Stoichiometry

High School Chemistry Rapid Learning Series - 12

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Stoichiometry

HS Ch i t R id Learning Series
Chemistry Rapid L
i
S i
Wayne Huang, PhD
Kelly Deters, PhD
Russell Dahl, PhD
Elizabeth James, PhD

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© Rapid Learning Inc. All rights reserved.

© Rapid Learning Inc. All rights reserved.:: http://www.RapidLearningCenter.com

1


Learning Objectives
By studying this tutorial you will learn…
What stoichiometry is.
How dimensional analysis is
used to solve stoichiometry
problems.
How to gather equalities for
use in stoichiometry problems.
How to use Molarity.
How to use the molar volume
of a gas at STP
f
t STP.
How to find the limiting
reactant.

stoichiometry

How to find percent yield.

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Concept Map
Previous content
Chemistry

New content
Limiting
Reactant

Studies

Percent
Yield
Can lead to

Matter
Stoichiometry
Undergoes

Chemical
Equations

Shown in

Are investigated
with

Chemical
Changes

uses

Dimensional
Di
i
l
Analysis

Shows
Using

Mole Ratio

Molar
Mass

Molarity

Molar
Volume of a Gas

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2


Relationships of
Compounds in
Chemical Equations

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Mole Ratios in Chemical Equations
One of the most important pieces of information in
a chemical reaction is the mole ratio (shown by the
coefficients).
and 2 moles of H2O are
produced.

For every 2 moles
of H2…

2

2

2 H2 + O2

2 H2O

Coefficient = 1
1 mole of O2 is
need to react…
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3


Using Dimensional
Analysis in
Stoichiometry

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Definition: Stoichiometry
Stoichiometry – Using
t e o e at o o t e
the mole ratio from the
balanced equation and
information about one
compound in the
reaction to determine
information about
another compound in the
reaction.

CH4+O2

CO2 + 2H2O

The word stoichiometry derives from two Greek words: stoicheion
(meaning "element") and metron (meaning "measure").
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4


KUDOS & Dimensional Analysis
Use the KUDOS method and dimensional analysis to
solve stoichiometry problems.

K

Identify the known.

U

Identify the unknown.

D

Identify needed definitions.
For dimensional analysis, the “definitions” are the equalities:
Ratio from balanced equation (one compound ↔ another compound)
Molar mass (grams ↔ moles)
Molarity (grams ↔ liters of a solution)
Molar volume of a gas (moles ↔ liters of a gas)

O

Find the output - perform the dimensional analysis.

S

Substantiate your answer.

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Mole-Mole Problems
Example:

If 4.2 mole of H2 reacts completely with O2, how
many moles of O2 are needed?
2 H2 + O2

2 H2O
From balanced equation:
2 mole H2
1 mole O2

K

U
1

mole O2

2

4.2 mole H2

D

mole H2

= ________ mole O2
2.1

O
2.1 is a reasonable answer when 4.2 is given.
“mole O2” is the correct unit.
2 sig. fig. given
2 sig. fig. in answer.

S

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5


Molar Mass Review
Molar mass is needed to convert between
grams and moles:
1 Count the number of each type of
atom.

2

Find the atomic mass of each atom
on the periodic table.

3

Multiply the # of atoms × atomic mass for each atom.

4

Find the sum of all the masses.

Example: Find the molar mass for CaBr2
Ca

1 × 40.08 g/mole =

40.08 g/mole

Br

2 × 79.91 g/mole =

+ 159.82 g/mole
199.90 g/mole

1

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2

4

3
1 mole of CaBr2
molecules would
have a mass of
199.90 g.

