Chapter 02 Chemical Formulas & Composition Stoichiometry

1

CHAPTER 2

CHEMICAL FORMULAS & COMPOSITION STOICHIOMETRY

2

Chapter Goals1. Atoms and Molecules2. Chemical Formulas3. Ions and Ionic Compounds4. Names and Formulas of Some

Ionic Compounds5. Atomic Weights6. The Mole

3

Chapter Goals7. Formula Weights, Molecular Weights,

and Moles8. Percent Composition and Formulas of

Compounds9. Derivation of Formulas from Elemental

Composition10.Determination of Molecular Formulas11.Some Other Interpretations of

Chemical Formulas12.Purity of Samples

4

Atoms and MoleculesDalton’s Atomic Theory - 1808Five postulates

1. An element is composed of extremely small, indivisible particles called atoms.

2. All atoms of a given element have identical properties that differ from those of other elements.

3. Atoms cannot be created, destroyed, or transformed into atoms of another element.

4. Compounds are formed when atoms of different elements combine with one another in small whole-number ratios.

5. The relative numbers and kinds of atoms are constant in a given compound.

Which of these postulates are correct today?

5

Atoms and MoleculesA molecule is the smallest particle of an element that can have a stable independent existence.

Usually have 2 or more atoms bonded together

Examples of molecules H2

O2

S8

H2O CH4

C2H5OH

6

Chemical Formulas

Chemical formula shows the chemical composition of the substance. ratio of the elements present in the molecule

or compound

He, Au, Na – monatomic elementsO2, H2, Cl2 – diatomic elements

O3, S8, P4 - more complex elements

H2O, C12H22O11 – compounds Substance consists of two or more elements

7

Chemical Formulas

Compound 1 Molecule ContainsHCl 1 H atom & 1 Cl atomH2O 2 H atoms & 1 O atom

NH3 1 N atom & 3 H atoms

C3H8 3 C atoms & 8 H atoms

8

Ions and Ionic CompoundsIons are atoms or groups of atoms that possess an electric charge.Two basic types of ions:Positive ions or cations one or more electrons less than neutral Na+, Ca2+, Al3+

NH4+ - polyatomic cation

Negative ions or anions one or more electrons more than neutral F-, O2-, N3-

SO42-, PO4

3- - polyatomic anions

9

Ions and Ionic CompoundsSodium chloride table salt is an ionic compound

10

Names and Formulas of Some Ionic Compounds

Table 2-3 displays the formulas, charges, and names of some common ions You must know the names, formulas, and

charges of the common ions in table 2-3.

Some examples are: Anions - Cl1-, OH1-, SO4

2-, PO43-

Cations - Na1+, NH41+, Ca2+, Al3+

11


Formulas of ionic compounds are determined by the charges of the ions. Charge on the cations must equal the charge on

the anions. The compound must be neutral.

NaCl sodium chloride (Na1+ & Cl1-)KOH potassium hydroxide(K1+ & OH1-)CaSO4 calcium sulfate (Ca2+ & SO4

2-)

Al(OH)3 aluminum hydroxide (Al3+ & 3 OH1-)

12


Table 2-2 gives names of several molecular compounds. You must know all of the molecular

compounds from Table 2-2.

Some examples are: H2SO4 - sulfuric acid FeBr2 - iron(II) bromide C2H5OH - ethanol

13


You do it! (# 1)What is the formula of nitric acid?HNO3

What is the formula of sulfur trioxide?SO3

What is the name of FeBr3?

iron(III) bromide

14


You do it! (# 2)What is the name of K2SO3?

potassium sulfite What is charge on sulfite ion?SO3

2- is sulfite ion

What is the formula of ammonium sulfide?(NH4)2S

15


You do it!What is charge on ammonium ion?NH4

1+

What is the formula of aluminum sulfate?Al2(SO4)3

What is charge on both ions?Al3+ and SO4

2-

16

Atomic Weights

Weighted average of the masses of the constituent isotopes if an element. Tells us the atomic masses

of every known element. Lower number on periodic

chart.

How do we know what the values of these numbers are?

17

The Mole

A number of atoms, ions, or molecules that is large enough to see and handle.A mole = number of things Just like a dozen = 12 things One mole = 6.022 x 1023 things

Avogadro’s number = 6.022 x 1023 Symbol for Avogadro’s number is NA.

