Unit One

Notes One Unit One

• Two Classes of Elements

• What Are Stable Elements?

• Stabilizing Sodium

• Stabilizing Oxygen

• Sodium Loses electrons to Oxygen

• Oxidation Numbers

• Key Elements• Examples

Pages 158-165

Two Classes of ElementsWhat are the Two Main Classes of Elements?Metals and Nonmetals Noble Elements

What Makes Elements Stable?

(Lose e-1)

(Gain e-1)

Losing or Gaining e-1.Do metals Lose or Gain e-1?Do nonmetals Lose or Gain e-1?

0 +1 +2 +3 +4 +5 +6 +7 +8 +9 +10-10 -9 -8 -7 -6 -5 -4 -3 -2 -1

Oxidation

Reduction

Stabilizing Sodium• How many e-1 for Na?• 11e-1

• What is the noble element closest to Na?• Ne• How many e-1 for Ne?• 10e-1

• Sodium loses/gains how many electrons?• 1e-1

• Na Na+1 + e-1

• (protons) + (electrons)=charge (+11) (-10) +1

Oxidation or reduction?

Stabilizing Oxygen• How many e-1 for O?• 8e-1

• What is the noble element closest to O?• Ne• How many e-1 for Ne?• 10e-1

• Oxygen loses/gains how many electrons?• 2e-1

• O + 2e-1 O-2 Oxidation or reduction?• (protons) + (electrons)=charge

(+8) (-10) -2

Sodium Loses electrons to Oxygen• Na Na+1 + e-1 (Stable Like Neon) Ox or Red?• O + 2e-1 O-2 (Stable Like Neon) Ox or Red?• How many sodium atoms are needed to satisfy

oxygen’s electron hunger?• 2e-1 means • How many oxygen atoms are needed to satisfy

sodium’s electron loss?• 2e-1 means

• Na2O

two Na

one O

High Electronegativity

Low Electronegativity

Oxidation Numbers• All elements Lose or Gain e-1.

• Some have multiple loss or gain possibilities.

Fe+2 Fe+3 S-2 S+4 S+6

Key Elements

(99%) H+1 H-1

(99%) O-2 O-1

(Always) Li+1, Na+1, K+1, Rb+1, Cs+1, Fr+1

(Always) Be+2, Mg+2, Ca+2, Ba+2, Sr+2, Ra+2

(Always) Al+3

(with only a metal) F-1, Cl-1, Br-1, I-1

(NO3-1) ion is always +5

(SO4-2) ion is always +6

Example OneFinding Oxidation Numbers

2 (+3)+ 3(S) = Zero

sum of the oxidation #’s =

Find Ox #’s for Al2S3?

zero

2 (Al)+ 3(S) = Zero

S = -2

+3 -2

Example TwoFinding Oxidation Numbers

sum of the oxidation #’s = zero

+2 -2Find Ox #’s for Ca3(PO4)2?

3(Ca)+ 8(O) = Zero

+5

2(P)+

3(+2)+ 8(-2) = Zero2(P)+

= +5P

Finding Oxidation #’s for Compounds

+1 -2

+1+5 -2HH33POPO44

H2O

HNO3

+1+5 -2

H2SO4

+1 -2+6

Hg2SO4

+6+1 -2

Na2Cr2O7

+1 +6 -2

H2CO3

+1 -2+4

(NH4)2CO3

-3 +1+4 -2

Ca3(AsO4)2

+2 +5 -2

Fe2(SO4)3

+6+3 -2

Ba(ClO4)2

+2 +7 -2

Al2(CO3)3

+3 +4 -2

Formula of Water Lab A

• Water will be converted in to elements by passing and electric current through the water. Acid( contains + and -ions) is needed in order to pass an electric current through water.

Lab A Setup

Formula of water Lab A results• 7. What gasses are being produced in the tubes?

• 8. When the electrode-tube with the most gas is 2/3 full, switch the alligator clips and finish filling the tubes.

• Observations

• 9. What is the identity of the gas with greater volume?

Formula of water Lab A resultsQUESTIONS:• 1. What is the ratio of volumes for the gases collected • 2. What is the formula for water?

• 3. Does the result of your work confirm the ratio of elements in water's formula?

• 4. Explain how it does so.

• 5. What mixture of hydrogen to oxygen should give off the best reaction using the spark device?

• 6. Why was the sulfuric acid solution used in this demonstration?

