Download Mass Relationships in Chemical Reactions

Mass Relationships in
Chemical Reactions
Chapter 3
HW: Ch.3a 2, 5, 13-16, 20, 23, 26, 27, 39-53 odd (no 47)
Ch.3b 59, 66, 69,75, 78, 83, 86, 93-94
Micro-World
(atoms & molecules)
Macro-World
(grams)
Atomic mass: mass of an atom in atomic mass units (amu)
where 1 amu = 1.66 x 10-24 g
By definition and International agreement:
1 atom 12C “weighs” 12 amu
On this scale
1H = 1.008 amu - which is 8.400% as massive
16O = 16.00 amu - which is 133.33% as massive
2
The average atomic mass is the weighted average
of all of the naturally occurring isotopes of
the element.
12121212121212121213
12121212121212121212
12121212121212121212
12121212121212121212
12121212121212121212
12121212121212121212
12121212121212121212
12121212121212121212
12121212121212121212
12121212121212121212
Carbon average atomic mass =
0.9890 * 12 amu + 0.0110 * 13.00335 amu =
12.01 amu
*98.90% = 0.9890
Example: Average Atomic Mass Calculation #1
A random sample of Magnesium is composed of
• 79.0% Mg-24 , 10.0% Mg-25 , and 11.0% Mg-26
Find the weighted average for the mass of Magnesium
79.0% of 24 + 10.0% of 25 + 11.0% of 26 = Average atomic mass
*Do Not Divide by the total
Use the percentage in the decimal form (e.g. 75% = 0.75)
(0.790 x 24) + (0.100 x 25) + (0.11 x 26) = 24.3 amu
Example: Average Atomic Mass Calculation #2
A random sample of chlorine is
75.8%
35Cl
(34.97 amu) &
24.2%
37Cl
(36.97 amu)
• Find the weighted average for the mass of chlorine by
writing percentage as a decimal (e.g. 50% = 0.50)
(0.758 x 34.97) + (0.242 x 36.97) = 35.45 amu
Average atomic mass for relative abundant isotopes
Some BIG Numbers…
World population: ~7,000,000,000
Stars in our galaxy: ~300,000,000,000
Cells in our body: ~50,000,000,000,000
Avogadro’s Constant (number)
602,214,150,000,000,000,000,000
~0.6
Septillion
6.022 x 1023
1 mole = NA = 6.022 x
Jean Baptist Perrin
23
10
23
6.02*10
grains of sand could cover
Texas in almost 3 feet 268,820
of sand miles2
23
6.02*10
©
Starbursts
would cover the
surface of the entire
US stacked and be
24.2 miles tall
The Moon
7.3 × 1022 kg
6.02*1023
moles (mammal)
The mass of a single carbon atom in grams:
0.0000000000000000000000199
The Mole (mol): A grouping unit to count numbers of
particles Dozen = 12
Pair = 2
The mole (mol) is the amount of a substance that
contains as many elementary units as there
12
are atoms in exactly 12.00 grams of C
1 12C atom
12.00 amu
12.00 g
x
6.022 x 1023 12C atoms
1 amu = 1.66 x 10-24 g
1.66 x 10-24 g
=
1 amu
or 1 g = 6.022 x 1023 amu
M = molar mass in g/mol
NA = Avogadro’s number
Molar
eggs
mass is the mass of 1 mole of shoes in grams
atoms
1 mole 12C atoms = 6.022 x 1023 atoms = 12.00 g
1 12C atom = 12.00 amu
12.00 g 12C = 1 mole 12C atoms
65.39 g of Zn = 1 mole Zinc atoms
18.015 g of water = 1 mole water molecules
58.443 g of NaCl = 1 mole NaCl formula units
*Read the masses on the periodic table as grams/mole
1 mol = NA = 6.02 x
23
10
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 units of MgO
or... (Scale everything up by NA)
2 moles Mg + 1 mole O2 makes 2 moles MgO 
so (using the Periodic Table we have)
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
Ted-Ed: How big is a mole? www.youtube.com/watch?v=TEl4jeETVmg
One Mole of:
S
C
Mercury
Copper
Iron
MM: Molar mass (grams/mol)
Avogadro’s constant
NA = 6.02 x 1023
Particles
Atoms
Molecules
Formula units
Remember Train Track Unit Conversions
How
many inches are in 3 miles?
