Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill.

46
We measure ordinary objects either by counting or weighing them, depending on which method is more convenient Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill. http://weyume.com/wp-content/uploads/2011/04/rice.jpg http://farm1.static.flickr.com/21/90994367_5613e69fd9.jpg

description

We measure ordinary objects either by counting or weighing them, depending on which method is more convenient. Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill. http://weyume.com/wp-content/uploads/2011/04/rice.jpg - PowerPoint PPT Presentation

Transcript of Silberberg, M. 2010. Principles of General Chemistry. 2 nd ed. New York: McGraw-Hill.

Page 1: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

We measure ordinary objects either by counting or weighing them,

depending on which method is more convenient

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.http://weyume.com/wp-content/uploads/2011/04/rice.jpg

http://farm1.static.flickr.com/21/90994367_5613e69fd9.jpg

Page 2: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Dozen = 12

Pair = 2

Certain nouns can be used to define a collection of objects

Page 3: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

The mole

Page 4: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

The mole (n or mol) is the amount of matter that contains as many entities (atoms, molecules, ions, or other particles) as there are atoms in

exactly 12 g of the carbon-12 isotope (12C)

• The actual number of atoms in 12 g of carbon-12 was determined experimentally

• Avogadro’s number (NA)

NA = 6.02 x 1023

Brown, , E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd. Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Page 5: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Just as 1 dozen of oranges contains 12 oranges, 1 mole of matter contains 6.02 x 1023 entities

Brown, , E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd. Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

1 mole 12C atoms = 6.02 x 1023 12C atoms

1 mole H2O molecules = 6.02 x 1023 H2O molecules

1 mole NO3- ions = 6.02 x 1023 NO3

- ions

Page 6: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Each of these contains one mole of the substance

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

copper iron

carbon sulfur

mercury

Page 7: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

One mole (or an Avogadro’s number)is an extremely big number

• One mole of softdrink cans would cover the surface of the earth to a depth of over 300 kilometers

• If we were able to count the number of atoms at a rate of 10 million per second, it would take about 2 billion years to count a mole of atoms

Ateneo
from www.nasa.gov/worldbook/earth_worldbook.html:"Earth has a diameter of about 7900 miles (12700 kilometers)"to picture dimensions:1) one lap in the oval track over at Moro is 200 m2) one lap in the LS swimming pool is 25 m3) the distance from the main door of Schmitt to before you climb the stairs going to C-116 is about 25 m
Page 8: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Molar mass

Page 9: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

The molar mass (M) of a substance is the mass of one mole of its entities (atoms, molecules,

ions, or other particles) in units of g/mol

MH2O = 18.0 g/mol

(one mole of H2O molecule weighs 18.0 g)

MNO3- = 62.0 g/mol

(one mole of NO3- ion weighs 62.0 g)

MC = 12.01 g/mol

(one mole of C atom weighs 12.01 g)

Brown, T., E. LeMay, and B. Bursten. 2000. Chemistry: The Central Science. 8th ed. Phils: Pearson Education Asia Pte. Ltd. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-

Hill.

Page 10: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

The periodic table is indispensable for calculating the molar mass of a substance

• Elements

– M is the numerical value from the periodic table

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

MH = 1.008 g/mol

MO = 16.00 g/mol

Page 11: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

The periodic table is indispensable for calculating the molar mass of a substance

• Compounds

– M is the sum of the molar masses of the atoms of the elements in the formula

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

MSO2 = MS + (2 x MO)

= 32.07 g/mol + (2 x 16.00 g/mol)

= 64.07 g/mol

Page 12: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

The periodic table is indispensable for calculating the molar mass of a substance

• Compounds

– M is the sum of the molar masses of the atoms of the elements in the formula

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

MK2S = (2 x MK) + MS

= (2 x 39.10 g/mol) + 32.07 g/mol

= 110.27 g/mol

Page 13: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Interconverting moles, mass,and chemical entities

(atoms, molecules, ions, or other particles)

Page 14: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

The factor-label method is used to convert from one unit to another

1 peso = 4 25-centavos

1 peso

4 25-centavos= 1

1 peso

4 25-centavos= 1

unit factor

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Michelle Jose
mention that the use of unit factors does not violate any math-remember that any number multiplied by one is the number itself-we are only translating the same thing from one unit to another (i.e. whether we speak of one peso coin or 4 25-centavo coins, we are talking about the same amount of money. we are only changing the denomination)
Page 15: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Alexa bought 18 fresh chicken’s eggs. How many dozens of egg did she buy?

dozens of egg = x18 eggs 1 dozen egg

12 eggs

= 1.5 dozens of egg

unit factor

Page 16: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

In order to convert between moles, mass, and chemical entities, the factor label method is used

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

Michelle Jose
for more advanced students, use silberberg page 76 fig 3.4
Page 17: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Methane (CH4) is the principal component of natural gas. How many moles of methane

are present in 6.07 g of CH4?