Notes Two Unit One

• Naming Inorganic Salts

• Example One

• Example One Thinking

• Example Two

• Computer Assignment One

Pages 176-183

Naming Inorganic Salts• TWO parts to the name• 1) Cation• 2) Anion• Cation Examples• Ca+2 • Al+3 • Fe+2 • Na+1

• Anion Examples• Cl-1

• NO3-1

• SO4-2

• N-3

PositiveNegative

Example One• Name the formula Fe2(CrO4)3

• Step #1 Find The + Ion(s).

Iron(II) Fe+2

Iron(III) Fe+3

Example One• Step #2 Find The - Ion(s)

Chromate CrO4-2

Fe2(CrO4)3 Fe+2 Fe+3 CrO4-2Iron(II) ChromateIron(III)

Fe+2 CrO4-2

Iron(II) Chromate

(+2) (-2)Y+ = 0

X=1 Y=1

FeCrO4

X

(+2) (-2)1+ = 01

Fe+3 CrO4-2

Iron(III) Chromate

(+3) (-2)Y+ = 0

X=2 Y=3

X

(+3) (-2)3+ = 02

Fe2(CrO4)3

Example One

Al2(CO3)3 Al+3 CO3-2 CarbonateAluminum

Al+3 CO3-2

Aluminum Carbonate

(+3) (-2)Y+ = 0

X=2 Y=3

X

(+3) (-2)3+ = 02

Al2(CO3)3

Example Two

Computer Assignment One/Two

• NAMING IONIC COMPOUNDS LEVELs ONE AND TWO

Writing a Formula From a Name

HH33POPO44

LiNOLiNO33Lithium NitrateLithium Nitrate ( )_( )_LiLi+1+1 NONO33

-1-11111

Hydrogen PhosphateHydrogen Phosphate ( )_( )_HH+1+1 POPO44-3-3

1133

CaCa33(AsO(AsO44))22

(NH(NH44))22COCO33Ammonium carbonateAmmonium carbonate ( )_( )_NHNH44

+1+1 COCO33-2-2

1122

Calcium ArsenateCalcium Arsenate ( )_( )_CaCa+2+2 AsOAsO44-3-3

2233

HgHg22SOSO44

Fe(IOFe(IO44))33Iron(III) periodateIron(III) periodate ( )_( )_FeFe+3+3 IOIO44

-1-13311

Mercury(I) SulfateMercury(I) Sulfate ( )_( )_HgHg22+2+2 SOSO44

-2-22222

NaNa22CrCr22OO77

Ba(ClOBa(ClO44))22Barium PerchlorateBarium Perchlorate ( )_( )_BaBa+2+2 ClOClO44

-1-12211

Sodium DichromateSodium Dichromate ( )_( )_NaNa+1+1 CrCr22OO77-2-2

1122

Pb(SOPb(SO44))22Lead(IV) SulfateLead(IV) Sulfate ( )_( )_PbPb+4+4 SOSO44

-2-24422

(Cation+?)X(Anion-?)Y(+?) (-?)Y+ = 0X

Lowest Whole Number Ratio

If X or Y is 2 or greater...

and the ion is polyatomic.

BaBa+2 +2 CrCr22OO77-2 -2 HgHg22

+2 +2 PbPb+4+4

Notes Three Unit One• Standard Amounts

• One Gopher

• One Mole

• Formula mass

• Percent Composition

• Empirical Formula

Pages 286-297

Standard Amounts• How many dollars is…• A) 120 pennies?• 1.2 dollars• B) 2 quarters?• 0.5 dollars• C) 15 nickels?• 0.75 dollars• How many dozens is…• D) 48 eggs?• 4 dozen• E)18 apple fritters• 1.5 dozen

One Gopher• One Gopher equals 12

items

• What is the mass of one gopher of…

• A) white beads?

• 2.81g/G

• B) blue beads?

• 0.50g/G

• C) orange Beads?

• 1.67g/G

Eight Rows

Seven Rows

One Gopher(12 items)• In Six groups

• (1) How many gophers of beads are in…(2) How many beads are in…A) ___gB) ___gC) ___gD) ___gE) ___gF) ___g

2.00

3.49

2.51

3.75

1.75

5.82

4.00G

1.24G

1.50G

2.25G

3.50G

2.07G

48 beads

15 beads

18 beads

27 beads

42 beads

25 beads

One Mole• One mole is 6.022x10+23 items.• Each element on the period table has

a mass per mole.

