CHEM181-M15 Expt. No. 9 FINAL REPORT - MENDOZA

COLLEGE OF ENGINEERING AND ARCHITECTURE
Department of Engineering Mathematics, Physics and Chemistry
Mendoza Adrian A. 10/23/2021

Family Name First Name Middle Initial Date Submitted
BSCE – 1 M15 5 Mrs. Dangin

Course & Year Section Group Instructor
Number
Experiment No. 9 ︎
CIGARETTE SMOKING AND AIR POLLUTION
I. Objectives:
To measure the quantity of nitrogen monoxide and carbon monoxide in
cigarette smoke in terms of nitrous acid and carbonic acid using titration.
II. Apparatus:
3 - 250 mL Erlenmeyer flasks base burette
1 - 50 mL graduated cylinder iron stand
3 – empty plastic bottles with screw cap burette clamp
III. Materials:
1 stick cigarette distilled water
0.025 N sodium hydroxide phenolphthalein

IV. Procedure and Observations:
In preparation of the sample, there were 3 samples of cigarette smoke
that were collected in different plastic bottles for 5 minutes each, then closed
immediately. Then, 30 mL of water was added to each sample that contained the
smoke and closed directly. The sample was shaken for 3 minutes, and after
shaking, it was transferred to a 250 mL Erlenmeyer flask. Then, 2 drops of
phenolphthalein were added to each sample.
For the titration, a set-up of titration was prepared. The burette with
0.025N sodium hydroxide solution was rinsed. The burette was filled with
0.025N sodium hydroxide solution. The bubbles and the hanging drops were
removed. The initial reading of the sodium hydroxide solution inside the burette
was taken. The sample was titrated slowly with 0.025N NaOH until the end point
was reached. The end point was reached when the colorless solution turned to
the faintest shade of pink on a white background. Note: During titration, be
careful that each drop in the sample is fully mixed before adding the next drop
to the sample. After adding each drop into the sample, circulate the Erlenmeyer
flask that contained the sample.

V. Tables:
Trial 1 Trial 2 Trial 3

Initial reading 0 1.0 2.15
NaOH Final reading 1.0 2.15 3.35
Volume 1.0 mL 1.15 mL 1.2 mL
Normality 0.025 N 0.025 N 0.025 N
Volume 50 mL 50 mL 50 mL
Normality 0.0005 Eq/L 0.000575 Eq/L 0.0006 Eq/L
HNO2 for NO Molar mass 47 g 47 g 47 g

