CHE 106: General Chemistry CHAPTER THREE  1

Chem 106, Prof. J.T. Spencer

CHE 106: General Chemistry

1



CHAPTER THREE

Copyright © James T. Spencer 1995 - 1999

All Rights Reserved


Stoichiometry

2

Chapter Three


Stoichiometry

3

• Antoine Lavoisier (1734 - 1794)

– Law of Conservation of Mass - atoms are neither created nor destroyed in chemical reactions

– total number of atoms = total number of atoms after reaction before reaction

– Stoichiometry - quantitative study of chemical formulas and reactions

(Greek; “stoichion”= element, “metron” = measure)

• Chemical Equations - used to describe chemical reactions in an accurate and convenient fashion

2H

2

+ O reactants

2

2 H

2

O products

Chapt. 3.1


Chemical Equations

4

To Write and Balance: (Shorthand

Communication for a great deal of information)

(1) Know Reactants

(2) Know ALL Products

(3) Balance - Same Number and Kinds of atoms on each side



5

• Chemical Equations

– Must have equal numbers of atoms of each element on each side of the equation = BALANCED EQUATION

2 H

2

+ O

2

4 hydrogen

2 oxygen

2 H

2

O

4 hydrogen

2 oxygen

N

2

O

5(g)

+ H

2 nitrogen

6 oxygen

2 hydrogen

2

O 2 HNO

2 nitrogen

6 oxygen

2 hydrogen

3

NOTE

The coefficients in front of the formula for a compound refers to the number of molecules (intact) involved while a subscript refers to the ratio of atoms within the molecule

Chapt. 3.1



6

• Chemical Equations

– balancing equations often requires some trial and error of coefficients

PCl

3(l)

+ 3 H

2

O

(l)

6 hydrogen

3 oxygen

1 phosphorus

3 chloride

H

3

PO

3(aq)

6 hydrogen

+ 3 HCl

3 oxygen

1 phosphorus

3 chlorine

C

6

H

12(l)

+ 9 O

2(g)

6 carbon

18 oxygen

12 hydrogen

6 CO

2(g)

+ 6 H

2

6 carbon

18 oxygen

12 hydrogen

O

(l)

NOTE

Never change subscripts in formulas when balancing chemical reactions!

subscripts change compounds; coefficients change amounts

Chapt. 3.1



7

Sample exercise: Balance the following equations by providing the missing coefficients:

C

2

H

4

+ O

2

CO

2

+ H

2

O

Chapt. 3.1



8


C

2

H

4

C

+ O

2

CO

C

2

+ H

2

O

H

O

H

O

Chapt. 3.1



9


C

2

H

4

C 2

+ O

2

CO

2

C 1

+ H

2

O

H 4

O 2

H 2

O 3

Chapt. 3.1



10


C

2

H

4

C 2

+ O

2

2 CO

2

+ H

C (1)2= 2

2

O

H 4

O 2

H 2

O 3 5

Chapt. 3.1



11


C

2

H

4

C 2

+ O

2

2 CO

2

+ 2H

C (1)2= 2

2

O

H 4

O 2

H (2)2 = 4

O 3 5 6

Chapt. 3.1



12


C

2

H

4

C 2

+ 3O

2

2 CO

2

+ 2H

C (1)2= 2

2

O

H 4 H (2)2 = 4

O (2)3 = 6 O 3 5 6

Chapt. 3.1



13


Al + HCl AlCl

3

+ H

2

Chapt. 3.1



14


Al + HCl AlCl

Al Al

3

+ H

2

H

Cl

H

Cl

Chapt. 3.1



15


Al + HCl AlCl

3

Al 1 Al 1

+ H

2

H 1

Cl 1

H 2

Cl 3

Chapt. 3.1



16


Al + 3HCl AlCl

Al 1 Al 1

3

+ H

2

H (1)3 = 3

Cl (1)3 = 3

H 2

Cl 3

Chapt. 3.1



17


Al + 6HCl AlCl

Al 1 Al 1

3

+ 3H

2

H (1)6 = 6

Cl (1)6 = 6

H (2)3 = 6

Cl 3

Chapt. 3.1



18


Al + 6HCl 2AlCl

3

Al 1

+ 3H

Al (1)2 = 2

2

H (1)6 = 6

Cl (1)6 = 6

H (2)3 = 6

Cl (3)2 = 6

Chapt. 3.1



19


2Al + 6HCl 2AlCl

3

+ 3H

Al (1)2 = 2 Al (1)2 = 2

2

H (1)6 = 6

Cl (1)6 = 6

H (2)3 = 6

Cl (3)2 = 6

Chapt. 3.1


Chemical Reactivity

20

• Chemical Reactions

–The course of a chemical reaction can often

be predicted by recognizing general patterns of reactivity through similar reactions previously observed. Elements in same family (column of table) have similar reactions.

–The periodic table is helpful in predicting

products of reactions. Atoms like to assume electron configurations of the

Noble Gases.

Chapt. 3.2



21

• Chemical Reactions

Example, if you know that

2Li + 2H

2

0 2LiOH + H

2 then you should be able to predict the products from the reaction of Na, K and the other members of group 1 (alkali metals) with water. Thus a general reaction would be;

2 M + 2 H

2

O 2 MOH + H

2

Chapt. 3.2



22

•Combustion Reactions

•Combination Reactions

•Decomposition Reactions

•Metathesis Reactions



23

• Combustion Reactions

– Reactions with oxygen (usually from the air)

– The complete combustion of hydrocarbons yield carbon dioxide (CO

2

) and water (H

2

O)

C x

H y

+ (2x+y) /

2

O

2

X CO

2

+ Y /

2

H

2

O

Generally:

Balance Carbon Atoms First

Balance Hydrgoens

Balance Oxygen Atoms

Chapt. 3.2



24

• Combustion Reactions

– Reactions with oxygen (usually from the air)

– The complete combustion of hydrocarbons yield carbon dioxide (CO

2

) and water (H

2

O)

C x

H y

+ (2x+y) /

2

O

2

X CO

2

+ Y /

2

H

2

O

Examples

octane: C

8

ethanol: C

H

18

H

+ 25 /

2

O

2

5

OH + 3 O

2

glucose: C

styrene: C

8

6

2

H

H

8

12

O

6

+ 9 O

+ 10 O

2

2

8 CO

2

2 CO

2

6 CO

2

8 CO

2

+ 9 H

2

+ 3 H

2

+ 6 H

+ 4 H

2

2

O

O

O

O

Chapt. 3.2



25

Sample exercise: Write the balanced equation for the reaction that occurs when ethanol, C

2

H

5

OH(l) is burned in air.

