Chapter 2 notes – Flashcards

# protons in nucleus

_ZX

# protons + # neutrons

^AX

p = 29

n = 63 - 29= 34                      

e = 29

(a) ¹⁹⁷₇₉Au

p = 79

n = 197 - 79=118                   

e = 79

(b) ⁹⁰₃₈Sr

p = 38

n = 90 - 38=52                       

e = 38

¹₁H : p = 1

n = 1 - 1=0                       

e = 1                               

²₁H : p = 1

n = 2 - 1=1                       

e = 1                               

³₁H : p = 1

n = 3 - 1=2                       

e = 1                               

²³⁵₉₂U : p = 92

n = 235 - 92=143             

e = 92                            

²³⁸₉₂U : p =92

n = 238 - 92=146            

e = 92                           

atomic weight = ∑ [(%abundance/100) X (mass of isotope)]

(0.5184 X 106.90509amu) + (.4816 X 108.90476) = 107.9amu

⁶Li = x     ⁷Li = (1 - x)

Li = 6.941amu

6.941amu = (x)(6.0151amu) + (1 - x)(7.0160amu)

6.941amu = 6.0151x + 7.0160 + 7.0160x

   1.0009x = 7.0160 - 6.941

   1.0009x = .075

            x = 0.075

⁶Li = 7.5%

⁷Li = 92.5%

ion with a positive charge (metals that lose electrons)

Ex: Na -- -e^- --> Na⁺

    (11p, 11e)    (11p, 10e)

ion with a negative charge (nonmetals gain electrons)

Ex: F -- +e^- --> F^-

       (9p, 9e)      (9p, 10e)

p = 13

          e = 13 - 3=10

p = 34

           e = 34 + 2=36

shows the exact number of atoms of each element in the smallest unit of a substance

Ex: H₂O

     C₆H₁₂O₆

     O₃

     N₂H₄

shows the simplest whole-number ratio of the atoms in a substance

Ex: H₂O

     CH₂O

     O

     NH₂

1. ionic compounds consist of a combination of cations and anions

2. the formula is always the same as the empirical formula

3. the sum of the charges on the cation(s) and anion(s) in each formula unit must equal zero

Ex: the ionic compound of NaCl = Na⁺ Cl^-

(a) Group I A = (H⁺), Li⁺, Na⁺, K⁺, Rb⁺, Cs⁺

     Group II A = Be²⁺, Hg²⁺, Ca²⁺, Sr²⁺, Ba²⁺

      Group III A = Al³⁺, Ga³⁺

(b) Fe²⁺, Cu⁺, Cr²⁺, Co²⁺

     Fe³⁺, Cu²⁺, Cr³⁺, Co³⁺

exception which are not transition metals: Sn²⁺, Pb²⁺

                                                           Sn⁴⁺, Pb⁴⁺

(a) Group V A = N^3-, P^3-

     Group VI A = O^2-, S^2-

     Group VII A = F^-, Cl^-, Br^-, I^-

Mg²⁺ F^- = MgF₂   

Al³⁺ O^2- = Al₂O₃   

Na⁺ SO₄^2- = Na₂SO₄   

Ca²⁺ PO₄^3- = Ca₃(PO₄)₂

(a) Li⁺ = Lithium

     Mg²⁺ = Magnesium

(b) Cu⁺ = Copper I

     Cu²⁺ = Copper II

(c) NH₄⁺ = ammonium

rule: name the matal cation first, followed by the non-metal anion.

Na⁺ Cl^- = NaCl = sodium chloride

K⁺ CN^- = KCN = potassium cyanide

Na⁺ SO₄^2- = Na₂SO₄ = sodium sulfate

Mg²⁺ N^3- = Mg₃N₂ = magnesium nitride

Mg²⁺ PO₄^3- = Mg₃(PO₄)₂ = magnesium phosphate

K⁺ CO₃^2- = K₂CO₃ = potassium carbonate

Na⁺ O₂^2- = Na₂O₂ = sodium peroxide

rule: (a) write the name of the cation

       (b) write the charge on the cation as a roman numeral

       (c) write the name of the anion

FeCl₂ = Fe²⁺ Cl^- = Iron II Chloride

Cu₂O = Cu⁺ O^2- = Copper I Oxide

Cr₂O₃ = Cr³⁺ O^2- = Chromium III Oxide

SnO₂ = Sn⁴⁺ O^2- = Tin IV Oxide

PbI₂ = Pb²⁺I^- = Lead II Iodide

rule: (a) name the first element and add -ide to the stem of the name of the second element