Need conversion factors: 1 mile = 5,280 ft; 1 foot = 12 in
3 miles
1
5280 feet
1 mile
12 inches
1 foot
Multiply all numbers on the top
Divide all numbers on the bottom
3 x 5280 x 12
1 x 1x 1
= 190,080 inches
Chemistry Conversion Factors
MM = molar mass in grams/mole
Converts between mass (g) & moles
23
10
NA = Avogadro’s number = 6.02 x
Converts between moles & particles
(atoms / molecules / formula units)
Grams to Moles example
How many moles of Helium atoms
are in 6.46 g of He?
Molar mass is the conversion factor and is
found on the periodic table to be 4.003 g/mol.
This can be written as 1 mol He = 4.003 g He
6.46 g He
1 mol He
x
4.00 g He
= 1.61 moles He
More Practice: 45.5 grams of Gold (Au) to moles
Moles to Grams example
How many grams of zinc are in
0.356 mole of Zn?
The conversion factor is the molar mass of Zn.
From the periodic table we find that
1 mol Zn = 65.4 g Zn
0.356 mol Zn
65.4 gram Zn
x
1 mol Zn
Zinc
= 23.3 gram Zn
More Practice: 0.56 mol Potassium to grams
Moles to Atoms example
How many atoms of Carbon are in
2.54 moles of C?
The conversion factor is Avogadro’s Number
1 mole Carbon = 6.02 x 1023 atoms of Carbon
Carbon
2.54 mol C
6.02 x 1023 atoms of C
x
1 mol C
= 1.53 x 1024 atoms of C
More Practice: 0.074 mol U to atoms of U
Grams to Atoms Example
How many atoms are in 16.3
g of Sulfur?
Grams of S → moles of S → # atoms S
MM: 1 mol S = 32.07 g
16.3 g S
1 mol S
x
32.1 g S
1 mol S = 6.02 x 1023 S atoms
6.02 x 1023 atoms of S
x
1 mol S
= 3.06 x 1023 atoms of S
More Practice: 0.01 g of Carbon to atoms
Atoms to Grams Example
What would be the mass in grams
25
of 2.4 x 10 atoms of Copper?
# atoms of Cu
→
moles of Cu
6.02 x 1023 Cu atoms = 1 mol Cu
2.4 x 1025 atoms Cu
→
grams Cu
MM: 1 mol Cu = 63.55 g Cu
1 mol Cu
x
x
6.02 x 1023 atoms
63.55 g Cu
1 mol Cu
= 2.5 x 103 g = 2.5 kg Cu
More Practice: 9.8 x 1021 atoms of Uranium to grams
More Practice Conversions
Convert 24 grams Potassium to moles
24 grams K
1 mol K
x
39.1 grams K
=
0.61 moles K
Convert 1.25 moles Gold to grams
1.25 mol Au
197.0 g Au
x
1 mol Au
=
246 grams Au
Convert 16 grams Argon to atoms
16 g Ar
1 mol Ar
x
39.9 g Ar
6.02 x 1023 Ar atoms =
x
1 mol Ar
2.4 x 1023 Ar
atoms
Convert 3.4 x 1025 atoms of Boron to grams of Boron
Bozeman Science Mole conversion tutorial:
www.youtube.com/watch?v=xPdqEX_WMjo
Molecular mass (or molecular weight) is the sum
of the atomic masses (in amu) in a molecule.
