Mass Relationships inChemical Reactions
Chapter 3
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2
By definition:
1 atom 12C “weighs” 12 amu
On this scale
1H = 1.008 amu
16O = 16.00 amu
Atomic mass is the mass of an atom inatomic mass units (amu)
Micro World
atoms & molecules
Macro World
grams
3
The average atomic mass is the weighted
average of all of the naturally occurring
isotopes of the element.
4
Naturally occurring lithium is:
7.42% 6Li (6.015 amu)
92.58% 7Li (7.016 amu)
7.42 x 6.015 + 92.58 x 7.016
100
= 6.941 amu
Average atomic mass of lithium:
Calcium has 6 isotopes. Their masses & percentageabundances are given in the table below. Calculate the averageatomic mass or atomic weight of Ca.
Isotope
Isotope mass(amu)
PercentageAbundance
Ca-40
39.96259
96.941
Ca-42
41.95862
0.647
Ca-43
42.95877
0.135
Ca-44
43.95548
2.086
Ca-46
45.95369
0.004
Ca-48
47.95253
0.187
{(39.96259 x 96.941) + (41.95862 x 0.647) + (0.135 x42.95877) + (2.086 x 43.95548) + (.004 x 45.95369) +(47.95253 x 0.187)}amu/100
=4007.8amu/100
= 40.078 amu
6
Average atomic mass (6.941)
7
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)
Dozen = 12
Pair = 2
The Mole (mol): A unit to count numbers of particles
Molar mass is the mass of 1 mole of in grams
eggs
shoes
marbles
atoms
1 12C atom = 12.00 amu
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 mole 12C atoms = 12.00 g 12C
Average mass of 1 Li atom = 6.941 amu
Molar mass of Li = mass of 1 mole lithium atoms =6.941 g
For any element,
numerical value of atomic mass (amu) = numerical value of molarmass (grams)
3.2
9
One Mole of:
C
S
Cu
Fe
Hg
10
1 amu = 1.66 x 10-24 g or 1 g = 6.022 x 1023 amu
1 12C atom
12.00 amu
x
12.00 g
6.022 x 1023 12C atoms
=
1.66 x 10-24 g
1 amu
M
= molar mass in g/mol
NA = Avogadro’s number
1. 15.0 g Mercury has
a.How many moles of mercury? 0.0748 mol
b.How many atoms of Mercury? 4.50 x 1022 atoms
2. 5.16 x 1022 atoms of Fe is
a.How many moles of Fe? 0.0857 mol Fe
b.What is the corresponding mass? 4.79 g Fe
3. 15.0 g Zinc has
a.How many moles of Zinc?
b.How many atoms of Zinc?
4. 5.16 x 1022 atoms of Cu is
a.How many moles of Cu?
b.What is the corresponding mass?
12
x
6.022 x 1023 atoms K
1 mol K
=
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 K
39.10 g K
x
8.49 x 1021 atoms K
Molecular mass (or molecular weight) is the sum of
the atomic masses (in amu) in a molecule.
SO2
1S
32.07 amu
2O
+ 2 x 16.00 amu
SO2
64.07 amu
For any molecule
The numerical value of molecular mass in amu = thenumerical value of molar mass in grams
1 molecule of SO2 weighs 64.07 amu
1 mole of SO2 weighs 64.07 g
1 mole of SO2 =6.022 x 1023 molecules of SO2
3.3
You cannot go directly from grams of a compound to moles of ACONSTITUENT ATOM
You have to first calculate the moles of the compound and thenfrom that calculate the moles of a constituent atom:
Grams of compound Moles of Compound Moles of aconstituent atom
The molecular formula not only tells us how many atoms of eachtype are present in one molecule of a compound but also howmany MOLES of atoms of each type are present in 1MOLE OFTHE COMPOUND.
For example, the molecular formula of sugar, C12H22O11, tells usthat 1 mole of sugar has 12 moles of C atoms, 22 moles of Hatoms & 11 moles of O atoms besides the fact that 12 atoms of C,22 H atoms & 11 O atoms are present in 1 sugar molecule.
1.Consider a sample of 15.0 g Borazine (B3N3H6)
a.How many moles of borazine are present? 0.186 mol
b.How many molecules of borazine are present? 1.12 x 1023molecules
c.How many moles of N atoms are present? 0.558 mol N
d.How many N atoms are present? 3.36 x 1023 N atoms
2. A sample of ethanol, C2H5OH has 6.00 x 1023 H atoms.
a.How many ethanol molecules are present? 1.00 x 1023molecules
b.How many moles of ethanol are present?0.166 mol
c.What is the mass of the sample? 7.65 g
Complete the following conversions;
1.Consider 0.00500 g of H2O
Calculate the following:
a.The moles of H2O: 2.77 × 10-4
b.The molecules of H2O: 1.67 × 1020
c.The moles of H atoms: 5.55 × 10-4
d.The number of H atoms present: 3.34 × 1020
2. Na atoms in 3.50 mole NaOH
3. grams mass of 2.50 mole O2
4. carbon atoms in 10.5 g C6H12O6
5. A sample of benzene, C6H6 has 3.00 x 1022 H atoms.
a.How many benzene molecules are present? 5.00 × 1021
b.How many moles of benzene are present? 8.30 × 10-3
c.What is the mass of the sample? 0.649 g
Do You Understand Molecular Mass?
