Chapter 3
Stoichiometry of Formulas and Equations
3.1 The Mole
3.2 Determining the Formula of an Unknown Compound
3.3 Writing and Balancing Chemical Equations
3.4 Calculating Quantities of Reactant and Product
FoS-1&2 Chemistry, NYUAD
Chemistry is a quantitative science. Knowing the number of atoms/molecules in the
samples and their relative ratios (stoichiometry) that you are dealing with is crucial.
Do planes carry oxygen
cylinders?
C footprint in foods
Clean, safe reactions take a clean
understanding of the quantitative
aspects.
So we need a “counting unit.”
FoS-1&2 Chemistry, NYUAD
Counting Atoms: The Mole
Counting through weighing
It’s far easier to express the number of small items through it’s mass (average mass;
since not all particles of a certain item will have exactly the same mass)
e.g., 5000 g of jelly beans will contain 1000 beans, if the average
mass is 5 g.
The number of particles in a sample can be expressed through it’ average mass (which
is determined with a -------)
E.g., average mass of a C atom is 12.01 amu. Hence 12,010 amu of carbon will have
1000 C atoms.
But amu is such a small mass. We need mass in grams for practicality.
FoS-1&2 Chemistry, NYUAD
“Mole” relates number to mass in grams:
Define mole: A mole is the amount of a substance that contains the same number of
elementary entities as there are carbon-12 atoms in exactly 12 g of carbon-12.
The atomic mass of C-12 = 12 amu/atom;
The unit of mole is mol
(1 amu = 1.66054 x 10-24 g)
So, mass of one C-12 atom = 12 × 1.66054x10-24 g
∴ 12 g of C-12 sample contain how many C-12 atoms?
12g
12×1.66054×10
−
24
g
= 6.022x1023 of C-12 atoms
1 mol of C-12 is 12 g of C-12
In fact, one mole (1 mol) of
every substance contains
6.022x1023 entities
Since all masses are expressed relative to C-12, the mass of one mole of any substance in grams
is numerically equal to it’s atomic / molecular/ formula unit mass in amu. That makes calculations
easier…
FoS-1&2 Chemistry, NYUAD
Mole relates number to mass:
A dozen (12)
Gross (144)
How many molecules of H2O?
One mole (1 mol) of every substance contains 6.022x1023 entities
The term “entities” refers to atoms, ions, molecules, in fact, any type of particle.
This number is called Avogadro’s constant and is abbreviated as NA.
FoS-1&2 Chemistry, NYUAD
Size of a Mole
Helium (in balloon), 4.002g
Water
(18.01g)
The amount of 1 mole
of different substances
Sugar (Sucrose)
Aluminum Copper
(26.98g) (63.54g) Salt (58.44g) 342.29g
A mole is always the same number of particles, but 1-mol of different substances carry
different mass & volume
FoS-1&2 Chemistry, NYUAD
Molar Mass
The mass in grams of one mole of a substance is called its Molar mass (M). Molar
mass is expressed in g/mol or gmol-1
The numerical value of the mass in grams of one mole of atoms would be equal to its
atomic/molecular mass in amu (which is read from the periodic table/calculated from
the molecular formula).
Examples;
1 atom of Sulfur (S) has a mass of 32.06 amu, 1 mole of S is 32.06g
1 atom of Iron (Fe) has a mass of 55.85 amu, 1 mole of Fe is 55.85 g
1 molecule of H2O has a molecular mass of 18.02 amu, 1 mole of H2O is 18.02 g
1 molecule of NaCl has a formula unit mass of 58.44 amu, 1 mole of NaCl is 58.44 g
1 mol of nitrogen is how many grams of nitrogen?
Be very clear of the difference:
•
•
•
Molecular mass is expressed in amu, while molar mass is given in grams
Molar mass can be calculated for atomic or molecular samples
Molecular mass is an extremely small value, molar mass is often sizeable (depending
on the size of the particle)
FoS-1&2 Chemistry, NYUAD
What is the molar mass of ethanol, C2H6O?