Chapt. 3.2



26


2

H

5


C

2

H

5

OH + O

2

CO

2

+ H

2

O

Chapt. 3.2



27


2

H

5


C

2

H

5

OH + O

2

CO

2

+ H

2

O

C C

H

O

H

O

Chapt. 3.2



28


2

H

5


C

2

H

5

OH + O

2

CO

2

+ H

2

O

C 2 C 1

H 6

O 3

H 2

O 3

Chapt. 3.2



29


2

H

5


C

2

H

5

OH + O

2

2CO

2

+ H

2

O

C 2 C (1)2 = 2

H 6

O 3

H 2

O 3 5

Chapt. 3.2



30


2

H

5


C

2

H

5

OH + O

2

2CO

2

+ 3H

2

O

C 2 C (1)2 = 2

H 6

O 3

H (2)3 = 6

O 3 5 7

Chapt. 3.2



31


2

H

5


C

2

H

5

OH + 3O

2

2CO

2

+ 3H

2

O

C 2 C (1)2 = 2

H 6

O 3 7

H (2)3 = 6

O 3 5 7

Chapt. 3.2



32

• Combination Reactions

– two or more substances react to form a single product

– especially common in the reactions of pure elements

A + B C

Ni(s) + 4 CO(g)

BF

3

(g) + NH

3

(g)

Ni(CO)

4

(g)

BF

3

NH

3

(s)

Chapt. 3.2



33

• Decomposition Reactions

–when one compound reacts to form two

or more products (opposite of combination reactions)

C A + B

• (often heat required)

2 NaN

3

(s)

B(OH)

3

Air Bag Inflator (J. Chem. Ed. 1990, 67, 61)



(heat)

2 Na(s) + 3 N

2

(g)

HBO

2

+ H

2

O

Chapt. 3.2



34

• Metathesis Reactions

–when ionic “partners” switch

AB + CD AD + BC

• (often in aqueous solutions)

Ag(NO

3

) + KCl

BaCl

2

+ Na

2

SO

4

AgCl(s) + KNO

3

BaSO

4

(s) + 2 NaCl

Chapt. 3.2



35

Sample Exercise: Write balanced chemical equations for the following reactions:

Solid mercury (II) sulfide decomposes into its component elements when heated

.

Chapt. 3.2



36


Solid mercury (II) sulfide decomposes into its component elements when heated

.

Hg +2 S -2

HgS  Hg + S

Chapt. 3.2



37


The surface of aluminum metal undergoes a combination reaction with oxygen in air.

Chapt. 3.2



38



Al +3 O -2

Al + O

2

 Al

2

O

3

Chapt. 3.2



39



Al +3 O -2

4Al + 3O

2

 2Al

2

O

3

Chapt. 3.2



40


Si

2

H

6 burns when exposed to air

.

Chapt. 3.2



41


Si

2

H


. Hint, Si is in the same group as C, and therefore reacts similarly.

Si

2

H

6

+ O

2

 SiO

2

+ H

2

O

Chapt. 3.2



42


Si

2

H


. Hint, Si is in the same group as C, and therefore reacts similarly.

2Si

2

H

6

+ 7O

2

 4SiO

2

+ 6H

2

O

Chapt. 3.2


Atomic and Molecular Weights

43

•Chemical equations indicate exactly the

amounts of two reagents which will react to form an exact amount of products

•Atomic Mass Scale - based upon

12 C isotope.

This isotope is assigned a mass of exactly 12 atomic mass units (amu) and the masses of all other atoms are given relative to this standard.

•Most elements in nature exist as mixtures of

isotopes.

Chapt. 3.3



44

• Atomic Mass Scale - given the following;

100 g of water contains 11.1 g of H and 88.9 g of O and the formula for water is H

2

O then;

•water has 8 times more O than H by mass

( 88.9

/

11

= 8)

•if water has 2 H for 1 O then O atoms must weigh

16 time more than H atoms

•if H is assigned an atomic mass of 1 amu then O must weigh 16 amu (using the 12 C standard)

•1 amu = 1.66054 x 10 -24 g

OR 1 g = 6.02214 x 10 23 amu

Chapt. 3.3


Mass Spectrometer

45

• Direct methods of measuring (separating) mass.

• Sample molecules are ionized by e-beam to cations (+1 by

“knocking off” one electron) which are then deflected by magnetic field - for ions of the same charge the angle of deflection in proportional to the ion’s mass vacuum chamber accelerating grid (-) sample beam of pos. ions

N

Mass

Spectrum

Hg

Int.

ionizing e- beam focusing slits

S magnetic field detector

200 mass number (amu)

Chapt. 3.3



46

Mass

Spectrum

Cl

35

Int.

Mass

Spectrum

C

12

Int.

Mass

Spectrum

P

31

Int.

37 13 mass number (amu)

35 Cl: 75% abundant

37 Cl: 24% abundant mass number (amu) mass number (amu)

12 C: 98.9% abundant

13 C: 1.11% abundant

31 P: 100% abundant



47

Unknown white powdery substance injested by unconscious patient.

What do you do? Is it Heroin, Cocaine, Caffeine?

