chapter 6 chapter opening figure chemical...

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MA

Introductory Chemistry, 3rd Edition

Nivaldo Tro

Chapter 6

Chemical

Composition

2009, Prentice Hall

Chapter opening figure

Outline• 6.1 - How much Sodium?

• 6.2 - Counting Nails by the Pound.

• 6.3 – Counting Atoms by the Gram.

• 6.4 – Counting Molecules by the Gram.

• 6.5 – Chemical Formulas as Conversion Factors

• 6.6 – Mass Percent Composition of Compounds

• 6.7 – Mass Percent Composition and Chemical Formula

• 6.8 – Calculating Empirical Formulas for Compounds

• 6.9 – Calculating Molecular Formulas for Compounds

Tro's "Introductory Chemistry",

Chapter 6

2

6.1 - How much Sodium?


Chapter 6

3


Chapter 6

4

Why Is Knowledge of

Composition Important?

• It is important to know the fraction (or %) of

material in a substance.

• Some Applications:

sodium in foods for diet.

iron in iron ore for steel production.

hydrogen in water for hydrogen fuel.

chlorine in freon to estimate ozone depletion.

Figures of

Salt

Water

And

Freon

6.2 - Counting Nails by the

Pound.


Chapter 6

5


Chapter 6

6

Counting Nails by the Pound

• Nails are not counted individually.

• Consider nails 500 nails per lb.

• Need 1500 nails need 3.000 lb

• What if you but 0.750 lb. How

many nails?

• 500 nails/1 lb or 1 lb/500 nails can be

used as a conversion factor.

• 0.75 lb nails x 500 nails = 375 nails

1lb nails

nails


Chapter 6

7

Counting Nails by the Pound,

Continued• What if we bought a different size nail?

A smaller nail would have a different number of

nails per pound, for example 1500 nails per lb

0.750 lb of these smaller nails.

• 0.75 lb nails x 1500 nails = 1125 nails

1lb nails

• Figure 6.1 big nails and small nails

6.3 – Counting Atoms by the

Gram.


Chapter 6

8


Chapter 6

9

Counting Atoms by Moles

• In the same way that we count nails by the

pounds we count atoms by the mole

• 6.022 x 1023 atoms is equal to a mole.1 mole = 6.022 x 1023 things.Like 1 dozen = 12 things.

Avogadro’s number.


Chapter 6

10

Chemical Packages—Moles

• Mole = equal to the number of atoms in 12 g of C-12.

1 atom of C-12 weighs exactly 12 amu.

1 mole of C-12 weighs exactly 12 g.

• In 12 g of C-12 there are 6.022 x1023 C-12 atoms.

• Atomic mass in amu is numerically equal to molar mass in

grams

mole 1

atoms 10022.6 23atoms 10022.6

mole 123

Molar Mass and Avogadro’s

Number as Conversion Factors• Molar mass has units of g

mole

conversions grams moles

• Avagadro’s number units 6.022 x 1023 particles

1 mole

conversions particles moles

Particles may be atoms, molecules, ions, protons, etc.


Chapter 6

11

How Many mol are There in 10.0 g

of Al?

• g Al mol Al

• Atomic mass is the conversion factor

• 10.0 g Al x 1 mol Al = 0.371 mol Al

26.983 g Al


Chapter 6

12

How Many Al atoms are There in

10.0 g of Al?


Chapter 6

13

• g Al mol Al number Al atoms

•The moles was determine on the previous slide.

•0.371 mol Al x 6.022 x 1023 Al atoms = 2.23 x 1023 Al atoms

1 mol Al

Summary of conversions


Chapter 6

14

g of A mol of A # particles of A

÷ molar mass

x molar mass

x 6.022 x 1023

÷ 6.022 x 1023

Need 3.75 mol of S for a reaction.

How many g of S is this?

How many S atoms is this

• mol S g S and mol S number S atoms

• Atomic mass is the conversion factor

• 3.75 mol S x 32.07 g S = 120 g S

1 mol S3.75 mol S x 6.022 x 1023 S atoms = 2.26 x 1024 S atoms

1 mol S


Chapter 6

15

6.4 – Counting Molecules by the

Gram.


Chapter 6

16

Counting molecules by the gram

• We can count molecules by the gram

• The molar mass is the conversion factor


Chapter 6

17

Example counting molecules by

the gram

• Dihydrogren sulfide, H2S, is the smell of

rotten eggs

• 12.0 g of H2S is how many g and now many

molecules.

