chemistry chapter 7

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Chemistry Chapter 7 Chemistry Chapter 7 Chemical Formulas and Chemical Compounds Heart cell rhythm depends on the opening and closing of a complex series of valves on the cell membrane, called ion channels. Some valves let certain ions ike potassium (K+) flow out, others let different ions like sodium (Na+) flow in. There are also pumps that actively move ions one direction or another.

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Chemistry Chapter 7. Chemical Formulas and Chemical Compounds. - PowerPoint PPT Presentation

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Page 1: Chemistry Chapter 7

Chemistry Chapter 7Chemistry Chapter 7Chemical Formulas

and Chemical

CompoundsHeart cell rhythm depends on the opening and closing of a complex series of valves on the cell membrane, called ion channels. Some valves let certain ions ike potassium (K+) flow out, others let different ions like sodium (Na+) flow in. There are also pumps that actively move ions one direction or another.

Page 2: Chemistry Chapter 7

IonsIons• CationCation: A positive ion: A positive ion

• MgMg2+2+, NH, NH44++

• AnionAnion: A negative ion: A negative ion• ClCl, SO, SO44

22

• Ionic BondingIonic Bonding: Force of attraction : Force of attraction between oppositely charged ions.between oppositely charged ions.

Page 3: Chemistry Chapter 7

Predicting Ionic ChargesPredicting Ionic ChargesGroup 1Group 1::Lose 1 electron to form Lose 1 electron to form 1+1+ ions ions

HH++ LiLi++ NaNa++ KK++

Page 4: Chemistry Chapter 7

Predicting Ionic ChargesPredicting Ionic ChargesGroup 2Group 2::Loses 2 electrons to form Loses 2 electrons to form 2+2+ ions ions

BeBe2+2+ MgMg2+2+ CaCa2+2+ SrSr2+2+ BaBa2+2+

Page 5: Chemistry Chapter 7

Predicting Ionic ChargesPredicting Ionic ChargesGroup 13Group 13:: Loses 3 Loses 3 electrons to form electrons to form 3+3+ ions ions

BB3+3+ AlAl3+3+ GaGa3+3+

Page 6: Chemistry Chapter 7

Predicting Ionic ChargesPredicting Ionic ChargesGroup 14Group 14:: Lose 4 Lose 4 electrons or gain electrons or gain 4 electrons?4 electrons?

Neither! Neither! Group 13 Group 13 elements rarely elements rarely form ions.form ions.

Page 7: Chemistry Chapter 7

Predicting Ionic ChargesPredicting Ionic ChargesGroup 15Group 15:: Gains 3 Gains 3 electrons to form electrons to form 3-3- ions ions

NN3-3-

PP3-3-

AsAs3-3-

NitridePhosphideArsenide

Page 8: Chemistry Chapter 7

Predicting Ionic ChargesPredicting Ionic ChargesGroup 16Group 16:: Gains 2 Gains 2 electrons to form electrons to form 2-2- ions ions

OO2-2-

SS2-2-

SeSe2-2-

OxideSulfideSelenide

Page 9: Chemistry Chapter 7

Predicting Ionic ChargesPredicting Ionic ChargesGroup 17Group 17:: Gains 1 Gains 1 electron to form electron to form 1-1- ions ions

FF1-1-

ClCl1-1-

BrBr1-1-Fluoride

Chloride

Bromide

II1-1- Iodide

Page 10: Chemistry Chapter 7

Predicting Ionic ChargesPredicting Ionic ChargesGroup 18Group 18:: Stable Stable Noble gases Noble gases do do notnot form ions! form ions!

Page 11: Chemistry Chapter 7

Predicting Ionic ChargesPredicting Ionic ChargesGroups 3 - Groups 3 - 1212::

Many Many transitiontransition elements elements have more than one possible oxidation have more than one possible oxidation state.state.Iron(II) = Fe2+ Iron(III) = Fe3+

Page 12: Chemistry Chapter 7

Predicting Ionic ChargesPredicting Ionic ChargesGroups 3 - Groups 3 - 1212::

Some Some transitiontransition elements elements have only one possible oxidation state.have only one possible oxidation state.Zinc = Zn2+ Silver = Ag+

Page 13: Chemistry Chapter 7

Writing Ionic Compound Writing Ionic Compound FormulasFormulas

Example: Barium nitrate1. Write the formulas for the cation and anion, including CHARGES!