Mass Spectrum of Unknown Compound

Mass 25 50 75 100 125 150 175 200 225 250 275 300



48

MS Library other peaks at

327 and 369

43

204

215

268

94

146

Mass 25 50 75 100 125 150 175 200 225 250 275 300

194

Caffeine

67

109

42

55

82

MS of Unknown

Mass 25 50 75 100 125 150 175 200 225 250 275 300



49

MS

Library

82

182

303

42

122

150

272

Mass 25 50 75 100 125 150 175 200 225 250 275 300

194

Caffeine

67

109

42

55

82

MS of Unknown

Mass 25 50 75 100 125 150 175 200 225 250 275 300



50

194 MS Library

67

109

55

82

42

Mass 25 50 75 100 125 150 175 200 225 250 275 300

194

Caffeine

67

109

42

55

82 MS of Unknown

Mass 25 50 75 100 125 150 175 200 225 250 275 300



51

Unknown white powdery substance injested by unconscious patient.

What do you do?

Mass Spectrum

O

H

3

C

N

O

N

CH

3

Caffeine

N

CH

3

N

Mol. Wgt

= 194

Mass 25 50 75 100 125 150 175 200 225 250 275 300


Atomic Weights

52

• Average Atomic Mass (AW)- weighted average (by

% natural abundance) of the isotopes of an element.

•Example (1);

10

11

B is 19.78% abundant with a mass of 10.013 amu

B is 80.22% abundant with a mass of 11.009 amu therefore the average atomic mass of boron is;

(0.1987)(10.013) + (0.8022)(11.009) = 10.82 amu

Chapt. 3.3


Atomic Weights

53

• Average Atomic Mass (AW)- weighted average (by

% natural abundance) of the isotopes of an element.

•Example (2):

194 Pt is 33.90% abundant with a mass of 193.963 amu



198 Pt is 7.210% abundant with a mass of 197.968 amu therefore the average atomic mass of platinum is;

(0.3390)(193.963) + (0.3380)(194.965) + (0.2530)(195.965 ) +

(0.07210)(197.968)= 195.504 amu

Chapt. 3.3



54

Sample exercise: Three isotopes of silicon occur in nature: 28 Si (92.21%), which has a mass of 27.97693 amu; 29 Si (4.70%), which has a mass of 28.97659 amu; and 30 Si (3.09%), which has a mass of 29.97376 amu. Calculate the atomic weight of silicon.

Chapt. 3.3



55


27.97693(0.9221) + 28.97659(0.0470) + 29.97376(0.0309) =

28.0856 amu

Chapt. 3.3



56


27.97693(0.9221) + 28.97659(0.0470) + 29.97376(0.0309) =

28.0856 amu

* 3 sig figs 28.1 amu

Chapt. 3.3


Atomic Weights

57

• Sample Problem : When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown are obtained. Use these data to compute the average mass of natural copper.

[masses for 63 Cu = 62.93 amu and 65 Cu = 64.93 amu]

69.09%

30.91%

Given:

•Masses for 63 Cu and 65 C

•Relative abundance of 63 Cu and 65 Cu

Find:

•Average Mass of Cu

Mass No. 63 65


Atomic Weights

58

(.6909 atoms)(62.93 amu) + (.3091 atoms)(64.93amu) = 63.55 amu atom atom.

average mass per atom is;

6355 amu = 63.55 amu/atom

100 atoms

69.09%

30.91%

Given:

•Masses for 63 Cu and 65 C

•Relative abundance of 63 Cu and 65 Cu

Find:

•Average Mass of Cu

Mass No. 63 65


Molecular Weights

59

• Formula Weights (FW) - sum of the atomic weights of each atom in its chemical formula. (note AW is atomic weight)

• formula weight of NaN

3

= 3(AW of N) + 1(AW of Na)

3(14) + 1(23) = 65 amu for sodium azide

• Molecular Weights (MW) - sum of atomic weights of each atom in its molecular formula

• molecular weight of B

2

H

6

= 2(AW of B) + 6(AW of H)

2(10.8) + 6(1) = 27.6 amu for diborane

• Difference between Molecular and Formula Weights

• ionic compounds, with extended arrays, have no well defined molecules (and no molecular formulas) so we use the formula weights (i.e., NaCl = 58 amu )

Chapt. 3.3



60

• Calculate molecular/formula weights for the following:

– P

4

O

10

– BrCl

– Ca(NO

3

)

2

P =31; O = 16; Br = 80; Cl = 35.5; Ca = 40; N = 14



61

• Calculate molecular weights/formula for the following:

– P

4

O

10

4(31) + 10(16) = 284 amu

– BrCl

1(80) + 1(35.5) = 115.5 amu

– Ca(NO

3

)

2

1(40) + 2(14) + 6(16) = 164 amu

P =31; O = 16; Br = 80; Cl = 35.5; Ca = 40; N = 14


Percentage Composition

62

• Percentage Composition - percentage by mass contributed by each element in the substance. May be used to verify the purity or identity of a particular compound.

• 100 [ (atoms of an element in formula)(AW) /

FW

] = % comp. element

Percentage Composition of C

6

(FW = 180)

H

12

O

6

% C = 100 (6)(12)/ (180) = 40.0% carbon

% O = 100 (6)(16)/ (180) = 53.3% oxygen

% H = 100 (12)(1)/ (180) = 6.7% hydrogen

2 example calculations follows

Chapt. 3.3



63

Problem : In 1987, the first substance to act as a superconductor at a temperature above that of liquid nitrogen (77 K) was discovered. The approximate formula of the substance is YBa

2

Cu

3

O

7

. Calculate the percent composition by mass of this material.

M W of YBa

2

Cu

3

O

7

= (88.9) + 2(137.3) + 3(63.6) + 7(16)

= 666.0 amu

AW: Y = 88.9; Ba =137.3; Cu = 63.6; O = 16.0



64

M W of YBa

2

Cu

3

O

7

= 666.0 amu

Y = 1(88.9) = 100 (88.9) = 13.3 %

666.0

Ba = 2(137.3) = 100 (274.6) = 41.3 %

666.0

Cu = 3(63.5)

O = 7(16.0)

= 100 (190.5) = 28.6 %

666.0

= 100 (112)

666.0

= 16.8 %

AW: Y = 88.9; Ba =137.3; Cu = 63.6; O = 16.0



65

Sample exercise: Calculate the percentage of nitrogen, by mass, in Ca(NO

3

)

2

.



