Atomic Mass l Atoms are so small, it is difficult to discuss how much they weigh in grams. l Use atomic mass units. l an atomic mass unit (amu) is one.

Atomic Mass Atoms are so small, it is difficult to

discuss how much they weigh in grams.

Use atomic mass units. an atomic mass unit (amu) is one

twelth the mass of a carbon-12 atom.

The decimal numbers on the table are atomic masses in amu.

kgmuamu C27

12 1066053892.112

111

numberwhole

unpm

ukgm

ukgm

um

kgmuamu

C

n

p

C

C

12

27

27

12

2712

1067492735.1

1067262178.1

12

1066053892.112

111

They are not whole numbers

Because they are based on averages of atoms and of isotopes.

mass

10000 Cu atom

6909 Cu-63

6909x62.93=434783.37

u

3091Cu-65

3091x64.93=200698.63

u

Total mass= 635482 uAverage mass= total mass/10000=63.55 u

93.64100

91.3193.62

100

09.6955.63

......

...

...

3

2

1

isotope

isotope

isotope

MassAtAbunNat

MassAtAbunNat

MassAtAbunNatMassAtomicaverage

The Mole The mole is a number. A very large number, but still, just

a number. 6.022 x 1023 of anything is a mole A large dozen. The number of atoms in exactly 12

grams of carbon-12.

The Mole 1 mol of atoms of any element

weighs the number value of atomic weight in g.

1 mol Cu weighs 63.55 g 1 mol Ga weighs 69.723 g

Molar mass, M Mass of 1 mole of a substance. Often called molecular weight. To determine the molar mass of an

element, look on the table. To determine the molar mass of a

compound, add up the molar masses of the elements that make it up.

Find the molar mass of

CH4 : CH4 → C + 4 H

mCH4= mC+ mH

MCH4= MC + 4×MH

Mg3P2 : M = 3×MMg + 2×MP

Ca(NO3)3

Al2(Cr2O7)3

CaSO4 · 2H2O

avN

Nn

M

mn

Number of moles, n

Avogadro’s number Nav=6.02x1023 mol-1

gmol

molgN

NMm

M

m

N

Nn

avav

21123

1 1042.21002.6

6.243

Number of moles, number of atoms in 10 g Al?

2323

1

1023.21002.6371.0.

371.0.98.26

10

molmolNnN

molmolg

g

M

mn

av

Number of moles, mass of 5.00x1020 atom Co?

gmolgmolMnm

molmolN

Nn

av

414

4123

20

1089.4.93.581030.8.

1030.81002.6

1000.5

Number of molecules of isopentyl acetate in 1 ug?

Number of carbon atoms?

atomcarbonN

moleculeperatomscarbonofnomoleculesofnoN

moleculemolmolNnN

OHCmolmolg

g

M

mn

molgMOHC

C

C

av

1615

15239

21479

1

6

2147

105.37105

..

1051002.6108.

108.18.130

101

/18.13000.162008.11401.127

Mass of CO32- in 4.86 mol CaCO3?

gmol

gmolm

MnmM

mn

molmol

COCaCO

CO

COCOCO

29216312186.4

86.486.4

23

23

23

23

233

Percent Composition Percent of each element a compound is

composed of.

What is the percent composition of C2H5OH?

%100% sample

X

m

mX

%100%

%100%

%100%

52

52

52

OHHC

O

OHHC

H

OHHC

C

m

mO

m

mH

m

mC

Suppose we have 1 mol C2H5OH

OHCOHHC MMMM

molmolmolmol

OHCOHHC

162

1621

62

52

52

%73.34%10007.46

00.161%100%100%

%13.13%10007.46

008.16%100

6%100%

%14.52%10007.46

01.122%100

2%100%

5252

5252

5252

OHHC

O

OHHC

O

OHHC

H

OHHC

H

OHHC

C

OHHC

C

M

M

m

mO

M

M

m

mH

M

M

m

mC

• Sum of all percentages=100%

• 100 g of C2H5OH contains 52.14 g C, 13.13 g H and 34.73 g O.

Find the percent compositions of the following compounds:

Al2(Cr2O7)3

CaSO4 · 2H2O

%100% 2

2 sample

OH

m

mOH

From percent composition, you can determine the empirical formula.

Empirical Formula the lowest ratio of atoms in a molecule.

