c h a p t e rc h a p t e r c h a p t e rc h a p t e r chemistry, 5 th edition mcmurry/fay chemistry,...

C h a p t e rC h a p t e r

Chemistry, 5th EditionMcMurry/Fay

Chemistry, 5th EditionMcMurry/Fay

33Formulas, Equations,

and MolesFormulas, Equations,

and Moles

2

The Structure of AtomsThe Structure of Atoms

Atomic Mass Unit

1 amu = 1/12 of the mass of on atom of Carbon-12

1 amu = 1.6605 x 10-24 g

3

Mass: proton = 1.00728 amuneutron = 1.0086 amuelectron = 0.000548612C atom = 12.00000 amu13C atom = 13.00335 amu

Atomic and Molecular MassAtomic and Molecular Mass

1 amu mass of carbon 12 atom12

4


• The atomic masses as tabulated in the periodic table are the averages of the naturally occurring isotopes.

• Mass of C = average of 12C and 13C

= 0.9889 x 12 amu + 0.0111 x

13.0034 amu

= 12.011 amu

5


The mass of a molecule is just the sum of the masses of the atoms making up the molecule.

m(C2H4O2) = 2·mC + 4·mH + 2·mO

= 2·(12.01) + 4·(1.01) + 2·(16.00)

= 60.06 amu

6

Avogadro and the MoleAvogadro and the Mole

• One mole of a substance is the gram mass value equal to the amu mass of the substance.

• One mole of any substance contains 6.02 x 1023 units of that substance.

• Avogadro’s Number (NA, 6.022 x 1023) is the

numerical value assigned to the unit, 1 mole.

7


8


• Methionine, an amino acid used by organisms to make proteins, is represented below. Write the formula for methionine and calculate its molar mass. (red = O; gray = C; blue = N; yellow = S; ivory = H)

9


• The Mole: Allows us to

make comparisons

between substances

that have

different

masses.

10

Balancing Chemical EquationsBalancing Chemical Equations

• A balanced chemical equation represents the conversion of the reactants to products such that the number of atoms of each element is conserved.

reactants products

limestone quicklime + gas

Calcium carbonate calcium oxide + carbon dioxide

CaCO3(s) CaO(s) + CO2(g)

11


CaCO3(s) CaO(s) + CO2(g)

The letters in parentheses following each substance are

called State Symbols

(g) → gas (l) → liquid (s) → solid (aq) → aqueous

12


A balanced equation MUST have the same number of

atoms of each element on both sides of the equation.

H2 + O2 → H2O Not Balanced

H2 + ½O2 → H2O Balanced

2H2 + O2 → 2H2O Balanced

13


The numbers multiplying chemical formulas

in a chemical equation are called:

Stoichiometric Coefficients (S.C.)

2H2 + O2 → 2H2O Balanced

Here 2, 1, and 2 are stoichiometric coefficients.

14


Hints for Balancing Chemical Equations:

1) Save single element molecules for last.

2) Try not to change the S.C. of a molecule containing

an element that is already balanced.

3) If possible, begin with the most complex molecule

that has no elements balanced.

15


Hints for Balancing Chemical Equations:

4) Otherwise, trial and error!!

16


Example 1: CH4 + O2 → CO2 + H2O

Balance O2 last

C is already balanced

Start by changing S.C. of H2O to balance H

CH4 + O2 → CO2 + 2H2O

17


Example 1: CH4 + O2 → CO2 + 2H2O

Now C and H are balanced

Balance O by changing the S.C. of O2

CH4 + 2O2 → CO2 + 2H2O

BALANCED!

18


Example 2: B2H6 + O2 → B2O3 + H2O

Balance O last

B is already balanced

Start by changing S.C. of H2O:

B2H6 + O2 → B2O3 + 3H2O

19


Example 2: B2H6 + O2 → B2O3 + 3H2O

B and H are balanced

Balance O by changing S.C. of O2

B2H6 + 3O2 → B2O3 + 3H2O

BALANCED!

20


Example 3: MnO2 + KOH + O2 → K2MnO4 + H2O

Balance O last

Mn is already balanced

Change S.C. of KOH to balance K

MnO2 + 2KOH + O2 → K2MnO4 + H2O

21


Example 3: MnO2 + 2KOH + O2 → K2MnO4 + H2O

Mn, K, and H are balanced (H was balanced by chance)

Balance O

MnO2 + 2KOH + ½O2 → K2MnO4 + H2O

or

2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O

22


Example 4: NaNO2 + H2SO4→

NO + HNO3 + H2O + Na2SO4

Hard one (no single element molecules)

S is balanced

Start with NaNO2 to balance Na

2NaNO2 + H2SO4→ NO + HNO3 + H2O + Na2SO4

23


Example 4: 2NaNO2 + H2SO4→

NO + HNO3 + H2O + Na2SO4

S, Na, and N are balanced

Cannot balance H without changing S.C. for H2SO4!

Boo! Option 1: trial and error

Option 2: Go on to next problem!

24


• Balance the following equations:

C6H12O6 → C2H6O + CO2

Fe + O2 → Fe2O3

NH3 + Cl2 → N2H4 + NH4Cl

KClO3 + C12H22O11 → KCl + CO2 + H2O

26



C6H12O6 → 2C2H6O + 2CO2

Fe + O2 → Fe2O3



27



C6H12O6 → 2C2H6O + 2CO2

4Fe + 3O2 → 2Fe2O3 (balance O first)



28



C6H12O6 → 2C2H6O + 2CO2

4Fe + 3O2 → 2Fe2O3 (balance O first)


N:H is 1:3 on left, must get 1:3 on right!

