The Mole Atoms are so small, it is impossible to count them by
the dozens, thousands, or even millions. To count atoms, we use the
concept of the mole 1 mole of atoms = That is, 1 mole of atoms =
__________ atoms The mole is the SI unit for amount of substance.
602,213,673,600,000,000,000,000 atoms 6.02 x 10 23
Slide 4
If I had a mole of dollars, I could give every person on Earth
How Big is a Mole? about the size of a chipmunk, weighing about 5
oz. (140 g), and having a length of about 7 inches (18 cm). I
Meant, How Big is 6.022 x 10 23 ? BIG. 6.022 x 10 23 marbles would
cover the entire Earth (including the oceans) to a height of 2
miles. There are ~ 6,880,900,000 people on Earth. $87.5 trillion or
$87.5 x 10 12
Slide 5
Welcome to Mole Island 1 mole = 22.4 L @ STP 1 mol = molar mass
1 mol = 6.02 x 10 23 particles
Slide 6
Mole Island Diagram (gases) Mass Particles Volume Mole Mass
Volume Particles Substance A Substance B 1 mole = molar mass (g)
Use coefficients from balanced chemical equation 1 mole = 22.4 L @
STP 1 mole = 6.02 x 10 23 particles (atoms or molecules) Mole 1
mole = molar mass (g) 1 mole = 22.4 L @ STP 1 mole = 6.02 x 10 23
particles (atoms or molecules)
Slide 7
Mole Island Diagram Mass Particles Volume Mole Mass Volume
Particles Substance A Substance B 1 mole = molar mass (g) Use
coefficients from balanced chemical equation 1 mole = 22.4 L @ STP
1 mole = 6.022 x 10 23 particles (atoms or molecules) (gases) Mole
1 mole = molar mass (g) 1 mole = 22.4 L @ STP 1 mole = 6.022 x 10
23 particles (atoms or molecules)
Slide 8
Mole Island Diagram Substance A Substance B
Slide 9
White Board Overlay Mass Particles Volume Mole Mass Volume
Particles Known Unknown Substance A Substance B
Slide 10
2 __TiO 2 + __Cl 2 + __C __TiCl 4 + __CO 2 + __CO 1. How many
mol chlorine will react with 4.55 mol carbon? 3 mol C 4 mol Cl 2
4.55 mol C = 6.07 mol Cl 2 2. What mass titanium (IV) oxide will
react with 4.55 mol carbon? = 242 g TiO 2 43221 CCl 2 CTiO 2 3 mol
C 2 mol TiO 2 4.55 mol C 1 mol TiO 2 79.9 g TiO 2 Stoichiometry
Practice Problems
Slide 11
3. How many molecules titanium (IV) chloride can be made from
115 g titanium (IV) oxide? TiCl 4 TiO 2 () = 8.66 x 10 23 mc TiCl 4
115 g TiO 2 1 mol TiO 2 79.9 g TiO 2 () 2 mol TiO 2 2 mol TiCl 4 ()
1 mol TiCl 4 6.02 x 10 23 mc TiCl 4 Island Diagram helpful
reminders: 1 mol coeff. 1 mol 1. Use coefficients from the equation
only when crossing the middle bridge. The other six bridges always
have 1 mol = as a part of the conversion. 2. The middle bridge
conversion factor is the only one that has two different substances
in it. The conversion factors for the other six bridges have the
