Chapter 3. Atomic Mass amu = Average Atomic Mass Unit Based on 12 C as the standard. 12 C = exactly 12 amu The average atomic mass (weight) of.

Chapter 3Chapter 3

Atomic MassAtomic Mass

amu = Average Atomic Mass Unit amu = Average Atomic Mass Unit Based on Based on 1212C as the standard. C as the standard. 1212C = exactly 12 amuC = exactly 12 amu The The average atomic mass (weight)average atomic mass (weight) of of

an element is equal to the sum of the an element is equal to the sum of the products of each isotope’s mass (in amu) products of each isotope’s mass (in amu) multiplied by it’s relative abundance.multiplied by it’s relative abundance.

EXAMPLE OF AVERAGE EXAMPLE OF AVERAGE ATOMIC MASS PROBLEMATOMIC MASS PROBLEM Naturally occurring chlorine is Naturally occurring chlorine is

75.53% 75.53% 3535Cl which has an atomic Cl which has an atomic mass of 34.969 amumass of 34.969 amu,, and and 24.47% 24.47% 3737Cl, which has an atomic mass of Cl, which has an atomic mass of 36.966 amu.36.966 amu.

Calculate the average atomic mass Calculate the average atomic mass of chlorine.of chlorine.

EXAMPLE OF AVERAGE ATOMIC EXAMPLE OF AVERAGE ATOMIC MASS PROBLEM (CONT)MASS PROBLEM (CONT)

Average atomic Mass = [ (%/100) Average atomic Mass = [ (%/100) (Atomic Mass) ] (Atomic Mass) ]

Average atomic mass = Average atomic mass = (0.7553) (34.969 amuCl(0.7553) (34.969 amuCl3535) + (0.2447) (36.966 ) + (0.2447) (36.966

amuClamuCl3737))

= 26.41 amu + 9.045 amu= 26.41 amu + 9.045 amu

= 35.46 amu= 35.46 amu

Formula Weight Formula Weight

AKA: molar mass (g/mol)AKA: molar mass (g/mol) Sum of all atomic weights of each Sum of all atomic weights of each

atom in its chemical formula atom in its chemical formula

Ex: HEx: H22S0S044

1(2H) + 32(1S) + 16(4O) = 98 amu 1(2H) + 32(1S) + 16(4O) = 98 amu

MOLE MOLE Avogadro’s NumberAvogadro’s Number

1 mole of any 1 mole of any substance is substance is

6.02 x 106.02 x 102323 particles particles

Question Question

How many nitrogen How many nitrogen atoms are in 0.25 atoms are in 0.25 mol of Ca(NOmol of Ca(NO33))22

Answer Answer

6.02 x 106.02 x 102323 molec Ca(NO molec Ca(NO33))22 x_ x_2N___ 2N___ X X .25.25

1 mol molec Ca(NO1 mol molec Ca(NO33))22

= atoms= atoms

Law of Conservation of MassLaw of Conservation of Mass

Mass is never Mass is never created or created or destroyed destroyed

Reason for Reason for balancing balancing chemical chemical equationsequations

Lavosier Says :

2 Na + Cl 2

2NaCl

46.0 g 70.9 g 116.9 g

Balancing Equations Balancing Equations

1. Write the correct formulas for the reactants 1. Write the correct formulas for the reactants and the productsand the products

2.Chose the compound that has the greatest 2.Chose the compound that has the greatest number of atoms, then look to the element in number of atoms, then look to the element in that compound that has the greatest number of that compound that has the greatest number of atoms. atoms.

3.Balance this element first by placing a 3.Balance this element first by placing a coefficient in front of the corresponding coefficient in front of the corresponding compound on the other side of the equation. compound on the other side of the equation.

Balancing Equations Balancing Equations Cont.Cont.

4. Balance H then O 4. Balance H then O

5. Check all coefficients to see that they 5. Check all coefficients to see that they are in the lowest possible ratio.are in the lowest possible ratio.

Examples:Examples:

CC22HH6 6 + O+ O22 -> CO -> CO2 2 + H+ H22OO

CC22HH66OO + + OO2 2 -> CO-> CO2 2 + H+ H22OO

CaCOCaCO33 + H + H33POPO44 -> Ca -> Ca33(PO(PO44))22 + CO + CO22 + H + H22OO

ANSWERANSWER

22CC22HH6 6 + + 77OO22 -> -> 44COCO2 2 + + 66HH22OO

CC22HH66OO + + 33OO2 2 -> -> 22COCO2 2 + + 33HH22OO

33CaHCaH66OO33 + + 22HH33POPO44 -> Ca -> Ca33(PO(PO44))22 + + 33COCO22 + + 33HH22OO

Percent CompositionPercent Composition

We can describe composition in two waysWe can describe composition in two ways1. number of atoms 1. number of atoms 2. % (by mass) of its elements. 2. % (by mass) of its elements.