1S
SO2
2O
SO2
32.07 amu
+ 2 x 16.00 amu
64.07 amu
For any molecule
molecular mass (amu) = molar mass (grams)
1 molecule SO2 = 64.07 amu
1 mole SO2 = 64.07 g SO2
Formula mass is the sum of the atomic masses
(in amu) in a formula unit of an ionic compound.
x
NaCl
1Na
22.99 amu
1Cl
NaCl
+ 35.45 amu
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
Determining Molar Mass Examples
Calculate the molecular or formula masses (in g/mol) of the
following compounds:
(a) Water (H2O)
(b) Isopropanol (C3H8O1) (rubbing alcohol)
(c) (NH4)2S
(d) Mg(NO3)2
(e) caffeine (C8H10N4O2)
Example
Molar Mass Solutions
a) H2O: 2*(1.01 g) + 1*(16.00 g) = 18.02 grams
b) C3H8O1: 3*(12.01 g) + 8*(1.01 g) + 1*(16.00 g) = 60.05 grams
c) (NH4)2S : 2*(14.01 g) + 8*(1.01 g) + 1*(32.07 g) = 68.17 grams
d) Mg(NO3)2: 1*(24.31 g) + 2*(14.01 g) + 6*(16.00 g) = 148.33 grams
e) C8H10N4O2: 8*(12.01 g) + 10*(1.01 g) + 4*(14.01 g) + 2*(16.00 g)
= 194.20 grams
Each Molar mass can be read with units “grams/mole”
Compound Mass to Mole Example
Methane (CH4) is the principal
component of natural gas.
How many moles of CH4 are present in
6.07 g of CH4?
Molar mass of CH4 = 12.01 g + 4*(1.008 g) = 16.04 g
6.07 g CH4
1 mol CH4
x
= 0.38 moles CH4
16.0 g CH4
More Practice: 32 grams of Isopropanol (C3H8O) to moles
Atoms in a Compound
Example
urea
How many hydrogen atoms are present in
25.6 g of urea [(NH2)2CO]
The molar mass of urea is 60.06 g.
*Note there are 4 H atoms per Urea molecule
So there are 4 moles of H per 1 mole urea
grams of urea → moles of urea → moles of H → atoms of H
25.6 g of urea
x
1 mol of urea
60.06 g urea
x
4 mol of H atoms
1 mol of urea
x
6.02 x 1023 H atoms
1 mole of H
= 1.03 × 1024 H atoms
More Practice: How many Fluorine atoms in 4.5 grams of AlF3?
Bell Ringer Practice Conversions
How many atoms of Phosphorous (P)
are found in 6.30 g of Mg3(PO4)2?
-Invented by Francis Aston in the 1920’s
-Separates particles by e/m (charge/mass)
-First demonstration of isotopes (Ne 20 & 22)
-Led to precise determination of atomic masses
and relative isotope abundancies.
1922 Nobel Prize in Chemistry
Mass Spectrum of Ne
Heavy
Light
Heavy
Sample gets
ionized
Light
Mass Spectrometer
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 mole of the
compound
%C =
%H =
%O =
Ethanol: C2H5OH
Molecular weight: 46.07 g
2 x (12.01 g)
46.07 g
6 x (1.008 g)
46.07 g
1 x (16.00 g)
46.07 g
x 100% = 52.14%
x 100% = 13.13%
x 100% = 34.73%
52.14% + 13.13% + 34.73% = 100.0%
Example Percent Composition #2
Phosphoric acid (H3PO4) is a colorless, syrupy
liquid used in detergents, fertilizers, toothpastes,
and in carbonated beverages for a “tangy” flavor.
Calculate the percent composition by mass of H, P, and O
Molar Mass of H3PO4
H: 3 x 1.008
P : 1 x 30.97
O: 4 x 16.00
= 97.99 g
More Practice: What is the % composition of Aluminum in Galaxite (MnAl2O4)?
How many grams Al in 35.6 g of Galaxite?
Example Using Percent Compositions #1
Chalcopyrite (CuFeS2) is a
principal mineral of copper.
Calculate the number of
kilograms of Cu in 3.71×103 kg of
chalcopyrite using percent
compositions.
Chalcopyrite.
How do we calculate mass percent of an element?
mass % =
(Element Mass)*(Quantity) x 100%
Compound Mass
Example
#1 Solution
The molar masses of Cu and CuFeS2 are 63.55 g and 183.5 g,
respectively. The percent composition of Cu is therefore:
To calculate the mass of Cu in a 3.71 × 103 kg sample of CuFeS2, we
need to convert the percentage to a fraction (that is, convert 34.63 percent
to 34.63/100, or 0.3463) and write:
mass of Cu in CuFeS2 = 0.3463 × (3.71 × 103 kg)
3
= 1.28 × 10 kg
Using Percent Compositions Example #2
Talc is a mineral composed of hydrated magnesium
silicate with the chemical formula H2Mg3(SiO3)4
Determine the number of grams of pure Magnesium
we could extract from 45.0 grams of the talc
mineral using percent composition.
