO2

1 Carbon 1 oxygen 1 carbon dioxide atom (C) molecule (O2) molecule (CO2)

In a balanced chemical equation, the number of atoms for left side equals the number for the right side.

Coefficients indicate the relative number of each unit involved.

+ Cu (s) O2 (g) CuO (s)

2 Copper oxygen 2 copper oxide

atoms (Cu) molecule (O2) molecules (CuO)


Chemists use the term formula unit when referring to the smallest unit of an ionic compound.

+ Cu (s) O2 (g) CuO (s)

2 Copper oxygen 2 copper oxide

atoms (Cu) molecule (O2) molecules (CuO)


Classwork

Answer questions 1-5 in Section C.2 pg 155

C.2 – Accounting for Atoms

1) Methane burning with oxygen

CH4 + 2 O2 CO2 + 2 H2O

Reactants ProductsC CH HO O

C.2 – Accounting for Atoms(continued)

2) Hydrobromic acid reacting with magnesium

HBr + Mg H2 + MgBr2

Reactants ProductsH HBr BrMg Mg


3. Hydrogen sulfide and metallic silver react

4 Ag + 4 H2S + O2 2 Ag2S + 4 H2O

Reactants Products

Ag AgH HS SO O


4. Cellulose burns to form carbon dioxide and water vapor.

C6H10O5 + 6 O2 6 CO2 + 5 H2O

Reactants ProductsC CH HO O


5. Nitroglycerin decomposes explosively to form nitrogen, oxygen, carbon dioxide and water vapor.

2 C3H5(NO3) 3 3 N2 + O2 + 6 CO2 + 5 H2O

Reactants ProductsC CH HN NO O

HW 2

Read & take notes on Section C.3Address & answer questions 1-6 in section C.4

C.3 – Nature’s Conservation: Balancing Chemical Equations

If polyatomic ions (examples NO3-, CO3

2-) appear as both reactants and product treat them as units.

If water is involved, balance the hydrogen and oxygen atoms last.

Recount all atoms after you think an equation is balanced.

C.4 – Writing Chemical EquationsWriting to balance the chemical equations…

Methane Chlorine Chloroform Hydrogen chloride

__ CH4 + __ Cl2 __ CHCl3 + __ HClReactants Products

C

H

Cl

C

H

Cl

C.4 – Writing Chemical Equations(continued)

1a. __ C + __ O2 __ COReactants Products

C

O

C

O


1b. __ Fe2O3 + __ CO __ Fe + __ CO2

Reactants Products

C

Fe

O

C

Fe

O


2. __ CuO + __ C __ Cu + __ CO2

Reactants Products

C

Cu

O

C

Cu

O


3. __ O3 __ O2

Reactants Products

O

O


4. __ NH3 + __ O2 __ NO2 + __ H2OReactants Products

NHO

NHO


5. __ Cu + __ AgNO3 __ Cu(NO3)2 + __ Ag

Reactants Products

CuAgNO

CuAgNO


6. __ C8H18 + __ O2 __ CO2 + __ H2O

Reactants Products

C

H

O

C

H

O

HW 3


C.5 – Introducing the Mole Concept

Chemist have created a counting unit for elements called the mole (symbolized mol).

One mole of ANY element or molecule contains

602 000 000 000 000 000 000 000 particles

6.02 x 1023

C.5 – Introducing the Mole Concept(continued)

Furthermore, the atomic weight of elements can be used to find the molar mass of a substance.

One mole of boron atoms (6.02 x 1023) would have a molar mass of 10.81 g


More examples…

One mole of carbon atoms (6.02 x 1023) would have a molar mass of _______ g

C.5 – Introducing the Mole Concept(continued)More examples…

One mole of copper atoms (6.02 x 1023) would have a molar mass of ______ g

One mole of silver atoms (6.02 x 1023) would have a molar mass of ______ g

One mole of gold atoms (6.02 x 1023) would have a molar mass of ______ g


(Curve ball) How about the molar mass of oxygen gas (O2)?

One mole of oxygen gas (O2) molecules (6.02 x 1023) would have a molar mass of _______ g


(Curve ball 2) How about the molar mass of water (H2O)?

One mole of water (H2O) molecules (6.02 x 1023) would have a molar mass of _______ g

C.6 – Molar Masses

HW Questions 1-4,6,8

pg 163

C.6 – Molar Masses1. One mole of nitrogen (N) atoms (6.02 x 1023)

would have a molar mass of _______ g

2. One mole of nitrogen (N2) molecules (6.02 x 1023) would have a molar mass of _______ g

C.6 – Molar Masses(continued)

3. One mole of table salt (NaCl) molecules (6.02 x 1023) would have a molar mass of _______ g


4. One mole of table sugar (C12H22O11) molecules (6.02 x 1023) would have a molar mass of _______ g


6. One mole of magnesium phosphate Mg3(PO4)2 molecules (6.02 x 1023) would have a molar mass of ______ g


8. One mole of calcium hydroxyapatite Ca10(PO4)6(OH)2 molecules (6.02 x 1023) would have a molar mass of ______ g

HW 5


C.7 – Equations and Molar RelationshipsLet’s revisit copper-refining…

2 CuO(s) + C(s) 2 Cu(s) + CO2(g)

Alternatively stated…

2 mol CuO + 1 mol C 2 mol Cu + 1 mol CO2

In this example, for every two moles of CuO that react, one mole of CO2 is produced.