We can find % mass from formula mass, We can find % mass from formula mass, by comparing by comparing eacheach element present in 1 element present in 1 mole of compound to the mole of compound to the total masstotal mass of 1 of 1 mole of compound mole of compound

Example of % Comp Example of % Comp

Calculate the percentage of nitrogen in Calculate the percentage of nitrogen in Ca(NOCa(NO33))22

Answer Answer

% N = % N = # N atoms(m.w N# N atoms(m.w N) X 100%) X 100%

m. w Ca(NOm. w Ca(NO33))22

% N = % N = 2(14.02 N amu2(14.02 N amu) X 100%) X 100%

164.12 Ca(NO164.12 Ca(NO33))22amuamu

= 17%= 17%

Inter-converting Inter-converting

Grams -> moles -> molec -> atoms Grams -> moles -> molec -> atoms

How many oxygen atoms are present in How many oxygen atoms are present in 4.20 grams NaHCO4.20 grams NaHCO33??

Answer Answer

4.20 g NaHCO4.20 g NaHCO3 3 (1mole NaHCO(1mole NaHCO33) ) ( (6.02e 6.02e 2323molec)molec) 3 Oxygen atoms3 Oxygen atoms

84 g NaHCO84 g NaHCO3 3 1 mol1 mol 1 molec NaHCO1 molec NaHCO33

= 9.033 x 10 = 9.033 x 10 22 22 atoms of Oxygen in 4.20 grams NaHCO atoms of Oxygen in 4.20 grams NaHCO33

Question Question

Determine the mass in grams of Determine the mass in grams of

3.00 x 103.00 x 1020 20 NN22 molecules molecules

Answer Answer

3.00 x 103.00 x 1020 20 molecmolec NN2 2 (1 mol)(1 mol) (28g N(28g N22) )

6.02e6.02e2323 molecmolec 1 mol N 1 mol N22

= 0.0140 g = 0.0140 g

Determining empirical Determining empirical formula from mass percent formula from mass percent

Recall: Empirical formula: simplest whole Recall: Empirical formula: simplest whole # ratio of atoms in a compound.# ratio of atoms in a compound.

Example: Vitamin C is composed of Example: Vitamin C is composed of 40.92% C, 4.58% H, and 54.50% O by 40.92% C, 4.58% H, and 54.50% O by mass. What is the empirical formula?mass. What is the empirical formula?

Answer: CHOAnswer: CHO

1.1. Convert mass % into grams (assume 100g) Convert mass % into grams (assume 100g) 2.2. Convert grams to moles Convert grams to moles 3.3. Divide each mol by the smallest number of moles Divide each mol by the smallest number of moles

present. You may round to nearest whole #present. You may round to nearest whole #

40.92 g C 40.92 g C 1mol C1mol C = 3.4 moles C = 3.4 moles C /3.4 = 1 C/3.4 = 1 C 12 g C 12 g C

54.40 g O 54.40 g O 1mol O 1mol O = 3.4 moles O = 3.4 moles O / 3.4 = 1 O/ 3.4 = 1 O 16g O 16g O

4.48 g H 4.48 g H 1mol H 1mol H = 4.48 moles H = 4.48 moles H/ 3.4 = 1 H/ 3.4 = 1 H 1 g H 1 g H

Determine Molecular Determine Molecular formula from Empirical formula from Empirical FormulaFormula

Recall: Molecular formula: the exact formula of a molec, giving Recall: Molecular formula: the exact formula of a molec, giving types of atoms and the number of each type. types of atoms and the number of each type.

1.1. Using mass % and molar mass, determine mass of each Using mass % and molar mass, determine mass of each element in 1 mole of compoundelement in 1 mole of compound

2.2. Determine number of moles of each element in 1 mole of Determine number of moles of each element in 1 mole of compound compound

3.3. The integers from the previous step represent the The integers from the previous step represent the subscripts in the molecular formula subscripts in the molecular formula

Let’s look back at our work Let’s look back at our work

40.92 g C 40.92 g C 1mol C 1mol C = 3.4 moles C= 3.4 moles C /3.4 = 1 C /3.4 = 1 C 12 g C 12 g C

54.40 g O 54.40 g O 1mol O 1mol O = = 3.4 moles O3.4 moles O / 3.4 = 1 O / 3.4 = 1 O 16g O 16g O

4.48 g H 4.48 g H 1mol H 1mol H = = 4.48 moles H/4.48 moles H/ 3.4 = 1.3 H 3.4 = 1.3 H 1 g H 1 g H

CC33HH44OO33 = molecular formula= molecular formula

Shortcut Shortcut

n = n = Molecular WeightMolecular Weight

empirical Formula Weight empirical Formula Weight

The molecular weight of butyric acid is The molecular weight of butyric acid is

88 amu. If the empirical formula is C88 amu. If the empirical formula is C22HH44O. O.