1 mol contains:
2 mol  C (12.01g C/1 mol) = 24.02 g C
6 mol  H (1.01g H/1 mol)
= 6.06 g H
1 mol  O (16.00g O/1 mol) = 16.00 g O
TOTAL = molar mass
= 46.08 g/mol
(Molecular mass of ethanol is 46.08 amu)
FoS-1&2 Chemistry, NYUAD
Interconverting Mass, Moles, and Number of Atoms
The mole concept is the cornerstone of quantitative chemistry and provides a
bridge between mass and number of particles.
It’s very important to master these conversions
FoS-1&2 Chemistry, NYUAD
PROBLEM: How many moles of Mg is represented by 0.200
g? How many atoms?
Molar mass of Mg 24.3050 g/mol.
So 1 mole of Mg = 24.3050 g of Mg, Then 0.200g of Mg is;
1 mol
0.200 g •
= 8.23 x 10-3 mol
24.31 g
How many atoms in this piece of Mg?
We know that 1 mole of Mg has 6.023×1023 atoms, So
6.022 x 10 23 atoms
-3
8.23 x 10 mol •
1 mol
= 4.95 x 1021 atoms Mg
FoS-1&2 Chemistry, NYUAD
1) Calculate the number of constituent particles in 9.005 g of water
1) Calculate the number of H atoms in 0.350 mol of C6H12O6
FoS-1&2 Chemistry, NYUAD
Can you leave out salt in this recipe here?
Not really, in a molecular sense!. Food is a after all a molecular composite.
¾ cup of sugar (≈ 150 g)
1 teaspoon salt (≈ 10 g)
Do your own calculations and decide.
FoS-1&2 Chemistry, NYUAD
Percent Composition
Percent composition or mass percentage of a compound is the mass
contributed by each element in the substance towards it’s molar mass.
It can easily be calculated ; 1) from the chemical formula.
e.g., each Oxygen atom contributes 50% (16 amu/16gmol-1) to the total mass (32
amu/gmol-1) of an oxygen molecule (O2)
Or, 2) experimentally If the compound is unknown (later slides)
The sum of the wt. % of each element in the compound must sum to be 100%.
FoS-1&2 Chemistry, NYUAD
Percent Composition
• A pure compound always consists of the same elements combined in the
same proportions by weight (law of constant composition).
• Therefore, we can express molecular composition as mass percentage
(which can then lead to empirical & molecular formula)
Ethanol, C2H6O
52.13% C
13.15% H
34.72% O
FoS-1&2 Chemistry, NYUAD
Percent Composition
1) What is the mass percent of each element in sodium chlorite, NaClO2?
molar mass = 22.990 + 35.453 + 2(15.999) = 90.441 g mol-1
mass of Na in 1 mol NaClO2
Molar mass of NaClO2
%Na =
=
22.990 g
90.441 g
x 100 % = 25.42%
mass of O …
Molar mass of NaClO2 …
%O =
=
2(15.999) g
90.441 g
x 100 %
x 100 %
x 100 % = 35.38%
FoS-1&2 Chemistry, NYUAD
Percent Composition
%Cl =
=
mass of Cl …
x 100 %
Molar mass of NaClO2 …
35.453 g
90.441 g
x 100 %
= 39.20%
Check your work:
%Na + %O + %Cl = 25.42 + 35.38 + 39.20 = 100%
FoS-1&2 Chemistry, NYUAD
Chemical formula from Mass percent
Mass information can be converted to mol ratio
E.g., The mass & of N and H in hydrazine is 87.5 % and 12.5% respectively.
⸫ 100g sample has 87.5 g N and 12.5 g hydrogen.
Number of mols of N = 87.5 g / 14.01 gmol-1 = 6.25 mol
Number of mols of H = 12.5 g / 1.008 gmol-1 = 12.50 mol
A chemical formula is a actually the mol ratio of the atoms of the molecule.
