Molar mass of alka seltzer tablet

Chem Post Lab

Experiment 4

Post Lab

Plop, Plop, Fizz, Fizz: The Mass Percent of NaHCO3

Name: Rawan Douar Date: 10/04/2016

Instructor: Daniel Otis Section: 691

Introduction:

In this experiment we are going to learn the importance of the mass percent and how to use it in order to determine mass of the ingredients in any tablet. Also we will learn that the tablet contains small amount of medication and the rest is just binders and flavors. This experiment is very important in our life especially for those who will specialize in Pharmacy; they will be able to create a new similar tablet of any kind by themselves. We will specifically concentrate on Alka-Seltzer tablets and create similar tablets with different concentration and compare it with the manufacturer’s mass percentage.

Materials:

-100mL Beaker

-Graduated cylinder

-Small watch glass

- Weighting machine

-De-ionized distilled water

-Kim wipes

-Alka-Seltzer

-Paper Towels

-Vinegar

Procedures:

1.) Weigh 100 ml empty beaker, clean it and dry it. Write down the weight in the lab notebook to the closest 0.001g

2.) Get a graduated cylinder and place 35 ml of de-ionized distilled (DI) water in the weighted 100 ml beaker.

3.) Get the beaker and make sure it’s clean and re-weight it including the water and then recorded it. (Clean the beaker with Kim wipes every time before weighing it)

4.) Break Alka-Seltzer tablet into two halves and weight each half carefully.

5.) Add one half of the tablets to the beaker that contains the distilled water. Then add the pre weighted watch glass on top of it in an inverted position to deflect any splatter. Mix the contents and make sure it dissolves completely in the solution.

6.) When the reaction finished and the bubbles disappear weigh all of the materials again separately (Beaker, solution and the watch glass). Record the new masses

7.) Empty the content of the beaker in the appropriate sink and clean it with distilled water. In the end make sure to dry the beaker and the watch glass using Kim wipes and paper towels.

8.) Again using a graduated cylinder, place a 5 ml of vinegar in the 100 ml beaker. Then pour water into the cylinder until it reaches 35 mL. Place the solution into the weighted beaker used at the beginning of the experiment.

9.) You must reweigh the beaker including the solution and make sure the beaker is clean and dry. Make sure you record the weigh

10.) Now add the 2nd half of the tablet to the beaker including the distilled water and cover it with the watch glass. Make sure you mix the solution until the tablet completely dissolves in the solution

11.) When the reaction is complete reweight the beaker, the solution, watch glass and write down the weight in the lab notebook.

12.) Over again do the experiment using 10, 15, 20, 25, 30, and 35 ml of vinegar.

Data Table 1

Figure 1: Lab Calculations

Volume

Mass Before Reaction (g)

Mass after Reaction (g)

CO2

NaHCO3

RUN

VINEGAR

WATER

½ TABLET

BEAKER

Beaker and Solution

Watch Glass

Beaker and Solution and Watch Glass Mass

Total Mass

Beaker and Solution and Watch Glass

GRAMS

MOLES

MOLES

GRAMS

MASS %

1

0

35

1.839

51.79059

86.068

27.9727

114.04

115.62

113.781

0.259

0.00589

0.00589

0.495

26.92

2

5

30

1.42

51.79059

86.510

27.9727

114.48

115.68

114.260

0.22

0.0050

0.0050

0.42

29.58

3

10

25

1.527

51.79059

85.708

27.9727

113.68

114.75

113.223

0.457

0.0104

0.0104

0.874

57.24

4

15

20

1.688

51.79059

86.008

27.9727

113.98

115.29

113.602

0.378

0.0859

0.0859

0.722

42.77

5

20

15

1.6443

51.79059

85.288

27.9727

113.26

114.57

112.927

0.333

0.00757

0.00757

0.636

38.71

6

25

10

1.566

51.79059

86.148

27.9727

114.12

115.35

113.784

0.336

0.0076

0.0076

0.641

40.93

7

30

5

1.854

51.79059

86.098

27.9727

114.07

115.54

113.686

0.384

0.00873

0.00873

0.733

39.54

8

35

0

1.389

51.79059

86.54

27.9727

114.512

115.57

114.184

0.328

0.00745

0.00745

0.626

45.17

Calculations

1) Run 1 :

Mass of CO2 = Total Mass of Beaker and Solution and Watch Glass Mass (Before the reaction) – the total mass after the Reaction (Beaker and Solution and Watch Glass Mass) = 114.04 – 113.781= 0.259 grams

Molar mass of CO2= 12.01+ (16x2) = 44.01g/mol

Molar mass of NaHCO3 = 22.98+1.008+12.01+(16x3)= 84 g/mol

Moles of CO2 = 0.259 g x 1 mole / 44.01 g/mol

= 0.00589 moles

0.00589 moles = grams of NaHCO3 g x 1 mole / 84 g/mol

= 0.495 grams of NaHCO3

Mass % = mass of NaHCO3/ mass of Tablet x 100%

= 0.495 / 1.839 x 100% = 26.92 %

2) Run 2 :