BaBa2+2+ NONO33--2. Check to see if charges are

balanced.

3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion.

Not balanced!

( )( ) 22

Page 14: Chemistry Chapter 7

Writing Ionic Compound Writing Ionic Compound FormulasFormulas

Example: Ammonium sulfate1. Write the formulas for the cation and anion, including CHARGES!

NHNH44++ SOSO44

2-2-2. Check to see if charges are balanced.

3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion.

Not balanced!

( )( )22

Page 15: Chemistry Chapter 7

Writing Ionic Compound Writing Ionic Compound FormulasFormulas

Example: Iron(III) chloride1. Write the formulas for the cation and anion, including CHARGES!

FeFe3+3+ClCl--2. Check to see if charges are balanced.

3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion.

Not balanced!

33

Page 16: Chemistry Chapter 7

Writing Ionic Compound Writing Ionic Compound FormulasFormulas

Example: Aluminum sulfide1. Write the formulas for the cation and anion, including CHARGES!

AlAl3+3+ SS2-2-2. Check to see if charges are balanced.

3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion.

Not balanced!

22 33

Page 17: Chemistry Chapter 7

Writing Ionic Compound Writing Ionic Compound FormulasFormulas

Example: Magnesium carbonate1. Write the formulas for the cation and anion, including CHARGES!

MgMg2+2+ COCO332-2-2. Check to see if

charges are balanced.

They are balanced!

Page 18: Chemistry Chapter 7

Writing Ionic Compound Writing Ionic Compound FormulasFormulas

Example: Zinc hydroxide1. Write the formulas for the cation and anion, including CHARGES!

ZnZn2+2+ OHOH--2. Check to see if charges are balanced.

3. Balance charges , if necessary, using subscripts. Use parentheses if you need more than one of a polyatomic ion.

Not balanced!

( )( ) 22

Page 19: Chemistry Chapter 7

Writing Ionic Compound Writing Ionic Compound FormulasFormulas

Example: Aluminum phosphate1. Write the formulas for the cation and anion, including CHARGES!

AlAl3+3+ POPO443-3-2. Check to see if charges are

balanced.

They ARE balanced!

Page 20: Chemistry Chapter 7

Naming Ionic CompoundsNaming Ionic Compounds• 1. 1. Cation first, then anionCation first, then anion

• 2. Monatomic cation = name of the 2. Monatomic cation = name of the elementelement

• CaCa2+2+ = calcium = calcium ionion

• 3. Monatomic anion = 3. Monatomic anion = rootroot + + -ide-ide• ClCl = chlor = chlorideide

• CaClCaCl22 = calcium chlor= calcium chlorideide

Page 21: Chemistry Chapter 7

Naming Ionic CompoundsNaming Ionic Compounds(continued)(continued)

• -- some some metal forms more than one metal forms more than one cationcation• -- use use Roman numeralRoman numeral in namein name

• PbClPbCl22

• PbPb2+2+ is cationis cation

• PbClPbCl22 = lead( = lead(IIII) chloride) chloride

Metals with multiple oxidation Metals with multiple oxidation statesstates

Page 22: Chemistry Chapter 7

Naming Binary CompoundsNaming Binary Compounds• -- Compounds between two Compounds between two nonmetalsnonmetals• -- First element First element in the formula is in the formula is named firstnamed first..• -- Second element Second element is named as if it were an is named as if it were an

anionanion..• -- Use prefixesUse prefixes• -- Only use Only use monomono on second element - on second element -

PP22OO55 ==COCO22 = =

CO =CO =NN22O =O =

didiphosphorus phosphorus pentpentoxideoxidecarbon carbon didioxideoxidecarbon carbon monmonoxideoxidedidinitrogen nitrogen monmonoxideoxide

Page 23: Chemistry Chapter 7

Calculating Formula MassCalculating Formula MassCalculate the formula mass of magnesium Calculate the formula mass of magnesium carbonate, MgCOcarbonate, MgCO33..