A sample is 59.53% C, 5.38%H, 10.68%N, and 24.40%O, what is its empirical formula?

C H N O

sample mass=100 g 59.53 g 5.38 g 10.68 g 24.4 g

n=m/M 59.53/12.01

5.38/1.008

10.68/14.00

24.4/16.00

n 4.96 5.34 0.76 1.53

Divide by smallest value

6.53 7.03 1 2.01E.F. C13H14N2O4

Empirical Formula and Molecular Formula

H2O2 EF: HO

Such compound doesn’t exist! MF = n × EF H2O2 = 2 (HO)

MMF=n × MEF 34 = 2 × 17

2×MH + 2×MO = 2 (MH+MO)

Alkenes: C2H4 , C3H8 , C4H8 , C5H10 …

EF : CH2

Cl C H

sample mass=100 g 71.65 g 24.27 g 4.07 g

n=m/M 71.65/35.45

24.27/12.01

4.07/1.008

n 2.021 2.021 4.04

Divide by smallest value

1 1 2

E.F. CH2Cl

2

)48.49(96.98

)45.351008.1201.121(96.98

n

n

n

MnM EFMF

M.F. C2H4Cl2

Elemental Analysis

A substance is known to be composed of C, H and nitrogen. When 0.1156 g of this substance is reacted with oxygen, 0.1638 g CO2 and 0.1676 g H2O. Find the empirical formula.

Assumptions:1) All C in sample converted into

CO2.

ggmm

m

mC

M

MC

mm

COC

CO

CCO

CO

CCO

sampleCCOC

04470.02729.01638.0%100

%29.27

%29.27%100%

%29.27%10001.44

01.12%100%

?

2

2

2

2

2

2

22

22

2%100%100%

CO

C

CO

C

CO

C

CO

CCO

m

m

M

M

m

m

M

MC

2) All H in sample converted into H2O.

ggmm

m

mH

M

MH

mm

OHH

OH

HOH

OH

HOH

sampleHOHH

01875.01119.01676.0%100

%19.11

%19.11%100%

%19.11%100016.18

008.12%100

2%

?

2

2

2

2

2

2

OH

H

OH

H

OH

H

OH

HOH

m

m

M

M

m

m

M

MH

22

22

2

2

%100%1002

%

gm

m

mmmm

N

N

NHCsample

0521.0

01875.004470.01156.0

C H N

mass 0.04470 g 0.01875 g 0.0521 g

n=m/M 0.04470/12.01

0.01875/1.008

0.0521/14.00

n 0.003722 0.01860 0.003721

Divide by smallest value 1 5 1

E.F. CH5N

Chemical Equations Are sentences that describe what

happens in a chemical reaction. Reactants ® Products

CH4 + O2 ® CO2 + H2O Redistribution of bonds, atoms don’t

change! Equations should be balanced: Have the

same number of each kind of atoms on both sides.

Law of conservation of mass. CH4 + 2O2 ® CO2 + 2 H2O

Balancing by inspection.

Abbreviations (s) ¯ (g) (l) (aq) heat

D

catalyst

Practice Ca(OH)2 + H3PO4 ® H2O +

Ca3(PO4)2

Cr + S8 ® Cr2S3

KClO3(s) ® Cl2(g) + O2(g) Solid iron(III) sulfide reacts with

gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas.

Fe2O3(s) + Al(s) ® Fe(s) + Al2O3(s)

Stoichiometric Calculations

What mass of oxygen will react completely with 96.1 g of propane?

C3H8 + O2 ® CO2 + H2O Balance the equation!

C3H8 + 5 O2 ® 3 CO2 + 4 H2O

Language of moles, not grams!

molmolg

g

M

mn

HC

HCHC 18.2

/1.44

1.96

83

83

83

C3H8 + 5 O2 ® 3 CO2 + 4 H2O

1 52.18 ?

molnO 90.101

518.22

2349/3290.10

.222

2

2

2

Ogmolgmol

Mnm

M

mn

OOO

O

OO

Grams of CO2 produced?

C3H8 + 5 O2 ® 3 CO2 + 4 H2O

1 3

2.18 ?

molnCO 54.61

318.22

2288/01.4454.6

.222

2

2

2

COgmolgmol

Mnm

M

mn

COCOCO

CO

COCO

Mass of CO2 absorbed by 1.00 kg LiOH(s)?