same substance in both the numerator and denominator. 3. The units
on the islands at each end of the bridge being crossed must appear
in the conversion factor for that bridge
Slide 12
How many molecules titanium (IV) chloride can be made from 115
g titanium (IV) oxide? TiCl 4 TiO 2 () = 8.7 x 10 23 mc TiCl 4 115
g TiO 2 1 mol TiO 2 79.9 g TiO 2 () 2 mol TiO 2 2 mol TiCl 4 () 1
mol TiCl 4 6.02 x 10 23 mc TiCl 4 Island Diagram helpful reminders:
1. Use coefficients from the equation only when crossing the middle
bridge. The other six bridges always have 1 mol = as a part of the
conversion. 1 mol coeff. 1 mol
Slide 13
How many molecules titanium (IV) chloride can be made from 115
g titanium (IV) oxide? TiCl 4 TiO 2 () = 8.7 x 10 23 mc TiCl 4 115
g TiO 2 1 mol TiO 2 79.9 g TiO 2 () 2 mol TiO 2 2 mol TiCl 4 () 1
mol TiCl 4 6.02 x 10 23 mc TiCl 4 Island Diagram helpful reminders:
2. The middle bridge conversion factor is the only one that has two
different substances in it. The conversion factors for the other
six bridges have the same substance in both the numerator and
denominator. 1 mol coeff. 1 mol
Slide 14
3. The units on the islands at each end of the bridge being
crossed must appear in the conversion factor for that bridge How
many molecules titanium (IV) chloride can be made from 115 g
titanium (IV) oxide? TiCl 4 TiO 2 () = 8.7 x 10 23 mc TiCl 4 115 g
TiO 2 1 mol TiO 2 79.9 g TiO 2 () 2 mol TiO 2 2 mol TiCl 4 () 1 mol
TiCl 4 6.02 x 10 23 mc TiCl 4 Island Diagram helpful reminders: 1
mol coeff. 1 mol
Slide 15
2 Ir + Ni 3 P 2 3 Ni + 2 IrP 1. If 5.33 x 10 28 mcules nickel
(II) phosphide react w/excess iridium, what mass iridium (III)
phosphide is produced? Ni 3 P 2 IrP = 3.95 x 10 7 g IrP 1 mol IrP
223.2 g IrP 1 mol Ni 3 P 2 2 mol IrP 1 mol Ni 3 P 2 6.02 x 10 23 mc
Ni 3 P 2 5.33 x 10 28 mc Ni 3 P 2 2. How many grams iridium will
react with 465 grams nickel (II) phosphide? = 751 g Ir 1 mol Ir
192.2 g Ir 1 mol Ni 3 P 2 2 mol Ir 1 mol Ni 3 P 2 238.1 g Ni 3 P 2
465 g Ni 3 P 2 Ni 3 P 2 Ir
Slide 16
2 mol Ir 3. How many moles of nickel are produced if 8.7 x 10
25 atoms of iridium are consumed? IrNi = 220 mol Ni 3 mol Ni 1 mol
Ir 6.02 x 10 23 at. Ir 8.7 x 10 25 at. Ir 2 Ir + Ni 3 P 2 3 Ni + 2
IrP iridium (Ir)nickel (Ni)
Slide 17
Zn 4. What volume hydrogen gas is liberated (at STP) if 50 g
zinc react w/excess hydrochloric acid (HCl)? __ Zn + __ HCl __ H 2
+ __ ZnCl 2 1211 H2H2 = 20 L H 2 1 mol H 2 22.4 L H 2 1 mol Zn 1
mol H 2 1 mol Zn 65.4 g Zn 50 g Zn 50 gexcess? L
Slide 18
5. At STP, how many mcules oxygen react with 632 dm 3 butane (C
4 H 10 )? C 4 H 10 O2O2 = 1.10 x 10 26 mc O 2 1 mol O 2 2 mol C 4 H
10 13 mol O 2 1 mol C 4 H 10 632 dm 3 C 4 H 10 22.4 dm 3 C 4 H 10
