Practice Exercise

Write a set of quantum numbers for each

of the electrons with an n of 4 in a Se
atom.
(4, 0, 0, +1/2)
(4, 0, 0, -1/2)
(4, 1, -1, +1/2)
(4, 1, -1, -1/2)
(4, 1, 0, +1/2
(4, 1, 1, -+/2)
2
Stoichiometry
 • Avogadro’s number
Topic • The mole concept
Outline • Percent composition and chemical
formulas
• Chemical reactions and chemical
equations
• Mass relationships in chemical
reactions
Think of these:
o A pair of shoes
o A dozen donuts
o A ream of papers

5
1. MOLE
 ⊙ is the SI unit for the amount of substance.
MOLE

⊙ the mole is a convenient way for chemists to

express the quantity of substances containing
very large number of atoms.
3. Moles
and Mass
 ⊙ Molar mass is the mass in grams of 1 mole of a
substance.
Number of
moles and
⊙ 1 mole of carbon-12 has a mass of 12 g and contains
mass 6.022 × 1023 carbon-12 atoms. The molar mass of
carbon-12 therefore is 12 g.

⊙ Molar mass of an element (g) = atomic mass (amu)

 ⊙ The molecular mass also called molecular
weight is the sum of the atomic masses (amu) in
Molecular and the molecule.
Formula Mass
How to calculate molecular mass?
⊙ In general, we need to multiply the atomic mass
of each element by the number of atoms of that
element present in the molecule and get the
sum for all of the elements.
 Example: Calculate the Molecular Mass of H2O
Molecular Mass H = 2 X 1.01 amu
O = 16.00 amu

H2O = (2 X 1.01 amu) + 16.00 amu = 18.02 amu

The molar mass of water therefore is 18.02

grams/mole
 ⊙ The formula mass is the sum of the atomic masses (in
amu) in a formula unit of an ionic compound.
⊙ The procedure for getting the formula mass of an
Formula Mass ionic compound is similar to the procedure in
computing for the molecular mass.

Example: Computing the Formula mass of NaCl

Na- 22.99 g/mol

Cl- 35.45 g/mol
NaCl = 22.99 + 35.45 = 58.44 g/mol

1 mole of NaCl has a mass of 58.44 grams.

Empirical &
Molecular
Formula
 Percent The percent composition by mass of a compound is the
Composition percent by mass of each element in a compound.

𝑝𝑒𝑟𝑐𝑒𝑛𝑡 𝑐𝑜𝑚𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎𝑛 𝑒𝑙𝑒𝑚𝑒𝑛t

𝑛 ×𝑚𝑜𝑙𝑎𝑟 𝑚𝑎𝑠𝑠 𝑜𝑓 𝑒𝑙𝑒𝑚𝑒𝑛𝑡

= ×100
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑐𝑜𝑚𝑝𝑜𝑢𝑛𝑑
Sample Computations

Determine the percent composition of each element in the compound glucose.

(C6H12O6)

6 ×12.01 𝑔
%𝐶 = ×100 = 𝟑𝟗. 𝟗𝟗 %
180.18 𝑔

12 ×1.01 𝑔
%𝐻 = ×100 = 𝟔. 𝟕𝟑 %
180.18 𝑔

6 ×16.00 𝑔
%𝑂 = ×100 = 𝟓𝟑. 𝟐𝟖 %
180.18 𝑔
 1. How many moles are in 0.661 g of Potassium (K)?
Review 2. Compute the weight, in grams of 100 moles NaCl
Empirical ⊙the simplest chemical formula is called the empirical
Formula formula.

⊙the empirical formula shows the simplest whole-

number ratio of the atoms present in a compound.

⊙the empirical formula of a compound can be

determined using the percent composition
Procedure for Determining Empirical Formula

Percent by Convert into grams

mass and divide by molar
mass

Number of Divide by the smallest

Moles of the number of moles
element

Mole Ratios Change to integer

of the subscripts
elements

EMPIRICAL
FORMULA
Empirical What is the empirical formula of a compound containing
Formula 70.19 % Pb, 8.14 % C, and 21.67% O?