Total + charge 1 1 1
Molarity 0.0005 mol/L 0.000575 mol/L 0.0006 mol/L
Concentration in mg/L 23.5 mg/L 27.025 mg/L 28.20 mg/L
Volume 50 mL 50 mL 50 mL
Normality 0.0005 Eq/L 0.000575 Eq/L 0.0006 Eq/L
H2CO3 for CO2
Molar mass 62 g 62 g 62 g
Total + charge 2 2 2
Molarity 0.00025 mol/L 0.0002875 mol/L 0.0003 mol/L
Concentration in mg/L 15.50 mg/L 17.825 mg/L 18.60 mg/L
VI. Discussion of Results:
The term "cigarette" was usually referring to a tobacco cigarette, but it
could also refer to similar devices that have been used to smoke other substances,
such as cannabis. A cigarette is distinguished from a cigar by its smaller size, use
of processed leaf, and paper wrapping, which is generally white but can also be
other colors and tastes. Cigars are typically manufactured fully from whole tobacco
leaves (Coursehero).
Each plastic bottle has put cigarette smoke, then was added by water and
shaken. After shake for about 3 minutes, the smoke was faded. It was transferred
to the Erlenmeyer flask and was added two drops phenolphthalein. Titration of the
solution was made and resulted that when we drop a NaOH solution the color of
the solution inside the Erlenmeyer flask was changed and turned into pink and
was shaken until the end point is reached from colorless to faintest shade of pink
was appeared. Then, calculated the given value with the NO and CO.
In the solution NaOH, each trial has a different initial and final reading. To
get the value of its volume, the final and initial reading was subtracted. For trial
1, its final reading was 1.0, subtracted from the initial was 0, so the volume in
trial 1 was 1.0 mL. For trial 2, the same process on how to get the volume was
followed, where the final was 2.15 mL subtracted from the initial 1.0, and the
volume was 1.15 mL. For trial 3, same process, the final was 3.35, subtracted to
2.15, where it is the initial reading, to have the value of its volume, which was 1.2
mL. The normality of each trial was 203 N.
The HNO2 for NO, the volume was 50 mL because it is the amount of water
that was used. The molar mass of HNO2 was 47 g/mol. To get its molar mass,
substitute the atomic mass of its element where H=1, N=14, and O=16, so
1+14+16(2) = 47 g/mol. The total + charge of each trial was 1. For the normality
of trial 1, used the normality which is 0.025N x 1 mL (volume)/50 mL (n) = 0.0005
Eq/L. For trial 2, 0.025N x 1.5 mL/50 mL = 0.000575 Eq/L. Then for trial 3, 0.025N
x 1.2 mL/50 mL =0.0006 Eq/L. It was substituted to the normality/total + charge
formula in the molarity, where trial 1 was 0.0005/1 = 0.0005 mol/L, trial 2 was
0.000575/1 = 0.000575 mol/L, and trial 3 was 0.0006/1 = 0.0006mol/L. Last for
the concentration, molar mass x molarity x 1000 mg/g, where trial 1 was 47 g/mol
x 0.0005 mol/L x 1000 mg/g = 23.50 mg/L, for trial 2 was 47 g/mol x 0.000575
mol/L x 1000 mg/g = 27.025 mg/L, and for trial 3 was 47 g/mol x 0.0006 mol/L
x 1000 mg/g = 28/20 mg/L.
For the H2CO3 for CO2, each trial has a volume of 50 mL, a molar mass of
62g, a total + charge of 2, and a normality for trial 1was 0.0005 Eq/L, trial 2 of
0.000575 Eq/L, and for trial 3 of 0.0006 Eq/L. The molarity of trial 1 was 0.0005/2
= 0.00025 mol/L, for trial 2 was 0.000575/2 = 0.0002875 mol/L, and for trial 3
was 0.0006/2 = 0.0003 mol/L. The concentration for trial 1 was 62 g/mol x
0.00025 mol/L x 1000 mg/g = 15.50 mg/L, the concentration for trial 2 was 62
g/mol x 0.0002875 mol/L x 1000 mg/g = 17.825 mg/L, and the concentration for
trial 3 was 62 g/mol x 0.0003 mol/L x 1000 mg/g = 18.60 mg/L.
VII. Significance of the experiment:
The significance of this experiment was to answer the objective that was
stated above, namely to measure the quantity of nitrogen monoxide and carbon
monoxide in cigarette smoke in terms of nitrous acid and carbon acid using
titration. It was calculated using the given value between the get the missing value
of the two solutions. This experiment can be useful because it gives knowledge,
especially to the people who were using cigarettes, and could make people aware
of the disadvantages of using cigarettes. It can also affect the air pollution in the
environment.
The disadvantage was that cigarettes can destroy our health, especially
when used very often. It could lead to a risk factor for chronic lung disease and
lead to conditions of such as asthma, bronchitis, emphysema, and increased risk
of lung cancer. Cigarette smoke can contribute to air pollution (Hitti, 2004). Air
pollution is a mixture of solid particles and gases in the air that is referred to as
air pollution (Health Topics).
VII. Reference
 https://www.coursehero.com/file/23728257/cigarette/
 https://www.webmd.com/lung/news/20040823/smoking-worse-than-
exhaust-forair-pollution
 https://medlineplus.gov/airpollution.html#:~:text=Air%20pollution%20is
%20a%2 0mixture,Some%20air%20pollutants%20are%20poisonous

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COLLEGE OF ENGINEERING AND ARCHITECTURE

Department of Engineering Mathematics, Physics and Chemistry

Mendoza Adrian A. 10/23/2021

BSCE – 1 M15 5 Mrs. Dangin

CIGARETTE SMOKING AND AIR POLLUTION

3 - 250 mL Erlenmeyer flasks base burette

1 - 50 mL graduated cylinder iron stand

3 – empty plastic bottles with screw cap burette clamp

1 stick cigarette distilled water

0.025 N sodium hydroxide phenolphthalein

In preparation of the sample, there were 3 samples of cigarette smoke

shaking, it was transferred to a 250 mL Erlenmeyer flask. Then, 2 drops of

phenolphthalein were added to each sample.

the faintest shade of pink on a white background. Note: During titration, be

flask that contained the sample.

Trial 1 Trial 2 Trial 3

HNO2 for NO Molar mass 47 g 47 g 47 g

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