How many H atoms are in 72.5 g of C3H8O ?
moles of C3H8O = 72.5 g / 60.095 g/mol = 1.21 mol
1 mol of H atoms is 6.022 x 1023 H atoms
5.82 x 1024 H atoms
3.3
1 mol C3H8O molecules contains 8 mol H atoms
72.5 g C3H8O
1 mol C3H8O
60 g C3H8O
x
8 mol H atoms
1 mol C3H8O
x
6.022 x 1023 H atoms
1 mol H atoms
x
=
Steps: 1. Convert grams of C3H8O to moles of C3H8O.
2. Convert moles of C3H8O to moles of H atoms.
3. Convert moles of H atoms to number of H atoms.
18
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
1Na
22.99 amu
1Cl
+ 35.45 amu
NaCl
58.44 amu
For any ionic compound
formula mass (amu) = molar mass (grams)
1 formula unit NaCl = 58.44 amu
1 mole NaCl = 58.44 g NaCl
NaCl
Do You Understand Formula Mass?
What is the formula mass of Ca3(PO4)2 ?310.18 amu
3.3
1 formula unit of Ca3(PO4)2
3 Ca
3 x 40.08 g/mol
2 P
2 x 30.97 g/mol
8 O
+ 8 x 16.00 g/mol
310.18 g/mol = molar mass
Units of grams per mole are the most practical
for chemical calculations!
1.You have a 10.0 g of Ca3(PO4)2. (Molar mass = 310.18 g/mol)
a.How many moles are present? 0.0322mol
b.How many moles of Ca+2 ions are present? 0.0966 mol
c. How many phosphate ions are present? =0.0644 mol = 3.88 x1022 ions.
2. You have a 10.0 g of (NH4)3PO4 (molar mass = 149.12g/mol)
a.How many moles are present? 0.0671 mol
b.How many moles of phosphate ions are present? 0.0671 mol
c. How many ammonium ions are present? 1.21 × 1023
21
Heavy
Heavy
Light
Light
Mass Spectrometer
Mass Spectrum of Ne
22
Percent composition of an element in a compound =
n x molar mass of element
molar mass of compound
x 100%
n is the number of moles of the element in 1 moleof 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%
Calculate percentage composition of
a.Borazine (B3N3H6; molar mass = 80.52g/mol) 40.28%B,52.20% N, 7.52% H
b.Ca3(PO4)2 (Molar mass = 310.18 g/mol)
38.76 % Ca, 19.97 % P, 41.27 % O
24
Percent Composition and Empirical Formulas
Determine the empirical formula of acompound that has the followingpercent composition by mass:K 24.75, Mn 34.77, O 40.51 percent.
nK = 24.75 g K x
= 0.6330 mol K
1 mol K
39.10 g K
nMn = 34.77 g Mn x
= 0.6329 mol Mn
1 mol Mn
54.94 g Mn
nO = 40.51 g O x
= 2.532 mol O
1 mol O
16.00 g O
To begin, assume for simplicity that you have 100 g of compound!
25
Percent Composition and Empirical Formulas
K :
~
~
1.0
0.6330
0.6329
Mn :
0.6329
0.6329
= 1.0
O :
~
~
4.0
2.532
0.6329
nK = 0.6330, nMn = 0.6329, nO = 2.532
KMnO4
Determine the empirical formula of the compound whosepercentage composition is
62.1% C, 5.21% H, 12.1% N, and 20.7% O
Step 1: assume 100 g: 62.1 g C, 5.21 g H, 12.1 g N, 20.7 g O
Step 2: (62.1/12.01)mol C = 5.17 mol C; (5.21/1.01) mol H = 5.16mol H; (12.1/14.01)mol N = 0.864 mol N; (20.7/16.00) mol O =1.29 mol O
Step 3: C:H:N:O = 5.17 : 5.16 : 0.864 : 1.29
= (5.17/0.864) : (5.16/0.864) : 1 : (1.29/0.864) =
= 5.98 : 5.97 : 1: 1.50
= 6 : 6 : 1 : 1.50 (DO NOT ROUND OFF NUMBERS WITH .5;MULTIPLY BY 2 BEFORE ROUNDING OFF)
= 12 : 12 : 2 : 3; C12H12N2O3
A sample of an organic compound was found to contain 9.01g C,1.01g H, 12.0 g O. Calculate the empirical formula of thiscompound.
