1

By definition:1 atom 12C “weighs” 12 amu

On this scale1H = 1.008 amu16O = 16.00 amu

Atomic mass is the mass of an atom in atomic mass units (amu)

Micro Worldatoms & molecules

Laboratory scale measurements

No matter what its source, a particular chemical compound is composed of the same elements in the same parts (fractions) by mass.

CaCO3

40.08 amu

12.00 amu3 x 16.00 amu

1 atom of Ca

1 atom of C3 atoms of O

100.08 amu40.08 amu100.08 amu = 0.401 parts Ca

12.00 amu100.08 amu = 0.120 parts C

48.00 amu100.08 amu = 0.480 parts O

Law of Definite (or Constant)Composition:

2

The Atomic Basis of the Law of Multiple Proportions

3

1. All matter consists of atoms.

2. Atoms of one element cannot be converted into atoms of another element.

3. Atoms of an element are identical in mass and other properties and are different from atoms of any other element.

4. Compounds result from the chemical combination ofa specific ratio of atoms of different elements.

DaltonDalton’’s Atomic Theorys Atomic Theory

The mole (mol) is the amount of a substance that contains as many elementary entities as there

are atoms in exactly 12.00 grams of 12C

1 mol = NA = 6.0221367 x 1023

Avogadro’s number (NA)

4

Water18.02 g

Figure 3.2

One Mole of Common Substances

CaCO3100.09 g

Oxygen32.00 g

Copper63.55 g

Molar mass is the mass of 1 mole of in grams

Na atomsPb atomsKr atomsLi atoms

1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g

1 mole 12C atoms = 12.00 g 12C

1 mole lithium atoms = 6.941 g of Li

For any elementatomic mass (amu) = molar mass (grams)

5

Mass (g) = no. of moles x no. of grams

1 mol

No. of moles = mass (g) xno. of grams

1 mol

No. of entities = no. of moles x6.022x1023 entities

1 mol

No. of moles = no. of entities x 6.022x1023 entities

1 mol

Interconversions of Moles, Mass, and Number of Chemical Entities

Molar MassAvogadro’s Number

6

How many atoms are in 0.551 g of potassium (K) ?

1 mol K = 39.10 g K

1 mol K = 6.022 x 1023 atoms K

0.551 g K 1 mol K39.10 g K

x x 6.022 x 1023 atoms K1 mol K

=

8.49 x 1021 atoms K

7

Molecular mass (or molecular weight) is the sum ofthe atomic masses (in amu) in a molecule.

SO2

1S 32.07 amu2O + 2 x 16.00 amuSO2 64.07 amu

For any moleculemolecular mass (amu) = molar mass (grams)

1 molecule SO2 = 64.07 amu1 mole SO2 = 64.07 g SO2

Calculating the Mass and the Number of Atoms in a Given Number of Moles of an Element

PROBLEM:

amount(mol) of Ag

mass(g) of Ag

(a) Silver (Ag) is used in jewelry and tableware but no longer in U.S. coins. How many grams of Ag are in 0.0342mol of Ag?

(b) Iron (Fe), the main component of steel, is the most important metal in industrial society. How many Fe atoms are in 95.8g of Fe?

(a) To convert mol of Ag to g we have to use the #g Ag/mol Ag, the molar mass M.

(b) We first have to find the #mols of Fe and then convert mols to atoms.

multiply by M of Ag (107.9g/mol)

0.0342mol Ag xmol Ag

107.9g Ag = 3.69g Ag

mass(g) of Fe

amount(mol) of Fe

atoms of Fe

95.8g Fe x55.85g Fe

mol Fe 6.022x1023atoms Fe

mol Fe

= 1.04x1024 atoms Fe

divide by M of Fe (55.85g/mol)

multiply by 6.022x1023

atoms/mol

8

Determination of Empirical Formula from Mass %

1. Assume 100 grams of material2. Determine moles of each element in

compound3. Divide all subscripts in empirical formula

by lowest number because the number of atoms must be an integer

Percent composition of an element in a compound =

n x molar mass of elementmolar mass of compound x 100%

n is the number of moles of the element in 1 mole of the compound

C2H6O

%C = 2 x (12.01 g)46.07 g x 100% = 52.14%

%H = 6 x (1.008 g)46.07 g x 100% = 13.13%

%O = 1 x (16.00 g)46.07 g x 100% = 34.73%

52.14% + 13.13% + 34.73% = 100.0%

3.5

9

Determining the Empirical Formula from Masses of Elements

PROBLEM:

PLAN:

SOLUTION:

Elemental analysis of a sample of an ionic compound gave the following results: 2.82g of Na, 4.35g of Cl, and 7.83g of O. What are the empirical formula and name of the compound?

