Introductory Chemistry , 2 nd Edition Nivaldo Tro

Roy KennedyMassachusetts Bay Community College

Wellesley Hills, MA

Introductory Chemistry, 2nd EditionNivaldo Tro

Chapter 6Chemical

Composition

2006, Prentice Hall

Tro's Introductory Chemistry, Chapter 6

Why is Knowledge of Composition Important?

• everything in nature is either chemically or physically combined with other substances

• to know the amount of a material in a sample, you need to know what fraction of the sample it is

• Some Applications: the amount of sodium in sodium chloride for diet the amount of iron in iron ore for steel production the amount of hydrogen in water for hydrogen

fuel the amount of chlorine in freon to estimate ozone

depletion


3

Counting Nails by the Pound

• I want to buy a certain number of nails for a project, but the hardware store sells nails by the pound!

• How do I know how many nails I am buying when I buy a pound of nails?

• AnalogyHow many atoms in a given

mass of an element?


4


A hardware store customer buys 2.60 pounds of nails. A dozen of the nails has a mass of 0.150 pounds. How many nails did the customer buy?

Solution map:


5


• The customer bought 2.60 lbs of nails and received 208 nails. He counted the nails by weighing them!

nails 208 doz. 1

nails 12

lbs. 0.150

nails doz. 1lbs. 2.60


6


• What if he bought a different size nail? Would the mass of a dozen be 0.150 lbs?Would there be 208 nails in 2.60 lbs? How would this effect the conversion factors?


7

Counting Atoms by Moles

• If we can find the mass of a particular number of atoms, we can use this information to convert the mass of a element sample to the number of atoms in the sample.

• The number of atoms we will use is 6.022 x 1023 and we call this a mole1 mole = 6.022 x 1023 things

Like 1 dozen = 12 things

http://wps.prenhall.com/wps/media/objects/1053/1078985/media/AACXJQT0.html


8

Chemical Packages - Moles• mole = number of particles equal to the

number of atoms in 12 g of C-12 1 atom of C-12 weighs exactly 12 amu1 mole of C-12 weighs exactly 12 g

• The number of particles in 1 mole is called Avogadro’s Number = 6.0221421 x 1023 1 mole of C atoms weighs 12.01 g and has

6.022 x 1023 atomsthe average mass of a C atom is 12.01 amu

Example 6.1Converting Between Moles and

Number of Atoms


10

Example:• A silver ring contains 1.1 x 1022 silver atoms. How many

moles of silver are in the ring?


11

Example:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?

• Write down the given quantity and its units.

Given: 1.1 x 1022 Ag atoms


12

• Write down the quantity to find and/or its units.

Find: ? moles

Information

Given: 1.1 x 1022 Ag atomsExample:A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring?


13

• Collect Needed Conversion Factors:

1 mole Ag atoms = 6.022 x 1023 Ag atoms

Information


Find: ? moles



14

• Write a Solution Map for converting the units :

Information


Find: ? moles

Conv. Fact.: 1 mole = 6.022 x 1023


atoms Agatoms Ag moles Agmoles Ag

atoms Ag10022.6

Ag mole 123


15

• Apply the Solution Map:

moles atoms Ag 106.022

Ag mole 1atoms Ag 101.1

2322

= 1.8266 x 1022 moles Ag

= 1.8 x 1022 moles Ag

• Sig. Figs. & Round:

InformationGiven: 1.1 x 1022 Ag atomsFind: ? molesConv. Fact.: 1 mole = 6.022 x 1023

Sol’n Map: atoms mole



16

• Check the Solution:

1.1 x 1022 Ag atoms = 1.8 x 10-2 moles Ag

The units of the answer, moles, are correct.The magnitude of the answer makes sense

since 1.1 x 1022 is less than 1 mole.

