The Mole &Chemical Composition

Section 1Avogadro’s Number &

Molar Conversions

For Review

MoleSI unit for amount# of atoms in 12g of carbon-12

Avogadro’s Number# of particles in a mole6.022 x 1023 - # of (?) in 1.000 moleUsed to count any kind of particle

The Mole is a Counting Unit

Ex. – 1 dozen = 12

The mole is used to count out a given number of particles, whether they are atoms, molecules, formula units, ions, or electrons.

Amount in Moles converted to # of Particles

6.022 x 1023 particles = 1 mol

1mol 1

10022.6 23

1

10022.6

mol 123

Choose a conversion factor that cancels the given units.

Example #1

Find the number of molecules in 2.5 mol of sulfur dioxide.

224

23

2 SO molecules 105.1mol 1

10022.6SO mol 5.2

Homework

Practice A p. 228#’s 2 - 4

Example #2

A sample contains 3.01 x 1023 molecules of sulfur dioxide, SO2. Determine the amount in moles.

2

223

22

23

SO mol 500.0

SO molecules 106.022

SO mol 1SO molecules 1001.3

Homework

Practice Bp. 229#’s 2 – 4#5 c – g

Molar Mass Relates Moles to Grams

The mass in grams of 1 mole of substance

= atomic mass of monatomic elements & formula mass of compounds & diatomic elements

ExamplesCarbon = 12g/molO2 = 16+16 = 32g/mol

CH4 = 12+1+1+1+1 = 16g/mol

Example #3

Find the mass in grams of 2.44 x 1024 atoms of carbon, whose molar mass is 12.01 g/mol.

C g7.48mol 1

C g01.12

106.022

mol 1atoms 1044.2

2324

Homework

Practice Cp. 231 #’s 2-4

Example #4

Find the number of molecules present in 47.5 g of glycerol, C3H8O3. The molar mass of glycerol is 92.11g/mol.

molecules 1011.3

mol 1

molecules 10022.6

OHC g 11.92

mol 1OHC g 5.47

23

23

383383

Homework

Practice Dp. 232#’s 2, 3

Quiz.7.1 Answer List

57.41 g695 g1.4 x 1024 atm1.195 mol0.0206 mol

The Mole &Chemical Composition

Section 2Relative Atomic Mass &

Chemical Formulas

For Review

Isotope – atoms of the same element with different #’s of neutrons (different mass #’s)

Average Atomic Mass – weighted average of atomic masses of elements isotopes

Calculating Average Atomic Mass

Need to know % abundance to calculate avg. atomic mass

Native copper is a mixture of two isotopes. Copper-63 contributes 69.17% of the atoms, and copper-65 the remaining 30.83%.

Example #1

The mass of Cu-63 atom is 62.94 amu, and that of a Cu-65 atom is 64.93 amu. Using the data for the previous figure, find the average atomic mass of Cu.

Isotope % Decimal

Contribution

Copper-63

69.17%

0.6917 62.94 x 0.6917

Copper-65

30.83%

0.3083 64.93 x 0.3083 amu 55.63)3083.0amu 93.64()6917.0amu 94.62(

Practice #1

1. Calculate the average atomic mass for gallium if 60.00% of its atoms have a mass of 68.926 amu and 40.00 % have a mass of 70.925 amu.

2. Calculate the average atomic mass of oxygen. Its composition is 99.76% of atoms with a mass of 15.99 amu, 0.038% with a mass of 17.00 amu, and 0.20% with a mass 18.00 amu.

To Understand

Chemical formula1. Which elements2. How much of each

Ionic compounds Show simplest ratio of cations & anionsKBr – 1:1 – 1 K+ & 1 Br-

Molar Mass of Compound

= sum of masses of all atoms in g/mol

Ex. H2O - H → 2 x 1.00 = 2.00

- O → 1 x 16.00 = 16.00 18.00

g/mol

Examples

ZnCl2

(NH4)2SO4

g/mol 29.136

90.7045.352Cl

39.6539.651Zn

g/mol 0.132

64164O

32321S

818H

28142N

Practice – Homework

Practice F p. 239 - 240

#’s 1 - 4

The Mole &Chemical Composition

Section 3Formulas & Percent

Composition

Definitions

Percent composition – percentage by mass of each element in a compound

Empirical formula – shows simplest ratio

a chemical formula that shows the composition of a compound in terms of relative #’s & kinds of atoms

Example – Find the empirical formula.C – 60.0%H – 13.4% O – 26.6%

1. Convert mass to moles.

Assume you have a 100g sample.

O mol 66.1O g 16

O mol 1 26.6g

H mol 4.13H g 1

H mol 113.4g

C mol 00.5C 12g

C mol 10.60

g

Example – Find the empirical formula.2. formulas are written using whole

#’s(to convert divide by smallest)

3. Write formula

C3H8O

166.1

66.107.8

66.1

4.1301.3

66.1

00.5

Practice – In Class

1. A compound is 63.52% Fe and 36.48% S.

2. 26.58% K, 35.35% Cr, 38.07% O

3. 32.37% Na, 22.58% S, 45.05% O

Practice – Homework

Practice G p. 243

#’s 1 - 4

Definitions

Molecular formulas – whole # multiple of empirical formula

Ex. Empirical Molecular

– CH2O x 1 = CH2O formaldehyde

x 2 = C2H4O2 acetic acid

x 6 = C6H12O6

glucose

Example- Find the molecular formula.

The empirical formula for a compound is P2O5. Its experimental molar mass is 284 g/mol. Determine the molecular formula of the compound.

g/mol 94.141

80165O

94.6197.302 P

00.294.141

284

10452 OP)OP(2

Practice – In ClassWhat is the molecular formula?1. Experimental MM = 232.41

g/molEmpirical formula = OCNCl

2. Experimental MM = 32.06 g/molEmpirical formula = NH2

Practice - Homework

Practice Hp. 245

#’s 1 - 3

Percent Composition

Calculate the percent composition of copper (I) sulfide, a copper ore called chalcocite.

g/mol 2.159SCu

32.132.071 S

127.163.552Cu

2

Cu %8.791002.159

1.127 S %2.20100

2.159

1.32

Check that everything adds up to 100%

Practice – In ClassDetermine the %

Composition1. NaClO

2. H2SO3

3. C2H5COOH

Answers to Practice

Na 1 x 23 = 23Cl 1 x 35.5 = 35.5

O 1 x 16 = 16

NaClO 74.5 g/mol

Na Cl O(23/74.5) x 100

= 30.9%(35.5/74.5) x 100 = 48.0%

(16/74.5) x 100 = 21.5%

Answers to Practice

H 2 x 1 = 2S 1 x 32.1 = 32.1

O 3 x 16 = 48

H2SO3 82.1 g/mol

H S O(2/82.1) x 100

= 2.5%(32.1/82.1) x 100 = 39.1%

(48/82.1) x 100 = 58.5%

Answers to Practice

C 3 x 12 = 36H 6 x 1 = 6

O 2 x 16 = 32

C2H5COOH 74.0 g/mol

C H O(36/74.0) x 100

= 48.6%(6/74.0) x 100

= 8.1%(32/74.0) x

100 = 43.2%

Qz.7.3Calculate the percent composition1. SrBr2

2. CaSO4

3. Mg(CN)2

4. Pb(CH3COO)2

Practice - Homework

Practice Ip. 248 #’s 1- 4