(Mg
Mass)*(3)
% Mg =
Total Mineral Mass
x 100%
72.93
g
Mg
% Mg =
=
19.23%
Mg
x
100%
379.29 g Mineral
g Mg = % Mg 45.0 gram compound
= 0.1923 x 45.0 g = 8.65 g Mg
More Practice: What is the % composition of Aluminum in Galaxite (MnAl2O4)?
How many grams Al in 35.6 g of Galaxite?
Percent Composition and Empirical Formulas
Step 1 & 2
Step 3
Step 4
Early analysis of many
molecules originally only
yielded % composition
Elemental analysis of
compounds reveals the mass
% of each element
This allows us to derive the
empirical formula
(atom to atom ratio)
Determining Empirical Formulas
When the element % composition is known:
1. Assume 100g sample and change element percents to grams
2. Convert each to moles by dividing by molar mass of each atom
3. Divide each number by the smallest number
If all Numbers are integers (or close) you are done; Put in formula
4. If not, multiply all numbers by the SAME smallest whole number so that
all numbers are whole.
We don’t expect numbers to be perfect integers due to some experimental
error, Generally round only if within one tenth.
Example: Empirical Formula Determination #1
An analysis of an unknown gas has determined that it contains
72.55% Oxygen and 27.45% Carbon by mass. What is the
empirical formula of the compound? C#O#
72.55 g O 1 mol O
16 g O
27.45 g C 1 mol C
12 g C
C(1 mol)O(2 mol)
= 4.53 mol O
4.53/2.29 = 1.98 mol
~ 2 mol O
2.29/2.29 = 1 mol C
= 2.29 mol C
C(1 atom)O(2 atom)
CO2
Example: Empirical Formula #2
Determine the empirical formula of Ascorbic acid
(vitamin C: cures/prevents scurvy).
It is composed of 40.92% C, 4.58% H, and 54.50% O by mass.
To determine the empirical formula, we will first assume a 100 g
sample of Ascorbic Acid
2nd: convert each mass to mole
40.92 g C
4.58 g H
54.50 g O
Example: Empirical Formula #2 Solution
3rd:We then take the smallest value of moles and divide each number by
that same smallest value
• This gives CH1.33O as the formula for ascorbic acid.
• If any # is more than 0.1 away from whole, it is too far to round.
• Multiply all values by integers until all values are whole numbers
Because 1.33 × 3 gives us an integer (4), we
1.33 × 1 = 1.33
multiply all the subscripts by 3 and obtain
1.33 × 2 = 2.66
C3H4O3 as the empirical formula for
1.33 × 3 = 3.99 ~ 4
ascorbic acid.
More Practice: CxHyNz : 42.1% C, 8.8% H & 49.1% N
Combust 11.5 g ethanol
Collect 22.0 g CO2 and 13.5 g H2O
“desiccant” SiO2
“scrubber” LiOH
g C = % C x 22.0 g CO2
27.3% x 22.0 g = 6.0 g C
6.0 g C = 0.5 mol C
g H = % H x 13.5 g H2O
11.1% x 13.5 g = 1.5 g H
1.5 g H = 1.5 mol H
g O = (11.5 g sample) – (6 g C + 1.5 g H) =
Empirical formula C0.5H1.5O0.25
Divide by smallest subscript (0.25)
Empirical formula C2H6O
4.0 g O = 0.25 mol O
Example: Molecular Formula
The molecular formula can be determined
with both % Composition and Molar Mass
A sample of a compound contains 30.46% Nitrogen and
69.54% Oxygen by mass
In a separate experiment, the molar mass of the compound is
estimated to be between 90 g and 95 g.
Determine the molecular formula and the accurate molar mass
of the compound.
Example: Molecular Formula Solution
We first assume 100 g and convert % to grams. We then convert
each element from grams to moles.