Once we find the relative number of moles of each element, we can divide by the lowest mol amount to find the relative mol ratios (empirical formula).

2.82g Namol Na

22.99g Na= 0.123 mol Na

4.35g Clmol Cl

35.45g Cl= 0.123 mol Cl

7.83g Omol O

16.00 O= 0.489 mol O

Na1 Cl1 O3.98 NaClO4Na1 Cl1 O3.98 NaClO4

NaClO4 is sodium perchlorate.

0.123 mol Na

0.123= 1.00

0.123= 1.000.123 mol Cl

0.123= 3.98

0.489 mol O

Two Compounds with Molecular Formula C2H6O

Property Ethanol Dimethyl Ether

M(g/mol)

Color

Melting Point

Boiling Point

Density at 200C

Use

Structural formulas and space-filling model

46.07

Colorless

-1170C

78.50C

0.789g/mL(liquid)

intoxicant in alcoholic beverages

46.07

Colorless

-138.50C

-250C

0.00195g/mL(gas)

in refrigeration

C CH

H

H

H

H

O H C OH

H

H

C HH

H

Structural Formulas vs. Molecular Formulas

10

Determination of Molecular Formula from Empirical Formula

olarMassEmpiricalMlarMassMoleculeMoMultiplier =

Empirical formula only can be determined by mass percent

3 ways of representing the reaction of H2 with O2 to form H2O

Chemical Equationsreactants products

11

Balancing Chemical Equations

1. Write the correct formula(s) for the reactants on the left side and the correct formula(s) for the product(s) on the right side of the equation.

Ethane reacts with oxygen to form carbon dioxide and water

C2H6 + O2 CO2 + H2O

2. Change the numbers in front of the formulas (coefficients) to make the number of atoms of each element the same on both sides of the equation. Do not change the subscripts.

2C2H6 NOT C4H12

Balancing Chemical Equations

3. Start by balancing those elements that appear in only one reactant and one product.

C2H6 + O2 CO2 + H2O start with C or H but not O

2 carbonon left

1 carbonon right

multiply CO2 by 2

C2H6 + O2 2CO2 + H2O

6 hydrogenon left

2 hydrogenon right multiply H2O by 3

C2H6 + O2 2CO2 + 3H2O

12

Balancing Chemical Equations

4. Balance those elements that appear in two or more reactants or products.

2 oxygenon left

4 oxygen(2x2)

C2H6 + O2 2CO2 + 3H2O

+ 3 oxygen(3x1)

multiply O2 by 72

= 7 oxygenon right

C2H6 + O2 2CO2 + 3H2O72

remove fractionmultiply both sides by 2

2C2H6 + 7O2 4CO2 + 6H2O

Balancing Chemical Equations

5. Check to make sure that you have the same number of each type of atom on both sides of the equation.

2C2H6 + 7O2 4CO2 + 6H2O

Reactants Products4 C

12 H14 O

4 C12 H14 O

4 C (2 x 2) 4 C12 H (2 x 6) 12 H (6 x 2)14 O (7 x 2) 14 O (4 x 2 + 6)

13

Chemical Equations

2 Mg + O2 2 MgO

2 atoms Mg + 1 molecule O2 makes 2 formula units MgO

2 moles Mg + 1 mole O2 makes 2 moles MgO

48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO

IS NOT2 grams Mg + 1 gram O2 makes 2 g MgO

2S (s) + 3O2(g) ->2 SO3(g)

Reactant Mixture Product Mixture

SO

14

1. Write balanced chemical equation

2. Convert quantities of known substances into moles

3. Use coefficients in balanced equation to calculate the number of moles of the sought quantity

4. Convert moles of sought quantity into desired units

Mass Changes in Chemical Reactions

Memorize these steps:

Mass react x 1 mol react x mol product x product g = massgrams react mol react mol prod.

molar massof reactant

molar massof productmol ratio from

balanced equation

15

Volumereactant

Grams reactant

Molesreactant

Molarity ofreactant

Atoms or moleculesOf reactant

* Moles product

Grams ofproduct

Volume ofproduct*

Molarity ofproduct

Atoms or moleculesOf products

moles productmoles reactant

a,b,c or dFrom balanced

equation

Density(g/mL) (mol/g)

1/Na

Volume(L)

Volume-1

(L-1) (g/mol)1/density(mL/g)

Na

Stoichiometric Relationships

a Reactant A + b Reactant B c Product C + d Product D

a Product A + b Product B c Reactant C + d Reactant Dor

© 2004 Randal Hallford

Methanol burns in air according to the equation2CH3OH + 3O2 2CO2 + 4H2O

If 209 g of methanol are used up in the combustion, what mass of water is produced?

grams CH3OH moles CH3OH moles H2O grams H2O

molar massCH3OH

coefficientschemical equation

molar massH2O

209 g CH3OH1 mol CH3OH32.0 g CH3OH

x4 mol H2O

2 mol CH3OHx

18.0 g H2O1 mol H2O

x =

235 g H2O

16

I have found the pot at the end of the rainbow!!