InformationGiven: 1.1 x 1022 Ag atomsFind: ? molesConv. Fact.: 1 mole = 6.022 x 1023

Sol’n Map: atoms mole



17

Relationship Between Moles and Mass

• The mass of one mole of atoms is called the molar mass

• The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu


18

Mole and Mass RelationshipsSubstance Pieces in 1 mole Weight of 1 mole hydrogen 6.022 x 1023 atoms 1.008 g

carbon 6.022 x 1023 atoms 12.01 g

oxygen 6.022 x 1023 atoms 16.00 g

sulfur 6.022 x 1023 atoms 32.06 g

calcium 6.022 x 1023 atoms 40.08 g

chlorine 6.022 x 1023 atoms 35.45 g

copper 6.022 x 1023 atoms 63.55 g

1 moleSulfur32.06 g

1 moleCarbon12.01 g

Example 6.2Converting Between

Grams and Moles of Atoms


20

Example:• Calculate the number of moles of sulfur in 57.8 g of

sulfur


21

Example:Calculate the number of moles of sulfur in 57.8 g of sulfur


Given: 57.8 g S


22


Find: ? moles S

Information

Given: 57.8 g SExample:Calculate the number of moles of sulfur in 57.8 g of sulfur


23


1 mole S atoms = 32.06 g

Information

Given: 57.8 g S

Find: ? moles S



24


Information

Given: 57.8 g S

Find: ? moles S

Conv. Fact.: 1 mole S = 32.06 g

g Sg S moles Smoles S

S g 2.063

S mole 1



25


S moles S g 2.063

S mole 1S g 57.8

= 1.80287 moles S

= 1.80 moles S


Information

Given: 57.8 g S

Find: ? moles S


Sol’n Map: g moles



26


57.8 g sulfur = 1.80 moles sulfur

The units of the answer, moles, are correct.The magnitude of the answer makes sense

since 57.8 g is more than 1 mole.

Information

Given: 57.8 g S

Find: ? moles S


Sol’n Map: g moles



Grams and Number of Atoms


28

Example:• How many aluminum atoms are in an aluminum can with

a mass of 16.2 g?


29

Example:How many aluminum atoms are in an aluminum can with a mass of 16.2 g?


Given: 16.2 g Al


30


Find: ? atoms Al

Information

Given: 16.2 g AlExample:How many aluminum atoms are in an aluminum can with a mass of 16.2 g?


31


1 mole Al atoms = 26.98 g Al

1 mole = 6.022 x 1023

Information

Given: 16.2 g Al

Find: ? atoms Al



32


g Al mol Al atoms Al

Al g 26.98

Al mol 1

Al mol 1

Al atoms 10.0226 23

Information

Given: 16.2 g Al

Find: ? atoms Al

C F: 1 mol Al = 26.98 g

1 mol = 6.022 x 1023



33

Al mole 1

Al atoms 10022.6

Al g 26.98

Al mole 1Al g 6.21

23


= 3.6159 x 1023 atoms Al

= 3.62 x 1023 atoms Al


Information

Given: 16.2 g Al

Find: ? atoms Al

CF: 1 mol Al = 26.98 g

1 mol = 6.022 x 1023

SM: g mol atoms



34


16.2 g Al = 3.62 x 1023 atoms Al

The units of the answer, atoms, are correct.The magnitude of the answer makes sense since 16.2 g is less than the mass of 1 mole, 26.98 g.

Information

Given: 16.2 g Al

Find: ? atoms Al

CF: 1 mol Al = 26.98 g

1 mol = 6.022 x 1023

SM: g mol atoms



35

Molar Mass of Compounds

• the relative weights of molecules can be calculated from atomic weights

Formula Mass = 1 molecule of H2O

= 2(1.01 amu H) + 16.00 amu O = 18.02 amu

• since 1 mole of H2O contains 2 moles of H and 1 mole of O

Molar Mass = 1 mole H2O

= 2(1.01 g H) + 16.00 g O = 18.02 g


Grams and Moles of Compound


37

Example:• Calculate the mass (in grams) of 1.75 mol of water


38

Example:Calculate the mass (in grams) of 1.75 mol of water


Given: 1.75 mol H2O


39


Find: ? g H2O

Information

Given: 1.75 mol H2OExample:Calculate the mass (in grams) of 1.75 mol of water


40


Molar Mass H2O = 2(atomic mass H) + 1(atomic mass O)

