St.

Louise de Marillac College of Bogo

Member: Daughters of Charity-St. Louise de Marillac Educational System General Chemistry 1 Module

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The St. Louise de Marillac College of Bogo employs FLEx (Flexible Learning Experience) Delivery Mode for
SY 2020 – 2021. The chosen delivery mode operates on four basic principle: (1) Principle of Individuality and
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MODULE 3: STOICHIOMETRY

Foreword

You pour vinegar into a glass of water containing baking soda and bubbles
form. You heat sugar in a pan, and it turns brown. The bubbles and color
change are visual evidence that something is happening. To an experienced
eye, these visual changes indicate a chemical change or chemical reaction.
In this module, we begin to explore some important aspects of chemical
change. Our focus will be both on the use of chemical formulas to represent
reactions and on the quantitative information we can obtain about the
substances involved in reactions.

In this module, you will learn the following:

Most Essential Learning Christian- Sustainable
Lesson No. Topics Competencies Vincentian Developmental
Values Goals
Lesson # 1 PERCENT ● Calculate the
COMPOSITION empirical formula from SDG#4:
AND CHEMICAL the percent Flexibility Quality
FORMULA composition of a and Education
compound Adaptability
(STEM_GC11PCIf-32)
● Calculate molecular
formula given molar Co-
mass (STEM_GC11PCIf- responsibility
32)
Lesson # 2 CHEMICAL ● Write and balance
REACTIONS chemical reactions
(STEM_GC11RIf-g-37)
● Construct mole or

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 mass ratio for a
reaction in order to
calculate the amount
of reactant needed or
product formed in
terms of moles or mass
(STEM_GC11MRg-h-38)
● Calculate % yield and
theoretical yield of the
reaction
(STEM_GC11MRIg-h-
39)
● Explain the concept of
limiting reagent in a
chemical reaction,
identify the excess
reagent
(STEM_GC11MRIg-h-
40)
● Determine mass
relationship in a
chemical reaction
((STEM_GC11MRIg-h-
42)

LESSON 1: PERCENT COMPOSITION AND CHEMICAL FORMULA

REVIEW ASSESSMENT:
Identify which of the following is an atom, a molecule or an ion.
molecule 1. Oxygen gas atom 4. Helium
ion 2. Na+ ion 5. O-2
molecule 3. Salt ion 6. fluoride

PRE-ACTIVITY:
Read about stoichiometry in your book and based on your reading,
define the following terms:
1. mole- collective units of all atoms.
2. molar mass- it is the total mass of the components in the molecule.__
3. molecular weight- the molecular weight id the mass of one mole of a
substance._____
4. dimensional analysis- is the process of converting between units.
5. conversion factor- a number used to change one set of units to another,
by multiplying or dividing.___
6. chemical formula- chemical formula is a way of presenting information
about the chemical proportions of atoms that constitute a particular
chemical compound or molecule.___
7. molecular formula- a chemical formula that gives the total number of
atoms of each element in each molecule of a substance.___
8. empirical formula- a chemical formula showing the simplest ratio of
elements in a compound rather than the total number of atoms in the
molecule. ___
9. limiting reagent - the substance which are entirely consumed in the
completion of a chemical reaction.____
10. theoretical yield-is a measure of the quantity of moles of a product
formed. ____

Stoichiometry is the quantitative relationship of reactants and products

in a chemical reaction. Stoichiometry makes use primarily the concept of
moles, mass and volume (for gases) in showing the relationship of the
reactants and products.
What is a mole? Imagine this paper. We know this paper is composed
of particles in it, in the form of atoms. If we have the chance to look closely
to the tiniest part of this paper, we can see millions of atoms or molecules. So,
instead of counting all the atoms in this paper, it is collectively expressed in
the unit of mole. According to Avogadro’s number, 1 mole is equivalent to
6.02 x 1023 particles ( i.e., atoms, ions, formula units or molecules).

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Example 1: How many atoms are present in 2 moles of oxygen?
Solution: To find the number of atoms in oxygen given the number of
moles, we make use of the Avogadro’s number (conversion factor).