To convert to whole numbers we divide each mole by the
smaller subscript (2.174). After rounding off, we obtain NO2 as
the empirical formula.
The Molecular formula must have a Molar Mass that is a
multiple of the Empirical formula’s Molar Mass
(NO2, N2O4, N3O6, N4O8…)
Example: Molecular Formula Solution
Empirical molar mass = 14.01 g + 2(16.00 g) = 46.01 g
Next, we determine the ratio between the molar mass and the empirical
molar mass
The molar mass is twice the empirical molar mass. This means that
there are two NO2 units in each molecule of the compound, and the
molecular formula is N2O4.
More practice: Find the Molecular formula PxOy : 43.7% P & 56.3% O
with a molar mass between 280 – 290 g/mol.
Bell Ringer: Molecular Formula
Aconitic acid has a mass composition of:
41.4% Carbon
3.4% Hydrogen
55.2% Oxygen
It also has a molar mass of 174 grams/mole
What is the molecular formula (CxHyOz)of Aconitic acid?
Chemical Reaction: A process in which one or more
substances is changed into one or more new substances.
A chemical equation uses chemical symbols to show what
happens during a chemical reaction:
reactants
products
3 ways of representing the reaction of H2 with O2 to form H2O
Equations Symbols
Symbol
Meaning
+
“and”
“yields” or “produces”
→
↔
Reversible
or
(g)
Gas
Vapor, H2O(g)
(l)
Liquid
Water, H2O(l)
(s)
Solid
Ice, H2O(s)
(aq)
Aqueous (dissolved in H2O)
NaCl(s) + H2O(l)  Na+(aq) + Cl-(aq) + H2O(l)
Phase State :
represent what
state of matter a
substance is in.
How to “Read” Chemical Equations
2 Mg + O2
2 MgO
2 atoms Mg + 1 molecule O2 makes 2 formula units MgO
Or
2 moles Mg + 1 mole O2 makes 2 moles MgO 
Or
48.6 grams Mg + 32.0 grams O2 makes 80.6 g MgO
NOT
2 grams Mg + 1 gram O2 makes 2 g MgO
We can not compare mass to mass,
it must be in particles or Moles
In 1 mole of HBr there are 80 grams of Br to 1 gram H,
Yet still 1 atom Br to 1 atom H
Law of Conservation of Matter/Mass
• Chemical reactions only rearrange bonded atoms
with no atoms created or destroyed
• Every atom from the reactants must be equal in
the products
Fire converts/Not destroys
2C2H6 + 7O2 → 4CO2 + 6H2O
Reactants
4 Carbons
12 Hydrogens
14 Oxygens
=
Products
4 Carbons
12 Hydrogens
14 Oxygens
Balancing Chemical Equations
1. Write the correct formula(s) for the reactants on the left
side and the product(s) on the right side of the equation.
Ethane reacts with oxygen to form carbon dioxide and water
C2H6 + O2
CO2 + H2O
We only change the coefficients (number in front) to make the
number of atoms of each element the same on both sides.
Never change the subscript (that would change the chemicals)
2C2H6

NOT
C4H12
Balancing Chemical Equations
2. Count the number of atoms on each side.
Check to see if it’s already balanced.
LiBr + KNO3 → LiNO3 + KBr
• On the reactant side, there is 1 Li, 1 Br, 1
K,1 N, and 3 O atoms.
• On the product side, there is 1 Li, 1 Br, 1
K,1 N, and 3 O atoms.
• This equation is balanced.
Balancing Chemical Equations
3. Leaving Hydrogen and Oxygen till last,place
a coefficient to balance one element.
Continue balancing the others elements with
coefficients
__H2SO4 + __NaOH
→ __Na2SO4 + __H
2
2 2O
Reactants
1 Sulfur
21 Sodium
43 Hydrogen
65 Oxygen
=
Products
1 Sulfur
2 Sodium
2 Hydrogen
4
5 Oxygen
6
Combustion of Glucose
6 2 → __CO
6
6
__C6H12O6 + __O
+
__H
2
2O
The balanced equation shows it take 6 molecules of O2 to
react with 1 molecule glucose to produce 6 molecules of
OR
CO2 and H2O each.
