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Ch. 6 – Chemical QuantitiesThe Mole
• What is a mole? It is a unit for _________in chemistry.
• It is similar to a dozen, except instead of 12 things, it’s
602 billion trillion things… (602,000,000,000,000,000,000,000)
• ___________ (in scientific notation)
• This number is named in honor of AmedeoAmedeo _________ (1776 – (1776 –
1856)1856), who studied quantities of gases and discovered that equal
volumes of gases under the same conditions of temperature and
pressure contained the same number of particles.
Avogadro
6.02 x 10 23
counting
• A mole is a term for a certain ______________ of objects.
1 mole = 6.02 x 1023 objects
*Other words that represent quantities:
1 pair = __ objects; 1 dozen = __ objects
1 gross = ____ objects; 1 _____ = 24 cans of soda
• Since this value is so huge, it is used to measure very small objects like ___________ and _______________.
The Mole Concept
number
2 12
144 case
atoms molecules
Just How Big is a Mole?Just How Big is a Mole?
• It’s enough soft drink cans to cover the surface of the earth to a depth of over 200 miles.
• If you had Avogadro's number of un-popped popcorn kernels, and spread them across the United States of America, the country would be covered in popcorn to a depth of over 9 miles.
• If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole.
• 1 dozen cookies = ___ cookies• 1 mole of cookies = ___________ cookies
• 1 dozen cars = ___ cars• 1 mole of cars = __________ cars
• 1 dozen Al atoms = ___ Al atoms• 1 mole of Al atoms = __________ atoms
• Mole is abbreviated ______ .
The MoleThe Mole12
6.02 x 1023
126.02 x 1023
126.02 x 1023
mol
Gram-Formula Mass
• The # of grams that 6.02 x 1023 particles, (or ___ mole), weighs is called the gram-formula mass. The mass is found from the atomic masses of the elements on the ____________ __________.
• This quantity has other nicknames too!
________ ______or _________ ________ or Formula Weight
1
periodic table
Practice Problems:
Calculate the gram-formula mass of each compound.a) CaCO3 b) (NH4)2SO4
c) N2O5
Molar Mass Formula Mass
Ca = 40.1C = 12.03 O’s =3 x 16.0 = 48.0
Add them up!
100.1 g/mol2 N’s = 2 x 14.0 = 28.08 H’s = 8 x 1.0 = 8.0S = 32.14 O’s = 4 x 16.0 = 64.0
Add them up!
132.1 g/mol
2 N’s = 2 x 14.0 = 28.05 O’s = 5 x 16.0 = 80.0
Add them up!
108.0 g/mol
3) Convert 835 grams of SO3 to moles.
4) How many molecules of CH4 are there in 18 moles?
5) How many grams of helium are there in 5.6 x 1023 atoms of helium?
6) How many molecules are there in 3.7 grams of H2O?
80.1 g SO3
1 mole SO3835 g SO3
x = 10.4 moles of SO3
1 mole CH4
6.02 x 1023 molecules CH418 moles CH4 x = 108 x 1023 molecules CH4
4.0 grams He5.6 x 1023 atoms He
x 3.72 grams He
18.0 grams H2O3.7 grams H2O
x = 1.24 x 1023 molecules H2O
or 1.08 x 1025 molecules CH4
6.02 x 1023 atoms He=
6.02 x 1023 molecules H2O
Calculating Percent Composition by MassStep 1: Find the molar mass of the compound by adding the
individual masses of the elements together.
Step 2: Divide each of the individual masses of the elements by the molar mass of the compound.
Step 3: Convert the decimal to a % by multiplying by 100.
Practice Problems:
(1) Find the % composition of the elements in each compound.
a) Na3PO4 b) SnCl4
3 Na’s = 3 x 23.0 = 69.0
P = 31.0
4 O’s = 4 x 16.0 = 64.0+164
÷ 164
÷ 164
÷ 164
= 0.421 = 42.1%
= 0.189 = 18.9%
= 0.390 = 39.0%
Sn = 118.7
4 Cl’s = 4 x 35.5 = 142.0+
260.7
÷ 260.7
÷ 260.7
= 45.5%
= 54.5%
Elements in the Universe: % Composition by Mass
Earth’s Crust: % Composition by Mass
Entire Earth (Including Atmosphere): % Composition by Mass
Human Body: % Composition by Mass
Determining the Empirical Formula for a Compound
• The empirical formula for a compound is the simplest __________ number __________ of the atoms in the compound.
Examples: H2O is the empirical formula for water.