__ C 4 H 10 + __ O 2 __ CO 2 + __ H 2 O 145 2810 13 6.02 x 10 23 mc
O 2 Suppose the question had been,How many ATOMS of oxygen 1.10 x
10 26 mc O 2 1 mc O 2 2 atoms O = 2.20 x 10 26 at. O
Slide 19
1 mol CH 4 A balanced eq. gives the ratios of moles-to-moles
Energy and Stoichiometry CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g)
+ 891 kJ 1. How many kJ of energy are released when 54 g methane
are burned? AND moles-to-energy. = 3.0 x 10 3 kJ 1 mol CH 4 891 kJ
54 g CH 4 16.04 g CH 4 EE CH 4
Slide 20
10,540 kJ 1 mol H 2 O 3. What mass of water is made if 10,540
kJ are released? CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O(g) + 891
kJ 2. At STP, what volume oxygen is consumed in producing 5430 kJ
of energy? 2 mol O 2 = 273 L O 2 1 mol O 2 22.4 L O 2 5430 kJ 891
kJ E 2 mol H 2 O = 426.2 g H 2 O 18.02 g H 2 O 891 kJ E O2O2 EE
H2OH2O
Slide 21
Limiting Reactants A balanced equation for making a Big Mac
might be: 3 B + 2 M + EE BM 30 B and excess EE 30 M excess M and
excess EE 30 B excess B and excess EE 30 M one can make andWith 15
BM 10 BM 10 BM
Slide 22
50 P A balanced equation for making a big wheel might be: one
can make and With 50 S + excess of all other reactants excess of
all other reactants 50 S excess of all other reactants 50 P 25 Bw 3
W + 2 P + S + H + F Bw 50 Bw 25 Bw
Slide 23
Solid aluminum reacts w/chlorine gas to yield solid aluminum
chloride. 2 Al(s) + 3 Cl 2 (g) 2 AlCl 3 (s) If 125 g aluminum react
w/excess chlorine, how many g aluminum chloride are made? = 618 g
AlCl 3 2 mol Al 1 mol AlCl 3 1 mol Al 26.98 g Al 125 g Al AlAlCl 3
133.34 g AlCl 3 2 mol AlCl 3 If 125 g chlorine react w/excess
aluminum, how many g aluminum chloride are made? = 157 g AlCl 3 3
mol Cl 2 1 mol AlCl 3 1 mol Cl 2 70.91 g Cl 2 125 g Cl 2 Cl 2 AlCl
3 133.34 g AlCl 3 2 mol AlCl 3
Slide 24
2 Al(s) + 3 Cl 2 (g) 2 AlCl 3 (s) If 125 g aluminum react w/125
g chlorine, how many g aluminum chloride are made? 157 g AlCl 3
(Were out of Cl 2 ) limiting reactant (LR): the reactant that runs
out first Any reactant you dont run out of is an excess reactant
(ER). amount of product is limited by the LR when pouring liquid
into a funnel, it doesnt matter how much you pour into the top, the
bottom of the funnel limits how much you get out. Think of this
analogy
Slide 25
How to Find the Limiting Reactant For the generic reaction R A
+ R B P, assume that the amounts of R A and R B are given. Should
you use R A or R B in your calculations? 1. Using Stoichiometry,
calculate the amount of product possible from both R A and R B (2
separate calculations) 2. Whichever reactant produces the smaller
amount of product is the LR 3. The smaller amount of product is the
maximum amount produced