MCH4 = MC + (4 x MH)

= 12.01 g/mol + (4 x 1.01 g/mol)

= 16.05 g/mol

Report final answer with the correct number of significant figures!

nCH4 = x6.07 g CH4 1 mol CH4

16.05 g CH4

= 0.378 mol CH4

Michelle Jose
when solving problems,1. what are the given?2. what are you supposed to look for?3. what will be your strategy?4. show solutionfor multi-step problems, round off only in the end
Page 18: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

How many molecules of methane are present 6.07 g of CH4?

molecules CH4 = x 6.07 g CH4 1 mol CH4

16.05 g CH4

= 2.28 x 1023 molecules CH4

x6.02 x 1023 molecules CH4

mol CH4

Page 19: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Glucose (C6H12O6), also known as blood sugar, is used by the body as energy source.

How many moles of glucose are present in 1.75 x 1022 molecules of glucose?

nC6H12O6 = x 1.75 x 1022 molecules C6H12O6 1 mol C6H12O6

6.02 x 1023 molecules C6H12O6

= 0.0291 mol C6H12O6

Michelle Jose
why are hospital patients intravenously fed with glucose (aka dextrose) and not any other sugar? -see explanation in the comment box
Page 20: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

How many grams of glucose are present in 1.75 x 1022 molecules of glucose?

MC6H12O6 = (6 x MC) + (12 x MH) + (6 x MO)

= (6 x 12.01 g/mol) + (12 x 1.01 g/mol)

+ (6 x 16.00 g/mol)

= 180.18 g/mol

Page 21: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

How many grams of glucose are present in 1.75 x 1022 molecules of glucose?

nC6H12O6 = x 1.75 x 1022 molecules C6H12O6 1 mol C6H12O6

6.02 x 1023 molecules C6H12O6

180.18 g C6H12O6x

1 mol C6H12O6

= 5.24 g C6H12O6

Page 22: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

1. Urea [(NH2)2CO] is used in animal feeds, as fertilizer and in the manufacture of polymers.

a. Draw the Lewis structure of area. C is surrounded by O and the N’s. (Where are the H’s connected?)

b. Calculate its molar mass.

c. Consider 25.6 g of urea. How many moles of urea present?

d. How many moles of N are present?

e. How many moles of C are present?

f. How many molecules of urea are present?

g. How many atoms of N are present?

Page 23: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

2. Vitamin C, ascorbic acid, is often sold as sodium ascorbate.

a. Calculate its molar mass.

b. Consider a 500.-mg tablet. How many moles of sodium ascorbate are present?

c. How many moles of C are present?

d. How many moles of Na are present?

e. How many formula units of sodium ascorbate are present?

f. How many atoms of Na are present?

OO

OH

-O

HOOH

Na+

Page 24: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Chemical reactions and chemical equations

Page 25: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

A chemical reaction shows the process in which a substance (or substances) is

changed into one or more new substances

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Page 26: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

A chemical equation uses chemical symbols to show what happens during a chemical reaction

reactants product

(g) (g) (l)

“Two molecules of hydrogen react with one molecule of oxygen to yield two moles of water”

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Page 27: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

The Law of Conservation of Mass states that matter is neither created nor destroyed

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Page 28: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

To conform with the Law of Conservation of Mass, there must be the same number of each type of atom on both sides of the arrow. Hence, we balance the equation by

adding coefficients before each chemical symbol

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill.

Page 29: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Calculating the amounts of reactant and product

Page 30: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Stoichiometry a double cheeseburger

2 bun slices + 2 cheese slices + 2 burger patties =

Page 31: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

In a balanced equation, the number of moles of one substance is equivalent to the number of

moles of any of the other substances

2CO(g) + O2(g) 2CO2(g)

2 mol CO = 1 mol O2

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

2 mol CO

1 mol O2

= 11 mol O2

2 mol CO= 1

Page 32: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

In a balanced equation, the number of moles of one substance is equivalent to the number of

moles of any of the other substances

2CO(g) + O2(g) 2CO2(g)

2 mol CO = 2 mol CO2

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

2 mol CO

2 mol CO2

= 12 mol CO2

2 mol CO= 1

Page 33: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

In a balanced equation, the number of moles of one substance is equivalent to the number of

moles of any of the other substances

2CO(g) + O2(g) 2CO2(g)

1 mol O2 = 2 mol CO2

Chang, R. 2002. Chemistry 7th ed. Singapore: McGraw-Hill. Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

1 mol O2

2 mol CO2

= 12 mol CO2

1 mol O2

= 1

Page 34: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

The amount of one substance in a reaction is related to that of any other

Silberberg, M. 2010. Principles of General Chemistry. 2nd ed. New York: McGraw-Hill.