= 2(1.01) + 1(16.00)

= 18.02 g/mol

1 mol H2O = 18.02 g H2O

Information

Given: 1.75 mol H2O

Find: ? g H2O



41


Information

Given: 1.75 mol H2O

Find: ? g H2O

C F: 1 mole H2O = 18.02 g H2O

mol H2Omol H2O g H2Og H2O

OH mol 1

OH g 18.02

2

2



42


OH g OH mol 1

OH g 18.02OH mol 1.75 2

2

22

= 31.535 g H2O

= 31.5 g H2O


Information

Given: 1.75 mol H2O

Find: ? g H2O

C F: 1 mole H2O = 18.02 g H2OSol’n Map: mol g



43


1.75 mol H2O = 31.5 g H2O

The units of the answer, g, are correct.The magnitude of the answer makes sense

since 31.5 g is more than 1 mole.

Information

Given: 1.75 mol H2O

Find: ? g H2O

C F: 1 mole H2O = 18.02 g H2OSol’n Map: mol g



Grams and Number of Molecules


45

Example:

• Find the mass of 4.78 x 1024 NO2 molecules?


46

Example:Find the mass of 4.78 x 1024 NO2 molecules


Given: 4.78 x 1024 NO2 molecules


47


Find: ? g NO2

Information

Given: 4.78 x 1024 molec NO2



48


Molar Mass NO2 = 1(atomic mass N) + 2(atomic mass O)

= 1(14.01) + 2(16.00)

= 46.01 g/mol

1 mole NO2 molec = 46.01 g NO2

1 mole = 6.022 x 1023

Information


Find: ? g NO2



49


molec NO2 mol NO2 g NO2

2

2

NO mol 1

NO g 01.46

223

2

NO molec 10.0226

NO mol 1

Information

Given: 4.78 x1024 NO2 molec

Find: ? g NO2

C F: 1 mol NO2 = 46.01 g NO2

1 mol = 6.022 x 1023



50

2

2

223

22

23

NO mole 1

NO g 6.014

NO molec 10022.6

NO mole 1NO molec 10022.6


= 365.21 g NO2

= 365 g NO2 • Sig. Figs. & Round:

Information


Find: ? g NO2

CF: 1 mol NO2 = 46.01 g

1 mol = 6.022 x 1023

SM: molec mol g



51


4.78 x 1024 molecules NO2 = 365 g NO2

The units of the answer, g, are correct.The magnitude of the answer makes sense since

4.78 x 1024 molecules is more than 1 mole.

Information


Find: ? g NO2

CF: 1 mol NO2 = 46.01 g

1 mol = 6.022 x 1023

SM: molec mol g



52

Chemical Formulas as Conversion Factors

• 1 spider 8 legs• 1 chair 4 legs• 1 H2O molecule 2 H atoms 1 O atom


53

Mole Relationships inChemical Formulas

• since we count atoms and molecules in mole units, we can find the number of moles of a constituent element if we know the number of moles of the compound

Moles of Compound Moles of Constituents

1 mol NaCl 1 mole Na, 1 mole Cl

1 mol H2O 2 mol H, 1 mole O

1 mol CaCO3 1 mol Ca, 1 mol C, 3 mol O

1 mol C6H12O6 6 mol C, 12 mol H, 6 mol O


Grams of a Compound and Grams of a Constituent Element


55

Example:

• Carvone, (C10H14O), is the main component in spearmint oil. It has a pleasant odor and mint flavor. It is often added to chewing gum, liquers, soaps and perfumes. Find the mass of carbon in 55.4 g of carvone.


56

Example:Find the mass of carbon in 55.4 g of carvone, (C10H14O).


Given: 55.4 g C10H14O


57


Find: ? g C

Information

Given: 55.4 g C10H14OExample:Find the mass of carbon in 55.4 g of carvone, (C10H14O).