24.31 g + 12.01 g + 3(16.00 g) 24.31 g + 12.01 g + 3(16.00 g) ==

84.32 g84.32 g

Page 24: Chemistry Chapter 7

Calculating Percentage Calculating Percentage CompositionComposition

Calculate the percentage composition of Calculate the percentage composition of magnesium carbonate, MgCOmagnesium carbonate, MgCO33..From previous slide:From previous slide:24.31 g + 12.01 g + 3(16.00 g) = 24.31 g + 12.01 g + 3(16.00 g) = 84.32 g84.32 g 24.31 100 28.83%

84.32Mg

12.01 100 14.24%84.32

C 48.00 100 56.93%84.32

O 100.00

Page 25: Chemistry Chapter 7

FormulasFormulas

molecular formula = (empirical molecular formula = (empirical formula)formula)nn [ [nn = integer] = integer]

molecular formula = Cmolecular formula = C66HH66 = (CH) = (CH)66

empirical formula = CHempirical formula = CH

Empirical formula: the lowest whole number ratio of atoms in a compound.Molecular formula: the true number of atoms of each element in the formula of a compound.

Page 26: Chemistry Chapter 7

FormulasFormulas (continued)(continued)

Formulas for Formulas for ionic compoundsionic compounds are are ALWAYSALWAYS empirical (lowest whole empirical (lowest whole number ratio).number ratio).Examples:Examples:

NaCl MgCl2 Al2(SO4)3 K2CO3

Page 27: Chemistry Chapter 7

FormulasFormulas (continued)(continued)

Formulas for Formulas for molecular compoundsmolecular compounds MIGHTMIGHT be empirical (lowest whole be empirical (lowest whole number ratio).number ratio).

Molecular:Molecular:

H2O

C6H12O6 C12H22O11

Empirical:

H2O

CH2O C12H22O11

Page 28: Chemistry Chapter 7

Empirical Formula Empirical Formula DeterminationDetermination

1.1. Base calculation on 100 grams of Base calculation on 100 grams of compound. compound.

2.2. Determine moles of each element in 100 Determine moles of each element in 100 grams of compound.grams of compound.

3.3. Divide each value of moles by the Divide each value of moles by the smallest of the values.smallest of the values.

4.4. Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.

Page 29: Chemistry Chapter 7

Empirical Formula Empirical Formula DeterminationDetermination

Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

49.32 14.107

12.01

g C mol Cmol C

g C

6.85 16.78

1.01

g H mol Hmol H

g H

43.84 12.74

16.00

g O mol Omol O

g O

Page 30: Chemistry Chapter 7

Empirical Formula Empirical Formula DeterminationDetermination

(part 2)(part 2)

4.107 1.502.74

mol Cmol O

6.78 2.472.74

mol Hmol O

2.74 1.002.74

mol Omol O

Divide each value of moles by the smallest Divide each value of moles by the smallest of the values.of the values.

Carbon:Carbon:

Hydrogen:Hydrogen:

Oxygen:Oxygen:

Page 31: Chemistry Chapter 7

Empirical Formula Empirical Formula DeterminationDetermination

(part 3)(part 3)Multiply each number by an integer to Multiply each number by an integer to obtain all whole numbers.obtain all whole numbers.

Carbon: 1.50Carbon: 1.50 Hydrogen: 2.50Hydrogen: 2.50 Oxygen: 1.00Oxygen: 1.00x 2 x 2 x 2

33 55 22Empirical formula:C3H5O

2

Page 32: Chemistry Chapter 7

Finding the Molecular Finding the Molecular FormulaFormulaThe empirical formula for adipic acid The empirical formula for adipic acid

is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

1. Find the formula mass of 1. Find the formula mass of CC33HH55OO22

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

Page 33: Chemistry Chapter 7

Finding the Molecular Finding the Molecular FormulaFormulaThe empirical formula for adipic acid The empirical formula for adipic acid

is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g

2. Divide the molecular mass by 2. Divide the molecular mass by the mass given by the emipirical the mass given by the emipirical formula.formula.

146 273

Page 34: Chemistry Chapter 7

Finding the Molecular Finding the Molecular FormulaFormulaThe empirical formula for adipic acid The empirical formula for adipic acid

is Cis C33HH55OO22. The molecular mass of . The molecular mass of adipic acid is 146 g/mol. What is the adipic acid is 146 g/mol. What is the molecular formula of adipic acid?molecular formula of adipic acid?

3(12.01 g) + 5(1.01) + 2(16.00) = 3(12.01 g) + 5(1.01) + 2(16.00) = 73.08 g73.08 g146 273

3. Multiply the empirical formula by 3. Multiply the empirical formula by this number to get the molecular this number to get the molecular formula.formula.

(C(C33HH55OO22) x 2 ) x 2 ==

CC66HH1010OO44