C.7 – Equations and Molar Relationships(continued)

Sample Problem: A refiner needs to convert 955.0 g CuO to pure Cu. What mass of C is needed for this reaction?

Using . . .


We can reason that . . .

1 mol CuO

79.55 g CuO

X mol CuO

955.0 g CuO=


Sample Problem: A refiner needs to convert 955.0 g CuO to pure Cu. What mass of C is needed for this reaction? (continued)

Solving for X . . . 1 mol CuO

79.55 g CuO

X mol CuO

955.0 g CuO=

955.0 g CuO X 1 mol CuO = 79.55 g CuO X X mol CuO

X mol CuO955.0 g CuO X 1 mol CuO

79.55 g CuO=

NOTICE . . .



Solving for X . . .

1 mol CuO

79.55 g CuO

X mol CuO955.0 g CuO X 1 mol CuO

79.55 g CuO=

12.01 mol CuO = XThis all started with . . .

A proportion we created called a conversion factor, both referring to the same number of particles.



So . . .

The refiner knows there are 12.01 mol CuO in 955.0 g.

Remembering the equation we started with . . .


12.01 mol CuO X 1 mol C

2 mol CuO= 6.005 mol C



So . . . Once we know

The refiner knows we need 6.005 mol C to refine 955 g of CuO we can calculate the actual mass of C needed . . .

Appropriate conversion factors . . .

6.005 mol C X 12.01 g C

1 mol C = 72.12 g C

C.8 – Molar Relationships

HW 6 Questions 1-4 pg 166


1. CuO(s) + 2 HCl(aq) CuCl2(aq) + H2O (l)

Molar mass of each…

C.8 – Molar Relationships(continued)

2. Mass (in grams) of:

a. 1.0 mol HCl

b. 5.0 mol HCl

c. 0.50 mol CuO

C.8 – Molar Relationships (continued)

3. # of moles represented by

a. 941.5 g CuCl2

b. 201.6 g CuCl2

c. 73.0 g HCl


4. CuO(s) + 2 HCl(aq) CuCl2(aq) + H2O (l)

a. How many moles of CuO are needed to react with 4 mol HCl?

a. How many moles of HCl are needed to react with 4 mol CuO?

Mole Quiz (explained)Start with 6 K + B2O3 3 K2O + 2 B

Please write out the conversion factor for the reactants and products of the above chemical equation.

1) 1 mol K

-------------------

39.098 g K

2) 1 mol B2O3

-------------------

69.619 g B2O3

3) 1 mol K2O

-------------------

94.194 g K2O

4) 1 mol B

-------------------

10.811 g B

6 K + B2O3 3 K2O + 2 BA processor needs to convert 955.0 g B2O3 to pure B. What mass of K is needed for this reaction? (show all your work)

Mole Quiz (continued)

3219 g K

16 Al + 3 S8 8 Al2S3

A processor plans to create 1000. g of Al2S3. What mass of pure S is needed for this reaction? (show all your work)

Mole Quiz (continued)

640.8 g S8

HW 7


C.9 – Compositions of MaterialsThe percent mass of each material

found in an item is called the percent composition.

Hint: remember solution concentration

Example: post-1982 penny has a mass of 2.500 g has 2.4375 g zinc & 0.0625 g copper. What is the percent composition?

2.4375 g zinc

2.500 g total

X 100% =

97.50 % zinc

C.9 – Compositions of Materials(continued)

Example: post-1982 penny has a mass of 2.500 g has 2.4375 g zinc & 0.0625 g copper. What is the percent composition?

0.0625 g copper

2.500 g total

X 100% =

2.50 % copper

C.9 – Compositions of Materials(continued)

Why are the ideas of molar mass & percentage composition so important?

Mass of copper

Mass of Cu2S

X 100% =

% copper

Some Copper-containing Minerals

Common Name Formula

Chalcocite Cu2S

Chalcopyrite CuFeS2

Malachite Cu2CO3(OH)2

It helps us determine which is more profitable to mine.

C.10 – Percent Composition

HW 8 – Questions 1-2 on pg 168

C.10 – Percent Composition

1. An atom inventory for Cu3(CO3)2(OH)2

There are : 3 Cu atoms

2 C atoms

8 O atoms

2 H atoms

C.10 – Percent Composition(continued)

2. Percent copper in ? a. Chalcopyrite CuFeS2

63.55 g + 55.85 g + (2 x 32.07g)

Start with . . .

= 183.54 g

63.55 g

183.54 gx 100 % = 34.62 % Cu

(2 x 63.55 g) + 12.01 g + (5 x 16.00 g) + (2 x 1.008)

C.10 – Percent Composition(continued)

2. Percent copper in ? b. malachite Cu2CO3 (OH)2

c. Chalcocite , at 79.85% copper, would be the most profitable to mine.

= 221.13 g(2 x 63.55 g)

221.13 gx 100 % = 57.48 % Cu

C.11 – Retrieving Copper

HW 9 – please pre-read the lab.

HW 10


C.12 – Conservation in the CommunityResources…

RenewableFresh water,Air,Fertile soil , Plants, andAnimals

EVENTUALLY replenished by natural processes.

NonrenewableMetals,Natural gas, coal and petroleum

CANNOT be readily replenished.

C.13 – Rethinking, Reusing, Replacing & Recycling

HW 11 Questions 1 & 2 on pg. 176

C.14 – The Life Cycle of Material

Not covered

C.15 – Copper Life-Cycle Analysis

Documents

Unit 2 – Section C