9.01g C = (9.01/12.01) mol C = 0.750 mol C
1.01 g H = (1.01/1.01)mol H = 1.00 mol H
12.0 g O = (12.0/16.00) mol O = 0.750 mol O
C : H : O = 0.750 : 1.00 : 0.750
= (0.750/0.750) : (1.00/0.750) : (0.750/0.750)
= 1.00 : 1.33 : 1.00 = 1 : 1.33 : 1
DO NOT ROUND OFF ANY NO. WITH 0.3 OR 0.33 OR 0.333 OR0.5 OR ANY THAT CAN BE ROUNDED OFF TO 0.5.
MULTIPLY THOSE WITH .3, .33 etc. WITH 3 (REMEMBER TOMULTIPLY ALL THE NUMBERS IN THE RATIO)
MULTIPLY THOSE WITH .5 WITH 2 (REMEMBER TO MULTIPLYALL THE NUMBERS IN THE RATIO)
C : H : O = 1 : 1.33 : 1 = 3 x (1 : 1.33 : 1) = 3 : 3.99 : 3 = 3: 4 : 3
Empirical formula = C3H4O3
28
g CO2
mol CO2
mol C
g C
g H2O
mol H2O
mol H
g H
g of O = g of sample – (g of C + g of H)
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
6.0 g C = 0.5 mol C
1.5 g H = 1.5 mol H
4.0 g O = 0.25 mol O
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
Determination of Molecular Formulas from Empirical Formulas
1.Determine the ratio of the molecular weight to empiricalformula weight (the molecular weight HAS to be provided):molecular weight/empirical formula weight
2.Multiply all subscripts in the empirical formula by the aboveratio to obtain the molecular formula.
1.A hydrocarbon has 12.01g C, 3.03g H. Its molecular weight =30.08 amu. Calculate the molecular formula of the compound.
12.01 g C, 3.03 g H = (12.01/12.01) mol C, (3.03/1.01) mol H =1.000 mol C, 3.00 mol H = 1 mol C, 3 mol H
Empirical formula =CH3
Empirical formula weight = (12.01) +(3 x 1.01) amu = 15.04 amu
Molecular weight/empirical formula weight = 30.08/15.04 = 2
Therefore molecular formula : C(1 x 2)H(3 x 2) = C2H6
30
3 ways of representing the reaction of H2 with O2 to form H2O
A process in which one or more substances is changed into oneor more new substances is a chemical reaction
A chemical equation uses chemical symbols to show whathappens during a chemical reaction
reactants
products
Balance equations ONLY by putting numbers IN FRONT OFformulas.
Such numbers are called COEFFICIENTS
NEVER CHANGE SUBSCRIPTS – doing so will change theidentity of reactants, products and your equation will no longerrepresent the reaction that you are considering
32
How to “Read” 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
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
1. Write the correct formula(s) for the reactants on the left sideand the correct formula(s) for the product(s) on the right side ofthe equation.
2. Change the numbers in front of the formulas (coefficients) tomake the number of atoms of each element the same on bothsides of the equation. Do not change the subscripts.
3. Start by balancing those elements that appear in only onereactant and one product.
4. Balance those elements that appear in two or more reactants orproducts.
5. Coefficients should be whole numbers
6. Check to make sure that you have the same number of eachtype of atom on both sides of the equation.
Balancing Chemical Equations
34
Balancing Chemical Equations
1.Write the correct formula(s) for the reactants onthe left side and the correct formula(s) for theproduct(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 ofeach element the same on both sides of theequation. Do not change the subscripts.
2C2H6
NOT
C4H12
35
Balancing Chemical Equations
3.Start by balancing those elements that appear inonly one reactant and one product.
C2H6 + O2
CO2 + H2O
start with C or H but not O
2 carbon
on left
1 carbon
on right
multiply CO2 by 2
C2H6 + O2
2CO2 + H2O
6 hydrogen
on left
2 hydrogen
on right
multiply H2O by 3
C2H6 + O2
2CO2 + 3H2O
36
Balancing Chemical Equations
4.Balance those elements that appear in two ormore reactants or products.
2 oxygen
on left
4 oxygen
(2x2)
C2H6 + O2
2CO2 + 3H2O
+ 3 oxygen
(3x1)
multiply O2 by
7
2
= 7 oxygen
on right
C2H6 + O2
2CO2 + 3H2O
7
2
remove fraction
multiply both sides by 2
2C2H6 + 7O2
4CO2 + 6H2O
37
Balancing Chemical Equations
5.Check to make sure that you have the samenumber of each type of atom on both sides of theequation.
