Download Chapter 6: Moles, Molar Mass, Percent Composition and Formulas

Chapter 6: Moles, Molar Mass, Percent Composition and
Formulas
6.1 Atoms and Moles
1) Avogadro’s Number
a) Memorize: 6.022 x 1023
b) Avogadro discovered that there are 6.022 x 1023 atoms
in 1 gram of hydrogen.
2) Be able to explain and use the concept of the “mole”
a) Avogadro called his number a “mole” of matter.
b) The word “mole” is just like the word “dozen”. Dozen means “12”. You can have
a dozen of anything. You can also have a mole of anything.
3) How big is a mole?
a) Here it is written out: 602,200,000,000,000,000,000,000
b) That’s 602 billion groups of a trillion!
c) Let’s just do an example with paper clips.
d) If you have a mole of paper clips and made them into a chain, how many times
could you go to the moon and back with your chain? (You don’t need to do this)
4) Convert moles to # of atoms
a) How many atoms are in 3.2 mol potassium (K)?
b) Remember: 1 mol = 6.02 x 1023 atoms
c) This can be written as a conversion factor:

atoms 
  19.264 atoms of K
1 mol

3.2 mol K  6.02  10

 6.02  1023 atoms 


1 mol


23
5) How do we use the “Mole” in chemistry?
a) The atomic mass of an element is the grams of 1 mole of that atom
b) Why do chemists use moles?
Chapter 6 Notes
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Updated 6/27/2017
i) It’s fun.
ii) It’s impossible to count atoms with your hands.
iii) Numbers of moles are smaller and easier to do math with than big numbers of
atoms and molecules.
6) Convert moles of an atom to grams
a) I need 2.0 moles of copper (Cu) for an experiment. How many grams is that?
b) Atomic mass of Cu = 63.55 g/mol (round to 2 decimals)
c) “mol” is the abbreviation of “Mole”… I know it’s only one letter different…
chemists!!!
2.0 mol Cu  63.55 g   127.01g
 1 mol 
 130g Cu (2 sig figs! )
7) Converting grams to moles
a) I have 302 grams of silver (Ag). How many atoms of silver do I have?
b) Step 1: Atomic weight of Ag = 107.8682 g/mol
c) Step 2: Calculate
302g Ag 
1 mol 
  2.79966 mol
 107.87 g 
 2.80 mol (3 sig figs)
6.2 Molar Mass and Percent Composition
1) Atomic Mass = mass of one mole of an atom
2) Molar Mass= mass of one mole of a substance
3) Calculate Molar Mass
a) Example: Calculate the Molecular Weight (MW) of RbI2
Step 1: Assume you have 1 mole of this molecule and determine how much each element
weighs from the periodic table.
Rb is atomic # 37. How much does each mole of Rb weigh?
85.47 grams/mol Rb
Chapter 6 Notes
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I is atomic # 53. How much does it weigh?
126.90 g/mol I
Step 2: Determine how many of each element you have
Look at the formula:
RbI2
We have 1 “Rb” atom and 2 “I” atoms
Step 3: Add all the masses together
You will need to show this work:
(85.47 g/mol Rb)  (1 Rb)  85.47 g/mol
plus
(126.90 g/mol)  (2 I)  253.80 g/mol
Because the units are the same we can add these two numbers together, so…
253.80 g/mol + 85.47 g/mol = 339.27 g/mol
339.27 g/mol is the “molar mass”
4) Converting from moles of a compound to grams
a) Example: I need 3.00 mol NaCl for an experiment. How many grams is that?
Step 1: Find the molar mass
Molar mass = 22.09g/mol + 35.45g/mol
= 57.54 g/mol
Step 2: Use the molar mass like a conversion factor.
3.00 mol  57.54 g   172.62 g
 1 mol 
 173g NaCl
5) Converting from grams of a compound to moles
a) Example: How many moles are in 10.0 g of Na2SO4?
Step 1: Find the molar mass.
Molar mass = 142.1 g/mol
Chapter 6 Notes
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Step 2: Use the molar mass like a conversion factor. You need “grams” on the
bottom of the fraction.
 1 mol 
2
moles of Na 2SO4 = 10.0 g  
  7.04 10 mol
 142.1 g 
6) Calculate “percent composition”
a) Just like any other %
 stuff


 x 100  Percent Compositio n
total
stuff


b) Stuff = grams of elements
7) Calculate “percent composition”
a) Ex: calculate % of Cu and S in Cu2S
Stuff = grams Cu
(63.55 g/mol Cu)(2 mol Cu) = 127.1g Cu
Total stuff = grams Cu + grams S
= 127.1 g + 32.07 g = 159.17 g = 159.2 g
 127.1 g 

 x 100  79.84%
 159.2 g 
6.3 Formulas of Compounds
1) Identify an “empirical formula” and a “molecular formula”
a) Empirical formula – simplest ratio of atoms of each element in a compound
(whole numbers only)!!
b) Molecular formula – actual # of atoms of each element in a compound
Chapter 6 Notes
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2) Using % composition to determine a formula
a) Law of Definite Composition – Any amount of a pure compound will always
have the same ratio of masses for the elements that make up that compound
b) Ex: H2O is always 88.9% O and 11.1% H by mass
i) Only the simplest formula (ratio) can be found… in other words, you can only
find empirical formulas
c) Process is as follows:
i) Calculate % by mass of each element
ii) Determine mass of each element
1. Easy if you use 100 g of the chemical
2. Use mass to find the # of moles of each element
3. Find the smallest ratio of the atoms
4. ÷ the number of moles of each element by the element with the smallest # of moles
5. Round to the nearest whole #
d) A molecule is 75% C & 25% H. Calculate the empirical formula.
i) Using 100g total = 75g C and 25 g H
ii) Calculate moles of each =
75g C  1 mol

  6.2 mol
 12.01g 
25 g H  1 mol   25mol
 1.01g 
iii) Ratio = 6.24C : 24.8H, simplify by ÷ each by 6.24 Whole number only!!
C
Chapter 6 Notes
6.2
1
6.2
H
25
4
6.2
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Updated 6/27/2017
iv) Final ratio ≈ 1C : 4H so CH4
v) When you solve this problem, make a table AND show your work:
Percent
100 g total
Moles
C
H
75%
75g
25%
25g
75g C 
Ratio
1 mol 
 
 12.01g 
25 g H  1 mol  
 1.01g 
6.2 mol
25 mol
6.2
=
6.2
25
=
6.2
1
4
Formula
CH4
3) Find the molecular formula
a) Ex: C3H6O2 is an empirical formula for a chemical. The molar mass of the
compound is 148 g/mol.
What is the molecular formula of the compound??
i) Point: The ratio of C:H:O will always be what the empirical formula shows
b) Steps
1.
Calculate the empirical formula mass
2.
Calculate molar mass/empirical formula mass
3.
Multiply your subscripts by that #.
4) Example:
1. Calculate the empirical formula mass:
C3H6O2 mass = (3)(12.01) + (6)(1.01) + (2)(16.00)
= 148.09 g/mol
2. Calculate (molar mass)/(empirical formula mass). Round to a WHOLE number.
 148 

2
 74 
3. Multiply the subscripts of the empirical formula by that number.
i) C3x2H6x2O2x2
Chapter 6 Notes
= C6H12O4
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