• Need the molar mass

• (2 x H) + (1 x S)

• (2 x 1.008 g) + (1 x 32.07 g) = 34.09 g/mol


Chapter 6

18

Example counting molecules by

the gram (cont.)• 12.0 g H2S x 1 mol H2S = 0.352 mol H2S

34.09 g H2S0.352 mol H2S x 6.022 x 1023 H2S molecules = 2.12 x 1023 H2S

1 mol H2S molecules


Chapter 6

19

6.5 – Chemical Formulas as

Conversion Factors


Chapter 6

20

Chemical Formulas can be Used

to Convert to Amount of an

Element


Chapter 6

21

•1 molecule H2O

•2 atoms of H

•1 atom of O

•1 mole H2O

•2 mole H

•1 mole O

•1 mole C3H6O3

•3 mole C

•6 mole H

•3 mole O

In 25.0 g of glucose, C6H12O6,

How Many g of Oxygen is there?

• g glucose mol glucose mol O g O

• Molar mass glucose = 180.16 g/mol• 25.0 g glucose x 1 mol glucose = 0.139 mol glucose

180.16 g glucose

• 0.139 mol glucose x 6 mol O = 0.834 mol O

1 mol glucose

• 3.75 mol O x 16.00 g O = 13.3 g O

1 mol O

•


Chapter 6

22

6.6 – Mass Percent Composition

of Compounds


Chapter 6

23

Definition of Mass Percent

• Mass % mass of element in compound

of = X 100

element total mass of the compound


Chapter 6

24

6.7 – Mass Percent Composition

and Chemical Formula


Chapter 6

25

Mass % from formula

• Mass % may be calculated conveniently

from the formula of a compound.

• Mass % mass of element in one mole

of = X 100

element molar mass of compound


Chapter 6

26

What is the mass % of each

element in Na2CrO4

• Molar mass = (2 x Na) + (1 x Cr) + (4 x O)• (2 x 22.99) + (52.00) + (4 x 16.00) g = 161.98 g

(45.98 g Na) (64.00 g O) mol Na2CrO4

• % Na = 45.98 g Na x 100 = 28.39 % Na

161.98 g total

• % Cr = 52.00 g Cr x 100 = 32.10 % Cr

161.98 g total

• % Na = 64.00 g O x 100 = 39.51 % O

161.98 g total

Note the sum of the % is equal to 100.00 %27

6.8 – Calculating Empirical

Formulas for Compounds


Chapter 6

28


Chapter 6

29

Empirical Formulas

• The simplest, whole-number ratio of atoms in a molecule is called the empirical formula.

Can be determined from percent composition or combining masses.

• The molecular formula is a whole number multiple of the empirical formula. (usually called n)

% A mass A (g) moles A100g MMA

% B mass B (g) moles B100g MMB

moles A

moles B

30

Empirical Formulas, Continued

Benzene

Molecular formula = C6H6

Empirical formula = CH

Glucose

Molecular formula = C6H12O6

Empirical formula = CH2O

Hydrogen Peroxide

Molecular formula = H2O2

Empirical formula = HOMolecules shown

On page 179


Chapter 6

31

Practice—Determine the Empirical

Formula of Benzopyrene, C20H12.


Chapter 6

32

Practice—Determine the Empirical Formula of

Benzopyrene, C20H12, Continued

• Find the greatest common factor (GCF) of the

subscripts:

20 and 12 are both divisible by 4.

• Divide each subscript by 4 to get the empirical

formula.

C20H12 = (C5H3) x4

Empirical formula = C5H3

33

Finding an Empirical Formula

from Experimental Data1. Convert the percentages to grams. (assume 100 g to

make the math easy.

a. Skip if already grams.

2. Convert grams to moles.

a. Use molar mass of each element.

3. Write a pseudoformula using moles as subscripts.

4. Divide all by smallest number of moles At least the smallest one will be a whole number.

5. Multiply all mole ratios by number to make all whole numbers, if necessary.

a. If in the ratio fractional part is ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all by 3, etc.

b. Skip if already whole numbers after Step 4.

Examples from Experimental

Data•An oxide of Nickel, Ni, is analyzed and found to

contain 0.2636 g of Ni and 0.0718 g of O.