2C2H6 + 7O2
4CO2 + 6H2O
Reactants
Products
4 C
12 H
14 O
4 C
12 H
14 O
4 C (2 x 2)
4 C
12 H (2 x 6)
12 H (6 x 2)
14 O (7 x 2)
14 O (4 x 2 + 6)
38
1.Write balanced chemical equation
2.Convert quantities of known substances into moles
3.Use coefficients in balanced equation to calculate thenumber of moles of the sought quantity
4.Convert moles of sought quantity into desired units
Amounts of Reactants and Products
39
Methanol burns in air according to the equation
2CH3OH + 3O2 2CO2 + 4H2O
If 209 g of methanol are used up in the combustion, whatmass of water is produced?
grams CH3OH
moles CH3OH
moles H2O
grams H2O
molar mass
CH3OH
coefficients
chemical equation
molar mass
H2O
209 g CH3OH
1 mol CH3OH
32.0 g CH3OH
x
4 mol H2O
2 mol CH3OH
x
18.0 g H2O
1 mol H2O
x
=
235 g H2O
40
Limiting Reagent:
2NO + O2 2NO2
NO is the limiting reagent
O2 is the excess reagent
Reactant used up first inthe reaction.
Limiting Reactant: Is that reactant that is first used up. When it isused up, the reaction stops because all reactants have to bepresent for the reaction to occur.
Excess Reactant: A reactant that is remaining even after thereaction has stopped
Theoretical Yield: The maximum mass of a product that isCALCULATED to form when all of the LIMITING REACTANT isconsumed. (The limiting reactant predicts the smallest yield)
Actual Yield: An EXPERIMENTAL value: the mass of a productACTUALLY formed when the experiment is performed
Actual yield ≤ Theoretical Yield
Percentage Yield = (Actual/Theoretical) x 100; ≤ 100%
Do You Understand Limiting Reactants?
In a reaction, 124 g of Al are reacted with 601 g of Fe2O3.
2 Al + Fe2O3 Al2O3 + 2 Fe
a.Calculate the thoeretical mass of Al2O3 that can form in grams.
b.What mass of excess reactant remains?
c.If the actual yield of Al2O3 is 220g, what is the percent yield
3.9
1.Balanced reaction: Done.
2.Moles of “given” reactants.
Moles of Al = 124 g / 26.9815 g/mol = 4.60 mol
Moles of Fe2O3 = 601 g / 159.6882 g/mol = 3.76 mol
Solution to a:
Fabulous Four Steps
3. Moles of “desired” product, Al2O3.
3.9
Moles of Al2O3 = 4.60 mol Al X 1 mol Al2O3 = 2.30 mol Al2O3
based on Al 1 2 mol Al
2 Al + Fe2O3 Al2O3 + 2 Fe
Moles of Al2O3 = 3.76 mol Fe2O3 X 1 mol Al2O3 = 3.76 mole Al2O3based on Fe2O3 1 1 mol Fe2O3
Keep the smaller answer! Al is the limiting reactant.
4. Grams of Al2O3.
Grams of Al2O3 = 2.30 mol X 101.9612 g/mol = 235 g
In a certain experiment 2.50 g of NH3 and 2.85 g of O2 aretaken. The following reaction occurs:
4NH3 + 5 O2 4NO + 6 H2O
1. Determine the theoretical yield of NO? 2.14 g not 4.40g
Limiting reactant = O2, excess reactant: NH3
2. How much excess reactant remains
after the liming reactant is completely consumed? 1.29 g
3. If the actual yield is 2.00g of NO, what is the percentageyield? 93.5%
If only 1 reactant amount is given, it is implied that that reactantis the limiting reactant. There is no need for any furtheridentification.
If only 1 reactant amount is given, it is implied that that reactant isthe limiting reactant. There is no need for any further identification.
Consider the reaction between benzene, C6H6 and bromine, Br2,forming bromobenzene, C6H5Br:
C6H6 + Br2 , C6H5Br + HBr
1.If 30.0g of C6H6 is used. What is theoretical yield of C6H5Br?60.30g
2. If the actual yield of C6H5Br is 56.7 g, what is the % yield? 94.0%
46
Theoretical Yield is the amount of product that would
result if all the limiting reagent reacted.
Actual Yield is the amount of product actually obtained
from a reaction.
% Yield =
Actual Yield
Theoretical Yield
x 100%
Reaction Yield
47
Chemistry In Action: Chemical Fertilizers
Plants need: N, P, K, Ca, S, & Mg
3H2 (g) + N2 (g) 2NH3 (g)
NH3 (aq) + HNO3 (aq) NH4NO3 (aq)
2Ca5(PO4)3F (s) + 7H2SO4 (aq)
3Ca(H2PO4)2 (aq) + 7CaSO4 (aq) + 2HF (g)
fluorapatite