Ø Convert the percent by mass into grams by assuming the

sample is 100 g. Divide the mass of each element by
their respective molar masses.
Empirical and Molecular Formula

Pb = 70.19 g, C = 8.14 g, O = 21.67 g

1 𝑚𝑜𝑙 𝑃𝑏
𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑃𝑏 = 70.19 𝑔 𝑃𝑏× = 0.3388 𝑚𝑜𝑙 𝑃𝑏
207.2 𝑔 𝑃𝑏

1 𝑚𝑜𝑙 𝐶
𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝐶 = 8.14 𝑔 𝐶× = 0.678 𝑚𝑜𝑙 𝐶
12.01 𝑔 𝐶

1 𝑚𝑜𝑙 𝑂
𝑛𝑜. 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑂 = 21.67 𝑔 𝑂× = 1.354 𝑚𝑜𝑙 𝑂
16.00 𝑔 𝑂
Empirical and Molecular Formula

Divide the computed number of moles of each element by the smallest

number of moles among the elements.

0.3388 𝑚𝑜𝑙 0.678 𝑚𝑜𝑙 1.354 𝑚𝑜𝑙

𝑷𝒃: =1 𝑪: ≈2 𝑶: ≈ 4.00
0.3388 𝑚𝑜𝑙 0.3388 𝑚𝑜𝑙 0.3388 𝑚𝑜𝑙

Change into integer subscripts and write the complete empirical formula.
PbC2O4
Exercise:

Determine the empirical formula of a compound that has the following

percent composition by mass:
K=24.74%
Mn=34.76%
O=40.50%
 ⊙The molecular formula shows the exact number of
atoms of each element in a molecule. It represents
Molecular the actual formula of a molecule.

Formula
⊙The empirical formula may or may not be the same
as the molecular formula

⊙The molecular formula is a whole number multiple

of the empirical formula

⊙ (Empirical formula)n = molecular formula

 Determine the empirical formula of the compound

Molecular
Formula

d
Calculate the empirical formula mass

Compute the value of n, where

d
𝑚𝑜𝑙𝑒𝑐𝑢𝑙𝑎𝑟 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠
𝑛=
𝑒𝑚𝑝𝑖𝑟𝑖𝑐𝑎𝑙 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 𝑚𝑎𝑠𝑠

Derive the molecular formula,

d
(Empirical formula)n = molecular formula

24
Determination of Molecular Formula
⊙ Caffeine found in coffee and tea contains 49.5% C, 5.20% H, 28.8 % N,
and 16.5% O by mass.

⊙ Its molecular mass is 194.1 g/mol. Find the molecular formula of

caffeine.
Determination of Molecular Formula
1 𝑚𝑜𝑙𝑒 𝐶
𝒏𝒐 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑪: 49.5 𝑔 𝐶 × = 4.12 𝑚𝑜𝑙
12.01 𝑔 𝐶
1 𝑚𝑜𝑙𝑒 𝐻
𝒏𝒐 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑯: 5.20 𝑔 𝐻 × = 5.15 𝑚𝑜𝑙
1.01 𝑔 𝐶
1 𝑚𝑜𝑙𝑒 𝑁
𝒏𝒐 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑵: 28.8 𝑔 𝑁 × = 2.06 𝑚𝑜𝑙
14.01 𝑔 𝑁
1 𝑚𝑜𝑙𝑒 𝑂
𝒏𝒐 𝒐𝒇 𝒎𝒐𝒍𝒆𝒔 𝒐𝒇 𝑶: 16.5 𝑔 𝑂 × = 1.03 𝑚𝑜𝑙
16.00 𝑔 𝑂

4.12 𝑚𝑜𝑙 5.15 𝑚𝑜𝑙 2.06 𝑚𝑜𝑙 1.03 𝑚𝑜𝑙

𝑪: =4 𝑯: = 5 𝑵: = 2 𝑶: = 1
1.03 𝑚𝑜𝑙 1.03 𝑚𝑜𝑙 1.03 𝑚𝑜𝑙 1.03 𝑚𝑜𝑙
ü Empirical Formula: C4H5N2O
Determination of Molecular Formula

ü Calculate the molar mass of the empirical formula.

4 12.01 𝑔 + 5 1.01 𝑔 + 2 14.01 + 16.00 𝑔 = 𝟗𝟕. 𝟏𝟏 𝒈
ü Compute for n. (Empirical formula)n = molecular formula
!"#$% !$'' "( !"#)*+#$% ("%!+#$
Where 𝑛 = !"#$% !$'' "( )!,-%-*$# ("%!+#$

194.1 𝑔/𝑚𝑜𝑙
𝑛= ≈2
97.11 𝑔/𝑚𝑜𝑙
ü Write the molecular formula
Molecular Formula: (C4H5N2O)2 = C8H10N4O2
 Determine the empirical formula and molecular formula of a
compound that is found to have 68.54% C, 8.63% H and 22.83%
O. The molecular weight of the compound is known to be 140
g/mol.
Exercise
Chemical Reactions
and Chemical
Equations

Introduction to Mass Relationships

29
Guide Questions

1. What is Chemical Reaction?

2. What is Chemical Equation?
3. What are the symbols used in
making chemical equation?

30
Chemical ⊙ A chemical reaction is a process in which one or
Reaction more substances is changed into one or more
new substances.
V/S
Chemical
⊙ A chemical equation is the symbolic
Equation representation of a chemical reaction. It uses
symbols and chemical formulas to illustrate
what happens in a chemical reaction.
Chemical ⊙ Reactants are the starting materials in a
Equation chemical reaction.

