we measure ordinary objects either by counting or weighing them, depending on which method is more...

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

Dozen = 12

Pair = 2

Certain nouns can be used to define a collection of objects

The mole

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.

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

Each of these contains one mole of the substance

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

copper iron

carbon sulfur

mercury

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

Molar mass

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.

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


• Compounds

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


MSO2 = MS + (2 x MO)

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

= 64.07 g/mol


• Compounds

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


MK2S = (2 x MK) + MS

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

= 110.27 g/mol

Interconverting moles, mass,and chemical entities

(atoms, molecules, ions, or other particles)

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


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)

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

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


Michelle Jose

for more advanced students, use silberberg page 76 fig 3.4

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

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

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

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

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

Chemical reactions and chemical equations

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

changed into one or more new substances


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”


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


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


Calculating the amounts of reactant and product

Figure on slices of bread + ham to make sandwich. Use this to

relate to stoich

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



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

2 mol CO = 2 mol CO2


2 mol CO

2 mol CO2

= 12 mol CO2

2 mol CO= 1



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

1 mol O2 = 2 mol CO2


1 mol O2

2 mol CO2

= 12 mol CO2

1 mol O2

= 1

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


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

alkali metal hydroxide

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

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


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


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

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)

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

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

Ch 2 F

• No meeting this Friday

• Lab discussion moved to March 2

• 1:30-3:30 pm

• SOM 201

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

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