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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
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
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
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
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
Chang, R. 2002. Chemistry 7th ed. Singapore: 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
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.
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
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
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
Chang, R. 2002. Chemistry 7th ed. Singapore: 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.
The Law of Conservation of Mass states that matter is neither created nor destroyed
Chang, R. 2002. Chemistry 7th ed. Singapore: 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.
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
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
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
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.
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
2Li(s) + 2H2O(l) 2LiOH(aq) + H2(g)
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
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