Chp 6Chp 6

The MoleThe Mole

Counting by WeighingCounting by Weighing

Every item has a certain average mass.Every item has a certain average mass. We can use that mass to calculate how We can use that mass to calculate how

much we would expect 50, 1000 or 1000000 much we would expect 50, 1000 or 1000000 of that item to weigh.of that item to weigh.

Then we could weigh out that amount rather Then we could weigh out that amount rather than count out 1000000 of that item.than count out 1000000 of that item.– This is how we can count atoms.This is how we can count atoms.– The atomic mass is the amount that 1 mole of The atomic mass is the amount that 1 mole of

that element weighs.that element weighs.

The MoleThe Mole

Just like the term “a dozen” represents the Just like the term “a dozen” represents the number 12, a mole represents the number number 12, a mole represents the number 6.023 x 106.023 x 102323..– That’s 602,300,000,000,000,000,000,000!!!That’s 602,300,000,000,000,000,000,000!!!– A number that big is hard to comprehend so a A number that big is hard to comprehend so a

few facts:few facts: If you distributed a mole of pennies between If you distributed a mole of pennies between

everyone alive on Earth, we’d each get a trillion $.everyone alive on Earth, we’d each get a trillion $. If you stacked a mole of paper, it would reach the If you stacked a mole of paper, it would reach the

moon and back 80 billion times.moon and back 80 billion times. A mole of marshmallows would completely cover the A mole of marshmallows would completely cover the

Earth in a pile 12 miles deep.Earth in a pile 12 miles deep.

The Practical ApplicationThe Practical Application

So 12.01 g of carbon contains 6.023 x 10So 12.01 g of carbon contains 6.023 x 102323 atoms of carbon.atoms of carbon.

Therefore 24.02 g of carbon contains 1.2046 Therefore 24.02 g of carbon contains 1.2046 x 10x 102424 atoms (2 times 6.023 x 10 atoms (2 times 6.023 x 102323).).

We can use this in dimensional analysis to We can use this in dimensional analysis to figure out fractions of this:figure out fractions of this:– Atomic mass = 1 moleAtomic mass = 1 mole– 1 mole = 6.023 x 101 mole = 6.023 x 102323 atoms atoms

An ExampleAn Example

How many moles are in 0.5 g of hydrogen?How many moles are in 0.5 g of hydrogen?– 1 mole H = 1.007 g1 mole H = 1.007 g

How many atoms is this?How many atoms is this?– 1 mole = 6.023 x 101 mole = 6.023 x 102323 atoms atoms

0.5 g H

1.007 g H

1 mole H= 0.496 mol H

0.496 mol H = 2.99 x 1023 atoms H

6.023 x 1023 atoms H1 mol H

What About Compounds?What About Compounds?

We can do the same thing for a compound We can do the same thing for a compound by first adding the individual masses to get by first adding the individual masses to get the molar mass of the compound.the molar mass of the compound.– Ex. CHEx. CH44

C = 12.01 g/molC = 12.01 g/mol

H = 1.007 g/mol x 4H = 1.007 g/mol x 4

So CHSo CH44 = 16.04 g/mol = 16.04 g/mol

– How many moles are in 0.5 g of CHHow many moles are in 0.5 g of CH44??0.5 g CH4

16.04 g CH4

1 mol CH4 = 0.03 mol CH4

Mass PercentMass Percent How much of a particular element makes up a How much of a particular element makes up a

compound.compound. Mass % = Mass % = mass elementmass element x 100 x 100

molar mass of compoundmolar mass of compound Ex. What is the mass % of C in CHEx. What is the mass % of C in CH44??

12.01 12.01 x 100 = 74.9% x 100 = 74.9%16.0416.04

What about H?What about H?4.0284.028 x 100 = 25.1% x 100 = 25.1%16.0416.04

** Should add to 100%** Should add to 100%

Empirical FormulasEmpirical Formulas

The simplest, whole number ratio of atoms The simplest, whole number ratio of atoms in a compound.in a compound.

Rules:Rules:1.1. Get the mass of each element present.Get the mass of each element present.2.2. Convert grams to moles.Convert grams to moles.3.3. Divide all answers by the smallest number Divide all answers by the smallest number

from step 2.from step 2.4.4. Multiply by an integer, if necessary, to get Multiply by an integer, if necessary, to get

whole numbers.whole numbers.5.5. Those numbers are the subscripts of the Those numbers are the subscripts of the

empirical formula.empirical formula.

An ExampleAn Example

What is the empirical formula of a compound What is the empirical formula of a compound containing 4.151 g Al and 3.692 g O?containing 4.151 g Al and 3.692 g O?

1.1. Masses are listed above for us.Masses are listed above for us.2.2. 4.151 g Al4.151 g Al1 mol Al1 mol Al

26.98 g Al26.98 g Al

3.692 g O3.692 g O 1 mol O1 mol O15.99 g O15.99 g O

= 0.1539 mol Al

= 0.2308 mol O

An ExampleAn Example3.3. 0.1539 is smallest so:0.1539 is smallest so:

0.15390.1539 = 1 mol Al = 1 mol Al0.15390.1539

0.23080.2308 = 1.5 mol O = 1.5 mol O0.15390.1539

4.4. Since we don’t have whole #s, we need to Since we don’t have whole #s, we need to multiply by an integer. Let’s try 2.multiply by an integer. Let’s try 2.

1 mol Al x 2 = 2 mol Al1 mol Al x 2 = 2 mol Al1.5 mol O x 2 = 3 mol O1.5 mol O x 2 = 3 mol O

5.5. Now we have whole numbers, so the empirical Now we have whole numbers, so the empirical formula isformula isAlAl22OO33. .

Molecular FormulasMolecular Formulas The actual ratio of atoms in a compound.The actual ratio of atoms in a compound. May be the same as the empirical or a May be the same as the empirical or a

multiple of it.multiple of it.– Ex. CHEx. CH4 4 is an empirical formulais an empirical formula

CHCH44, C, C22HH88, C, C33HH1212 could all be molecular could all be molecular

formulas based on that E.F.formulas based on that E.F.

Found by dividing the actual molar mass by Found by dividing the actual molar mass by the molar mass of the empirical compound. the molar mass of the empirical compound. If the answer isn’t 1, you multiply the E.F. If the answer isn’t 1, you multiply the E.F. subscripts by that number.subscripts by that number.

An ExampleAn Example

What is the molecular formula of a What is the molecular formula of a compound with a molar mass of 283.33 compound with a molar mass of 283.33 g/mol if the empirical formula is Pg/mol if the empirical formula is P22OO55??

The mass of PThe mass of P22OO5 5 is:is:

2 x 30.97 = 61.94 g2 x 30.97 = 61.94 g5 x 15.99 = 5 x 15.99 = 79.95 g79.95 g

141.89 g/ mol 141.89 g/ mol

An ExampleAn Example

So, now we divide the real M.M. by the M.M. So, now we divide the real M.M. by the M.M. of the empirical formula.of the empirical formula.

283.33 g283.33 g = 2 = 2141.89 g141.89 g

Now we multiple the subscripts of the Now we multiple the subscripts of the empirical formula by that 2.empirical formula by that 2.

2 (P2 (P22OO55) = P) = P44OO10 10 is our answer!is our answer!