Note that in the solution, the mole is in the denominator to be able to

cancel the unit of moles of oxygen that is given.

Example 2: How many moles of water will you have if you have 18.06 x 1023
molecules of water?
Solution: In this example we are given the molecules, so in the Avogadro’s
number, the molecules must be placed in the denominator.

However, there are no instruments that immediately measure moles. In

this case, the amount of the substance first is measured using the unit of
mass. We can measure the mass of a substance using a digital balance, a
weighing scale and any other measuring instruments.
How can we convert mass to moles and vice versa?

The same with the conversion of moles to atoms and vice versa, we
need to make use of a conversion factor. This process of conversion is called
dimensional analysis. The main idea is to cancel out same units appearing in
both the numerator and the denominator until you are left with the target
unit.
In the conversion from mass to moles and vice versa, we need the
molar mass of the substance. Molar mass is in the unit of grams (g) per
mole(mol). This value is the sum of all the atomic masses of the element
present in a compound.
Example 3: Calculate the mass, in grams, of 0.433 moles of calcium nitrate.
Solution:
Step 1: Analyze the problem, see what is given and what is
asked.
Identify the chemical formula of calcium nitrate. From
the previous module, we know that calcium ion is Ca 2+
and the nitrate ion is , thus, calcium nitrate is
Ca(NO3)2.
Step 2: Identify what conversion factor is needed to relate the
mass and moles. Identify the molar mass of Ca(NO3)2.
Ca = 40 g /mol x 1 =40 g /mol
N = 14 g/mol x 2 = 28 g/mol
O = 16 g/mol x 3(2)= 96 g/mol
Total : 164 g/ mol
Step 3: Start with the given information to solve the problem.
0.433 moles of Ca(NO3)2 = _____grams Ca(NO3)2
Step 4: Solve for the mass of calcium nitrate.

Reading Activity

Read more examples in the book page 91-93

Figure 1. Conversion of units

Figure 1 presented the summary of the conversion factors needed to

convert one unit to the other.

PRACTICE EXERCISE NO. 1

Solve the follow problems. Write your answers in a separate sheet
of paper.

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1. What is the mass (g) of 7.38 x 10 23 molecules of acetic acid (CH3 COOH)?
MOLE:
nCH3COOH=7.38x1023 molecules of CH3COOHx
nCH3COOH=1.23 moles CH3COOH
MOLES TO MASS:
mCH
3COOH=1.23 moles CH3COOHx m CH COOH=73.8 g CH3COOH
3

2. Find the mass of 3 mol of potassium perchlorate.

3 mol x = 215.65 g

3. How many atoms are there in 1.50 mol of Hg?

1.50 moles x = 9.05x10²³ atoms

4. Determine the number of molecules in 2.50 mol of O 2.

2.50 moles x = 1.05x10²4 molecules

5. How many atoms of hydrogen can be found in 24.6 g of ammonia?

mol NH3= =1.45
N=1.45x6.02x10²³
=8.729x10²³ atoms

CHEMICAL FORMULA
A chemical formula is a chemical symbol of a substance used to
illustrate the composition of a compound. Chemical formula can be further
categorized into molecular formula, empirical formula and formula unit.
Molecular formula (MF) is the actual chemical formula of a compound
that reflects the composition of a molecule. Empirical formula (EF) is the
chemical formula that shows the simple whole ratios among the atoms of
elements in the compound. Formula unit, on the other hand, is the chemical
formula of an ionic compound. Table 1 shows the difference between
molecular formula and empirical formula.
 Table 1. Molecular Formula vs. Empirical Formula
Compound Molecular Formula Empirical Formula
Water H2O H2O
Hexane C6H14 C3H7
glucose C6H12O6 CH2O

You can see that the empirical formula is jus like the simplest form of a
molecular formula.

Note: The numerical subscripts in the chemical formula are always expressed
as whole numbers. You will never encounter a compound with fractional
subscripts because fractions of atoms cannot form compounds.