18

The Mole

How do we know when we have a mole? count it out weigh it out

Molar mass - mass in grams numerically equal to the atomic weight of the element in grams.H has an atomic weight of 1.00794 g 1.00794 g of H atoms = 6.022 x 1023 H atoms

Mg has an atomic weight of 24.3050 g 24.3050 g of Mg atoms = 6.022 x 1023 Mg atoms

19

The Mole

Example 2-1: Calculate the mass of a single Mg atom in grams to 3 significant figures.

Mgg ?

20

The Mole


atom Mg1 Mgg ?

21

The Mole


atoms Mg 106.022

atoms Mg mol 1atom Mg 1Mg g ?

23

22

The Mole

Example 2-1: Calculate the mass of a single Mg atom, in grams, to 3 significant figures.

Mg g 104.04atoms Mg mol 1

Mgg24.30

atoms Mg 106.022

atoms Mg mol 1atom Mg 1Mg g ?

23

23

23

The Mole

Example 2-2: Calculate the number of atoms in one-millionth of a gram of Mg to 3 significant figures.

atoms Mg?

24

The Mole


Mgg 24.30

Mg mol 1 Mgg 101.00atoms Mg? 6

25

The Mole


atoms Mgmol 1

atoms Mg106.022

Mgg 24.30

Mg mol 1 Mgg 101.00atoms Mg?

23

6

26

The Mole


atoms Mg102.48atoms Mgmol 1

atoms Mg106.022

Mgg 24.30

Mg mol 1 Mgg 101.00atoms Mg?

1623

6

27

The Mole

Example 2-3. How many atoms are contained in 1.67 moles of Mg?

atoms Mg ?

28

The Mole


Mg mol 1.67atoms Mg ?

29

The Mole


Mg mol 1

atoms Mg 106.022 Mg mol 1.67atoms Mg ?

23

30

The Mole


atoms Mg 101.00

Mg mol 1

atoms Mg 106.022 Mg mol 1.67atoms Mg ?

24

23

31

The Mole


atoms Mg101.00

Mgmol 1

atoms Mg106.022Mg mol 1.67atoms Mg?

24

23

32

The Mole

Example 2-4: How many moles of Mg atoms are present in 73.4 g of Mg?

You do it!You do it!

33

The Mole


Mg g 4.73Mg mol ?

34

The Mole


Mg g 24.30

atoms Mg mol 1 Mg g 4.73Mg mol ?

35

The Mole


Mg mol 02.3

Mg g 24.30

atoms Mg mol 1 Mg g 4.73Mg mol ?

IT IS IMPERATIVE THAT YOU KNOWHOW TO DO THESE PROBLEMS

36

Formula Weights, Molecular Weights, and Moles

How do we calculate the molar mass of a compound? add atomic weights of each atom

The molar mass of propane, C3H8, is:

amu 44.11 mass Molar

amu 8.08 amu 1.01 8H 8

amu 36.03amu 12.01 3C 3

37


The molar mass of calcium nitrate, Ca(NO3)2 , is:


38


amu 164.10 massMolar

amu 96.00 amu 16.006 O6

amu 28.02 amu 14.012 N2

amu 40.08 amu 40.081 Ca1

39


One Mole of Contains Cl2 or70.90g 6.022 x 1023 Cl2 molecules

2(6.022 x 1023 ) Cl atoms

C3H8 You do it!You do it!

40


One Mole of Contains Cl2 or 70.90g 6.022 x 1023 Cl2 molecules 2(6.022 x 1023 ) Cl atoms C3H8 or 44.11 g 6.022 x 1023 C3H8 molecules 3 (6.022 x 1023 ) C atoms 8 (6.022 x 1023 ) H atoms

41


Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane. 8383 HC g 74.6molecules HC ?

42


Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.

83

83

8383

HC g 44.11

HC mole 1

HC g 74.6molecules HC ?

43


Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.3 8 3 8

233 8 3 8

3 8 3 8

? C H molecules 74.6 g C H

1 mole C H 6.022 10 C H molecules

44.11 g C H 1 mol C H

44


Example 2-5: Calculate the number of C3H8 molecules in 74.6 g of propane.

molecules 10 02.1

HC g 44.11

molecules HC 106.022

HC g 44.11

HC mole 1

HC g 74.6molecules HC ?