6 mole of O2 reacts with 1 mole glucose to produce 6 moles
of CO2 and H2O each.
Reactants
6 Carbon
12 Hydrogen
188 Oxygen
=
Products
61 Carbon
122 Hydrogen
13
183 Oxygen
Balance these Chemical equations:
•
H3PO4 + NaOH → Na3PO4 + H2O
•
Mg + O2 → MgO
•
NH4NO2 → N2 + H2O
•
P2O5 + H2O → H3PO4
•
NaHCO3 → Na2CO3 + CO2 + H2O
•
Li +
H2O
→
LiOH + H2
Balancing Difficult Chemical Equations
Equations can be difficult if the reactants don’t react in a simple
1:2, 1:3, 1:4 etc. ratio.
C2H6 + O2
CO2 + H2O
C2H6 + O2
2CO2 + H2O
C2H6 + O2
2CO2 + 3H2O
2 oxygen atoms
*multiply CO2 by 2
*multiply H2O by 3
7 oxygen atoms
Because it’s O2, no whole number coefficient can give us 7 atoms
on the reactants (only multiples of 2)
Balancing Chemical Equations
C2H6 + _O2
2CO2 + 3H2O
7 oxygen atoms
Balance using the smallest fraction possible that would give the
desired quantity
*multiply O2 by 7 to balance
2
7
C 2H6 +
O2
2
2CO2 + 3H2O
Because each particle must be an integer, remove the fraction by
multiplying the whole reaction by the denominator
x2
2C2H6 + 7O2
4CO2 + 6H2O
Example
3.12
When aluminum metal is exposed to air, a protective layer of
aluminum oxide (Al2O3) forms on its surface. This layer prevents
further reaction between Al and O2, and it is why Al does not corrode.
Write a balanced equation for the formation of Al2O3.
An atomic scale image
of aluminum oxide.
Use the simplest fraction to balance Oxygen
Example
3.12
Multiplying both sides of the equation by 2 gives whole-number
coefficients.
Check
The equation is balanced. Also, the coefficients are reduced to the
simplest set of whole numbers.
More Practice: _C3H6O2 + _O2 → _CO2 + _H2O
Stoichiometry is the quantitative
relationship found between reactants and/or
products in a reaction.
• 2 hydrogen molecules and 1 oxygen
molecules produce 2 water molecules
Multiply all by
Avogadro's #
• 2 moles hydrogen and 1 mole oxygen
produce 2 moles water
Stoichiometry is the "recipe of a reaction".
From the equation we see that 1 mole CH4  2 mole O2
 = “Stoichiometric equivalents”
1x
2x
2x
1x
Mole Ratio
1C2H6O + 3O2 → 2CO
2CO2 + 3H2O
• The coefficients in a balanced reaction tell
us the mole ratio.
• It takes 1 mole of ethanol to react with 3
moles of oxygen. This produces 2 moles of
carbon dioxide and 3 moles of water.
• The mole ratio will act as our conversion
factor for calculations.
Combustion of Ethanol (C2H6O)
1C2H6O + 3O2 → 2CO2 + 3H2O
• If we have 4 moles of ethanol, how many moles
oxygen are needed for the reaction?
3
mol
oxygen
4 mol ethanol x
= 12 mol oxygen
1 mol ethanol
Mole ratio conversion factor
• How many moles CO2 are made from reacting
0.6 moles ethanol?
0.6 mol ethanol x
2 mol CO2
1 mol ethanol
= 1.2 mol CO2
Mole ratio conversion factor
Calculations can go forward or backward, always
looking at the mole to mole ratio
2C2H6 + 7O2 → 4CO2 + 6H2O
How many moles of oxygen does it take to produce
3.5 moles of water (with excess ethane)?
3.5 mol water
x
7 mol oxygen
6 mol water
= 4.1 mol oxygen
How many moles of ethane (C2H6) does it take to
produce 3.5 moles of water (with excess oxygen)?