58


Molar Mass C10H14O = 10(atomic mass C) + 14(atomic mass H) + 1(atomic mass O)

= 10(12.01) + 14(1.01) + (16.00)

= 150.2 g/mol

1 mole C10H14O = 150.2 g C10H14O

1 mole C10H14O 10 mol C

1 mole C = 12.01 g C

Information


Find: g C



59


gC10H14O

C mol 1

C g 0112.

OHC g 50.21

OHC mol 1

1410

1410

Information


Find: g C

CF: 1 mol C10H14O = 150.2 g

1 mol C10H14O 10 mol C

1 mol C = 12.01 g


molC10H14O

molC

gC

OHC mol 1

C mol 10

1410


60

C mole 1

C g 2.011

OHC mol 1

C mol 10

OHC g 50.21

OHC mole 1OHC g 5.45

14101410

14101410


= 44.2979 g C

= 44.3 g C • Sig. Figs. & Round:

Information

Given: 55.4 g C10H14OFind: g C

CF: 1 mol C10H14O = 150.2 g

1 mol C10H14O 10 mol C 1 mol C = 12.01 g

SM: g C10H14O mol C10H14O mol C g C



61


55.4 g C10H14O = 44.3 g C

The units of the answer, g C, are correct.The magnitude of the answer makes sense since

the amount of C is less than the amount of C10H14O.

Information

Given: 55.4 g C10H14OFind: g C

CF: 1 mol C10H14O = 150.2 g

1 mol C10H14O 10 mol C 1 mol C = 12.01 g

SM: g C10H14O mol C10H14O mol C g C



62

Percent Composition• Percentage of each element in a compound

By mass

• Can be determined from 1. the formula of the compound2. the experimental mass analysis of the

compound• The percentages may not always total to 100%

due to rounding

100%wholepart

Percentage


63

Mass Percent as a Conversion Factor

• the mass percent tells you the mass of a constituent element in 100 g of the compoundthe fact that NaCl is 39% Na by mass means that

100 g of NaCl contains 39 g Na

• this can be used as a conversion factor100 g NaCl 39 g Na

Na g NaCl g 100

Na g 39 NaCl g NaCl g

Na g 39

NaCl g 100 Na g


64

Example - Percent Composition from the Formula C2H5OH

1. Determine the mass of each element in 1 mole of the compound

2 moles C = 2(12.01 g) = 24.02 g6 moles H = 6(1.008 g) = 6.048 g1 mol O = 1(16.00 g) = 16.00 g

2. Determine the molar mass of the compound by adding the masses of the elements

1 mole C2H5OH = 46.07 g


65

Sample - Percent Composition from the Formula C2H5OH

3. Divide the mass of each element by the molar mass of the compound and multiply by 100%

52.14%C100%46.07g24.02g

13.13%H100%46.07g6.048g

34.73%O100%46.07g16.00g


66

Example – Percent Composition of Carvone if a 30.0 g sample contains 24.0 g of C, 3.2 g

O and the rest H?

1. Determine the masses of all the elements in the sample

C = 24.0 g, O = 3.2 g

H = 30.0 g – (24.0 g + 3.2 g) = 2.8 g

67

Example – Percent Composition of Carvone if a 30.0 g sample contains 24.0 g of C, 3.2 g

O and the rest H?2. Divide the mass of each element by the

total mass of the sample then multiply by 100% to give its percentage

80.0%C100%30.0g24.0g

11%O100%30.0g3.2g

9.3%H100%30.0g2.8g


68

Empirical Formulas• The simplest, whole-number ratio of atoms in a

molecule is called the Empirical Formulacan be determined from percent composition or