Mole-Mass Problems
Example:

How many grams of AgCl will be precipitated if
0.45 mole AgNO3 is reacted as follows:
2 AgCl + 2 Ca(NO3)2
2 AgNO3 + CaCl2
2 mole AgCl
2 mole AgNO3

K

U
2

mole AgCl

2

0.45 mole AgNO3

D

Molar Mass of AgCl:
1 mole AgCl = 143.35 g

mole AgNO3

143.35 g AgCl
1

= ________ g AgCl
65

mole AgCl

O
65 is a reasonable answer for grams.
“g AgCl” is the correct unit.
2 sig. fig. given

S

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6


Mass-Mass Problems
Example:

How many grams Ba(OH)2 are precipitated from 14.5
g of NaOH in the following reaction:
2 NaOH + BaCl2
Ba(OH)2 + 2 NaCl
q
2 mole NaOH
1 mole Ba(OH)2
Molar Mass of NaOH:
1 mole NaOH = 40.00 g

D

Molar Mass of Ba(OH)2:
1 mole Ba(OH)2 = 171.35 g

K
14.5 g NaOH

1 mole NaOH
40.00 g NaOH

1

mole Ba(OH)2 171.35 g Ba(OH)2
( )

2

mole NaOH

1

mole Ba(OH)2

U

31.1
= ________ g Ba(OH)2
31.1 is a reasonable answer for grams.
“g Ba(OH)2” is the correct unit.
3 sig. fig. given

S

O

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Moles,
Moles Volumes of
Solutions and
Concentrations

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7


Definition: Concentration
The substance
being dissolved in
a homogeneous
mixture (solution).
(solution)

Concentration – How much solute is in the
solution.
Molarity (M) – A
concentration unit,
t ti
it
mol/L.

15/32

Definition: Molarity

Molarity =

moles (solute)
L (solution)

Molarity gives the number of
moles of the solute that are
in 1 liter of the solution.

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8


Using Molarity
Molarity is used to convert between moles and liters.
Example:

If 0.85 moles NaOH are needed and you have a 1.5 M
solution, how many liters of the solution do you need?
From concentration:
1.5 moles NaOH = 1 L

K

D

U

0.85 mol NaOH

1

L

0.57
= ________ L

1.5 mol NaOH

O
0.57 is a reasonable answer for L.
“L” is the correct unit.
2 sig. fig. given
2 sig. fig. in answer. S
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Mass-Volume Problems (Solutions)
Example:

If you need to precipitate 15.7 g of Ba(OH)2, how
many liters of 2.50 M NaOH solution is needed?
Ba(OH)2 + 2 NaCl
2 NaOH + BaCl2
q
2 mol NaOH
1 mol Ba(OH)2
Concentration of NaOH:
2.50 mole NaOH = 1 L

D

Molar Mass of Ba(OH)2:
1 mol Ba(OH)2 = 171.35 g

K
15.7 g Ba(OH)2
( )

1

mol Ba(OH)2

2

mol NaOH

171.35 g Ba(OH)2

1

mol Ba(OH)2

1

L NaOH

2.50 mol NaOH

U

0.0733
= ________ L NaOH
0.0733 is a reasonable answer for L (73.3 mL).
“L NaOH” is the correct unit.
3 sig. fig. given

O

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9


Molar Volume
of a Gas

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Definition: Molar Volume of a Gas
Standard Temperature
and Pressure (STP) – 1
atm (760 mm Hg) and 273
K (0°C).
Molar Volume of a Gas – At
STP, 1 mole of any gas =
22.4 liters.

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10


Mass-Volume Problems (Gases)
Example:

If you need react 1.5 g of zinc completely, what
volume of gas will be produced at STP?
2 HCl (aq) + Zn (s)
ZnCl2 (aq) + H2 (g)
q
1 mole Zn
1 mole H2
Molar volume of a gas:
1 mole H2 = 22.4 L

D

Molar Mass of Zn:
1 mole Zn = 65.39 g

K
1.5 g Zn

1

mole Zn

1

mole H2

65.39 g Zn

1

mole Zn

22.4
1

L H2
mole H2

U

0.51
= ________ L H2
0.51 is a reasonable answer for L.
“L H2” is the correct unit.
2 sf given
2 sf in answer.