The elements in the compound are stoichiometrically equivalent. This

relationship is used in solving chemical calculation problems. Take aluminum
chloride (AlCl 3) as an example. This compound has an equivalent ratios of Al
and Cl in terms of atoms and moles.

See Example 4 of page 96

Percentage Composition
Percent composition refers to the relative amounts of the components
of a compound or mixture expressed in terms of percentage. This percent
composition can be used to verify the purity of a compound by comparing
the calculated percentage composition of the substance with that found
experimentally. Forensic chemist, for example, will measure the percentage
composition of an unknown white powder and compare it to the
percentage composition for sugar, salt or cocaine to identify powder.
Calculating percent composition is a straightforward matter if the
chemical formula is known. It can be calculated using the formula:

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Example 4
Calculate the percentage of carbon, hydrogen and oxygen (by mass) in
C12H22O11.
Solution:
We can use the formula of % mass of element, use the periodic table for the
atomic mass of each component.
Step 1: Based on the chemical formula, we can see that there are 12
moles of carbon (C), 22 moles of hydrogen (H) and 11 moles of
oxygen (O) in the glucose compound.
Step 2: Solve for the molar mass of C12H22O11 for the mass of the substance.

( ) ( )

Step 3: Solve for the % composition of the component:

% C:

% H:

% O:

To check your answers, you add the % composition of each and it must have
a total of 100%

See Examples 5-7 of the book page 98-99

PRACTICE EXERCISE NO. 2
Solve the following problems. Write your answers in a separate sheet of
paper.

1. Compute the percentage of H, C and O in carbonic acid (H 2CO3).

( )ᐩ ( )ᐩ ( )=
62 g

% H= =3.2%

% C= =19.4%

% O= = 77.4%

2. Compute the percent composition of vanillin (C 8H8O3) extracted from

vanilla beans.
( ) ( )ᐩ

( )=152 g

% C= = 63.2%

% H= = 5.3%

% O= = 31.5%
3. Compute the mass of chromium contained in 35.8 g (NH4)2Cr2O7?
Nitrogen: 14 g/mol Hydrogen: 1 g/mol Chromium: 52 g/mol Oxygen:
16 g/mol
+ + + = 252 g/mol
4. Compute and compare the percentage of Cu present in the following
copper ores:
a. Chalcopyrite (CuFeS2)
Cu=1x64=64
Fe=1x56=56
S=2X32=64
=184
% Cu= x 100=34.8%
% Fe= x 100=30.4%

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 % S= x 100=34.8%
b. Tennantite (Cu12As4S13)
Cu=12x64=768
As=4x75=300
S=13X32=416
=1484
% Cu= x 100=51.8%
% As= x 100=20.2%
% S= x 100=28%
c. Cuprite (Cu2O)
Cu=2x64=128
O=1x16=16
=144

% Cu= x 100=88.9%
% O= x100=11.1%
How do we determine the chemical formula from the percent composition?

Figure 2 summarizes the steps on how to make use of the percent

composition to determine the empirical formula of the substance.

Figure 2. Steps for determining empirical formula from percent composition

Example 5
Ascorbic acid (Vitamin C) contains 40.92% , 4.58% H, and 54.50 % O
by mass. What is the empirical formula of ascorbic acid?
Solution
Step 1: Assume 100 gram sample. Thus, the mass of each element is:
Mass of C (m C) = 40.92 grams C
Mass of H (m H) = 4.58 grams H
 Mass of O (m O) = 54.5 grams O
Step 2: Convert each mass to moles.

Step 3: Determine the simplest whole number ratio of moles by

dividing each number of moles by the smallest number of
moles, in this case, we have 3.41 mols;

Thus, the ratio then is C:H:O= 1:1.33:1

Step 4: We make the ratio in to a whole number since numerical
subscripts of each element must be a whole number. The ratio
for H is too far from 1 to attribute the difference to
experimental error; however it is close to 1 . This suggests that is
we multiply the ratio by 3, we will obtain whole numbers.