combining masses

• The Molecular Formula is a multiple of the Empirical Formula

% A mass A (g) moles A100g MMA

% B mass B (g) moles B100g MMB

moles Amoles B


69

Empirical FormulasHydrogen PeroxideMolecular Formula = H2O2

Empirical Formula = HOBenzeneMolecular Formula = C6H6

Empirical Formula = CHGlucoseMolecular Formula = C6H12O6

Empirical Formula = CH2O


70

Finding an Empirical Formula1) convert the percentages to grams

a) skip if already grams

2) convert grams to molesa) use molar mass of each element

3) write a pseudoformula using moles as subscripts4) divide all by smallest number of moles5) multiply all mole ratios by number to make all whole

numbersa) if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply

all by 3, etc. b) skip if already whole numbers

http://wps.prenhall.com/wps/media/objects/165/169519/MolFormulaDetermin.html

Example 6.11Finding an Empirical Formula

from Experimental Data


72

Example:• A laboratory analysis of aspirin determined the following

mass percent composition. Find the empirical formula.

C = 60.00%

H = 4.48%

O = 35.53%


73

Example:Find the empirical formula of aspirin with the given mass percent composition.


Given: C = 60.00%

H = 4.48%

O = 35.53%

Therefore in 100 g of aspirin there are 60.00 g C,

4.48 g H and 35.53 g O


74


Find: empirical formula, CxHyOz

Information

Given: 60.00 g C, 4.48 g H, 35.53 g O



75


1 mole C = 12.01 g C

1 mole H = 1.01 g H

1 mole O = 16.00 g O

Information

Given: 60.00 g C, 4.48 g H, 35.53 g O




76

• Write a Solution Map:

gC, H, O

molC, H, O

molratio

empiricalformula

Information

Given: 60.00 g C, 4.48 g H, 35.53 g O


CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 g


77

• Apply the Solution Map:calculate the moles of each element

Information

Given: 60.00 g C, 4.48 g H, 35.53 g O


CF: 1 mol C = 12.01 g; 1 mol H = 1.01 g; 1 mol O = 16.00 gSM: g C,H,O mol C,H,O

mol ratio empirical formula


C mol 9964C g 12.01

C mol 1C g 0060 ..

H mol 444H g 1.01

H mol 1H g .484 .

O mol 2212O g 16.00

O mol 1O g 5.533 .


78

• Apply the Solution Map:write a pseudoformula

Information

Given: 4.996 mol C, 4.44 mol H,

2.221 mol O





C4.996H4.44O2.221


79

• Apply the Solution Map: find the mole ratio by dividing by the smallest number of moles

Information

Given: C4.996H4.44O2.221





122.25

2.221

2.221

2.221

4.44

2.221

4.996

OHC

OHC


80

• Apply the Solution Map:multiply subscripts by factor to give whole number

Information

Given: C2.25H2O1





122.25 OHC{ } x 4

C9H8O4


81

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


82

All these molecules have the same Empirical Formula. How are the

molecules different?

Name Molecular

Formula

Empirical

Formula

Molar

Mass, g

glyceraldehyde C3H6O3 CH2O 90

erythrose C4H8O3 CH2O 120

arabinose C5H10O5 CH2O 150

glucose C6H12O6 CH2O 180


83

Molecular Formulas

• The molecular formula is a multiple of the empirical formula

• To determine the molecular formula you need to know the empirical formula and the molar mass of the compound

Molar Massreal formula

Molar Massempirical formula

= factor used to multiply subscripts


84

Example – Determine the Molecular Formula of Cadinene if it has a molar mass of

204 g and an empirical formula of C5H8

1. Determine the empirical formula• May need to calculate it as previous

C5H8

2. Determine the molar mass of the empirical formula

5 C = 60.05 g, 8 H = 8.064 g

C5H8 = 68.11 g


85

3. Divide the given molar mass of the compound by the molar mass of the empirical formula

Round to the nearest whole number

3g 11.68

g 204




86

4. Multiply the empirical formula by the factor above to give the molecular formula

(C5H8)3 = C15H24



Introductory Chemistry , 2 nd Edition Nivaldo Tro

Documents

Transcript of Introductory Chemistry , 2 nd Edition Nivaldo Tro