24

83

8323

83

83

8383

45


Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.


46


Example 2-8. Calculate the number of O atoms in 26.5 g of Li2CO3.

atoms O 106.49

COLi unit formula 1

atoms O 3

COLi mol 1

COLi unitsform.106.022

COLi g 73.8

COLi mol 1COLi g 26.5atoms O ?

23

3232

3223

32

3232

47


Occasionally, we will use millimoles. Symbol - mmol 1000 mmol = 1 mol

For example: oxalic acid (COOH)2 1 mol = 90.04 g 1 mmol = 0.09004 g or 90.04 mg

48


Example 2-9: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2.


49


Example 2-9: Calculate the number of mmol in 0.234 g of oxalic acid, (COOH)2.

22

2

22

(COOH) mmol 2.60(COOH) g 0.09004

(COOH) mmol 1

(COOH) g 234.0(COOH) mmol ?

50

Percent Composition and Formulas of Compounds

% composition = mass of an individual element in a compound divided by the total mass of the compound x 100%Determine the percent composition of C in C3H8.

81.68%

100%g 44.11

g 12.013

100%HC mass

C mass C %

83

51


What is the percent composition of H in C3H8?


52


What is the percent composition of H in C3H8?

81.68%100%18.32%

or

%18.32100%g 44.11

g 1.018

100%HC

H8

100%HC mass

H massH %

83

83

53


Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 significant figures.

You do it!

54


Example 2-10: Calculate the percent composition of Fe2(SO4)3 to 3 sig. fig.

100% Total

O 48.0% 100%g 399.9

g 16.012 100%

)(SOFe

O12 O %

S 24.1% 100%g 399.9

g 32.13 100%

)(SOFe

S3 S %

Fe 27.9% 100%g 399.9

g 55.82 100%

)(SOFe

Fe2Fe %

342

342

342

55

Derivation of Formulas from Elemental Composition

Empirical Formula - smallest whole-number ratio of atoms present in a compound

CH2 is the empirical formula for alkenes No alkene exists that has 1 C and 2 H’s

Molecular Formula - actual numbers of atoms of each element present in a molecule of the compound

Ethene – C2H4

Pentene – C5H10

We determine the empirical and molecular formulas of a compound from the percent composition of the compound.

percent composition is determined experimentally

56


Example 2-11: A compound contains 24.74% K, 34.76% Mn, and 40.50% O by mass. What is its empirical formula?Make the simplifying assumption that we have 100.0 g of compound. In 100.0 g of compound there are: 24.74 g of K 34.76 g of Mn 40.50 g of O

57


1 mol K? mol K 24.74 g K 0.6327 mol K

39.10 g K

58


Mn mol 0.6327Mn g 54.94

Mn mol 1Mn g 34.76 Mn mol ?

K mol 0.6327K g 39.10

K mol 1K g 24.74 K mol ?

59


1 mol K? mol K 24.74 g K 0.6327 mol K

39.10 g K

1 mol Mn? mol Mn 34.76 g Mn 0.6327 mol Mn

54.94 g Mn

1mol O? mol O 40.50 g O 2.531 mol O

16.00 g O

obtain smallest whole number ratio

60


Mn 10.6327

0.6327Mnfor K 1

0.6327

0.6327Kfor

rationumber wholesmallest obtain

O mol 2.531O g 16.00

O mol1O g 40.50 O mol ?

Mn mol 0.6327Mn g 54.94

Mn mol 1Mn g 34.76 Mn mol ?

K mol 0.6327K g 39.10

K mol 1K g 24.74 K mol ?

61


1 mol K? mol K 24.74 g K 0.6327 mol K

39.10 g K

1 mol Mn? mol Mn 34.76 g Mn 0.6327 mol Mn

54.94 g Mn

1mol O? mol O 40.50 g O 2.531 mol O

16.00 g O

obtain smallest whole number ratio

0.6327for K

4

0.63271 K for Mn 1 Mn

0.6327 0.63272.531

for O 4 O 0.6327

thus the chemical formula is KMnO

62


Example 2-12: A sample of a compound contains 6.541g of Co and 2.368g of O. What is empirical formula for this compound?


63



ratio number wholesmallest find

O mol 0.1480O g 16.00

O mol1O g 2.368O mol ?