3 [C:H:O= 1:1.33:1]=3:4:3
Thus, the empirical formula of ascorbic acid is C3H4O3.

How to determine the molecular formula from the empirical formula?

We obtain the molecular formula from the empirical formula if we are

given the molecular weight or molar mass of the substance. The subscripts in
the molecular formula of a substance are always a whole number multiple of
the corresponding subscripts in its empirical formula. This multiple can be
found by comparing the empirical formula weight with the molecular formula
weight.

Example 6:

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 Determine the molecular formula of the ascorbic acid given in Example
5. The molecular weight of an ascorbic acid is 176 g/mol.
Solution
Step 1: Solve for the molecular weight (MW) of the empirical formula of the
ascorbic acid. Recall that the empirical formula is C3H4O3

( ) ( )

Step 2: Solve for the whole number multiple

Step 3: Lastly multiply this whole number to the subscripts in the empirical
formula to get the molecular formula.
C(3x2)H(4x2)O(3 x2)
Thus, the molecular formula of ascorbic acid is C6H8O6.

See Example 8 in page 100-101 for another example.

PRACTICE EXERCISE NO. 3

Solve the following problems. Write your answers in a separate

sheet of paper.

1. Eugenol, a chemical substance with the flavor of cloves , consists of

73.19% C, 19.49% O and 7.37%H. What is its EF?
73.119% C, 19.49%, 7.37% H

moles C (nc)=73.19 g C x =6 mol C

moles O (no)=19.49 g O x =1.22 mol O

moles H (nH)=7.37 g H x =7.37 mol H

C= =5

O= =1

H= =6

C5H6O

2. The laboratory analysis of an unknown sample shows that it contains

164.9 g C, 27.5 g H, and 220.1 g O. What is the empirical formula of the
unknown sample?

164.9 g C, 27.5 g H, 220.1 g O

moles C=164.9 g C x = =13.74 mol C
moles H= 27.5 g H x = =27.5 mol H
moles O= 220.1 g O x = =13.8 mol O

C= =1
H= =2
O= =1
CH2O

3. An alcohol contains 64.81% C, 13. 60% H, and 21.59 % O. The molecular

mass of this compound is 74 g/mole. Determine its EF and MF.

64.81% C, 13.60 % H, 21.59% O

moles C= 64.81g C x =5.4 mol C
moles H= 13.60 g H x =13.60 mol H
moles O= 21.59 g O x =1.35 mol O
C= =4 ( )+( )+( )

H= =10 48 + 10 + 16 = 74
O= =1 MF: C4H10O
EF: C4H10O

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Christian Vincentian Value: Flexibility and Adaptability

Just like a chemical formula, our identity can be expressed as the

simplified version of us (empirical formula) or the complete version (molecular
formula) of us. But regardless of how we express ourselves, our composition
does not change. We tend to be flexible and adaptable to the situations but
remained the same person.
When was the last time you become flexible and adaptable? What
version of yourself did you show? Why?
When the Covid 19 arrived in the Philippines, I became more adaptable and
flexible. I am acclimating to the new situation. I became a more caring
person and became more aware of the situations that make me stronger and
more secure.

-PROBLEM SET 1

1.Determine the molar mass of each of the following compounds:

Final answer must be enclosed in a box.(3 points: 2 point solution, 1
point final answer)
a) Nitric acid
HNO3= 1 H + 1 N + 3(O)
= 1 (1) + 1 (14) + 3 (16)
= 1 = 4 + 48
= 63 g/mol
 b) KMnO4
= 1K + 1 Mn + 4 (o)
= 1 (39) + 1 (55) + 4 (16)
= 39 + 55 + 64
= 158 g/mol

c.) SiO4
= 1 S + 4 (O)
= 1 (28) + 4 (16)
= 28 + 64
= 92 g/mol
d.) butane
C4H10= 4 C + 10 H
= 4 (12) + 10 (1)
= 48 + 10
= 58 g/mol
e.) sucrose
C12H22O11= 12 C + 22 H + 11 (O)
= 12 (12) + 22 (1) + 11 (16)
= 144 + 22 + 16
= 342 g/mol

