AP chemistryCCHS, Mr. Bartelt’s class

Moles and Stoichiometry

Mass spectroscopyInformation found on pg 82

What’s amuAmu stands for atomic mass unit.1 amu is defined as 1/12th of the mass of the

carbon—12 nucleus.If you recall from nuclear chemistry, as the size

of the nucleus increases some of the mass is converted into energy to keep the nucleus from being blown apart by the repulsive force between the protons.

This causes the mass number (protons and neutrons) to not correspond perfectly with amus. Don’t worry too much about this. Amus will be given when they are needed.

Calculating atomic massesLook on pg83-85Chlorine has two natural isotopes:

Chlorine—35 (75.77% in nature)Chlorine—37 (24.23% in nature)

Determining the molar mass is simple. It’s a mater of performing a weighted average. You do it ever semester when you calculate your final grade. LookSemester average 86.4 (80% of final grade)Final exam 96.0 (20% of final grade)

So you plug: 86.4*.8+96.0*.2 into your calculator and find your final average to be 88.32. Sorry, you got a “B”

Calculating atomic massesSo think of chlorine just like your gradeChlorine has two natural isotopes:

Chlorine—35 (75.77% in nature)Chlorine—37 (24.23% in nature)

So you plug: 35*.7577+.37*.2423 into your calculator and find the atomic weight to be 35.48. Which is close to the value on the periodic table 35.453

NOTE: If you use mass number’s you be a little off because amu and mass number are not the same thing. Amus for the different isotopes will be given if you need them

The molePages 90-93The molar mass that you look up on the periodic

table is the mass in grams that one mole will weigh.

A mole is defined as 6.022*1023 (it has no units)A mole is like a dozen (12) in that it’s a counting

unit. A mole however is significantly larger.If you gave away a million dollars every second

for the last 4.5 billion years (since the earth formed) you would have only given a way ¼ of a mole of dollars.

A mole is HUGE

ConversionsHow many moles are in 1.204*1023

molecules of O2?

How much would that weigh in grams?First calculate the molar mass of O2

Next find the mass

223223 O mol 2000.

molecules 10*022.6

mol 1*O molecules 10*204.1

22 O g 400.6mol 1

g 2.003*O mol 0002.

g/mol 32.00 g/mol 16.00*2

g/mol 16.00 O of massMolar

Sample problems (work in groups)Find the mass of 1.5*1019 atoms of argon. How many moles of sugar (C12H22O11) are in

200.0 grams of sugar?How many alcohol (C2H5OH) molecules are

in 14.5 pounds?What weighs more 400. atoms of krypton

or 900. molecules of N2?

% compositionThis is easy (93-95)I’ll show you. Here’s how you find the %

composition of cocaine (C17H21NO4)First, find the mass contribution of all the

elements and the total mass

molg /4.303 Total

g/mol 21.17 21 * g/mol 1.008Hydrogen

g/mol 64.00 4 * g/mol 16.00Oxygen

g/mol 14.01 1 * g/mol 14.01Nitrogen

g/mol 204.2 17 * g/mol 12.01Carbon

% composition continuedSecond, find the percent by dividing each

part by the whole

That’s it

%98.6100*0698.0303.4g/mol

g/mol 21.17Hydrogen

%1.21100*211.0303.4g/mol

g/mol 64.00Oxygen

%62.4100*0462.0303.4g/mol

g/mol 14.01Nitrogen

%3.67100*673.0303.4g/mol

g/mol 204.2Carbon

PracticeFind the % composition of octane (C8H18)Find the % composition of MDMA

(C11H15NO2)Find the % composition of prozac

(C17H18F3NO)

Empirical formula from % compositionPg 96-102It is possible to determine the empirical

formula from % composition data.A compound is found to be 20.00% C, 53.30%

O, 23.34% N, 3.36% H by mass. What is the empirical formula?

This seems like a difficult problem but it’s easy once you learn the technique for solving it.

First assume that you have a 100.0 gram sample. This causes the % signs to become mass in grams

Empirical formula from % composition

20.00g C, 53.30g O, 23.34g N, 3.36g HNext, convert grams into moles

H mol 3.331 1.008g

mol 1 * H g .363

N mol 1.666 14.01g

mol 1 * N g 3.342

O mol 3.331 16.00g

mol 1 * O g 30.53

C mol 1.666 12.01g

mol 1 * C g 00.20

Empirical formula from % compositionNow you have a mole ratio. The easiest

way to simplify this ratio is to divide each mole amount by the lowest value.

This means that the empirical formula is:CO2NH2

The numbers in the simplified ratios are the subscripts in the empirical formula

21.666 H mol 3.331

11.666 N mol 1.666

21.666 O mol 3.331

11.666 C mol 1.666

Sample problems1.63.50% Ag, 28.25% N, 8.25% N2.12.63% Li, 58.21% O, 29.17%

S3.27.93% Fe, 48.01% O, 24.06%

S4. 29.28% C, 52.01% O, 17.08 N, 1.64% HThe last one’s a little tricky. Ask if you need

help

Molecular formulaPg 100-101As you saw in the bellwork, many different

molecules can have the same empirical formula. In order to determine the molecular formula you need more than just the % composition. You also need the molar mass.