Slide 26
For the Al / Cl 2 / AlCl 3 example: 2 Al(s) + 3 Cl 2 (g) 2 AlCl
3 (s) If 125 g aluminum react w/excess chlorine, how many g
aluminum chloride are made? = 618 g AlCl 3 2 mol Al 1 mol AlCl 3 1
mol Al 26.98 g Al 125 g Al AlAlCl 3 133.34 g AlCl 3 2 mol AlCl 3 If
125 g chlorine react w/excess aluminum, how many g aluminum
chloride are made? = 157 g AlCl 3 3 mol Cl 2 1 mol AlCl 3 1 mol Cl
2 70.91 g Cl 2 125 g Cl 2 Cl 2 AlCl 3 133.34 g AlCl 3 2 mol AlCl 3
LR ER
Slide 27
2 Fe(s) + 3 Cl 2 (g) 2 FeCl 3 (s) 223 g Fe 179 L Cl 2 Which is
the limiting reactant: Fe or Cl 2 ? 1 mol Fe 55.85 g Fe 223 g Fe 1
mol Cl 2 22.4 L Cl 2 179 L Cl 2 How many g FeCl 3 are produced? =
648 g FeCl 3 2 mol FeCl 3 2 mol Fe 162.21 g FeCl 3 1 mol FeCl 3 2
mol FeCl 3 3 mol Cl 2 162.21 g FeCl 3 1 mol FeCl 3 = 864 g FeCl 3
648 g FeCl 3 *Remember that the LR limits how much product can be
made! Limiting Reactant Practice
Slide 28
2 H 2 (g) + O 2 (g) 2 H 2 O(g) 13 g H 2 80 g O 2 Which is the
limiting reactant: H 2 or O 2 ? 1 mol H 2 2.02 g H 2 13 g H 2 1 mol
O 2 32.00 g O 2 80 g O 2 How many g H 2 O are produced? = 120 g H 2
O 2 mol H 2 O 2 mol H 2 18.02 g H 2 O 1 mol H 2 O 2 mol H 2 O 1 mol
O 2 18.02 g H 2 O 1 mol H 2 O = 90 g H 2 O 90 g H 2 O *Notice that
the LR doesnt always have the smaller amount (13 v. 80)
Slide 29
How many g O 2 are left over? How many g H 2 are left over?
zero; O 2 is the LR and therefore is all used up We know how much H
2 we HAD (i.e. 13 g) To find how much is left over, we first need
to figure out how much was USED in the reaction. H2H2 O2O2 HAD 13
g, USED 10 g Start with the LR and relate to the other 1 mol O 2
32.00 g O 2 80 g O 2 2 mol H 2 1 mol O 2 2.02 g H 2 1 mol H 2 10 g
H 2 USED = 3 g H 2 left over
Slide 30
181 g Fe 96.5 L Br 2 Which is the limiting reactant: Fe or Br 2
? 1 mol Fe 55.85 g Fe 181 g Fe 1 mol Br 2 22.4 L Br 2 96.5 L Br 2
How many g FeBr 3 are produced? = 958 g FeBr 3 2 mol FeBr 3 2 mol
Fe 295.55 g FeBr 3 1 mol FeBr 3 2 mol FeBr 3 3 mol Br 2 295.55 g
FeBr 3 1 mol FeBr 3 = 849 g FeBr 3 849 g FeBr 3 2 Fe(s) + 3 Br 2
(g) 2 FeBr 3 (s)
Slide 31
How many g of the ER are left over? FeBr 2 181 g Fe 96.5 L Br 2
2 Fe(s) + 3 Br 2 (g) 2 FeBr 3 (s) HAD 181 g, USED 160.4 g 1 mol Br
2 22.4 L Br 2 96.5 L Br 2 2 mol Fe 3 mol Br 2 55.85 g Fe 1 mol Fe
160. g Fe USED = 21 g Fe left over
Slide 32
101.96 g Al 2 O 3 Percent Yield molten sodium solid aluminum
oxide molten aluminum solid sodium oxide Find mass of aluminum
produced if you start w /575 g sodium and 357 g aluminum oxide. +
Al 2 O 3 (s) Al(l) 6 Na(l) + Na 2 O(s) 123 Na + O 2 Al 3+ O 2 Al 1
mol Na 22.99 g Na 575 g Na 1 mol Al 2 O 3 357 g Al 2 O 3 = 225 g Al
