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)
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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