       (b) indicate the number of each element by the use of Greek prefixes

N₂O₃ = dinitrogen trioxide

P₄S₃ = tetraphosphorous trisulfide

CO₂ = carbon dioxide

CO = carbon monoxide

PCl₅ = phosphorous pentachloride

SF₆ = sulfur hexafluoride

Cl₂O₇ = dochlorine heptoxide

CCl₄ = carbon tetrachloride

compounds that have a specific number of water molecules attatched to them

BaCl₂ • 2H₂O = barium chloride dihydrate

LiCl • H₂O = litium chloride monohydrate

MgSO₄ • 7H₂O = magnesium sulfate heptahydrate

Sr(NO₃)₂ • 4H₂O = strontium nitrate tetrahydrate

(a) NH₄NO₃ (NH₄⁺ NO₃^-)

(b) CoSO₄ (Co²⁺ SO₄^2-)

(c) Ba(CH₃CO₂)₂ (Ba²⁺ CH₃CO₂^-)

(d) V₂O₃ (V³⁺ O^2-)

(f) lithium carbonate

(g) copper (II) chloride

(a) CO₂

(b) PI₃

(c) SCl₂

(d) tetraphosphorous decaoxide

(e) sulfur tetraflouride

(f) dinitrogen tetraflouride

the mass of 1 mole of a substance in grams

;

(M = molar mass in g/mol;;;; N_A = Avagadro's number)

g --m/M--> mol --nN_A--> # atoms

g <--nM-- mol <--N/N_A-- # atoms

g-->mol

mass(g) X 1mol/molar mass(g) = mol

87.3g Mg X 1mol Mg/24.30g = 3.59mol

g-->mol-->atoms

0.551g K X (1mol K/39.10g K) X (6.022 X 10²³atom K/1mol K) = 8.49 X 10²¹atoms K

the sum of the atomic masses (amu) in a molecule

SO₂:

1 S = 32.07amu    +

2 O = 2 X 16.00amu

= 64.07amu   

the sum of the atomic masses (amu) in a formula unit of an ionic compound

NaCl:

1 Na = 22.99amu   

1 Cl = 35.45amu +

= 58.44amu 

(C₆H₁₂O₆ = 180.0g/mol)

mol --> g

3.25mol C₆H₁₂O₆ X (180.0g/1mol C₆H₁₂O₆₎ = 585g C₆H₁₂O₆

(CHCl₃ = 1119.5g/mol)

g --> mol

198g CHCl₃ X (1mol CHCl₃/119.5g CHCl₃) = 1.66mol CHCl₃

(HCl = 36.5g/mol)

g --> mol --> molecules

3.46g HCl X (1mol HCl/36.5g) X (6.022 X 10²³ molec HCl/1mol HCl) = 5.71 X 10²² HCl molec

(C₃H₈O = 60.0g/mol)

g --> mol --> molecules --> atoms

72.5g C₃H₈O X (1mol C₃H₈O/60.0g C₃H₈O) X (6.022 X 10²³ molec C₃H₈O/1mol C₃H₈O) X (8 atom H/1molec C₃H₈O) = 5.82 X 10²⁴ atoms H

(C₂H₆O = 46.07g/mol)

%C = [(2 X 12.01g)/46.07g] X 100 = 52.14% C

%H = [(6 X 1.008g)/46.07g] X 100 = 13.13% H

%O = [(1 X 16.00g)/46.07g] X 100 = 34.73% O

(C₆H₁₂O₆ = 180.0g/mol)

%C = 40.04%

%H = 6.72% +

%O = 53.33% +

 = 100%

(C₁₂H₂₂O₁₁ = 342.3g/mol)

%C = [(12 X 12.01g)/342.3g] X 100 = 42% C

61.8g X 0.42 = 26.0g C

2.1g H X (1mol H/1.008g) = 2.08mol/1.02 = 2

32.6g S X (1mol S/32.0g S) = 1.02mol/1.02 = 1

65.3g O X (1mol O/16.0g O) = 4.08/1.02 = 4

H₂SO₄

49.32g C X (1mol C/12.01g C) = 4.107mol/2.74 = 1.5

43.84g O X (1mol O/16.00g O) = 2.74/2.74 = 1

6.85g H X (1mol H/1.008g H) = 6.796mol/2.74 = 2.5

C₃H₅O₂

_{DO NOT ROUND SUBSCRIPTS!!!}

49.48g C X (1mol C/12.01g C) = 4.12mol/1.03 = 4

5.19g H X (1mol H/1.008g H) = 5.15mol/1.03 = 5

28.85g N X (1mol N/14.01g N) = 2.06mol/1.03 = 2

16.48g O X (1mol O/16.0g O) = 1.03mol/1.03 = 1

Empirical = C₄H₅N₂O

integral factor = (194.19/97.1) = 2

Molecular=

C₈H₁₀N₄O₂

amount of oxygen = 1.68g - 1.25g = 0.43g

1.25g Ga X (1mol Ga/69.72g Ga) = .0179mol/.0179 = 1

0.43g O X (1mol O/16.00g O) = .0269mol/.0179 = 1.5

Ga₂O₃