66


3

)

2

.

Formula Mass:

1(40.1) = 40.1

2(14.0) = 28.0

6(16.0) = 96.0

164.1 amu



67


3

)

2

.

Formula Mass: part x 100

1(40.1) = 40.1 total

2(14.0) = 28.0

6(16.0) = 96.0

164.1 amu



68


3

)

2

.


1(40.1) = 40.1 total

2(14.0) = 28.0

6(16.0) = 96.0 28.0 x 100 =

164.1 amu 164.1



69


3

)

2

.


1(40.1) = 40.1 total

2(14.0) = 28.0

6(16.0) = 96.0 28.0 x 100 = 17.1%

164.1 amu 164.1



70

• Example ; determine the elemental percent composition of strychnine N

•Molecular formula of strychnine = C

21

H

22

N

2

O

2

•Percent composition;

% C = 100 (21)(12)/ (334) = 75.4% carbon

% N = 100 (2)(14)/ (334) = 6.59% nitrogen

% O = 100 (2)(16)/ (334) = 9.59% oxygen

O

N

O

•Molecular weight (MW)

MW = 21(12) + 22(1) + 2(14) + 2(16) = 334 amu

% H = 100 (22)(1)/ (334) = 8.38% hydrogen

Chapt. 3.3

Mole


71

• Very small macroscopic samples contain VERY many atoms, molecules, etc... (e.g. 1 tsp. H

2

O contains 2 x 10 23 mol). [Need convenient counting unit]

• known no. of H atoms in 1 g of H = no. of atoms of O in

16 g of O = no. of C atoms in 12 g of C = etc... (based upon atomic weights)

• Def. = The number of carbon atoms in 12 g of 12 C is called Avogadro’s number . One Mole (latin “mole”

= a mass) is the amount of material that contains

Avogadro’s number

Chapt. 3.4

•Note: a mole refers to a fixed number of any type of particles!

•Avogadro’s number = 6.023 x 10 23


Avagadro’s Number and the Mole

72

• 1 mole of 12 C atoms = 6.02 x 10 23 atoms

• 1 mole of 11 B atoms = 6.02 x 10 23 atoms

• 1 mol of PCl

3 molecules = 6.02 x 10 23 molecules

• 1 mol of Na+ ions = 6.02 x 10 23 Na ions

• 1 mol of toasters = 6.02 x 10 23 toasters

• 1 mole of students = 6.02 x 10 23 students

Chapt. 3.4


Mole

73

•Examples

How many C atoms in 0.5 moles of Carbon?

C atoms = (0.5 moles C)(6.02 x 10 23 atoms) = 3.01 x 10 23 mole

How many C atoms are in 0.25 mol of C

6

H

12

O

6

?

C ato. = (0.25 mol C

6

H

12

O

6

)(6.02 x 10 23 molec) (6 C atoms) mol (1 C

6

H

12

O

6 molec)

C atoms = 1.5 x 10 23 atoms

Chapt. 3.4

Mole


74

Sample exercise: How many O atoms are in a) 0.25 moles Ca(NO

3

)

2

Chapt. 3.4

Mole


75


3

)

2

0.25 mol Ca(NO

3

)

2

Chapt. 3.4

Mole


76


3

)

2

0.25 mol Ca(NO

3

)

2

6 mole O

1 mole Ca(NO

3

)

2

Chapt. 3.4

Mole


77


3

)

2

0.25 mol Ca(NO

3

)

2

6 mole O

1 mole Ca(NO

3

)

2

1.5 mol O 6.02 x 10 23 atom O

1 mol O

Chapt. 3.4

Mole


78


3

)

2

0.25 mol Ca(NO

3

)

2

6 mole O

1 mole Ca(NO

3

)

2

1.5 mol O 6.02 x 10 23 atom O = 9.03 x 10 23

1 mol O

Chapt. 3.4

Mole


79

Sample exercise: How many O atoms are in a) 1.50 moles sodium carbonate

Chapt. 3.4

Mole


80


Na +1 CO

3

-2  Na

2

CO

3

Chapt. 3.4

Mole


81


Na +1 CO

3

-2  Na

2

CO

3

1.50 mol Na

2

CO

3

Chapt. 3.4

Mole


82


Na +1 CO

3

-2  Na

2

CO

3

1.50 mol Na

2

CO

3

3 mol O

1 mol Na

2

CO

3

Chapt. 3.4

Mole


83


Na +1 CO

3

-2  Na

2

CO

3

1.50 mol Na

2

CO

3

3 mol O

1 mol Na

2

CO

3

4.50 mol O 6.02 x 10 23 atom O

1 mol O Chapt. 3.4

Mole


84


Na +1 CO

3

-2  Na

2

CO

3

1.50 mol Na

2

CO

3

3 mol O

1 mol Na

2

CO

3

4.50 mol O 6.02 x 10 23 atom O = 2.71 x 10 24

Chapt. 3.4

1 mol O


Molar Mass

85

• Example - since one 12 C atoms weighs 12 amu and a 24 Mg atom weighs 24 amu (twice as massive) and since a mole always contains the

same number of particles, a mole of 24 Mg must weigh twice as much as a mole of 12 C.

1 12 C atom weighs 12 amu; 1 mol 12 C weigh 12 g

1 24 Mg atom weighs 24 amu; 1 mol 24 Mg weigh 24 g

1 238 U atom weighs 238 amu; 1 mol 238 U weighs 238 g

Molar Mass - (in grams) of any substance is always numerically equal to its formula weight (in amu).

Chapt. 3.4


Molar Relationships

86

Name Formula Formula Mass of 1 mol Number and kind weight of form units of particles in 1 mol atomic nitrogen N molec. nitrogen N

2

14.0

28.0

14.0

28.0

6.02 X 10 23 N atoms

6.02 X 10 23 N

2 molec.