Co mol 0.1110Cog 58.93

Co mol 1Co g 6.541Co mol ?

64



3 4

0.1110 0.1480for Co 1 Co for O 1.333 O

0.1110 0.1110multipy both by 3 to turn fraction to whole number

1 Co 3 3 Co 1.333 O 3 4 O

Thus the compound's formula is:

Co O

65

Determination of Molecular Formulas

Example 2-13: A compound is found to contain 85.63% C and 14.37% H by mass. In another experiment its molar mass is found to be 56.1 g/mol. What is its molecular formula? short cut method

H of g 8.100.1437g 56.1

C of g 48.00.8563g 56.1

H is 14.37% and C is 85.63%

g 56.1 contains mol 1

66

Determination of Molecular Formulas

4 8

convert masses to moles

1 mol C48.0 g of C 4 mol C

12.0 g C

1 mol H8.10 g of H 8 mol H

1.01 g H

Thus the formula is:

C H

67

Some Other Interpretations of Chemical Formulas

Example 2-16: What mass of ammonium phosphate, (NH4)3PO4, would contain 15.0 g of N?

4 3 4molar mass of (NH ) PO 149.0 g/mol

1 mol N? mol N 15.0 g of N 1.07 mol N

14.0 g N

68



434434

434

PO)(NH mol 0.357N mol 3

PO)(NH mol 1N mol 1.07

N mol 1.07N g 14.0

N mol 1N of g 15.0N mol ?

g/mol 149.0PO)(NH of massmolar

69



434434

434434

434434

434

PO)(NH g 53.2PO)(NH mol 1

PO)(NH g 149.0PO)(NH mol 0.357

PO)(NH mol 0.357N mol 3

PO)(NH mol 1N mol 1.07

N mol 1.07N g 14.0

N mol 1N of g 15.0N mol ?

g/mol 149.0PO)(NH of massmolar

70

Purity of Samples

The percent purity of a sample of a substance is always represented as

impurities includes sample of mass

%100sample of mass

substance pure of mass =purity %

71

Purity of SamplesExample 2-18: A bottle of sodium phosphate, Na3PO4, is 98.3% pure Na3PO4. What are the masses of Na3PO4 and impurities in 250.0 g of this sample of Na3PO4?

sample g 100.0

PONa g 98.3factor unit 43

72


43

4343

43

PONa g 246=

sample g 100.0

PONa g 98.3sample g 0.250PONa g?

sample g 100.0

PONa g 98.3factor unit

73


3 4

3 43 4

3 4

3 4

98.3 g Na POunit factor

100.0 g sample

98.3 g Na PO?g Na PO 250.0 g sample

100.0 g sample

=246 g Na PO

?g impurities = 250.0 g sample - 246 g Na PO

= 4.00 g

impurities

74

Synthesis Problem

In 1986, Bednorz and Muller succeeded in making the first of a series of chemical compounds that were superconducting at relatively high temperatures. This first compound was La2CuO4 which superconducts at 35K. In their initial experiments, Bednorz and Muller made only a few mg of this material. How many La atoms are present in 3.56 mg of La2CuO4?

75

426

42

42

42

42

CuOLa mol 1078.8CuOLa g 405.3

CuOLa mol 1

mg 1000

g 1CuOLa mg 3.56

g/mol 405.3=CuOLa of massmolar

Synthesis Problem

76

atoms La 1006.1CuOLa molecule

atoms La 2

CuOLa mol 1

CuOLa molecules 10022.6)CuOLa mol 1078.8(

CuOLa mol 1078.8CuOLa g 405.3

CuOLa mol 1

mg 1000

g 1CuOLa mg 3.56

g/mol 405.3=CuOLa of massmolar

19

42

42

4223

426

426

42

42

42

42

Synthesis Problem

77

Group Activity

Within a year after Bednorz and Muller’s initial discovery of high temperature superconductors, Wu and Chu had discovered a new compound, YBa2Cu3O7, that began to superconduct at 100 K. If we wished to make 1.00 pound of YBa2Cu3O7, how many grams of yttrium must we buy?

78

End of Chapter 2

The mole concept and basic stoichiometry ideas introduced in this chapter are essential components for the remainder of this course.

Chapter 02 Chemical Formulas & Composition Stoichiometry

Documents

Transcript of Chapter 02 Chemical Formulas & Composition Stoichiometry