chapter 6: moles, molar mass, percent composition and formulas

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?

WwWWwwhhHHHhhyyYYYYyy???

6.1 Atoms and Moles

Avogadro’s Number 6.022 x 1023

Avogadro discovered that there are 6.022 x 1023 atoms in 1 gram of hydrogen.

Amedeo Avogadro

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”. Dozenmeans “12”. You can have a dozen of anything. You can also have a mole of anything.

Hmmm… I shall call 6.022 x 1023... a “mole”. Yes…that has a nice ring to it.

So How Big is a “MOLE”

Ummm… NO!Here it is written out602,200,000,000,000,000,000,000That’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?

Don’t be cruel now…

Aaaiiiee

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:

mole

cmx

mole

clipsxx

clip

cm 2423

10 806.1 1

1002.63

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 conversionThis of course is a “2-step conversion” because

both units have a prefix

kmxcmx 1924 10806.11000m

km 1

cm 100

1m 10806.1

I love conversions!

We’re almost done!!!

That’s 23 trillion trips!! Mole-tastic!Marshmallow example: A bed of

marshmallows covering the U.S. would be 776 miles deep

trips1036.2)764,342km

back and trip1( 10806.1 1319 xkmx

Convert moles to # of atoms

How many atoms are in 3.2 mol potassium (K)?Remember: 1 mol = 6.02 x 1023 atomsThis can be written as a conversion factor:

K of atoms x101.9

K of atoms10 x 9.2641 mol 1

10 6.02K mol 2.3

24

23 23

atoms

mol 1

10 6.02 23 atoms

How do we use the “Mole” in chemistry?

The atomic mass of an element is the grams of 1 mole 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!!!

)figs! sig (2Cu 130g

127.01 mol 1

g 55.63Cu mol 0.2

Converting grams to moles

figs) sig (3 mol 2.80

mol 2.79966 g 107.87

mol 1Ag g302

I have 302 grams of silver (Ag). How many moles of silver do I have?Step 1: Atomic mass of Ag = 107.87 g/molStep 2: Calculate

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 RbI2

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

Step 2: Determine how many of each element you have

Step 3: 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: RbI2

We have 1 “Rb” atom and 2 “I” atoms

Step 3: Add all the masses together

You will need to show this work:

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/mol339.27 g/mol is the “molar mass”

g/mol 253.80 I) (2 g/mol) 90.126(

g/mol 85.47 Rb) 1( Rb) g/mol 47.85(

plus

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.

NaCl 173g

g 62.172mol 1

g 54.57mol 00.3

Converting from grams of a compound to moles

Example: How many moles are in 10.0 g of Na2SO4?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.

22 4

1 molmoles of Na SO = 10.0 g 7.04 10 mol

142.1 g

6.3 Formulas of Compounds

Calculate “percent composition”Just like any other %

Stuff = grams of elements

nCompositioPercent 100 x stuff total

stuff

Calculate “percent composition”

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

nCompositioPercent 100 x stuff total

stuff

%84.79100 x g 159.2

g 127.1

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

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 compoundEx: H2O 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 formulaProcess 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 #

ExampleA molecule is 75% C & 25% H. Calculate the empirical

formula.Using 100g total = 75g C and 25 g HCalculate moles of each =

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

Final ratio ≈ 1C : 4H so CH4

mol251.01g

mol 1 H g 25 mol 2.6

12.01g

mol 1 C 75g

6.2

25 H

6.2

6.2 C 1 4

C H

12.01g

mol 1 C 75g

1.01g

mol 1 H g 25

6.2

6.26.2

25

C H

Percent 75% 25%

100 g total 75g 25g

Moles

6.2 mol 25 mol

Ratio = =

1 4

Formula CH4

Find the molecular formula

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??

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

Steps1. Calculate the empirical formula mass2. Calculate molar mass/empirical formula mass3. Multiply your subscripts by that #.

Steps1. 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.

3. Multiply the subscripts of the empirical formula by that number.

C3x2H6x2O2x2 = C6H12O4

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