6 green used up6 red left over

Limiting Reagents

17

2S (s) + 3O2(g) ->2 SO3(g)

Reactant Mixture Product Mixture

SO

Excess Reagent

Limiting Reagent

Limiting Reagent Problems

1. Balance chemical equation2. Determine limiting reagent

• Do two separate calculations for the amount of product each reactant would produce ifthey were the limiting reagent

• The reactant that gives the lower number is the limiting reagent.

3. The amount of product produced is the number calculated by limiting reagent

18

Limiting Reagent Problems

4. Determination of the amount of excess reagent left over

• Calculate the amount of excess reagent (ER) used in chemical reaction

• Subtract the ER used from original amount of ER.

Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant

PROBLEM: A fuel mixture used in the early days of rocketry is composed oftwo liquids, hydrazine(N2H4) and dinitrogen tetraoxide(N2O4), which ignite on contact to form nitrogen gas and water vapor. How many grams of nitrogen gas form when 1.00x102g of N2H4and 2.00x102g of N2O4 are mixed?

PLAN: We always start with a balanced chemical equation and find the number of mols of reactants and products which have been given.

In this case one of the reactants is in molar excess and the other will limit the extent of the reaction.

mol of N2 mol of N2

divide by M

molar ratio

mass of N2H4

mol of N2H4

mass of N2O4

mol of N2O4

limiting mol N2

g N2

multiply by M

19

Calculating Amounts of Reactant and Product in Reactions Involving a Limiting Reactant

SOLUTION: N2H4(l) + N2O4(l) N2(g) + H2O(l)

1.00x102g N2H4 = 3.12mol N2H4

mol N2H4

32.05g N2H4

3.12mol N2H4 = 4.68mol N23 mol N2

2mol N2H4

2.00x102g N2O4 = 2.17mol N2O4

mol N2O4

92.02g N2O4

2.17mol N2O4 = 6.51mol N23 mol N2

mol N2O4

N2H4 is the limiting reactant because it produces less product, N2, than does N2O4.

4.68mol N2mol N2

28.02g N2 = 131g N2

2 43

Limiting Reagents

124 g of Al are reacts with 601 g of Fe2O3 :

2Al + Fe2O3 Al2O3 + 2FeCalculate the mass of Al2O3 formed.

g Al mol Al mol Fe2O3 needed g Fe2O3 needed

ORg Fe2O3 mol Fe2O3 mol Al needed g Al needed

124 g Al1 mol Al27.0 g Al

x1 mol Fe2O3

2 mol Alx

160. g Fe2O3

1 mol Fe2O3x = 367 g Fe2O3

Start with 124 g Al need 367 g Fe2O3

Have more Fe2O3 (601 g); Al is limiting reagent

20

Use limiting reagent (Al) to calculate amount of product thatcan be formed.

g Al mol Al mol Al2O3 g Al2O3

124 g Al1 mol Al

27.0 g Alx

1 mol Al2O3

2 mol Alx

102. g Al2O3

1 mol Al2O3x = 234 g Al2O3

2Al + Fe2O3 Al2O3 + 2Fe

Theoretical Yield is the amount of product that wouldresult if all the limiting reagent reacted.

Actual Yield is the amount of product actually obtainedfrom a reaction.

% Yield = Actual Yield

Theoretical Yieldx 100

21

Actual Yields < 100%

Calculating Percent Yield

PROBLEM: Silicon carbide (SiC) is an important ceramic material that is made by allowing sand(silicon dioxide, SiO2) to react with powdered carbon at high temperature. Carbon monoxide is also formed. When 100.0kg of sand are processed, 51.4kg of SiCare recovered. What is the percent yield of SiC in this process?