2 mol Al 6 mol Na 26.98 g Al 1 mol Al 2 mol Al 1 mol Al 2 O 3 26.98
g Al 1 mol Al = 189 g Al189 g Al
Slide 33
Now suppose that we perform this reaction and get only 172
grams of aluminum. Why? This amt. of product (______) is the
theoretical yield. amt. we get if reaction is perfect found by
calculation using Stoich 189 g couldnt collect all Al not all Na
and Al 2 O 3 reacted some reactant or product spilled and was lost
Actual yield
Slide 34
= 91.0% Find % yield for previous problem. Note: % yield should
never be > 100% Batting average FG % GPA
Slide 35
= 41,384 g Li 2 CO 3 On NASA spacecraft, lithium hydroxide
scrubbers remove toxic CO 2 from cabin. 1. For a seven-day mission,
each of four individuals exhales 880 g CO 2 daily. If reaction is
75% efficient, how many g Li 2 CO 3 will actually be produced? CO 2
(g) + 2 LiOH(s) Li 2 CO 3 (s) + H 2 O(l) CO 2 Li 2 CO 3 () 880 g CO
2 person-day x (4 p) x (7 d)= 24,640 g CO 2 percent yield 1 mol CO
2 24, 640 g CO 2 1 mol Li 2 CO 3 1 mol CO 2 73.9 g Li 2 CO 3 1 mol
Li 2 CO 3 44.0 g CO 2 theo yield A = 31,000 g Li 2 CO 3
Slide 36
2. Automobile air bags inflate with nitrogen via the
decomposition of sodium azide: 2 NaN 3 (s) 3 N 2 (g) + 2 Na(s) At
STP and a % yield of 85%, what mass sodium azide is needed to yield
74 L nitrogen? N2N2 NaN 3 percent yield act yield x = 87.1 L N 2
Theo yield 1 mol N 2 87.1 L N 2 2 mol NaN 3 3 mol N 2 65 g NaN 3 1
mol NaN 3 = 170 g NaN 3 22.4 L N 2
Slide 37
___ZnS + ___O 2 ___ZnO + ___SO 2 100 g100 g X g ? (assuming 81%
yield) Strategy: 1. 2. 3. Balance and find LR Use LR to calc. X g
ZnO (theo. yield) Actual yield is 81% of theo. yield 2322 ZnO 2 mol
ZnS 1 mol ZnS 97.5 g ZnS 100 g ZnS 1 mol O 2 100 g O 2 = 83.5 g ZnO
2 mol ZnO 81.4 g ZnO 1 mol ZnO 2 mol ZnO 3 mol O 2 81.4 g ZnO 1 mol
ZnO = 169.6 g ZnO 83.5 g ZnO 32 g O 2 x = 67.6 g ZnO
Slide 38
___Al + ___Fe 2 O 3 ___Fe + ___Al 2 O 3 X g? X g? 800. g needed
**Rxn. has an 80.% yield. 2112 FeAl FeFe 2 O 3 act yield theo =
1000 g Fe 2 mol Fe 1 mol Fe 55.85 g Fe 1000 g Fe = 480 g Al 2 mol
Al 26.98 g Al 1 mol Al 2 mol Fe 1 mol Fe 55.85 g Fe 1000 g Fe =
1400 g Fe 2 O 3 1 mol Fe 2 O 3 159.7 g Fe 2 O 3 1 mol Fe 2 O 3
Slide 39
Reaction that powers space shuttle is: 2 H 2 (g) + O 2 (g) 2 H
2 O(g) + 572 kJ From 100 g hydrogen and 640 g oxygen, what amount
of energy is possible? EE 1 mol H 2 = 14300 kJ 2 mol H 2 572 kJ 100
g H 2 2 g H 2 1 mol O 2 = 11440 kJ 1 mol O 2 572 kJ 640 g O 2 32 g
O 2 11440 kJ Review Questions
Slide 40
What mass of excess reactant is left over? 2 H 2 (g) + O 2 (g)
2 H 2 O(g) + 572 kJ H2H2 O2O2 Started with 100 g, used up 80 g 20 g
H 2 left over 1 mol O 2 640 g O 2 2 mol H 2 1 mol O 2 2 g H 2 1 mol
H 2 = 80 g H 2 32 g O 2 100 g 640 g