Mol ratio of N = 12.50/6.25 = 2
Mol ratio of H = 12.50/12.50 = 1
So the “chemical formula” of hydrazine you derive is NH2!
But the actual molecular formula of hydrazine is N2H4.
Mass percent information always provides the simplest whole number ratio of the
atoms. It may or may not be the same as actual mol ratio. See the next slide..
FoS-1&2 Chemistry, NYUAD
Empirical & Molecular Formula
Molecular formula provides the actual number ratio in a molecule
(or the actual mole ratio in a mole)
The molecular formula of hydrazine is N2H4
H
= 4:2
N
The smallest whole number ratio of the atoms in a compound is called the
empirical formula
The empirical formula of hydrazine is NH2
H
= 2:1
N
In many cases, where the such whole number reduction is not possible,
molecular formula is identical to the empirical formula. e.g., H2O, CH4, SO3
etc.
FoS-1&2 Chemistry, NYUAD
Why Empirical formula is important?
Many times you are presented with the task of experimentally
identifying an unknown compound. Breaking down the sample would
provide the mass percentage of the constituent elements and these
values leads to empirical formula (and not molecular formula)
Empirical formula leads to molecular formula through the following relation
Molecular formula = n (empirical formula) n = 1, 2,3,4…
The value of n is calculated by taking the ratio of molecular or molar mass to
empirical formula mass (in amu/in g)
n for hydrazine is =
molar mass
empirical formula mass
=
30.04 𝑔
15.2 𝑔
= 2
Molecular formula of hydrazine = 2 (NH2) = N2H4
FoS-1&2 Chemistry, NYUAD
Empirical formula calculations
Construction of empirical formula from mass %
Since the numbers in a formula are the mol ratios (and not mass ratios), the mass of each element
should be converted into moles. These mol numbers should then be converted mol ratios and into
whole numbers. The various steps are as follows;
a) Assume a 100.0 g sample.
• % becomes mass in grams
b) Divide each mass by its atomic mass.
• Gives the number of moles of each (in 100 g).
c) Divide each by the smallest answer found.
• The smallest integer ratio = empirical formula.
d) multiply all the subscripts by the smallest
integer to produce whole number subscripts
FoS-1&2 Chemistry, NYUAD
Empirical formula calculations
1) A sample of an unknown compound contains 0.21 mol of zinc, 0.14 mol of
phosphorus, and 0.56 mol of oxygen. What is its empirical formula?
Using the numbers of moles of each element given, we write the preliminary formula
Zn0.21P0.14O0.56
Next we divide each fraction by the smallest one; in this case 0.14:
0.21
= 1.5
0.14
0.14
= 1.0
0.14
0.56
= 4.0
0.14
This gives Zn1.5P1.0O4.0
We convert to whole numbers by multiplying by the smallest integer that gives whole
numbers; in this case 2:
1.5 x 2 = 3
1.0 x 2 = 2
4.0 x 2 = 8
This gives us the empirical formula Zn3P2O8
FoS-1&2 Chemistry, NYUAD
Empirical formula calculations
2) An orange compound is 26.6% K, 35.4% Cr and 38.0% O. Determine its
empirical formula.
Start with converting the grams to moles
26.6 g
×
35.4 g
×
38.0 g
×
mol
39.10 g
mol
52.00 g
1 mol O
16.00 g O
= 0.6803 mol K
= 0.6808 mol Cr
= 2.375 mol O
FoS-1&2 Chemistry, NYUAD
Empirical formula calculations
Empirical formula = smallest integer ratio.