If you found that the empirical formula of a molecule was C2O2NH2. That could be the molecular formula, but so could C4O4N2H4 or C6O6N3H6.

Molecular formulaIf you know that the molar mass of the

molecule with empirical formula C2O2NH2 is 144.1 g/mol then you can determine the molecular formula.

First find the molar mass of the empirical formula

molg /05.72 Total

g/mol 1.016 2 * g/mol 1.008Hydrogen

g/mol 32.00 2 * g/mol 16.00Oxygen

g/mol 14.01 1 * g/mol 14.01Nitrogen

g/mol 24.02 2 * g/mol 12.01Carbon

Molecular formulaNow divide144.1 g/mol by the molar mass of

the empirical formula, 72.05 g/mol.

Now distribute that number across the empirical formula and you have the molecular formula

And you’re done

2/05.72

/1.144

molg

molg

4244222 HNOCNHOC2

Sample problems You find a sample to be 44.45% C,

19.74% O, 34.57% N, 1.24% H by mass. You know that the molar mass of the molecule is 243.2 g/mol.

Find the empirical and molecular formulas

Balancing equationsPg 102-108Practice makes perfect (balance these)1.Zn + HCl ---> ZnCl2 + H2

2.C10H16 + Cl2 ---> C + HCl

3.C7H6O2 + O2 ---> CO2 + H2O

4.KClO3 ---> KClO4 + KCl

5.Al(OH)3 + H2SO4 ---> Al2(SO4)3 + H2O

6.Al2(SO4)3 + Ca(OH)2 ---> Al(OH)3 + CaSO4

7.NaOH + Cl2 ---> NaCl + NaClO + H2O

And much, much more

StoichiometryMost stoichiometry problems will give you

grams of one compound and ask for grams of another compound.

ExampleGiven the equation below (unbalanced), how many grams of sulfur hexaflouride will be produced if 85.23 grams of sulfur are reacted?

S8 + F2 SF6

A stoichiometry problem has 4 basic steps

Stoichiometry (4 steps)1) Balance the equation (sometimes this is

already done)S8 + F2 SF6 becomes S8 + 24F2 8SF6

2) Convert grams to moles (given: 85.23 grams of sulfur)

3) Perform mole ratioS8 + 24F2 8SF6

8

8

3322.056.256

1* 23.85

/56.2568*07.32S of massMolar

Smolmol

Sulfurg

molg

0.3322 X

6 658.28

3322.0

1SFmolx

x

Stoichiometry (Last step)4) Convert moles to grams

What you just found is the theoretical yield. That’s how much product would be produced in a “perfect world”. We don’t live in perfect world so we often times look at the % yield.

6

6

6

2.3771

07.146*658.2

/07.1466(19.00)32.07 SF of MassMolar

658.28

3322.0

1

SFgmol

gmol

molg

SFmolxx

Stoichiometry % yieldExample

Given the equation below (unbalanced), 155.23 grams of tetraphosphorous decoxide yields only 102.25 grams of phosporic acid, what’s the % yield?

P4O10 + H2O H3PO4

1) BalanceP4O10 + 6H2O 4H3PO4

2) Grams to moles

104104

104

OP 547.088.283

1*OP 23.155

/88.28310*00.164*97.30OP of massMolar

molg

molg

molg

3) Perform mole ratioP4O10 + 6H2O 4H3PO4

4) Convert moles to grams

5) Determine percent yield

Next

Yield lTheoreticaPO 34.2141

99.97*1872.2

/99.974(16.00)30.97 )008.1(3POH of MassMolar

POH 1872.24

54682.0

1

43

43

43

Hgmol

gmol

molg

molxx

%70.47100*(found) 214.34

)(25.102100*

yield lTheoretica

yield Actual yieldPercent

given

Stoichiometry (LR) Finding the limiting reagent is not very

hard. Example: Given the equation below, if

you have 100.0 grams of aluminum and 75.00 grams of oxygen, what will the limiting reagent be?

Al +O2 Al2O3

Balance4Al +3O2 2Al2O3

First convert both from grams to moles

2O mol 344.232.00g

mol 1*O2 00.75

Al mol 708.326.98g

mol 1*Al 0.100

g

g

Stoichiometry (LR)4Al +3O2 2Al2O3

Next, divide both by their coefficient

Oxygen is the limiting reagent. Solve for the theoretical yield using oxygen.

Forget about Al! The oxygen will run out and leave a little Al behind.

22 O mol 344.232.00g

mol 1*O 00.75

Al mol 708.326.98g

mol 1*Al 0.100

g

g

781.03

O mol 344.2

927.04

Al mol 708.3

2