What is the molecular formula?What is the molecular formula?

CC22HH44O = 12 + 12+ 1+1+1+1+16 = 44 amu O = 12 + 12+ 1+1+1+1+16 = 44 amu

n = n = 88 amu88 amu = 2 = 2

44 44

Molecular formula = (empirical)Molecular formula = (empirical)nn

(C(C22HH44O)O)22 = Molecular Formula = C = Molecular Formula = C44HH88OO2 2

StoichiometryStoichiometry: : mixing exactly mixing exactly enough chemical so that all is usedenough chemical so that all is used

Mass-Mass problems Mass-Mass problems

g given g given mol given mol given mol required mol required g required g required

(grams to moles to moles to grams) (grams to moles to moles to grams)

Silicon carbide is made by heating silicon dioxide Silicon carbide is made by heating silicon dioxide to high temperatures.to high temperatures.

SiOSiO22 (s) + 3C (s) (s) + 3C (s) SiC(s) + 2CO (g) SiC(s) + 2CO (g)

How many grams of CO are formed by complete How many grams of CO are formed by complete rxn of 5.00 g SiOrxn of 5.00 g SiO22? ?

AP EXAM HINT: always make sure your equation AP EXAM HINT: always make sure your equation is balanced first or mole ratios will be wrong. is balanced first or mole ratios will be wrong.

Given: 5.00 g SiOGiven: 5.00 g SiO22

Find : CO g Find : CO g

SiOSiO22 (s) + 3C (s) (s) + 3C (s) SiC(s) + SiC(s) + 2CO2CO (g) (g)

grams to moles to moles to grams grams to moles to moles to grams

5.00 g SiO5.00 g SiO2 2 1mol SiO1mol SiO22 2 mol CO2 mol CO 28 g CO 28 g CO = 4.6 g CO = 4.6 g CO

60 g SiO60 g SiO2 2 1 mol SiO1 mol SiO22 1 mol CO 1 mol CO

Mole ratio

How many moles of sulfuric acid would be How many moles of sulfuric acid would be needed to produce 4.80 moles of molecular needed to produce 4.80 moles of molecular iodine (Iiodine (I22) according to the following ) according to the following

balanced equation. balanced equation.

10HI + 2KMnO10HI + 2KMnO4 4 + 3H+ 3H22SOSO44 5I 5I22 + 2MnSO + 2MnSO4 4 + K+ K22SOSO44 + 8H + 8H22OO

4.80 mol I4.80 mol I22 3 mol H3 mol H22SOSO44 = 2.88 mol H= 2.88 mol H22SOSO44

5 mol I5 mol I22

Limiting reagent Limiting reagent This will be at least 1 AP question! This will be at least 1 AP question!

The number of products that can form is The number of products that can form is limited by the amount of reactant present.limited by the amount of reactant present.

The limiting reactant is the one that gives The limiting reactant is the one that gives the least amount of product. the least amount of product.

Reactants Reactants Products Products

When a mixture of silver and sulfur is heated, When a mixture of silver and sulfur is heated, silver sulfide is formed: silver sulfide is formed:

16 Ag (s) + S16 Ag (s) + S88 (s) (s) 8 Ag 8 Ag22S (s) S (s)

What mass of AgWhat mass of Ag22S is produced from a mixture S is produced from a mixture

of 2.0 g of Ag and 2.0 g of S?of 2.0 g of Ag and 2.0 g of S?

2.0 g Ag 12.0 g Ag 1mol Ag mol Ag 8 mol Ag8 mol Ag22S S 247.9 g Ag247.9 g Ag22SS

107.9 g Ag 16 mol Ag 1 mol Ag107.9 g Ag 16 mol Ag 1 mol Ag22SS

= 2.3 g Ag= 2.3 g Ag22S S

2.0 g S2.0 g S88 1 1mol mol SS8 8 8 mol Ag8 mol Ag22S S 247.9 g Ag247.9 g Ag22SS

256.8 S256.8 S88 1 mol S 1 mol S8 8 1 mol Ag 1 mol Ag22SS

= 15 g Ag= 15 g Ag22S S

Theoretical / Percent Theoretical / Percent yield yield

The amount of product that is calculated The amount of product that is calculated based on the limiting reactant. based on the limiting reactant.