Divide by the smallest number
K
0.6803 mol = 1.000
0.6803 mol
x2
2
Cr
0.6808 mol = 1.001
0.6803 mol
x2
2
O
2.375 mol = 3.491
0.6803 mol
x2
7
Choose a multiplier
to make integer
Empirical formula: K2Cr2O7
FoS-1&2 Chemistry, NYUAD
Conversion of Empirical formula to Molecular Formula
1) A compound is found to be 64.82 % carbon, 21.59 % oxygen and 13.59 %
hydrogen. What is the molecular formula for this compound? Molar mass of the
compound was determined to be 222.1 g/mol
a) The empirical formula calculated is C4H10O (check yourself)
b) The empirical formula weight for C4H10O is 74.12 g/mol (check yourself)
c) Determining the value of n
d) Determining the molecular formula
FoS-1&2 Chemistry, NYUAD
Mass percent of unknown compounds
Unknown samples (especially those containing C and H) are combusted with oxygen
and the mass of the resulting products (CO2 & H2O) provide the mols of C & H and
thereby the empirical formula. The mass of a third element if present, is calculated by
subtracting the mass of C & H from the total mass of the sample.
FoS-1&2 Chemistry, NYUAD
Determining a Molecular Formula from Combustion Analysis
PROBLEM:
When a 1.000 g sample of vitamin C (M = 176.12 g/mol) is placed in a
combustion chamber and burned, the following data are obtained:
mass of CO2 absorber after combustion = 85.35 g
mass of CO2 absorber before combustion = 83.85 g
mass of H2O absorber after combustion = 37.96 g
mass of H2O absorber before combustion = 37.55 g
What is the molecular formula of vitamin C?
PLAN: The masses of CO2 and H2O produced will give us the
masses of C and H present in the original sample. From
this we can determine the mass of O.
FoS-1&2 Chemistry, NYUAD
SOLUTION:
For CO2: 85.35 g - 83.85 g = 1.50 g
1.50 g CO2 x 12.01 g C = 0.409 g C
44.01 g CO2
For H2O: 37.96 g - 37.55 g = 0.41 g
0.41 g H2O x
2.016 g H
= 0.046 g H
18.02 g H2O
mass of O = mass of vitamin C – (mass of C + mass of H)
= 1.000 g - (0.409 + 0.046) g = 0.545 g O
FoS-1&2 Chemistry, NYUAD
Convert mass to moles:
0.409 g C
= 0.0341 mol C
12.01 g/mol C
0.545 g O
0.046 g H
= 0.046 mol H
1.008 g/mol H
= 0.0341 mol O
16.00 g/mol O
Divide by smallest to get the preliminary formula:
0.046
0.0341
= 1.3
=1
H
O
C
0.0341
0.0341
C1H1.3O1 = C3H3.9O3
0.0341
=1
0.0341
C 3 H 4 O3
Divide molar mass by mass of empirical formula:
176.12 g/mol
88.06 g/mol
= 2.000
C 6 H 8 O6
FoS-1&2 Chemistry, NYUAD
Chemical Equations
Reactants
Products
A chemical equation is a symbolic representation of a
transformation: of the substances, their physical state,
quantities, and the relative quantities (both in number and
mass) of the substances involved.
C25H52(s) + 38 O2(g)
hydrocarbon
in candle wax
25 CO2(g) + 26 H2O(g)
oxygen
carbon
dioxide
water
•Physical states are often listed:
(g)
(ℓ)
gas
liquid
(s) solid
(aq) aqueous (dissolved in water)
FoS-1&2 Chemistry, NYUAD
Chemical Equations
•Chemical equations must comply to the the law of conservation of
mass (Lavoisier 1789).This is ensured by balancing the equation with
necessary coefficients on either side. These stoichiometric numbers
stand for the relative number and moles of the reactants and products.