Page 35: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

All alkali metals react with water to produce hydrogen gas and the corresponding

alkali metal hydroxide

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

Page 36: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

How many moles of H2 will be formed by the complete reaction of 6.23 moles of Li with water?

nH2 = x6.23 mol Li 1 mol H2

2 mol Li

= 3.12 mol H2

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

Page 37: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

How many grams of H2 will be formed by the complete reaction of 80.57 g of Li with water?

2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)

mH2 = x x80.57 g Li 1 mol Li 1 mol H2

6.941 g Li 2 mol Li

2.016 g H2x

1 mol H2

= 11.70 g H2

Page 38: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

In a lifetime, the average American uses about 794 kg of copper in coins, plumbing, and wiring. Copper is obtained from sulfide ores (such as Cu2S) by a multistep process. After an initial

grinding, the first step is to “roast” the ore (heat it strongly with O2) to form Cu2O and SO2

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

Page 39: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

How many moles of oxygen are required to roast 10.0 mol of Cu2S?

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

nO2 = x10.0 mol Cu2S 3 mol O2

2 mol Cu2S

= 15.0 mol O2

Page 40: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

How many grams of SO2 are formed when 10.0 mol of Cu2S is roasted?

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

mSO2 = x x 10.0 mol Cu2S 2 mol SO2 64.07 g SO2

2 mol Cu2S 1 mol SO2

= 641 g SO2

Page 41: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Ch 2 F

• No meeting this Friday

• Lab discussion moved to March 2

• 1:30-3:30 pm

• SOM 201

Page 42: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

How many grams of O2 are required to form 2.86 kg of Cu2O?

2Cu2S(s) + 3O2(g) 2Cu2O(s) + 2SO2(g)

mO2 = x x2.86 kg Cu2O 1000 g Cu2O 1 mol Cu2O

1 kg Cu2O 143.10 g Cu2O

3 mol O2 32.00 g O2x x

2 mol Cu2O 1 mol O2

= 960 g O2

Page 43: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Limiting Reactants

The reactant used up first in a chemical reaction is called the limiting reactant. Excess reactants are present in quantities greater than necessary to react with the quantity of the limiting reactant.

A + B --- C + D

Given the amounts of A and B, which is the limiting reactant?

Page 44: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Urea is prepared by reacting ammonia with carbon dioxide:

2NH3(g) + CO2(g) -- (NH2)2CO(aq) + H2O(l)

In one process, 637.2 g of NH3 are allowed to react with 1142 g of CO2.

• Which is the limiting reactant?

• How much urea (in grams) is produced?

• How much of the excess reactant (in grams) is left at the end of the reaction?

Page 45: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

Strategy

• Convert mass of each reactant to moles

• Calculate the amount of product formed from the each of the reactants.

• The reactant the produces the less amount is the limiting reactant.

Page 46: Silberberg, M. 2010.  Principles of General Chemistry. 2 nd  ed.  New York: McGraw-Hill.

1. The reaction between aluminum and iron (III) oxide can generate temperatures around 3000⁰C and is used in welding metals:

2Al + Fe2O3 -- Al2O3 + 2Fe

In one process, 124 g of Al are reacted with 601 g of ferric oxide.

(a)Which is the limiting reactant?

(b)How much Al2O3 (in grams) is produced?

(c)How much of the excess reactant (in grams) is left at the end of the reaction?

2. Titanium is a strong & light metal used in rockets & aircrafts. It is prepared by the reaction between titanium (IV) chloride with molten magnesium at around 1000⁰C: TiCl4 + 2Mg -- Ti + 2MgCl2In a certain industrial operation, 3.54 x 107g of TiCl4 are reacted with 1.13 x 107 g of magnesium.

(a)Which is the limiting reactant?

(b)How much Ti (in grams) is produced?

(c)How much of the excess reactant (in grams) is left at the end of the reaction?