Determine the empirical fomula.g of mol of pseudoformuladivide by clear

elements elements smallest fractions

0.2636 g Ni x 1 mol Ni = 0.004491 mol Ni

58.69 Ni

0.0718 g O x 1 mol O = 0.00449 mol O

16.00 OTro's "Introductory Chemistry",

Chapter 6

34

Examples from Experimental

Data (cont)• A pseduoformula has the correct ratio of moles of

compound, but may not ne whole numbers. It is

not correct because a formula can not contain a

fraction of an atom

• Ni0.04491O0.0449

• Divide by smallest to at least convert one element

to a whole number 1

• Ni0.04491O0.0449 Ni1O1 NiO empirical

• formula

• 35

0.04490.0449

More Examples

• Cisplatin

65.02 % Pt, 9.34 % N, 2.02 % H, 23.63 % Cl

• Naproxen (active ingredient in Aleve)

73.03 % C, 6.13 % H, 20.85 % O,


Chapter 6

36

Cisplatin

• Cisplatin

65.02 % Pt, 9.34 % N, 2.02 % H, 23.63 % Cl

• We assume 100 g of compound to make the

conversion to g easy. In 100 g of Cisplatin there

are

• 65.02 % Pt

• 9.34 % N

• 2.02 % H

• 23.63 % Cl


Chapter 6

37

Cisplain (cont)

• Convert each of these mass to moles

• 65.02 g Pt x 1 mol Pt = 0.3333 mol Pt

195.1 g Pt

• 9.34 g N x 1 mol N = 0.667 mol Pt

14.01 g Pt

• 2.02 g H x 1 mol H = 2.00 mol Pt

1.008 g H

• 23.63 g Cl x 1 mol Cl = 0.6667 mol Pt

35.45 g ClTro's "Introductory Chemistry",

Chapter 6

38

Cisplain (cont)

• Pseudoformula

• Pt0.333N0.667H2.00Cl0.6667

• Divide by smallest 0.3333

• Pt0.3333N0.667 H2.00 Cl0.6667


Chapter 6

39

0.3333 0.3333 0.3333 0.3333

PtN2H6Cl2 empirical formula

Naproxen

• Naproxen (active ingredient in Aleve)

73.03 % C, 6.13 % H, 20.85 % O,

• Assume 100 g convert to moles

73.03 g C 6.08 mol C

6.13 g H 6.08 mol H

20.85 g O 1.302 mol O

• Pseudoformula C6.08H6.08O1.303


Chapter 6

40

Naproxen (cont)

• Dividc by the smallest

• C6.08H6.08O1.303 C4.67H4.67O1

• Not all are whole numbers but 4.67 = 4 and 2/3

• We can remove fractions by multiplying all

subscript by 3. --? C14H14O3 empirical formula


Chapter 6

41

1.303 1.3031.303

Fractions in empirical formula

calculations• When dividing by the smallest number of moles fractions often

occur

• Learn to recognize common fractions

• 0.5 = ½

• 0.33 = 1/3 0.67 = 2/3

• 0.25 = 1/4 0.75 = ¾

• 0.20 = 1/5 0.40 = 2/5 0.60 = 3/5 0.80 = 4/5

• In each case multiply by the denominator to

clear the fraction 42

6.9 – Calculating Molecular

Formulas for Compounds


Chapter 6

43


Chapter 6

44

All These Molecules Have the Same

Empirical Formula. How Are the

Molecules Different?

Name Molecular

Formula

Empirical

Formula

Glyceraldehyde C3H6O3 CH2O

Erythrose C4H8O4 CH2O

Arabinose C5H10O5 CH2O

Glucose C6H12O6 CH2O


Chapter 6

45

All These Molecules Have the Same Empirical

Formula. How Are the Molecules Different?,

ContinuedName Molecular

Formula

Empirical

Formula

Molar

Mass, g

Glyceraldehyde C3H6O3 CH2O

Empirical

Mass = 30

90 =30 x 3

Erythrose C4H8O4 CH2O 120 = 30 x 4

Arabinose C5H10O5 CH2O 150 =30 x 5

Glucose C6H12O6 CH2O 180 = 30 x 6

Notice the molar mass is a multiple of the empirical mass.

Molecular Formula

• Compound has empirical formula C3H2N

and molar mass 312.39. What is the

molecular formula?

• Empirical mass (3 x C) + (2 x H) + (1 x N)

• (3 x 12.01) + (2 x 1.008) + (14.01) = 52.06 g

• Molar mass = n 312.39 g = 6.00

Empirical mass 52.06 g

Molecular formula = C3H2N x 6 = C18H12N6

46

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