NOC

14.0g16.0g12.0g

6.022x10+23atoms6.022x10+23atoms6.022x10+23atoms

N

O

C

7.0g

4.0g

18.0g

=0.50m

=0.25m

=1.50m

• How many moles are in each?

=3.01x10+23atoms

=1.51x10+23atoms

=9.03x10+23atoms

÷14.0g/m

÷16.0g/m

÷12.0g/m

• How many atoms are in each?

x6.022x10+23atoms/m

x6.022x10+23atoms/m

x6.022x10+23atoms/m

Calculations Bases on Chemical Formulas

•Formula mass (Molecular Mass or Gram-Formula Mass)•Empirical Formula•Percent Composition

Rounding Atomic Mass

CFeO

12.01155.84715.9994

BiKAu

208.98083739.0983196.96654

OsMgNa

190.2324.305022.98968

12.055.816.0209.039.1197.0190.224.323.0

Formula Mass Example OneCalculate the formula mass for 1 mole of C6H12O6.

CHO

6 x12 x6 x

12.0 = 1.0 =16.0 =

72.012.096.0

180.0g/mol

E # Mass

How many molecules of C6H12O6 is 180.0g/mol?

6.022x10+23 molecules

12.0111.007915.9994

Empirical Formula Example One

3) Write the formula

What is the empirical (simplest) formula containing 36.8% N, 63.2% O?

X by 2 to get whole numbers

1) Calculate moles of each element.

NO

36.8 g ÷63.2 g ÷

14.0 =16.0 =

2.63 mol N 3.95 mol O

E Q Mass

2) Calculate the lowest ratio.

NO

2.63 mol N ÷3.95 mol O ÷

2.63 mol =2.63 mol =

1.00 1.50

E Moles Lowest Ratio

N2O3

14.006715.9994

Mass

Percent Composition Example OneCalculate the percentage composition of H2O.

HO

2 x1 x

1.0 =16.0 =

2.0 16.018.0g/mol

E # Mass

2) Divide each contribution by the total mass.

3) Add the percentages to check work.

1)Calculate the formula mass for 1 mole of H2O

HO

2.0 ÷16.0 ÷

18.0 =18.0 =

0.11 0.889

11% 88.9%100.%

1.007915.9994

( x 100) = 11%( x 100) = 88.9% Answer

End

Empirical Formula Example Two

3) Write the formula

What is the empirical (simplest) formula containing 69.58% Ba, 6.090% C, 24.32% O?

X by 1 to get whole numbers

1) Calculate moles of each element.

BaC

69.58 g ÷6.090 g ÷

137.33 =12.01 =

0.50666 mol Ba 0.50708 mol C

E Q Mass

2) Calculate the lowest ratio.

BaC

0.50666 mol ÷0.50708 mol ÷

0.50666 mol =0.50666 mol =

1.000 1.001

E Moles Lowest Ratio

BaCO3

O 24.32 g ÷ 16.00 = 1.520 mol O

O 1.520 mol ÷ 0.50666 mol = 3.00

Mass

Percent Composition Example TwoCalculate the percentage composition of Fe(ClO4)3.

ClO

3 x12 x

35.5 =16.0 =

106.5 192.0

354.3g/mol

E # Mass

2) Divide the each contribution by the total mass.

3) Add the percentages to check work.

1)Calculate the formula mass for 1 mole of Fe(ClO4)3.

ClO

106.5 ÷192.0 ÷

354.3 =354.3 =

0.30060.5419

30.1 % 54.2 %100.1%

Fe 1 x 55.8 = 55.8

Fe 55.8 ÷ 354.3 = 0.1575

15.8 %

55.84735.45315.9994

( x 100) = 15.8%( x 100) = 30.06%( x 100) = 54.19%

Answer

Formula Mass Example ThreeCalculate the formula mass for 1 mole of Al2O3

AlO

2 x3 x

27.0 =16.0 =

54.0 48.0102.0g/mol

E # Mass

How many molecules of Al2O3 is102.0g/mol?

6.022x10+23 molecules

26.9815415.9994

Formula Mass Example Two

Calculate the formula mass for 1 mole of CaCO3.

CaCO

1 x1 x3 x

40.1 =12.0 =16.0 =

40.112.0

48.0100.1g/mol

E # Mass

How many molecules of CaCO3 is100.1g/mol?

6.022x10+23 molecules

40.07812.01115.9994