⊙ Products are the substances formed as a result

of the chemical reaction. A chemical reaction
can have one or more products.
Symbols + this symbol means reacts with, is added to, or
used in combines with.
Chemical
→ this symbol means produces, yields, or forms.
Equation
(aq)- Aqueous solution, dissolved in water
(s)- Solid
(l) – liquid
(g) – gas
∆- heat
 Reading
Chemical
Equations ⊙ 1 mole of methane reacts with 2 moles of
oxygen to produce 1 mole of carbon dioxide and
2 moles of water
 Why do we balance chemical equations?
Balancing
Chemical ⊙ Chemical equations must be balanced in order
Equations to conform to the law of conservation of mass.

⊙ In chemical reactions, atoms are only

rearranged, therefore there must be the same
number of each type of atom on the reactant
and product side.
 Balancing Chemical Equations
Balancing
Chemical 1. Count the atoms of each element in the
Equations reactants and the products.
2. Begin balancing the equation by adjusting the
coefficients to make the number of atoms of
each element the same on both sides of the
equation.
3. Check if the equation is already balanced. Make
sure that you have the same total number of
each type of atoms on both sides of the
equation.
 Strategies
Balancing
Chemical Look for elements that appear only once on each
Equations side of the equation that have unequal numbers of
atoms.
Try this:
C3H8 + O2 ® CO2 + H2O
N2 + H2 ® NH3
Practice Exercise
Balance the following chemical equations:

1. Na + MgF2 ® NaF + Mg
2. Na + O2 ® Na2O
3. Na + HCl ® H2 + NaCl
4. Al(OH)3 + H2CO3 ® Al2(CO3)3 + H2O
 Mass
relationships
in a chemical
reaction
 The study of the quantitative relationships between
Stoichiometry the reactants and products in a chemical reaction.

⊙ “How much product will be formed from

specific amounts of reactants?”

⊙ “How much of the reactant/s must be used to

obtain a certain amount of product?”
Calculating Amount of Reactants and Products

⊙ Mole method of solving stoichiometry problems

⊙ In mole method, the stoichiometric coefficients in a balanced chemical
equation is interpreted as the number of moles of each substance.

N2 + 3H2 ® 2NH3
1 mole of N2 combines with 3 moles of H2 to form 2 moles of NH3 gas
Molar Ratios

⊙ In stoichiometric calculations, the coefficients are the considered as the

molar ratios.

⊙ The said relationships/ molar ratios can also be written as conversion factors.
1N2 + 3H2 ® 2NH3

2 𝑚𝑜𝑙 𝑁𝐻. 3 𝑚𝑜𝑙 𝐻/ 1 𝑚𝑜𝑙 𝑁/

1 𝑚𝑜𝑙 𝑁/ 2 𝑚𝑜𝑙 𝑁𝐻. 3 𝑚𝑜𝑙 𝐻/
Stoichiometry Sample Problems:

KClO3 ® KCl + O2
How many moles of O2 are produced by the decomposition of six moles of
potassium chlorate?

STRATEGY: Write the balanced chemical equation.

2KClO3 ® 2KCl + 3O2

Use the mole ratio from the balanced equation to calculate the number of moles of oxygen produced
from 6 moles of potassium chlorate.

3 𝑚𝑜𝑙 𝑂"
6.0 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂! × = 𝟗. 𝟎 𝐦𝐨𝐥𝐞𝐬 𝐨𝐟 𝐎𝟐
2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂!
Moles to Mass

2KClO3 ® 2KCl + 3O2

b) What is the mass of potassium chlorate needed to produce 2.50 moles of
oxygen gas?