2. While Julio puzzles over mole-to mass conversions, a friend heats water in a
copper kettle and makes him a cup of tea. He tells Julio to place 0.0120
mol of table sugar (sucrose: C12H22O11) in the tea. What mass of sugar has
he added? (3 points: organization of solution-2, final answer-1)

C- 12(12)

H- 22(1.008)

O- 11(16)

= 342 g/mol

1 mol C12H22O11 = 342 g/mol C12H22O11

x = 4.104 g sugar (C12H22O11)

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3. A substance named Agorca M5640(officially 5-nonyl salucylaldoxine) is
used for concentrating extracted copper ore. Its molecular formula is
C16H25NO2. It has a molar mass of 263.4 g/mol. If you have a 150-g sample
of Agorca M5640, how many molecules do you have? (3 points
:organization of solution-2, final answer-1)

C- 12(16)

H- 1.008(25)

N= 14(1)

O= 16(2)

= 263.4 g/mol

x x

= 3.43 x 10²³ molecules of C16H25NO2

4. The mineral malachite is a beautiful green color, often with swirling bands
of different intensity. It is often used in jewelry because of its attraction
appearance. An analysis of malachite gives the following composition:
57.48% Cu, 5.43% C, 0.91% H and 36.18 % O. Determine:

a. the empirical formula of the malachite.(5 points: organization of

solution-3, final answer-1, neatness-1)
moles Cu (nCu) = 57.48 g Cu x Cu = =2
= 0.91 mol Cu
moles C (nC) = 5.43 g C x C= =1
= 0.5 mol C
moles H (nH) = 0.91 g H x H= =2
= 0.91 mol H
moles O (nO) = 36.15 g O x O= =5
= 2.3 mol O
Cu:C:H:O = Cu2CH2O5
 b. molecular formula (given with the molar mass of malachite to be
221.12 g/mol) (4 points: organization of solution-2, final answer-1,
neatness-1)

Cu - 63.5(2)

C - 12.011(1)

H - 1.008(2)

O - 16(5)

=221.12 g/mol

=(2 moles Cu x ) + (1 mole C x ) + (2 moles H x ) + (5

moles O x )

= 63 g + 12 g + 2 g + 80 = 157 g

whole number multiple = =1

Cu(2x1)C(1x1)H(2x1)O(5x1) = Cu2CH2O5

Chemical composition often requires us to know the number of atoms,

molecules, or ions in a sample. Because these values are so large, it was
collectively called in terms of moles. This number of moles tells us the relative
amounts of the elements that are in a compound which has constant
composition.
Knowing these values and conversions is important to be able to understand
more the next lesson which is the chemical reactions- involve the

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rearrangement of atoms that converts one set of substances into another set
of substances.

LESSON 2: CHEMICAL REACTION

Activity 1

Directions: Read the given situation below and fill out the blanks below.

Go to your kitchen. List three chemical reactions that you can observe in your
kitchen and give their importance in our daily living.

1. Chemical Reaction: Combustion

Importance in your daily life: Combustion reactions are used to heat

everything that needs heat and generate our electricity.

2. Chemical Reaction: Washing Dishes

Importance in your daily life: to make our utensils clean and safe to use
again.

3. Chemical Reaction: Boiling of hard eggs

Importance in your daily life: excellent source of lean protein. They will
fill you up without packing too many calories, which is helpful if you
want to lose weight.

Chemical reactions are represented in a concise way by chemical

equations. Chemical equations are a shorthand for a chemical reactions and
its essential when planning an experiment.
When the gas hydrogen (H2)burns, for example, it reacts with oxygen (O2) in
the air to form water (H2O). We write the chemical equation for this reaction
as follows:

We read the + sign as “ reacts with” and the arrow (→) as “ produces”.
Symbols used in a chemical equation were presented in Table 5.1 of page
104 of the book.

Reading Activity

There are different types of chemical reactions. Read on each

type and see how they differ from each other. Read page 104 to 111 of the
book.