2(6.02 X 10 23 ) N atoms scandium ScCl

3

151.5

151.5

6.02 X 10 23 ScCl

3 units chloride 6.02 X 10 23 Sc 3+ ions

3(6.02 X 10 23 ) Cl- ions glucose C

6

H

12

O

6

180.0

180.0

Chapt. 3.4

6.02 X 10 23 gluc. molec.

6(6.02 X 10 23 ) C atoms

12(6.02 X 10 23 ) H atoms



87

(1) How many moles of phosphorus trichloride, PCl

50 g of the substance? (MW = 137.4 amu)

3

, are in

Moles of PCl

3

= 1mol PCl

3

137.4 g

(50 g PCl

3

) = 0.36 moles PCl

3

(2) How many molecules of PCl

3 molecules of PCl

3 are in 50 g?

= (0.36 moles )(6.023 x 10 23 molecules)

= 2.2 x 10 23 PCl

3

1 mole molecules

(3) How many grams of PCl

3 are in 0.75 moles?

x grams = (0.75 mole)(137.4 g PCl

1 mole PCl

3

3

) = 103 g of PCl

3

Chapt. 3.4

Grams



88 use molar mass use molar mass Moles use Avagadro’s number use Avagadro’s number

Items

(molecules, atoms, etc...)

Chapt. 3.4



89

Sample exercise: How many moles of

NaHCO

3 are present in 508 g of this substance?

Chapt. 3.4



90


NaHCO


508 g

Chapt. 3.4



91


NaHCO


508 g 1 mol

84 g

Chapt. 3.4



92


NaHCO


508 g 1 mol = 6.05 g NaHCO

3

84 g

Chapt. 3.4



93

Sample exercise: What is the mass, in grams, of a) 6.33 mol NaHCO

3

Chapt. 3.4



94

Sample exercise: What is the mass, in grams, of a) 6.33 mol NaHCO

3

6.33 mol NaHCO

3

84 g NaHCO

3

1 mol NaHCO

3

= 532 g NaHCO

3

Chapt. 3.4



95

Sample exercise: What is the mass, in grams, of b) 3.0 x 10 -5 mol sulfuric acid

Chapt. 3.4



96

Sample exercise: What is the mass, in grams, of b) 3.0 x 10 -5 mol sulfuric acid

3.0 x 10 -5 mol H

2

SO

4

98 g H

2

SO

1 mol H

2

4

SO

4

= 2.9 x 10 -3 g

Chapt. 3.4



97

Sample exercise: How many nitric acid molecules are in 4.20 g of HNO

3

?

Chapt. 3.4



98

Sample exercise: How many nitric acid molecules are in 4.20 g of HNO

3

?

4.20 g HNO

3

6.02 x 10 23 molec HNO

3

63 g HSO

3

= 4.01 x 10 22 molec

Chapt. 3.4


Empirical Formulas

99

• Empirical Formula - Relative number of each element in a compound.

• Using moles and percent weight (elemental analysis by chemical means), we can calculate an empirical formula

• Steps;

» assume a 100 g (convenient since working with % because the elements % can be thought of as g)

» calculate moles of element present in 100g sample

» find ratios of moles (approx) to lead to integral formula subscripts.

Chapt. 3.5


Empirical Formulas

100

Mass Percent of Elements assume

100 g sample

Grams of each Element

Use atomic weights

Empirical

Formula calculate mole ratio

Moles of each Elements

Chapt. 3.5


Empirical Formulas, Examples

101

Determine the empirical formula for a compound which contains 87.5 % N and 12.5% H by mass.

% g in 100g moles ratio

87.5 % N = 87.5 g N 1 mole = 6.25 moles N = 1

14 g

12.5% H = 12.5 g H 1 mole H = 12.5 moles H = 2

1 g

Empirical Formula = NH

2

Chapt. 3.5



102

Sample exercise: A 5.325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain

3.758 g of carbon, 0.316 g of hydrogen, and

1.251 g of oxygen. What is the empirical formula of this substance?

Chapt. 3.5



103




3.758 g x 100 = 70.58% C -> 70.58 g C 1 mol C = 5.88 mol C

5.325 g 12 g C

Chapt. 3.5



104




0.316 g x 100 = 5.93% H -> 5.93 g H 1 mol H = 5.93 mol H

5.325 g 1 g H

Chapt. 3.5



105




1.251 g x 100 = 23.49% O -> 23.49 g O 1 mol O = 1.47 mol O

5.325 g 16 g O

Chapt. 3.5



106




5.88 mol C ; 5.93 mol H ; 1.47 mol O

Chapt. 3.5



107




5.88 mol C ; 5.93 mol H ; 1.47 mol O

1.47 mol 1.47 mol 1.47 mol

4 4 1

Chapt. 3.5



108




5.88 mol C ; 5.93 mol H ; 1.47 mol O C

1.47 mol 1.47 mol 1.47 mol

4

H

4

O

4 4 1

Chapt. 3.5



109

A white unknown substance (mass spec. problem) found on an unconscious patient is suspected by a forensic chemist of being either cocaine or caffeine. Combustion of a 50.86 mg sample yielded 150.0 mg of CO

2 and 46.05 mg of water. Further analysis showed the compound contained 9.39% N by mass. The formula of cocaine is C

17

H

21

NO

4

. Can the substance be cocaine?

Unknown

% C, H, and O.

Percent Composition

Known

50.86 mg of cmpd gave 150.0 mg of CO

2

46.05 mg of H

2

O.

compound contains 9.39% N.

formula of Cocaine is C

17

H

21

NO

4

.

and

Chapt. 3.5



110

•Combustion Reactions

–Reactions with oxygen (usually from the air)

–The complete combustion of hydrocarbons yield carbon dioxide (CO

2

) and water (H

2

O)

C x

H y

+ (2x+y) /

2

O

2

X CO

2

+ Y /

2

H

2

O

Known


2

46.05 mg of H

2

O.