A stoichiometric
coefficient
2 C2H6(g) + 7 O2(g) → 4 CO2(g) + 6 H2O(ℓ)
Nanoscale 2 molecules
Macroscale 2 moles
MC2H6 (molar mass)
2(30.0)= 60.0 g
A chemical equation reflects
the stoichiometry in both
the microscopic as well as
the macroscopic world
7 molecules
4 molecules
6 molecules
7 moles
4 moles
6 moles
MCO2
MO2
MH2O
7(32.0)= 224.0 g → 4(44.0)= 176.0 g 6(18.0)= 108.0 g
284.0 g
→
284.0 g
FoS-1&2 Chemistry, NYUAD
Information Contained in a Balanced Equation
Viewed in
Terms of
Reactants
C3H8(g) + 5 O2(g)
Molecules 1 molecule C3H8 + 5 molecules O2
Amount (mol)
1 mol C3H8 + 5 mol O2
Mass (amu) 44.09 amu C3H8 + 160.00 amu O2
Mass (g)
Total Mass (g)
44.09 g C3H8 + 160.00 g O2
204.09 g
Products
3 CO2(g) + 4 H2O(g)
3 molecules CO2 + 4 molecules H2O
3 mol CO2 + 4 mol H2O
132.03 amu CO2 + 72.06 amu H2O
132.03 g CO2 + 72.06 g H2O
204.09 g
FoS-1&2 Chemistry, NYUAD
Calculating Quantities of Reactants and Products: Application of
Reaction coefficient.
1) Copper is obtained from copper(I) sulfide by roasting it in the presence of
oxygen gas to form powdered copper(I) oxide and gaseous sulfur dioxide.
How many moles of oxygen are required to roast 10.0 mol of copper(I)
sulfide? The balanced chemical equation is given below.
2Cu2S (s) + 3O2 (g) → 2Cu2O (s) + 2SO2 (g)
By looking at the molar ratio, it’s clear the 3 moles of O2 is required for consuming 2
moles of Cu2S
Hence, the amount of O2 required for roasting 10.0 mol of Cu2S is;
10.0 mol Cu2S x
3 mol O2
= 15.0 mol O2
2 mol Cu2S
FoS-1&2 Chemistry, NYUAD
Balancing Chemical Equations
1.
Write an unbalanced equation with correct formulas for all
substances.
2.
Balance the atoms of one of the elements.
i. Start with the most complex molecule.
ii. Change the stoichiometric coefficients.
iii. Do NOT alter the chemical formulas.
3. Balance the remaining elements.
FoS-1&2 Chemistry, NYUAD
Balancing Chemical Equations
1) Combustion of rocket fuel:
C2H8N2 + N2O4 → N2 + H2O + CO2
2C + 8H + 4N + 4O
2N + 2H + 3O + 1C
Not
balanced
• Balance C and H in C2H8N2 first:
•
C2H8N2 + N2O4 → N2 + 4 H2O + 2 CO2
2C + 8H + 4N + 4O
2N + 8H + 8O + 2C
• Still not balanced. Adjust N and O
Balanced
C2H8N2 + 2 N2O4 → 3 N2 + 4 H2O + 2 CO2
FoS-1&2 Chemistry, NYUAD
Balancing Chemical Equations
2) Balance : Al + Fe2O3 → Al2O3 + Fe
step 1
Al
1 Al
+ Fe2O3 →
Al2O3 +
(2Fe + 3O) (2Al + 3O)
Fe
1Fe
not balanced
balance Fe
from Fe2O3
not balanced
step 2
Al
1 Al
+ Fe2O3 →
Al2O3 + 2 Fe
(2Fe + 3O) (2Al + 3O)
2Fe
balanced
not balanced
step 3
2 Al + Fe2O3
2Al
→
Al2O3 + 2 Fe
(2Fe + 3O) (2Al + 3O)
2Fe
FoS-1&2 Chemistry, NYUAD
Balancing Equations: Practice
1
C3H8(g) +
5
3 CO (g)
2
2 B H (g)
4 10
+
11
→
O2(g)
+
4
H2O(g)
O2(g) →
4
B2O3(g) +
10 H2O(g)
FoS-1&2 Chemistry, NYUAD
LIMITING REACTANT
A stochiometric mixture is one that contain relative amounts of reagents as per the
balanced chemical equation. e.g., Suppose you are making ammonia as per the
balanced equation
The actual H:N ratio (15:5) is the same as the
ratio in the balanced equation. Clean reaction.