_______ is the empirical formula for glucose, C6H12O6.
Step 1: Divide the % composition data by the atomic mass of the element. This will give you a ratio of the # of atoms in the formula.
Step 2: Divide each of these answers by the smallest ratio.
Step 3: If there is still a decimal, multiply each answer by the denominator of the “freak”, (i.e. -- multiply all the ratios by the denominator of the ratio that is still a decimal.)
[1/2= 0.5 1/3≈ 0.33 2/3≈ 0.67 3/4= 0.75]
wholeratio
Helpful Rhyme: % to mass, mass to mole, divide by small, times ’til whole.
C1H2O1
Practice Problems: 1) An unknown compound is composed of 81.8% carbon and 18.2%
hydrogen. Determine the empirical formula for the compound.
2) An unknown compound is composed of 42.9% carbon and the rest of the compound is oxygen. Determine the empirical formula for the compound.
C = 81.8% = 81.8 gH = 18.2% = 18.2 g
(mass to moles) 81.8 g C ÷ 12.0 = 6.82 moles18.2 g H ÷ 1.0 = 18.2 moles
(÷ by small)C6.82H18.2
6.82 6.82
C1H2.67
(x ‘til whole)
x 3 x 3
C3H8
C = 42.9% = 42.9 gO = 57.1% = 57.1 g
(mass to moles) 42.9 g C ÷ 12.0 = 3.575 moles57.1 g O ÷ 16.0 = 3.569 moles
(÷ by small)C3.575O3.569
3.569 3.569
C1.0O1.0 CO
2) An unknown compound is composed of 42.9% carbon and the rest of the compound is oxygen. Determine the empirical formula for the compound.
C = 42.9% = 42.9 gO = 57.1% = 57.1 g
(mass to moles) 42.9 g C ÷ 12.0 = 3.575 moles C57.1 g O ÷ 16.0 = 3.569 moles O
(÷ by small)C3.575O3.569
3.569 3.569
C1.0O1.0 CO
To check your answer, you can simply find the % composition by mass of your formula…
% C = 12.0/28 = 42.9%
%O = 16.0/28 = 57.1%
CO = 28 g/mole
That matches up with the original problem!!
Determining the Molecular Formula for a Compound• The molecular formula for a compound is either the same as the
empirical formula ratio or it is a “_________ _________ of this ratio. It represents the true # of atoms in the molecule.
Examples: 1) H2O is the empirical & molecular formula for water. 2) CH2O is the empirical formula for sugar, ethanoic
acid, and methanol. The molecular formula for glucose is C6H12O6, (___times the empirical ratio!)
Step 1: Determine the empirical formula for the compound. (See the previous steps in the notes.)
Step 2: Calculate the empirical formula mass of the compound.
Step 3: Determine the “whole # multiple” by dividing the molecular formula mass (given in the problem) by the empirical
formula mass. Multiply each of the empirical ratios by this whole number.
whole # multiple
6
Practice Problems:
1) An unknown compound is composed of 58.8% carbon and 9.8% hydrogen and 31.4% oxygen. The molecular formula mass is 204 g. Determine the molecular formula for the compound.
C = 58.8% = 58.8 gH = 9.8% = 9.8 g
(mass to moles) 58.8 g C ÷ 12.0 = 4.9 moles C9.8 g H ÷ 1.0 = 9.8 moles H
(÷ by small)C4.9H9.8O1.96
1.96 1.96
C2.5H5O1
(x ‘til whole)
x 2 x 2 x 2
C5H10O2
O = 31.4% = 31.4 g 31.4 g O ÷ 16.0 = 1.96 moles O
1.96
Our formula mass = 102 g(Compare)
(204 ÷ 102 = 2)[C5H10O2 ]x 2 = C10H20O4
Practice Problems:
2) An unknown compound is composed of 40% carbon, 6.6% hydrogen, and 53.4% oxygen. Determine the molecular formula for the compound if the mass of one mole of the compound is 120 g.
C = 40% = 40 gH = 6.6% = 6.6 g
(mass to moles) 40 g C ÷ 12.0 = 3.3 moles C6.6 g H ÷ 1.0 = 6.6 moles H
(÷ by small)C3.3H6.6O3.3
3.3 3.3
C1H2O1
O = 53.4% = 53.4 g 53.4 g O ÷ 16.0 = 3.3 moles O
3.3
Our formula mass = 30 g(Compare)
(120 ÷ 30 = 4) [C1H2O1 ]x 4 = C4H8O4
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