17

H

21

NO

4

.

and

Chapt. 3.5



111

Compute % C and % H (from combustion).

C:

H:

N:

O:

0.150 g CO

2

= 0.00341 mol CO

2

= 0.00341 mol C = 40.9 mg C

0.04605 g H

2

O = 0.00256 mol H

2

O = 0.00512 mol H = 5.12 mg H

(0.0939)(50.86) = 4.77 mg N = 0.000341 mol N

(50.86 g sample)-(40.9 mg C + 5.12 mg H + 4.77 mg N) =

= 0.08 mg O = 0.000006 mol O

Known


2

46.05 mg of H

2

O.



17

H

21

NO

4

.

and

C:

H:

N:

O:



112 mg in sample % in sample calc’n

40.9 mg C 100 (40.9mg/50.86 mg)

5.12 mg H

4.77 mg N

0.08 mg O

100 (5.11mg/50.86mg)

100 (4.77mg/50.86 mg)

100 (0.08mg/50.86mg)

% sample % cocaine

80.5 % C 67.3% C

10.1% H

9.4% N

0.02% O

6.9% H

4.6% N

21.2% O

Known


2

46.05 mg of H

2

O.



17

H

21

NO

4

.

and

C:

H:

N:

O:



113 g in cocaine % in sample calc’n

(303 grams)

17(12) = 204 g C 100 (204/303)

21(1) g H 100 (21/303)

1(14) g N

4(16) g O

100 (14/303)

100 (64/303)


80.5 % C 67.3% C

10.1% H

9.4% N

0.02% O

6.9% H

4.6% N

21.2% O

Known: formula of Cocaine is C

17

H

21

NO

4

.

MW = 17(12) + 21(1) + 1(14) + 4(16) = 303 amu

C:

H:

N:

O:




40.9 mg C 100 (40.9mg/50.86 mg)

5.12 mg H

4.77 mg N

0.08 mg O

100 (5.11mg/50.86mg)

100 (4.77mg/50.86 mg)

100 (0.08mg/50.86mg)


80.5 % C 67.3% C

10.1% H

9.4% N

0.02% O

6.9% H

4.6% N

21.2% O

C:

H:

N:

O:




40.9 mg C 100 (40.9mg/50.86 mg)

5.12 mg H

4.77 mg N

0.08 mg O

100 (5.11mg/50.86mg)

100 (4.77mg/50.86 mg)

100 (0.08mg/50.86mg)


80.5 % C 67.3% C

10.1% H

9.4% N

0.02% O

6.9% H

4.6% N

21.2% O

Compound is Not Cocaine from analysis


Combustion Reactions

116

 Reaction of Hydrogen with Oxygen

[COMBUSTION] (note precautions)

» 2 H

2

(g) + O

2

(g) 2 H

2

O(g)



H = -232 kJ/mol H

2

O

» Ignition temperature = 580 ° - 590° C

» Explosive [“when stuff gets really big really fas t” Beakman’ World]

» The rapid release of energy [-232 kJ/mol

H

2

O] into the surrounding air causes the air to very quickly expand. the explosion from pure H

2 sound quiter because the air expansion is slower.

[Video No. 20-21; 4:42 +1:29 m]

[Video No. 22; 2:50 m] Chem 106, Prof. J.T. Spencer

Combustion Reactions

117

 Combustion of Alcohol (ethanol):

C

2

H

5

OH(g) + 3 O

2

(g) 2CO

2

(g) + 3 H

2

O(g)



H =1366.2 kJ mol -1

 Tesla coil produces a high voltage electric spark.

The spark is required to initiate this reaction.

 Conversion of chemical energy (PE stored in bonds) to mechanical energy.



Questions for After Demonstration

Are other types of energy are produced besides mechanical energy?

 Why can the reaction not be repeated without flushing the bottle with air first?

O

2 flow


Combustion Analysis

118

• Empirical Formula from reaction with oxygen

• Organic Compounds - C to CO

2 and H to H

2

O

• Use CO

2 and H furnace

2

O to determine the amount of C and H in original sample

H

2

O absorbant

(Mg(ClO

4

)

2

) sample contaminant catalyst (CuO); oxidizes traces of

CO and C to CO

2

CO

2 absorbant

(NaOH)

Chapt. 3.5



119

The combustion of 5.00g of an organic compound containing

C, H, and O yields 9.57 g CO

2 empirical formula?

and 5.87 g H

2

O. What is the

Chapt. 3.5



120

The combustion of 5.00g of an organic compound containing

C, H, and O yields 9.57 g CO

2 empirical formula?

and 5.87 g H

2

O. What is the

9.57 g CO

2

5.87 g H

2

= 0.217 mol CO

2

O = 0.326 mol H

2

= 0.217 mol C = 2.60 g C

O = 0.652 mol H = 0.652 g H whatever is left over must be the amt. of O originally present;

5.00 g - (2.60 g C + 0.652 g H) = 1.75 g of O = 0.109 mol O thus; 0.217 mol C

0.109 mol O

0.652 mol H divide each by 0.109

1.99

1.00 thus C

2

5.98

H

6

O

(ethanol)

Chapt. 3.5



121

The combustion of 0.596 g of a compound containing only B and H yields 1.17 g H

2

O and all the boron is recovered as B

2

O

What is the empirical formula?

3

.

Chapt. 3.5



122


2


2

O


3

.

(1) Chem Equation; B x

H y

+ O

2

H

2

O + B

2

O

3

Chapt. 3.5



123


2


2

O


3

.

(1) Chem Equation; B

(2) 1.17 g H

2 x

H y

+ O

2

O = 1.17g = 0.065 mol H

18 g/mol

2

O

H

2

O + B

2

O

3

(3) g H = (0.065 mol H

2

O) (1 g H

2

O) (2 H mol H) = 0.130 mol H

1 mol H 1 mol H

2

O

0.130 mol H = 0.130 g H

Chapt. 3.5



124


2



2

O

3

.