PLAN:

write balanced equation

find mol reactant & product

find g product predicted

percent yield

actual yield/theoretical yield x 100

SOLUTION:

SiO2(s) + 3C(s) SiC(s) + 2CO(g)

100.0kg SiO2mol SiO2

60.09g SiO2

103g SiO2

kg SiO2

= 1664 mol SiO2

mol SiO2 = mol SiC = 1664

1664mol SiC40.10g SiC

mol SiC

kg

103g= 66.73kg

x100 =77.0%51.4kg

66.73kg

22

Calculating the Molarity of a Solution

PROBLEM: Hydrobromic acid(HBr) is a solution of hydrogen bromide gas in water. Calculate the molarity of hydrobromic acid solution if 455mL contains 1.80mol of hydrogen bromide.

mol of HBr

concentration(mol/mL) HBr

molarity(mol/L) HBr

1.80mol HBr

455 mL soln

1000mL

1 L

divide by volume

103mL = 1L

= 3.96M

Calculating Mass of Solute in a Given Volume of Solution

How many grams of solute are in 1.75L of 0.460M sodium monohydrogen phosphate?

volume of soln

moles of solute

grams of solute

multiply by M

multiply by M

SOLUTION:

Molarity is the number of moles of solute per liter of solution.Knowing the molarity and volume leaves us to find the # moles and then the # of grams of solute. The formula for the solute is Na2HPO4.

1.75L 0.460moles

1 L

0.805mol Na2HPO4 141.96g Na2HPO4

mol Na2HPO4

= 0.805mol Na2HPO4

= 114g Na2HPO4

23

Preparing a Dilute Solution from a Concentrated Solution

“Isotonic saline” is a 0.15M aqueous solution of NaCl that simulates the total concentration of ions found in many cellularfluids. Its uses range from a cleaning rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80L of isotomic saline from a 6.0M stock solution?

The number of moles of solute does not change during the dilution but the volume does. The new volume will be the sum of the two volumes, that is, the total final volume.

volume of dilute soln

moles of NaCl in dilute soln = mol NaCl in concentrated soln

L of concentrated soln

multiply by M of dilute solution

divide by M of concentrated soln

MdilxVdil = #mol solute = MconcxVconc

SOLUTION:

0.80L soln

= 0.020L soln

0.15mol NaCl

L soln

0.12mol NaCl L solnconc

6mol

= 0.12mol NaCl

24

Calculating Amounts of Reactants and Products for a Reaction in Solution

A common antacid contains magnesium hydroxide, which reacts with stomach acid to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10M HCl to simulate the acid concentration in the stomach. How many liters of “stomach acid” react with a tablet containing 0.10g of magnesium hydroxide?

Write a balanced equation for the reaction; find the grams of Mg(OH)2; determine the mol ratio of reactants and products; use mols to convert to molarity.

mass Mg(OH)2

divide by M

mol Mg(OH)2

mol ratio

mol HCl

L HCl

divide by M

25

Calculating Amounts of Reactants and Products for a Reaction in Solution

Mg(OH)2(s) + 2HCl(aq) MgCl2(aq) + 2H2O(l)

0.10g Mg(OH)2mol Mg(OH)2

58.33g Mg(OH)2

2 mol HCl

1 mol Mg(OH)2

1L

0.10mol HCl

= 3.4x10-2 L HCl

Solving Limiting-Reactant Problems for Reactions in Solution

0.050L of 0.010M mercury(II) nitrate reacts with 0.020L of 0.10Msodium sulfide. How many grams of mercury(II) sulfide form?

As usual, write a balanced chemical reaction. Since this is a problem concerning a limiting reactant, we proceed as before and find the amount of product which would be made from each reactant. We then chose the reactant which gives the lesser amount of product.

26

SOLUTION:

Solving Limiting-Reactant Problems for Reactions in Solution

L of Na2S

mol Na2S

mol HgS

multiply by M

mol ratio

L of Hg(NO3)2

mol Hg(NO3)2

mol HgS

multiply by M

mol ratio

Hg(NO3)2(aq) + Na2S(aq) HgS(s) + 2NaNO3(aq)

0.050L Hg(NO3)2

x 0.010 mol/L

x 1mol HgS

1mol Hg(NO3)2

0.020L Hg(NO3)2

x 0. 10 mol/L

x 1mol HgS

1mol Na2S

= 5.0x10-4 mol HgS = 2.0x10-3 mol HgS

Hg(NO3)2 is the limiting reagent.

5.0x10-4 mol HgS232.7g HgS

232.7mol HgS= 0.12g HgS

Preparation of Stock Solutions

27

Volumereactant

Grams reactant

Molesreactant

Molarity ofreactant

Atoms or moleculesOf reactant

* Moles product

Grams ofproduct

Volume ofproduct*

Molarity ofproduct

Atoms or moleculesOf products

moles productmoles reactant

a,b,c or dFrom balanced

equation

Density(g/mL) (mol/g)

1/Na

Volume(L)

Volume-1

(L-1) (g/mol)1/density(mL/g)

Na

Stoichiometric Relationships

a Reactant A + b Reactant B c Product C + d Product D

a Product A + b Product B c Reactant C + d Reactant Dor

© 2004 Randal Hallford