3.5 mol water x
2 mol ethane
6 mol water
= 1.2 mol ethane
More Practice: 1) moles of CO2 from 1.5 moles C2H6
2) moles of H2O from 0.34 moles of O2
Amounts of Reactants and Products
1.
2.
3.
4.
Write balanced chemical equation
Convert quantities of known substances into moles
Use coefficients in balanced equation to calculate the
number of moles of the sought quantity
Convert moles of sought quantity into desired units
Stoichiometry Example #1
Mass to Mole to Mole to Mass
(Products)
to
(Reactants)
All alkali metals react in water to produce hydrogen gas
and the corresponding alkali metal hydroxide.
How many grams of Li are needed to produce 9.89
g of H2?
Lithium reacting with
water to produce
hydrogen gas.
Stoichiometry Example #1 Solution
? g of Li
9.89 g of H2
More Practice: Grams of LiOH produced from 2.9 grams Li
Stoichiometry Example #2
The food we eat is broken down, in our bodies to provide
energy. The overall equation for this complex process
represents the degradation of glucose (C6H12O6).
If 86 g of C6H12O6 is consumed by a person, what is the
mass of CO2 produced?
(180 g/mol) (1 mol C6H12O6  6 mol CO2)
86 g Glucose
x
(44 g/mol)
1 mol Glucose x 6 mol CO2 x 44 g CO2 = 126 g CO
2
180 g Glucose 1 mol Glucose 1 mol CO2
Stoichiometry: Chemistry for Massive Creatures - Crash Course Chemistry #6
www.youtube.com/watch?v=UL1jmJaUkaQ
Stoichiometry Example #3
How many grams of water are produced from
the combustion of 35 g of propane? (C3H8)
35 g C3H8 x 1 mol C3H8 x 4 mol H2O x 18 g H2O
=
57.3
44 g C3H8
1 mol C3H8 1 mol H2O
g H2O
How many grams of CO2 are produced from
the combustion of 35 g of propane?
35 g C3H8 x 1 mol C3H8 x 3 mol CO2 x 44 g CO2
44 g C3H8
1 mol C3H8 1 mol CO2
*Note: Different amounts of each product
are made by the same amount of reactant
= 105 g
CO2
More Practice
How many grams of Barium phosphate are produced
when adding 1.3 grams of Barium hydroxide to a
solution of phosphoric acid?
__Ba(OH)2 + __ H3PO4 → __ Ba3(PO4)2 + __H2O
1.3 grams
Excess
? grams
Practice problems
• How many grams are produced of each product in the
following decomposition reactions?
2.8 g 2KClO3 → 2KCl + 3O2
12.0 g CaCO3 → CaO + CO2
• How many grams of H2O are required for complete
reaction with 37 g of the other reactant?
K2O +
SiCl4 +
H2O →
4H2O →
2KOH
H4SiO4 +
4HCl
• How many grams of Aluminum are required to obtain
100 g Iron? 2Al +
3FeO → Al2O3 +
3Fe
? g of Al
100 g of Fe
Reaction Percent Yield
Not every reaction goes to 100% completion every time
Or we can’t always collect (purify) all of the product.
Theoretical Yield is the amount of product that would
result if all the limiting reagents reacted. This is
what you have already been calculating.
Actual Yield is the amount of product actually obtained
from a reaction. This is an observed value and you
must be told this amount.
Actual Yield
% Yield =
Theoretical Yield
x 100%
Example: Percent Yield
2Mg + O2 → 2MgO
If we react 10 grams of O2 with excess Magnesium, how much
MgO would be theoretically expect to obtain?
10 g O2 x 1 mol O2 x 2 mol MgO x 40.3 g MgO = 25.2 g MgO
32 g O2
1 mol O2
1 mol MgO
Theoretical yield = 25.2 g MgO
If we actually only ended up with 20.3 grams MgO, what is our
percent yield?
% Yield =
Actual Yield
Theoretical Yield
x100% =
20.3 g MgO
25.2 MgO
x 100% = 80.6%
We obtained a reaction yield of 80.6% MgO from this reaction.
Limiting Reagents
The reaction can only proceed as far as the limiting
reagent will allow it.
Once it is gone, the other reactant has nothing to react
with so the reaction stops.