Now consider another container;
(H:N = 9:5)
Unreacted nitrogen. Impure reaction mixture
The reagent in short supply (H here) “limits” the quantity of the product formed. The
unreacted regent is referred to as the excess reagent. Quantity of product must be
calculated based on the limiting reagent (if present).
Pics taken from
the textbook
of Zumdahl
FoS-1&2
Chemistry,
NYUAD
Limiting Reagent
•Given 10 slices of cheese and 14 slices of
bread. How many sandwiches can you
make?
“Balanced equation”
1 cheese + 2 bread
1 sandwich
Only 7 sandwiches can be made as the bread runs out (limiting reagent)
Limiting reagent is the one that produces the least amount of product
FoS-1&2 Chemistry, NYUAD
Limiting Reagent
1) Chlorine trifluoride, an extremely reactive substance, is formed as a gas
by the reaction of elemental chlorine and fluorine. The molecular scene
shows a representative portion of the reaction mixture before the reaction
starts. (Chlorine is green, and fluorine is yellow.)
Copyright © The McGraw-Hill Companies, Inc.
Permission required for reproduction or display.
(a) Find the limiting reactant.
(b) Write a reaction table for the process.
FoS-1&2 Chemistry, NYUAD
SOLUTION: The balanced equation is Cl2 (g) + 3F2 (g) → 2ClF3 (g)
As per the equation, 1 molecule of Cl2 is stoichiometrically equal to 3
molecules of F2 (1:3).
We have 3 molecules of Cl2 and 6 molecules of F2, the ratio is 1:2.
All the 6 F2 molecules would be consumed by the time 2 molecules of Cl2
reacts.
Since F2 limits the amount of product formed, it’s the limiting reagent.
Another way of figuring out the liming reagent is to see which reactant in the given molar
quantity produces the lowest amount of product. That will be the limiting reagent.
E.g., here 3 molecules/moles of Cl2 is capable of producing 6 molecules/moles of ClF3, while
6 molecules/moles of F2 can produce only 4 molecules/moles of ClF3. So F2 is the liming
reagent.
FoS-1&2 Chemistry, NYUAD
We use the amount of F2 to determine the “change” in the reaction table,
since F2 is the limiting reactant:
Molecules Cl2 (g)
+
3F2 (g) → 2ClF3 (g)
Initial
Change
3
-2
6
-6
0
+4
Final
1
0
4
The final reaction scene shows that all the F2 has reacted and that there is
Cl2 left over. 4 molecules of ClF3 have formed:
FoS-1&2 Chemistry, NYUAD
Reaction Yields
The theoretical yield is the amount of product calculated using the molar
ratios from the balanced equation.
Often theoretical amounts of products cannot be isolated in reactions (list a few
factors on why..).
The actual yield of product is given as a percentage of the theoretical yield. This
is called percent yield.
% yield =
actual yield
x 100
theoretical yield
FoS-1&2 Chemistry, NYUAD
Percent Yield
2.50 g of copper heated with an excess of sulfur made
2.53 g of copper(I) sulfide
•
16 Cu(s) + S8(s)
→ 8 Cu2S(s)
•What was the percent yield for this reaction? (Formula
mass of Cu is 63.5 gmol-1 , and that of Cu2S is 159 gmol-1)
nCu used:
1 mol
2.50 g
= 0.03934 mol Cu
63.55g
16 mol Cu used is equivalent to 8 mol Cu2S made
Theoretical yield in mols:
0.03934 mol Cu 8 Cu2S = 0.01967 mol Cu2S
16 Cu
FoS-1&2 Chemistry, NYUAD
Percent Yield
Theoretical yield in grams
159.2 g
= 0.01967 mol Cu2S 1 mol = 3.131 g Cu2S
Actual yield = 2.53 g Cu2S (in problem)
2.53 g
Percent yield =
x 100% = 80.8%
3.131 g
FoS-1&2 Chemistry, NYUAD