● Synthesis Reaction
● Decomposition Reaction
● Single Replacement Reaction
● Double Replacement Reaction
● Neutralization Reaction
● Combustion Reaction

BALANCING CHEMICAL EQUATIONS

We balance the chemical equation by determining the coefficients

that provide equal numbers of each type of atom on each side of the
equation. In most cases, a balanced chemical equation should contain the
smallest possible whole-number coefficients.

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 In balancing chemical equation, you need to understand the
difference between a coefficient in front of a chemical formula and a
subscript within a formula. You should never change the subscripts when
balancing an equation.In contrast, placing a coefficient in front of a formula
changes only the amount of the substance and not its identity.

Because atoms are neither created nor destroyed in any reaction, a

chemical equation must have an equal number of atoms of each element
on each side of the arrow. When this condition is met, the equation is said to
be balanced.

To illustrate the process of balancing an equation, consider the

combustion of reaction of methane.

CH4 + O2 → CO2 + H2O

Reactants Products

CH4 C =1
CO2 C =1
H =4 O =2

2 O2 O= 2 x 2 =4
2 H2O O = 1 x 2 =2
H = 2 x 2 =4

First we need to see the number of atoms of each element in the chemical
reaction as shown with the number (not bold) in the table above. In
balancing, we have to keep the number of moles of each element in the
reactant and product equal.

● We first look at the number of moles of carbon: 1 mole in the reactant

and 1 mole in the product, thus they are balanced.
● Next is we look at the number of moles of hydrogen: 4 moles in the
product but initially there were only 2 mole of hydrogen in the product.
So to balance, we multiply 2 to the hydrogen in the product. This makes
the number of hydrogen in the product equal to 4. They are now
balanced.
 ● Hydrogen in the product is with oxygen, thus we have to also multiply 2
to oxygen since they exist in the same water compound. Adding all the
oxygen in the product (from the CO2 and water), the total number of
oxygen is now 4. However, there are only 2 oxygen in the reactant. To
balance oxygen, we have to multiply 2 to the oxygen in the reactant.

The balance chemical equation is now:

CH4 + 2O2 → CO2 + 2H2O

PRACTICE EXERCISE NO. 4

Answer What Have I Learned so Far? page 112 numbers 1 & 2 of the
book. Write your answers in a separate sheet of paper.

1. a. 2P + 3Cl 2 = 2PCl3
P= 1 (2)=2 P= 1 (2)=2
Cl= 2 (3)=6 Cl= 3 (2)=6
b. K5 + O2 = K5C2
c. CaCo3 = Ca + CO3

d. I 2 + 2Cl = 2ICl

I= 2(1)=2 I=1(2)=2

Cl=1(2)=2 Cl=1(2)=2

e. Na2S + 2HCl = 2NaCl + H2S

Na= 2 Na= 1 x 2=2

S= 1 Cl= 1 x 2=2

H= 1 X 2=2 H= 2

Cl= 1 x 2=2 S=1

2. a. 4HC2H3O2 + Ba (OH)2 = 4H2O + Bd(2C2H3O2)2

H = 6 x 4=24 H = 8 x 4=24

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C = 2 x 4=8 C = 4 x 2=8

O = 4 x 4=16 O = 5 x 4=20

Ba = 1 Ba = 1

Neutralization Reaction

b. H2 + Cl2 = 2HCl

H =2 H =1 x 2=2

Cl =2 Cl =1 x 2=2

Combination Reaction

c. C6H14 + 9.5O2 = 6CO2 + 7H2O

C =6 x 1=6 C =1 x 6=6

H =14 H = 2 x 7=14

O = 2 (19/2) O = (2 x 6) (1 x 7) = 19

Neutralization Reaction

d. NaNO3 + HCl = NaCl + HNO3

Na= 1 Na= 1

N= 1 N= 1

O= 3 O= 3

H= 1 H= 1

Cl= 1 Cl= 1

Double Replacement Reaction

e. Zn + 2HCl = ZnCl2 + H2

Zn= 1 Zn= 1

H= 1 x 2=2 H= 2 x 1=2

Cl= 1 x 2=2 Cl= 2

Single Replacement Reaction

With the balanced chemical equation, we can already determine the
theoretical amount of a product that will be produced from a given reactant
or the amount of reactant needed to produce a desired amount of a
product. Figure 3 shows the steps needed to perform stoichiometry in a
balanced chemical reaction.