2 𝑚𝑜𝑙𝑒𝑠 𝐾𝐶𝑙𝑂. 122.55 𝑔

2.50 𝑚𝑜𝑙𝑒𝑠 𝑂/ × × = 𝟐𝟎𝟒 𝒈 𝑲𝑪𝒍𝑶𝟑
3 𝑚𝑜𝑙𝑒𝑠 𝑂/ 1 𝑚𝑜𝑙𝑒 𝐾𝐶𝑙𝑂.
Mass to Mass Calculations

C) Calculate the mass of oxygen gas (in grams) that can be obtained from 46.0 g
of KClO3.

1 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂. 3 𝑚𝑜𝑙 𝑂/ 32 𝑔

46.0 𝑔 𝐾𝐶𝑙𝑂. × × × = 𝟏𝟖. 𝟎 𝒈 𝑶𝟐
122.55 𝑔 𝐾𝐶𝑙𝑂. 2 𝑚𝑜𝑙 𝐾𝐶𝑙𝑂. 1 𝑚𝑜𝑙 𝑂/
 Given the reaction: N2 + H2 ® NH3
RECALL !
Suppose 16.0 g of H2 reacts completely with N2 to
form NH3. How many grams of NH3 will be formed?

Make sure to balance the equation!

 Given the chemical equation
Try this! C2H2 + O2 ® CO2 + H2O

a. Write the balanced chemical equation.

b. How many moles of CO2 can be produced from
3.60 mol C2H2?
c. What is the mass of O2 needed to produce 1.10
moles of CO2?
d. How many grams of H2O is produced when 113 g
of C2H2 is reacted?
 Limiting
Reagents &
Reaction
Yield
Limiting and
Excess
Reactants
 ⊙The limiting reactant is consumed first in a chemical
reaction.
Limiting ⊙It is called a limiting reactant because it limits the
extent of a chemical reaction.
Reagent
⊙ The maximum amount of product formed depends
on how much limiting reagent was originally present.

⊙The reactants present in quantities greater than

Excess necessary to react with the quantity of the limiting
Reagent reagent.
⊙These reactants are what is left after the limiting
reactant is fully consumed.
Determining Steps in determining the limiting reactant
Limiting
Reactant ⊙One way of finding the limiting reagent is by
calculating the amount of product that can be
formed by each reactant; the one that produces
less product is considered to be the limiting
reagent.
Determining NH3 + O2 ® NO + H2O
Limiting
Reactant In an experiment, 3.25 g of NH3 are allowed to react
with 3.50 g of O2.

a. Which reactant is the limiting reagent?

b. How many grams of NO are formed?
c. How much of the excess reactant remains in grams
after the reaction?
 The percent yield of a chemical reaction is the percent
Reaction ratio of the actual yield to the theoretical yield. It is
Yield calculated using the formula:

𝒂𝒄𝒕𝒖𝒂𝒍 𝒚𝒊𝒆𝒍𝒅
% 𝒚𝒊𝒆𝒍𝒅 = ×𝟏𝟎𝟎
𝒕𝒉𝒆𝒐𝒓𝒆𝒕𝒊𝒄𝒂𝒍 𝒚𝒊𝒆𝒍𝒅
Actual and The theoretical yield of a chemical reaction is the
amount of product (maximum) that would result if all
Theoretical the limiting reagent reacted.
Yield
The actual yield on the other hand is the amount of
product that was actually obtained from a chemical
reaction.
Actual and ⊙In the laboratory, percent yield is almost always less
than 100 percent; this means that the actual yield is
Theoretical less than the theoretical yield.
Yield
⊙In experiments, it is difficult to obtain 100 percent
yield because the limiting reactant is not always fully
consumed in most chemical reactions.
 NH3 + O2 ® NO + H2O
….
In an experiment, 3.25 g of NH3 are allowed to react
with 3.50 g of O2.

a. Which reactant is the limiting reagent?

b. How many grams of NO are formed?
c. How much of the excess reactant remains after the
reaction?
d. What is the % yield of the reaction if only 2.13 g of
NO was produced?
 Given the chemical reaction:
Reaction
Yield FePO4 + Na2SO4 ® Fe2(SO4)3 + Na3PO4

A. If 25 grams of iron (III) phosphate reacts with excess

sodium sulfate, what is the theoretical yield for the
chemical reaction?

B. Calculate the percent yield of the reaction if 18.5

grams of iron (III) sulfate was actually obtained from the
reaction.
 MnO2 + 4HCl → Cl2 + MnCl2 + 2H2O
Assignment
If 0.86 mole of MnO2 and 48.2 g of HCl react,

A. Which reagent will be used up first?

B. How many grams of Cl2 will be produced?
C. Calculate the reaction yield if only 20.1 g of Cl2 was
produced.