(1) Chem Equation; B

(2) 1.17 g H

2 x

H y

+ O

2

O = 1.17g = 0.065 mol H

18 g/mol

2

O

H

2

O + B

2

O

3

(3) g H = (0.065 mol H

2

O) (1 g H

2

O) (2 H mol H) = 0.130 mol H

1 mol H 1 mol H

2

0.130 mol H = 0.130 g H

O

(4) (0.596 g tot)-(0.130 g H) = 0.466 g B; 0.466 g B = 0.043 mol B

(5) B = 0.0431 mol = 1.00

10.8 g/ mol

H = 0.130 mol = 3.01

0.0431

BH

3

0.0431

Chapt. 3.5


Chemical Problem Solving

125

• Read and UNDERSTAND the problem - determine what is being given and what is required.

• Identify the Unknown and Given data.

• Set up the problem - determine what kinds of information bear upon the problem, what solution pathways may be available, what chemical principles should give guidance, etc...

• Solve the problem - Use the data given and the appropriate relationships or equations to work throught the problem.

• Check your work - not just the mathematical functions but ask if the answer makes sense and provides what is being asked for! (sig. figs)

Chapt. 3.5


Equations and the Mole

126

• Coefficients in a balanced chemical equation refer to both the relative number of molecules involved in a reaction and the relative number of moles.

Chapt. 3.6

• Stoichiometric equivalence - from coefficients in a chemical equation; B

2

H

6

+ 3 O

2

3 H

2

O + B

2

O

3

• 1 mol B

2

H

6 equiv. to 3 moles O

2 equiv. to 3 mol H

• Used to calculate quantities involved in a reaction

2

0, ...

grams of compound A grams of compound B use molar mass of A moles of compound A use coeff of A and B from balanced eqn.

use molar mass of B moles of compound B


Mole Calculations

127

• Given the reaction for the formation of B of LiH?

2

H

6

(diborane), how many grams of diborane can be prepared from 3.0 g

6 LiH + 8BF

3

[B

2

H

6

6 LiBF

4

+ B

2

H

MW = 27.6 and LiH MW = 7.9]

6

3.0 g LiH (1 mol LiH) = 0.38 mol LiH

7.9 g LiH

0.38 mol LiH (1 mol B

2

H

6

) 27.6 g B

2

H

6

6 mol LiH 1 mol B

2

H

6

= 1.7 g B

2

H

6

Chapt. 3.6



128

• Given the reaction for the formation of B

2

H

6 how many grams of BF

LiH ?

3

(diborane), are required to react with 3.0 g of

[B

2

H

6 LiH + 8BF

6

3

MW = 27.6, BF

3

6 LiBF

4

+ B

2

H

6

= 67.8 and LiH MW = 7.9]

3.0 g LiH (1 mol LiH) (8 mol BF

3

) (67.8 g BF

3

7.9 g LiH 6 mol LiH 1 mol BF

3

) = 34 g BF

3

Chapt. 3.6



129

Sample exercise: A common laboratory method for preparing small amounts of O involves the decomposition of KClO

3

:

2

2KClO

3

 2KCl + 3O

2

Chapt. 3.6



130


3

:

2

2KClO

3

 2KCl + 3O

2

How many grams of oxygen is produced from 4.50 g KClO

3

?

4.50 g KClO

3

Chapt. 3.6



131


3

:

2

2KClO

3

 2KCl + 3O

2


3

?

4.50 g KClO

3

1 mol KClO

3

122.6 g KClO

3

Chapt. 3.6



132


3

:

2

2KClO

3

 2KCl + 3O

2


3

?

4.50 g KClO

3

1 mol KClO

3

3 mol O

2

122.6 g KClO

3

2 mol KClO

3

Chapt. 3.6



133


3

:

2

2KClO

3

 2KCl + 3O

2


3

?

4.50 g KClO

3

1 mol KClO

3

3 mol O

2

32 g O

2

122.6 g KClO

3

2 mol KClO

3

1 mol O

2

Chapt. 3.6



134


3

:

2

2KClO

3

 2KCl + 3O

2


3

?

4.50 g KClO

3

1 mol KClO

3

3 mol O

2

32 g O

2

122.6 g KClO

3

2 mol KClO

3

1 mol O

2

= 1.76 g O

2

Chapt. 3.6



135

Sample exercise: Propane, C fuel used for cooking and home heating.

What mass of O

2

3

H

8

, is a common is consumed in the combustion of 1.00 g of propane?

Chapt. 3.6



136


What mass of O

2

3

H

8


C

3

H

8

+ O

2

 CO

2

+ H

2

O

Chapt. 3.6



137


What mass of O

2

3

H

8


C

3

H

8

+ 5O

2

 3CO

2

+ 4H

2

O

Chapt. 3.6



138


What mass of O

2

3

H

8


C

3

H

8

+ 5O

2

 3CO

2

+ 4H

2

O

1.00 g C

3

H

8

Chapt. 3.6



139


What mass of O

2

3

H

8


C

3

H

8

+ 5O

2

 3CO

2

+ 4H

2

O

1.00 g C

3

H

8

1 mol C

3

H

44 g C

3

H

8

Chapt. 3.6



140


What mass of O

2

3

H

8


C

3

H

8

+ 5O

2

 3CO

2

+ 4H

2

O

1.00 g C

3

H

8

1 mol C

3

H

44 g C

3

H

8

5 mol O

2

1 mol C

3

H

8

Chapt. 3.6



141


What mass of O

2

3

H

8


C

3

H

8

+ 5O

2

 3CO

2

+ 4H

2

O

1.00 g C

3

H

8

1 mol C

3

H

44 g C

3

H

8

5 mol O

2

1 mol C

3

H

8

32 g O

2

1 mol O

2

Chapt. 3.6



142


What mass of O

2

3

H

8


C

3

H

8

+ 5O

2

 3CO

2

+ 4H

2

O

1.00 g C

3

H

8

1 mol C

3

H

44 g C

3

H

8

5 mol O

2

1 mol C

3

H

8

32 g O

2

1 mol O

2

= 3.64 g O

2

Chapt. 3.6


Limiting Reagent

143

• Sometimes after one reagent is completely consumed in the reaction some of another reagent is left over. The reagent which is completely consumed limits the extent of the reaction = LIMITING REAGENT.