Figure 3. Steps for calculating amount of reactants or products in a reaction

Example 7
How many grams of water are produced in the oxidation of 1.00 g of
glucose, C6H12O6? The balanced chemical equation is:
C6H12O6 + 6O2 → 6 CO2 + 6H2O
SOLUTION
Step 1: Convert the given mass of glucose into moles. The molar mass
of glucose is 180 g/mol

Step 2: Convert this moles of glucose to moles of water based on the

balanced chemical equation. For every 1 mole of glucose, 6
moles of water is produced based on the balanced chemical
equation.

Step 3: Use the molar mass of water compute for the mass of water in

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grams. The molar mass of water is 18g/mol.

ANSWER: Thus, 0.6 g of water is produced for every 1 g of glucose.

See Examples 27-29 in page 114-116 of the book for more examples.

Read your book

Read and study on limiting reagents and theoretical yield page
117 - 120. Write your questions on points you were not able to
understand and ask these questions during our consultation time.

PRACTICE EXERCISE NO. 5

Answer What Have I Learned so Far? page 116 numbers 1 & 2 of
the book. Write your answers in a separate sheet of paper.
1. 2CO + O2 → 2CO2

Reactants Products

C = 1 (2) = 2 C = 1 (2) = 2

O=2+2=4 O = 2 (2) = 4
Given: M Co2 = 7.2 moles CO
Asked: O2
Solution: MCO = 7.2 mol CO x = 3.6 mol O2

2. Given: O2 = 30.2 g O2
Asked: CO2
Solution: MO2 = 30.2 g mol O2 x = O.94 mol O2

MCO2 = 0.94 mol O2 x = 1.88 mol CO2

Christian Vincentian Value: Co-responsibility

In stoichiometry we create new and more useful product out of 1 or

more reactants. But most of the time we react with others to create better
and useful products. We work together as one and are responsible of each
other.
As a student of SLMCB, what are your desired products of your current self?
How do you see yourself 10 years from now? What could be the limiting
reagents in achieving this desired product?
My current self’s desired outcome is to complete school and to be successful in
the future. A vision I’ve always had for myself in 10 years is to be a wealthy,
independent woman who can help my parents and anyone else who needs it
and to have my own happy family. And all I need to do to fulfill these desires of
mine is to apply passion, effort, determination, and understanding.

SDG#4: Quality Education

Create your formula of success. In your formula, write the needed reactants
(values and actions) to achieve your desired product (dreams).

Effort + Determination + Understanding = Success

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PROBLEM SET 2

1. Consider the following statement.

1.28 mols of sodium hydroxide reacts with 1 mol of carbon dioxide
produces sodium carbonate and water.
a. What are the reactants of the reaction? (2 points) 2N9OH
b. What type of chemical reaction is the given? COMBUSTION
REACTION
c. Write the balanced chemical equation of the statement. (3 points: 2
point solution, 1 point final answer)
2N9OH + CO2 = Na2CO3 + H2O
d. What is the limiting reactant of the given reaction? 0.64
e. How many moles of Na2CO3 can be produced?(3 points: 2 point
solution, 1 point final answer)
1.28 mol NaOH x = 0.64 mol Na2CO3
f. How many moles of the excess reactant will be left after the
reaction?(3 points: 2 point solution, 1 point final answer)
1.28 mol NaOH x = 0.64 mol O2

2. Benzene, C6H6 , is a known carcinogen that burns in air according to the

following equation:

2C6H6 (l) + 15 O2 (g) → 12CO2(g) + 6 H2O(g)

(a) What is the mole ratio of O2 to C6H6 ? (1 point)

(b) How many moles of O2 are required to react with each mole of
C6H6 ? ?(3 points: 2 point solution, 1 point final answer)
(c) How many moles of O2 are required to react with 0.38 mol of
C6H6 ? ?(3 points: 2 point solution, 1 point final answer)
 Vinegar and Baking Soda Stoichiometry Lab

Objective:

To predict the amount of carbon dioxide gas that should be produced

in a chemical reaction; then calculate the amount of CO 2 released,
the percent yield.