% Yield = % Yield = Actual yieldActual yield X 100% X 100% theoretical yieldtheoretical yield

\\

HWHW

7, 19, 21, 31,337, 19, 21, 31,33

35, 56,57, 58, 62,63, 67, 69. 71,74 35, 56,57, 58, 62,63, 67, 69. 71,74 81,82,83, 95, 98, 100a-b,10581,82,83, 95, 98, 100a-b,105

Types of reactions Types of reactions

Chemical Reactivity Chemical Reactivity

Combination/Synthesis Reaction:Combination/Synthesis Reaction:

2 or more substances react to form one new 2 or more substances react to form one new productproduct

A + B A + B C C

+ +

solid magnesium and oxygen gas react to solid magnesium and oxygen gas react to

produce solid magnesium oxideproduce solid magnesium oxide

22MgMg (s) + (s) + OO22(g) (g) 2 2MgMgOO (s) (s)

Metal nonmetal ionic compound Metal nonmetal ionic compound Diatomic Diatomic

2+ 2-2+ 2-

Decomposition Rxn Decomposition Rxn

One substance undergoes a reaction to One substance undergoes a reaction to produce two or more substances. produce two or more substances.

Typically occurs when things are heated.Typically occurs when things are heated.

AX AX A + X A + X

++

Solid calcium carbonate reacts to Solid calcium carbonate reacts to produce solid calcium oxide and produce solid calcium oxide and carbon dioxide gascarbon dioxide gas

CaCOCaCO33 (s) (s) CaO (s) + CO CaO (s) + CO22 (g) (g)

2+ (2-) 2+ 2- 4+ 2(2-)2+ (2-) 2+ 2- 4+ 2(2-)

Single displacement Single displacement

One element replaces a similar element in One element replaces a similar element in a compound a compound

A + BX A + BX AX + B AX + B

BX + Y BX + Y BY + X BY + X

+ + + +

Solid copper is dissolved in aqueous silver Solid copper is dissolved in aqueous silver nitrate to produce solid silver and aqueous nitrate to produce solid silver and aqueous copper nitrate.copper nitrate.

Cu(s)Cu(s) + Ag + AgNONO3 3 (aq)(aq) Ag(s) + Ag(s) + CuCu(NO(NO33))22 (aq) (aq)

Write the sentence for this reaction:Write the sentence for this reaction:

Fe (s)Fe (s) + Cu + Cu(NO(NO33))2 2 (aq)(aq) FeFe(NO(NO33))22 (aq)+ Cu (s) (aq)+ Cu (s)

Double Replacement Rxn/ Double Replacement Rxn/ Metathesis Metathesis The The ionsions of of two compoundstwo compounds exchange places exchange places

in an aqueous solution to form two new in an aqueous solution to form two new compounds. compounds.

AX + BY AX + BY AY + BX AY + BX

One of the compounds formed is usually a One of the compounds formed is usually a

precipitate, an insoluble gas that bubbles out precipitate, an insoluble gas that bubbles out of solution, or a molecular compound, of solution, or a molecular compound, usually water.usually water.

Double Replacement Rxn/ Double Replacement Rxn/ MetathesisMetathesis

AX + BY AX + BY AY + BX AY + BX

+ + + +

Write the sentence for these double replacement Write the sentence for these double replacement reactionsreactions

KKOOH (aq) + HH (aq) + H22SOSO44 (aq) (aq) KK22SOSO44 (aq) + H (aq) + H22O (l)O (l)

FeFeS (aq)S (aq) + H + HCl (aq) Cl (aq) FeFeClCl22 (aq) + H (aq) + H22S (aq)S (aq)

Combustion Reaction Combustion Reaction

A substance combines with oxygen, releasing a A substance combines with oxygen, releasing a large amount of energy in the form of light and large amount of energy in the form of light and heat. heat.

CC33HH8 8 (g)+ (g)+ 5O5O22 (g) (g) 3CO 3CO2 2 (g) + H(g) + H22O (g)O (g)

Usually COUsually CO22 (carbon dioxide) / CO (carbon monoxide) (carbon dioxide) / CO (carbon monoxide) and water are produced. and water are produced.

Reactive elements combine with Reactive elements combine with oxygenoxygen

PP44(s) + 5O2(g) (s) + 5O2(g) P P44OO10 10 (s)(s)

(This is also a synthesis reaction)(This is also a synthesis reaction)

The burning of natural gas, wood, The burning of natural gas, wood, gasolinegasoline

CC33HH88(g) + 5O(g) + 5O22(g) (g) 3CO 3CO22(g) + 4H(g) + 4H22O(g)O(g)