10.2The Mole-Mass Relationship

Transcript 10.2The Mole-Mass Relationship

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

From moles to mass and to the moon!

AMU (Atomic Mass Units)

 The mass of Carbon-12 is 12 AMU.

 But wait, when I look on the periodic table, the atomic mass is listed as 12.01078 AMU??? WHY? Why, cruel world?  WwWWwwhhH

HHhhyyYY

YYyy???

6.1 Atoms and Moles Amedeo Avogadro



Avogadro’s Number



6.022 x 10 23



Avogadro discovered that there are 6.022 x 10 23 atoms in 1 gram of hydrogen.

Count Lorenzo Romano Amedeo Carlo Avogadro di Quaregna e Cerreto

Be able to explain and use the concept of the “mole”



This number is called a “mole.”



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.

Hmmm… I shall call 6.022 x 10 to it.

23 ... a “mole”. Yes…that has a nice ring

So How Big is a “MOLE”



Ummm… NO!



Here it is written out 602,200,000,000,000,000,000,000 That’s 602 billion groups of a trillion!



Let’s just do an example with paper clips.



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?

 Assume a paper clip ( still folded) is about 3 cm long.

 To find the total distance of the paper clips we use the following equation: 3

cm clip x

  6 .

x

10 23 1

mole clips

   1 .

806

x

10 24

cm mole

Notice the unit “clips” cancels!!! Isn’t that Great… Anyone… Anyone see the greatness???

Man I love Conversions!

The moon is 382,171 km from Earth, so to the moon and back would be 764,342 km.



So we need to convert our cm into km… … oh how fun… this is a metric conversion



This of course is a “2-step conversion” because both units have a prefix

 1 .

806

10 24

    1m 100 cm       1 km 1000m     1 .

806

10 19

I love conversions!

 We’re almost done!!!

1 .

806

10 ( 1 trip and back 764,342km )



2 .

36

10 trips

 That’s 23 trillion trips!! Mole-tastic!

 Marshmallow example: A bed of marshmallows covering the U.S. would be 776 miles deep

Convert moles to # of atoms

 How many atoms are in 3.2 mol potassium (K)?

 Remember: 1 mol = 6.02 x 10 23 atoms  This can be written as a conversion factor:   6.02

 10 23 1 mol

atoms

   3 .

2 mol K    6.02

 10 23 1 mol

atoms

   1 9.264

x 10 23 atoms of K  1.9

x10 24 atoms of K

How do we use the “Mole” in chemistry?

 The atomic mass

grams of 1 mole

of an element is the of that atom  Why do chemists use moles?

 It’s fun.

 It’s impossible to count atoms with your hands.

 You can easily measure the mass (in grams) of a chemical.

Atomic mass = grams of 1 mole of this element, Cobalt

Convert moles of an atom to grams



I need 2.0 moles of copper (Cu) for an experiment. How many grams is that?



Atomic mass of Cu = 63.55 g/mol (round to 2 decimals)



“mol” is the abbreviation of “Mole”… I know it’s only one letter different… chemists!!!



2 .

0 mol Cu

   

63 1 .

55 mol g

   

127.01



130g Cu (2 sig figs!

)

Converting grams to moles



I have 302 grams of silver (Ag). How many moles of silver do I have?



Step 1: Atomic mass of Ag = 107.87 g/mol



Step 2: Calculate



302 g Ag

  

1 mol 107.87

g

  

2.79966

mol



2.80 mol (3 sig figs)

6.2 Molar Mass and Percent Composition

 

Atomic Mass

= mass of one mole of an atom

Molar Mass

= mass of one mole of a substance

Calculate Molecular Weights

Example: Calculate the Molecular Weight (MW) of RbI 2

Step 1 Step 2 Step 3 : Assume you have 1 mole of this molecule and determine how much each element weighs from the periodic table.

: Determine how many of each element you have : Add all the masses together

Step 1: Find 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



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: RbI 2 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 ”

Converting from moles of a compound to grams 

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 1 .

54 mol g

   

172 .

62 g



173g NaCl

Converting from grams of a compound to moles 

Example: How many moles are in 10.0 g of Na 2 SO 4 ?



Step 1: Find the molar mass.



Molar mass = 142.1 g/mol



Step 2: Use the molar mass like a conversion factor. You need “grams” on the bottom of the fraction.

moles of Na SO = 10.0 g 2 4     1 mol 142.1 g      2

6.3 Formulas of Compounds

 Calculate

“percent composition”

 Just like any other % stuff total stuff  x 100  Percent  Stuff = grams of elements Compositio n

Calculate

“percent composition”



Ex: calculate % of Cu and S in Cu 2 S

stuff total stuff  x 100  Percent Compositio n 

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

127.1

159.2 g g

= 159.17 g = 159.2 g

 

x 100



79 .

84 %

You should be able to…



Identify an “empirical formula” and a “molecular formula”



Empirical formula – simplest ratio of atoms of each element in a compound (whole #’s only)



Molecular formula – actual # of atoms of each element in a compound

Molecular

H 2 O 2 C 3 H 6 N 2 O 3 C 2 H 6 C 3 H 9

Empirical

HO CH 2 N 2 O 3 ?

Using % composition to determine a formula

 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  Ex: H 2 O is always 88.9% O and 11.1% H by mass  Only the simplest formula (ratio) can be found… in other words, you can only find empirical formulas

Using % composition to calculate the formula



Process is as follows: 1.

Calculate % by mass of each element 2.

Determine mass of each element



Easy if you use 100 g of the chemical 3.

Use mass to find the # of moles of each element 4.

Find the smallest ratio of the atoms



÷ the number of moles of each element by the element with the smallest # of moles



Round to the nearest whole #

Example

 A molecule is 75% C & 25% H. Calculate the empirical formula.

 Using 100g total = 75g C and 25 g H  Calculate moles of each =   75g C     1 mol 12.01g

   

6 .

2 mol  25 g H     1 mol 1.01g

  

 25

mol

Ratio = 6.2C : 25H, simplify by ÷ each by 6.2. Whole number only!!  C  6.2

6.2

 H  25 6.2

 Final ratio ≈ 1C : 4H so CH 4

Percent 100 g total Moles Ratio Formula C

75% 75g  75g C    1 mol 12.01g

   6.2 mol 6.2

6.2

= 1 CH 4

25% 25g  25 g H    1 mol 1.01g

   25 mol 25 6.2

= 4

Find the molecular formula



Ex: C 3 H 6 O 2 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??



Point: The ratio of C:H:O will always be what the empirical formula shows



Steps



1. Calculate the empirical formula mass



2. Calculate molar mass/empirical formula mass



3. Multiply your subscripts by that #.