LiOHmolmolg

g

M

mn

LiOH

LiOHLiOH 8.41

/95.23

1000

2 1

41.8 ?

molnCO 9.202

18.412

2920/01.449.20

.222

COgmolgmol

Mnm COCOCO

Antiacids

molmolg

g

M

mn

molmolg

g

M

mn

OHMg

OHMgOHMg

NaHCO

NaHCONaHCO

0171.0/32.58

00.1

0119.0/01.84

00.1

2

2

2

3

3

3

)(

)()(

1 1

0.0119 ?

molnHCl 0119.0

1 2

0.0172 ?molnHCl 0344.0

1

20172.0

Examples

One way of producing O2(g)

involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O2(g) are produced?

How many grams of potassium chloride?

How many grams of oxygen?

Examples A piece of aluminum foil 5.11 in x

3.23 in x 0.0381 in is dissolved in excess HCl(aq). How many grams of H2(g) are produced?

How many grams of each reactant are needed to produce 15 grams of iron form the following reaction?

Fe2O3(s) + Al(s) ®

Fe(s) + Al2O3(s)

Examples

K2PtCl4(aq) + NH3(aq) ®

Pt(NH3)2Cl2 (s)+

KCl(aq)

what mass of Pt(NH3)2Cl2 can be

produced from 65 g of K2PtCl4 ? How much KCl will be produced?

How much from 65 grams of NH3?

Limiting Reactant

31:Re toratiotricstoichiomeinareagents

mixturetricstoichiome

1 25 ?=10 mol NH3

3 29 ?=6 mol NH3

15 (not available)

H2 is LR.

3 mol nitrogen12 mol hydrogen

1 23 ?=6 mol NH3

3 212 ?=8 mol NH3

1

4 (not available)

N2 is LR.

Limiting Reactant Reactant that determines the

amount of product formed. The one you run out of first. Makes the least product.

N2 + H2 ® NH3

What mass of ammonia can be produced from a mixture of 100. g N2

and 500. g H2 ? How much unreacted material

remains?

molmolg

g

M

mn

molmolg

g

M

mn

H

HH

N

NN

01.248/016.2

500

57.3/00.28

100

2

2

2

2

2

2

1 23.57 ?=7.14 mol NH3

3 2248 ?=165.3 mol NH3

356.121/024.1714.7

.333

NHgmolgmol

Mnm NHNHNH

Reaction Yield The theoretical yield is the amount

you would make if everything went perfect.

The actual yield is what you make in the lab.

Percent Yield

%100% yieldltheoretica

yieldactualyield

68.5 kg CO is reacted with 8.60 kg H2. Calculate

the percent yield if the actual yield was 3.57x104 g CH3OH.

%100% yieldltheoretica

yieldactualyield

)(3)(2)( 2 lgg OHCHHCO

molmolg

g

M

mn

molmolg

g

M

mn

H

HH

CO

COCO

9.4265/016.2

106.8

6.2445/01.28

105.68

3

3

2

2

2

)(3)(2)( 2 lgg OHCHHCO

1 12445.6 ?=2445.6 mol CH3OH

2 14265.9 ?=2132.9 mol CH3OH

OHCHgmolgmol

Mnm OHCHOHCHOHCH

36.68343/042.329.2132

.333

%2.52%1006.68343

1057.3%100%

4

gyieldltheoretica

yieldactualyield

Examples Aluminum burns in bromine

producing aluminum bromide. In a laboratory 6.0 g of aluminum reacts with excess bromine. 50.3 g of aluminum bromide are produced. What are the three types of yield.

2 Al + 3 Br2 ® 2 AlBr3

Examples Years of experience have proven

that the percent yield for the following reaction is 74.3%

Hg + Br2 ® HgBr2

If 10.0 g of Hg and 9.00 g of Br2 are

reacted, how much HgBr2 will be

produced?

Examples Commercial brass is an alloy of Cu

and Zn. It reacts with HCl by the following reaction

Zn(s) + 2HCl(aq) ® ZnCl2 (aq) +

H2(g)

Cu does not react. When 0.5065 g of brass is reacted with excess HCl, 0.0985 g of ZnCl2 are eventually isolated. What is the composition of the brass (i.e. Zn% and Cu%)?