CH3COOH + NaHCO3 → NaCH3COO + H2O +CO2

Materials:

Baking Soda (NaHCO3), Vinegar (CH3COOH), 2 clear glass container and

weighing scale.

Procedure:

1. Obtain and record the mass of the first container. This is container A.
2. With container A still on the balance, add approximately 10.0 g of
baking soda to the cup. (The mass does not have to be exactly 10.0
g, as long as you record the mass accurately.)
3. Obtain and record the mass of another container. This is container B.
4. Place container B on the scale, weigh and record approximately 50.0
g of vinegar. (The mass does not have to be exactly 50.0 g, as long
as you record the mass accurately.
5. Performing the reaction:
a. Slowly add vinegar to container A until the reaction has
stopped.

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 b. DO NOT add all of the vinegar, just enough to complete the
reaction.
c. After the reaction is completed reweigh and record the mass of
both container A and B.
6. Calculate the mass of CO2 that escaped.

Useful Formula:

Data Collection:

Data Units
2.85 g
1 Mass of beaker A (empty)
10 g
2 Mass of Beaker A + Baking Soda
9.75 g
3 Mass of Baking Soda (2-1)
2.85 g
4 Mass of beaker B (empty)
3.35 g
5 Mass of Beaker B + Vinegar
18 g
6 Mass of Beaker B + Vinegar after reaction
2 g
7 Mass of Vinegar added to Beaker A (5-6)
11 g
8 Mass of Beaker A after reaction
4.1 g
9 Mass of product after the reaction (8-1)
4.2 g
10 Mass of Baking Soda + Vinegar (3+7)
1 g
11 Mass of Carbon Dioxide lost (10-9)

Guide Questions: (Write your answers in a separate sheet of paper)

 1. What are the reactants in this experiment? CH3COOH + NaHCO3
2. Which are the products in this experiment? NaCH3COO + H2O + CO2
3. Identify the limiting reactant: NaHCO3
4. Identify the excess reactant: CH3COOH
5. Using stoichiometry (i.e. mass of Baking Soda) calculate the theoretical
yield of carbon dioxide: (show your solution)
Solution:
9.75 g NaHCO3 x x x

Theoretical yield = 9.63 g NaCH3COO

6. What is the percent yield? 42.6%
7. What is the percent error? -57.4%
8. Matter cannot be created nor destroyed during a reaction. Does this
apply to this lab? (Yes or No) Explain your Answer.

Yes, because I believe the reactants in your chemical reaction are the
same as the product of your chemical reaction. The mass of the
reaction and the products remain unchanged, and no reactants were
destroyed when the product was released. As a result, I believe that
this statement applies to this lab.

Apodaca, D. C. (2020). General Chemistry 1 (2nd ed.). Makati: Diwa Learning

System Inc.

Bauer, R. C., Birk, J. P., & Marks, P. S. (2016). Introduction to Chemistry (4th
ed.). McGraw-Gill Education.

This learning kit is owned by and intended for students of St. Louise de Marillac College of Bogo.
Any unauthorized use and duplication is deemed subject to the violation of Copyright Laws Page 29 of 30
 Brown, T., LeMay, H., Bursten, B., Murphy, C., & Woodward, P. (2009).
Chemistry the Central Science. Singapore: Pearson Education South
Asia Pte Ltd.

Espinosa, A. A. (2016). General Chemistry 1. Makati City, Philippines: Diwa

Learning Systems Inc.

Written by: Checked by:

Engr. Florenei O. Rosaña Ingrid R. Quiliquite, LPT

Teacher Science Coordinator