Limiting Reagent

Chapt. 3.7

+


Limiting Reagent Calculations

144


2

H

6

5.0 g of LiH and 5.0 g of BF

3 which reagent would be left over?

(diborane), if were reacted how much of

[B

2

6 LiH + 8BF

H

6

3

MW = 27.6. BF

3

6 LiBF

4

+ B

2

H

6

= 67.8 and LiH MW = 7.9]

Know :

Quantities (g and moles) of starting materials

Molar ratios between all the starting materials and products.

Find :

Which reagent is completely consumed ( limiting reagent ) and which is left over

Chapt. 3.6



145


2

H

6




[B

2

6 LiH + 8BF

3

H

6

MW = 27.6. BF

3

6 LiBF

4

+ B

2

H

6

= 67.8 and LiH MW = 7.9]

5.0 g LiH (1 mol LiH) = 0.63 mol AND 5.0 g BF

3

(1 mol BF

3

) = 0.074 mol

7.9 g LiH 67.8 g BF

3

If all the LiH were consumed, then 0.84 mol BF

3 would be required

[(0.63 mol LiH)(8 mol BF

3

)] = 0.84 mol BF

6 mol LiH)

3

Chapt. 3.6



146


2

H

6




[B

2

6 LiH + 8BF

3

H

6

MW = 27.6. BF

3

6 LiBF

4

+ B

2

H

6

= 67.8 and LiH MW = 7.9]

5.0 g LiH (1 mol LiH) = 0.63 mol AND 5.0 g BF

3

(1 mol BF

3

) = 0.074 mol

7.9 g LiH

Since only 0.074 mol of BF

(all consumed).

0.074 mol BF

3

3 is available, BF

3

67.8 g BF

3

If all the LiH were consumed, then 0.84 mol BF

3 would be required is the limiting reagent

(6 mol LiH) = 0.056 mol LiH consumed

8 mol BF

3 therefore remaining LiH = (0.63 mol - 0.056 mol)(7.9 g/mol) = 4.53 g

Chapt. 3.6


Limiting Reagent Problems

147

• Equal weights (5.00 g) of Zn(s) and I together to form ZnI

2

2

(s) are mixed

. How much ZnI and which is the limiting reagent?

2 is formed? How much of each reactant remains at the end of the reaction

Zn (AW = 65.4 amu)

I

2

(MW = 253.8 amu)

Zn(s) + I

2

(s) ZnI

2

(s)

Chapt. 3.7


Limiting Reagent Problems

148

• Equal weights (5.00 g) of Zn(s) and I together to form ZnI

2

2

(s) are mixed

. How much ZnI and which is the limiting reagent?

2 is formed? How much of each reactant remains at the end of the reaction

Zn (AW = 65.4 amu)

I

2

Zn(s) + I

2

(s)

(MW = 253.8 amu)

Zn = 5.0 g Zn (1 mol Zn) = 0.076 mol Zn

ZnI

2

(s)

I

2

= 5.0 g I

2

I

2

65.4 g Zn

(1 mol I

2

253.8 g I

) = 0.020 mol I

2 is the limiting reagent.

2

Zn remaining = (0.076 Zn - 0.020 mol Zn) (64.5 g Zn) = 3.66 g Zn

1 mol Zn

Chapt. 3.7


Theoretical Yields

149

• Theoretical Yield - quantity of product calculated to form when all the limiting reagent is consumed (calculated from molar ratios).

• Actual Yield - the amount of product experimentally obtained from a reaction

• Percent Yield - describes relationship between theoretical and actual yields; percent yield = actual yield (100) theoretical yield

Chapt. 3.6


Percent Yields

150

• Given the reaction of 2.05 g of hydrogen sulfide with 1.84 g of sodium hydroxide, calculate how the theoretical yield of

Na

2

S. What is the percent yield if the amt. of Na

2 obtained was 3.65 g. [H

2

S (MW = 34.1); Na

2

S

S (MW = 78.1)]

H

2

S(g) + 2 NaOH(aq) Na

2

S(aq) + 2 H

2

O

(2.05 g H

2

S)(1 mol H

2

S)(1 mol Na

34.1 g H

2

S 1 mol H

2

2

S)(78.1 g Na

S 1 mol Na

2

S) = 4.70 g Na

2

S

2

S theoretical yield

% yield = 3.65 g (actual yield) (100) = 77.7 % yield

4.70 g (theoretical yield)

Chapt. 3.6


End of Chapter 3

151

Major Topics (not exhaustive list):

(1) Chemical Equations

(2) Periodic Table and Reaction

Types

(3) Atomic and Molecular

Weights (formula weights, % compositions, etc...)

(4) Molar Concepts

(5) Empirical Formulas

(6) Info from Balanced Eqns.

(7) Limiting Reagents

(8) Percent Yields

CHE 106: General Chemistry CHAPTER THREE  1

CHE 106: General Chemistry

CHAPTER THREE

Chapter Three

Chemical Equations

–The course of a chemical reaction can often

–The periodic table is helpful in predicting

•Combustion Reactions

•Combination Reactions

•Decomposition Reactions

•Metathesis Reactions

–when one compound reacts to form two

–when ionic “partners” switch

Atomic and Molecular Weights

•Chemical equations indicate exactly the

•Atomic Mass Scale - based upon

•Most elements in nature exist as mixtures of

Atomic and Molecular Weights

Atomic and Molecular Weights

Atomic and Molecular Weights

Atomic and Molecular Weights

4(31) + 10(16) = 284 amu

1(80) + 1(35.5) = 115.5 amu

1(40) + 2(14) + 6(16) = 164 amu

Compound is Not Cocaine from analysis

Combustion Reactions

Combustion Reactions

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CHE 106: General Chemistry CHAPTER THREE  1