29



N:H is 1:3 on left, must get 1:3 on right!

4NH3 + Cl2 → N2H4 + 2NH4Cl

30



C6H12O6 → 2C2H6O + 2CO2

4Fe + 3O2 → 2Fe2O3

4NH3 + Cl2 → N2H4 + 2NH4Cl

KClO3 + C12H22O11 → KCl + CO2 + H2O (tough!)

31




balance C

KClO3 + C12H22O11 → KCl + 12CO2 + H2O

balance H

KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O

balance O

8KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O

32



8KClO3 + C12H22O11 → KCl + 12CO2 + 11H2O

balance K (and hope Cl is balanced)

8KClO3 + C12H22O11 → 8KCl + 12CO2 + 11H2O

Balanced!

33


• Write a balanced equation for the reaction of element A (red spheres) with element B (green spheres) as represented below:

34


35

StoichiometryStoichiometry

• Stoichiometry: Relates the moles of products and reactants to each other and to measurable quantities.

36


Aqueous solutions of NaOCl (household bleach) are

prepared by the reaction of NaOH with Cl2:

2 NaOH(aq) + Cl2(g) NaOCl(aq) + NaCl(aq) + H2O(l)

How many grams of NaOH are needed to react with

25.0 g of Cl2?

37


2 NaOH + Cl2 → NaOCl + NaCl + H2O

25.0 g Cl2 reacts with ? g NaOH

22

22 353.0

90.70

10.25 Clmoles

Clg

ClmoleClg

NaOHgNaOHmole

NaOHg

Clmole

NaOHmoles

Clg

ClmoleClg 2.28

1

0.40

1

2

90.70

10.25

22

22

38


• Calculate the molar mass of the following:

Fe2O3 (Rust)

C6H8O7 (Citric acid)

C16H18N2O4 (Penicillin G)

• Balance the following, and determine how many moles

of CO will react with 0.500 moles of Fe2O3.

Fe2O3(s) + CO(g) Fe(s) + CO2(g)

39


Fe2O3 + CO → Fe + CO2

Balance (not a simple one)Save Fe for last

C is balanced, but can’t balance OIn the products the ratio C:O is 1:2 and can’t change

Make the ratio C:O in reactants 1:2

Fe2O3 + 3CO → 2Fe + 3CO2

40


Fe2O3 + 3CO → 2Fe + 3CO2

COmolesOFemole

COmoleOFemoles 50.1

1

3500.0

3232

41


42


• Aspirin is prepared by reaction of salicylic acid (C7H6O3)

with acetic anhydride (C4H6O3) to form aspirin (C9H8O4)

and acetic acid (CH3CO2H). Use this information to

determine the mass of acetic anhydride required to

react with 4.50 g of salicylic acid. How many grams of

aspirin will result? How many grams of acetic acid will

be produced as a by-product?

43


Salicylic acid + Acetic anhydride →

Aspirin + acetic acid

C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2H

C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2

Balanced!

Equal # moles for all

44


4.50 g Salicylic acid (C7H6O3) = ? moles

MW C7H6O3 = 7 x 12.01 + 6 x 1.008 + 3 x 16.00

= 138.12 g/mole

..0326.0..12.138

..1..50.4 ASmoles

ASg

ASmoleASg

45


Since all compounds have the same S.C., there must be

0.0326 moles of all 4 of them involved in the reaction.

g Aspirin (C9H8O4) = 0.0326 moles x MW Aspirin

= .0326 x [9x12.01 + 8x1.008 + 4x16.00]

=.0326 mole x 180.15 g/mole

5.87 g Aspirin

46


• Yields of Chemical Reactions: If the actual amount

of product formed in a reaction is less than the

theoretical amount, we can calculate a percentage

yield.

100% yieldproduct lTheoretica

yieldproduct Actual yield%

47


• Dichloromethane (CH2Cl2) is prepared by reaction

of methane (CH4) with chlorine (Cl2) giving

hydrogen chloride as a by-product. How many

grams of dichloromethane result from the reaction

of 1.85 kg of methane if the yield is 43.1%?

48


CH4 + Cl2 → CH2Cl2 + HCl

Balance

CH4 + 2Cl2 → CH2Cl2 + 2HCl

1.85 kg CH4 = ? moles CH4

49



1.85 kg CH4 = ? moles CH4

MW CH4 = 1x12.01 + 4x1.008 = 16.04 g/mole

44

44 115

4.16

1100085.1 CHmoles

CHg

CHmole

kg

gCHkg

50



115 moles CH4

in theory we should produce:

115 moles of CH2Cl2 and 230 moles of HCl

And use up 230 moles of Cl2

51



115 moles of CH2Cl2 = ? g

MW CH2Cl2 = 12.01 + 2x1.008 + 2x35.45 = 84.93

115 moles x (84.03 g/mole) = 9770 g

52



Expect 9770 g CH2Cl2

but the yield is 43.1%

So we produced just 0.431 x 9770 g

4.21 kg CH2Cl2

53



Suppose the reaction went to completion

(100% yield)

Is mass conserved?

54



Start with 115 moles CH4 and 230 moles Cl2

total mass = 115x16.04 + 230x70.90

= 1850 + 16300 = 18150

only 3 sig. figs. → 18.2 kg

55



End with 115 moles CH2Cl2 and 230 moles HCl

total mass = 115x84.93 + 230x36.46

= 9770 + 8390 = 18160

only 3 sig. figs → 18.2 kg

c h a p t e rc h a p t e r c h a p t e rc h a p t e r chemistry, 5 th edition mcmurry/fay chemistry,...

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