S

O

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Limiting
Reactants

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11


Planning a Meal
You go to the grocery store and you
buy 1 package of Brats (5 Brats), 1
package of cheese (16 slices) and 1
p
package of hot dog buns (8 buns).
g
g
(
)
If you use all of these…

You can make this many…

5 Brats

5 meals

16 slices of cheese

16 meals

8 hot dog buns

8 meals

So you have the possibility of making 5, 16 or 8 meals…which is it?
You’ll never get the chance to make 8 or 16 meals…you’ll run out of
Brats after 5.
Once you run out of one component, you have to stop making meals.
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Definition: Limiting Reactant
Limiting Reactant – The
reactant that runs out first
and causes the reaction to
stop.
In the previous example, the Brats were the
limiting reactant—once they were gone you had
gone,
to stop!
Once one of the reactants runs out, the reaction
stops…it can’t make any more product.
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12


Limiting Reactant Problems
Example:

How many moles of H2O is produced when 2.3
moles of O2 and 2.3 moles of H2 react?
2 H2 + O2
2 H2O
2 mol H2
2 mol H2O
1 mol O2
2 mol H2O

K

D

U
2.3 mol O2

2 mol H2O
1

2.3
2 3 mol H2

2 mol H2O
2

4.6
= ________ mol H2O

mol O2
2.3
= ________ moles H2O

mol H2

O

You’ll never get to 4.6 moles of
H2O because H2 runs out when
2.3 moles H2O is made.

2.3 is reasonable for moles.
“moles H2O” is the correct unit.
2 sig. fig. given

S

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Identifying the Limiting Reactant
Example:

What is the limiting reactant in the last problem?
2 H2 + O2
2 H2O …Calculate and compare the
amount of products produced by each reactant .

2.3 mol O2

2 mol H2O
1

2.3 mol H2

2 mol H2O
2

4.6
46
= ________ mol H2O
l

mol O2
2.3
= ________ mol H2O

mol H2

Less
product

The reactant that produces a lesser amount of product is the
limiting reactant.
H2 was the reactant that ran out (caused the smallest number
of products) therefore, H2 is the limiting reactant.

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Simple Limiting Reagent Finder Mnemonic: Find the moles of each reactant
and divide the moles of each reactant by its coefficient. The reactant with the
smallest number is the limiting reagent = “Smallest m/c is Limited”.


13


Percent Yield

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Definition: Percent Yield
Percent Yield – Compares how
much the reaction actually
produced (the “actual yield”)
to how much the stoichiometry
says you should get if things
react completely 100% (the
“theoretical yield”).
% yield =
i ld

28/32

actual yield
y
×100
theoretical yield

% yield is always calculated with
masses—not moles!


14


Percent Yield Problem
Example:

A student works out a stoichiometry problem and
determines that they should produce 1.56 g AgCl in the
lab. They complete the lab and find that they obtained
1.25 g AgCl. Find the percent yield of the lab.

K
U
D

Theoretical yield = 1.56 g AgCl
Actual yield = 1.25 g AgCl
% yield = ?

% yield =

% yield =

actual yield
×100
theoretical yield

O

1.25g AgCl
×100 = 80.1%
1.56 g AgCl

80.1 is a reasonable answer for percent yield.
“%” is the correct unit.
3 sig. fig. given
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Learning Summary
The limiting
reactant is the
reactant th t runs
t t that
out first and
causes the reaction
to stop.

Stoichiometry is
using the mole
ratio from the
balanced equation
to find information
about various
compounds.

Stoichiometry uses
dimensional analysis
and various equalities.

Percent yield
compares the
actual yield of a
reaction to the
theoretical yield
from stoichiometry.

Molarity is used to form
an equality between
lit b t
moles & liters of a
solution. The molar
volume of a gas is used to
find liters of a gas in
stoichiometry.

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15


Congratulations
You have successfully completed
the core tutorial

Stoichiometry

Chemistry :: Biology :: Physics :: Math

What’s N t
Wh t’ Next …

Step 1: Concepts – Core Tutorial (Just Completed)
Step 2: Practice – Interactive Problem Drill
Step 3: Recap – Super Review Cheat Sheet

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Stoichiometry

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