To determine, we must calculate the product yield from
both reactants and choose the smaller amount.
Limited by the number of car bodies so production stops
Limiting Reagents: Reactant used up first in the
reaction to result in its completion.
Unless all reactants are in exact stoichiometric equivalents, one
will limit the reaction (i.e. 2 moles of N2 and 3 moles of H2)
N2 + 3H2 → 2NH3
H2 is limiting
N2 Excess
Bozeman Science Stoichiometry Tutorial:
www.youtube.com/watch?v=LQq203gyftA
Example #1: Limiting Reagent and Excess
1P2O5 + 3H2O → 2H3PO4
If we react 44 grams of P2O5 with 25 grams of water, which is the
limiting reagent and how many grams of H3PO4 will be produced?
We first find out how many grams of product could be made in
each case?
44 g P2O5 x 1 mol P2O5 x 2 mol H3PO4 x 98 g H3PO4 = 61.2 g
141 g P2O5
1 mol P2O5
1 mol H3PO4
25 g H2O x 1 mol H2O x 2 mol H3PO4 x 98 g H3PO4 = 90.7 g
18 g H2O
3 mol H2O
1 mol H3PO4
We then choose the smaller amount of product: 61.2 g of H3PO4
P2O5 is the limiting reagent and there is excess water.
Example #2: Limiting Reagent and Excess
Urea [(NH2)2CO] is prepared by reacting ammonia with CO2:
63.7 g of NH3 are reacted with 114.2 g of CO2.
(a) Which of the two reactants is the limiting reagent?
(b) Calculate the mass of product (NH2)2CO formed.
(c) How much excess reagent (in grams) is left at the end of the
reaction?
Example #3: Limiting Reagent and Excess
40 g of NO are mixed with 8 g O2 to form NO2:
2NO + O2 → 2NO2
• Which reactant is limiting the reaction?
• How much NO2 is produced?
• How much of the other reactant is left?
• If only 18 g is recovered from the reaction,
what’s the percent yield?
Fe + CuSO4 → FeSO4 + Cu
How much Iron was added to Copper (II) Sulfate to actually
obtain 32.1 g Copper (a 73.1% reaction yield)?
% Yield =
Actual Yield
Theoretical Yield
x 100% =
32.1 g Cu
? g Cu
x 100% = 73.1%
The theoretical yield must have been 43.9 g Cu
43.9 g Cu x 1 mol Cu x 1 mol Fe x 55.8 g Fe = 38.6 g Fe
63.5 g Cu 1 mol Cu 1 mol Fe
38.6 grams of Iron was added to CuSO4
Fun Practice Problems
How many oxygen atoms are found in 20 g of CaCO3?
A compound analysis yields the following data:
44.4% C,
6.21% H,
39.5% Sulfur,
9.86% Oxygen
What is the molecular formula if the molar mass ~120?
Fun? Practice Problems
CO + O2 → CO2
Balance the above reaction and then calculate the mass of
oxygen needed to produce 27 g CO2 with excess CO
If 10 g methane CH4 is combusted (uses O2) to form CO2 and
H2O. How many grams of each product are formed by the
reaction and how many grams of oxygen were used?
40 g of NO are mixed with 8 g O2 to form NO2, which reactant
is limiting the reaction, how much NO2 is produced, and how
much of the other reactant is left? (Write equation first)
If only18 g is recovered from the reaction, what’s the % yield?
__H3PO4 + __NaOH → __Na3PO4 + __H2O
What is the theoretical yield of water if 12.5
grams of H3PO4 reacts with 18.4 grams of
NaOH?
The actual yield was 5.2 grams water, what is
the percent yield?
Mass Relationships Exam Review
1) Average Atomic Mass: 80.1% 11B and 19.9% 10B
2) How many Phosphorous atoms are in 2.4 g Mg3(PO4)2
3a) Find the Molecular formula PxOy : 43.7% P & 56.3% O
with a molar mass between 280 – 290 g/mol.
3b) 5.8 g of PxOy react with 15 g water to form H3PO4.
Balance Rxn and find the theoretical yield.
3c) What is the % yield if only 5.4 grams of product is
recovered?