Molar mass of CO2= 12.01+ (16x2) = 44.01g/mol

Molar mass of NaHCO3 = 22.98+1.008+12.01+(16x3)= 84 g/mol

Moles of CO2 = 0.220 g x 1 mole / 44.01 g/mol

= 0.00500 moles

0.00500 moles = grams of NaHCO3 g x 1 mole / 84 g/mol

= 0.420 grams of NaHCO3

Mass % = mass of NaHCO3/ mass of Tablet x 100%

= 0.420 / 1.420 x 100% = 29.58 %

3) Run 3 :

Molar mass of CO2= 12.01+ (16x2) = 44.01g/mol

Molar mass of NaHCO3 = 22.98+1.008+12.01+(16x3)= 84 g/mol

Moles of CO2 = 0.457 g x 1 mole / 44.01 g/mol

= 0.0104 moles

0.0104 moles = grams of NaHCO3 g x 1 mole / 84 g/mol

= 0.874 grams of NaHCO3

Mass % = mass of NaHCO3/ mass of Tablet x 100%

= 0.874 / 1.527 x 100% = 57.24 %

4) Run 4 :

Molar mass of CO2= 12.01+ (16x2) = 44.01g/mol

Molar mass of NaHCO3 = 22.98+1.008+12.01+(16x3)= 84 g/mol

Moles of CO2 = 0.378 g x 1 mole / 44.01 g/mol

= 0.00859 moles

0.00859 moles = grams of NaHCO3 g x 1 mole / 84 g/mol

= 0.722 grams of NaHCO3

Mass % = mass of NaHCO3/ mass of Tablet x 100%

= 0.722 / 1.688 x 100% = 42.77 %

5) Run 5 :

Molar mass of CO2= 12.01+ (16x2) = 44.01g/mol

Molar mass of NaHCO3 = 22.98+1.008+12.01+(16x3)= 84 g/mol

Moles of CO2 = 0.333 g x 1 mole / 44.01 g/mol

= 0.00757 moles

0.00757 moles = grams of NaHCO3 g x 1 mole / 84 g/mol

= 0.636 grams of NaHCO3

Mass % = mass of NaHCO3/ mass of Tablet x 100%

=0.636 / 1.644 x 100% = 38.71 %

6) Run 6 :

Molar mass of CO2= 12.01+ (16x2) = 44.01g/mol

Molar mass of NaHCO3 = 22.98+1.008+12.01+(16x3)= 84 g/mol

Moles of CO2 = 0.336 g x 1 mole / 44.01 g/mol

= 0.00763 moles

0.00763 moles = grams of NaHCO3 g x 1 mole / 84 g/mol

= 0.641 grams of NaHCO3

Mass % = mass of NaHCO3/ mass of Tablet x 100%

=0.641 / 1.566 x 100% = 40.93 %

7) Run 7 :

Molar mass of CO2= 12.01+ (16x2) = 44.01g/mol

Molar mass of NaHCO3 = 22.98+1.008+12.01+(16x3)= 84 g/mol

Moles of CO2 = 0.384 g x 1 mole / 44.01 g/mol

= 0.00873 moles

0.00873 moles = grams of NaHCO3 g x 1 mole / 84 g/mol

= 0.733 grams of NaHCO3

Mass % = mass of NaHCO3/ mass of Tablet x 100%

=0.733 / 1.854 x 100% = 39.54 %

8) Run 8 :

Molar mass of CO2= 12.01+ (16x2) = 44.01g/mol

Molar mass of NaHCO3 = 22.98+1.008+12.01+(16x3)= 84 g/mol

Moles of CO2 = 0.328 g x 1 mole / 44.01 g/mol

= 0.00745 moles

0.00745 moles = grams of NaHCO3 g x 1 mole / 84 g/mol

= 0.626 grams of NaHCO3

Mass % = mass of NaHCO3/ mass of Tablet x 100%

=0.626 / 1.386 x 100% = 45.17 %

Figure 2: Graph between Mass % of NaHCO3 and Acetic Acid (Vinegar)

Questions

1) The ratio between the mass of the Alka-Seltzer and the mass of the tablet are the same because it is a homogenous mixture. The homogenous mixture of Alka-Seltzer tablet components are equally mixed in the entire tablet. Therefore, the percent of the Alka-Seltzer is same in half mass of the tablet or even 1/4th mass of tablet. So, it remains constant. Law of constant composition.

2) The graph show that the relationship between the mass percentage and the acetic acid is directly proportion, so as the volume of vinegar increases the mass percentage increases so we can determine the amount of NaHCO3 in Alka-Seltzer by knowing the amount of vinegar needed.

Conclusion

It is obvious that there is a relationship between the mass of the CO2 and the NaHCO3 which is almost 1:2 for example in run number the mass of the CO2 is 0.333g and in the NaHCO3 is 0.636 which means that the half amount of NaHCO3 converted to CO2 after the reaction. Also the percentage mass is directly proportional to the mass of the NaHCO3 as shown in Figure number 1.

As the amount of Acetic acid increases the amount of NaHCO3 reacted increase so the minimum number of Acetic acid needed to react the entire amount of NaHCO3 can be calculated as follow:

Taking the last run for example:

35 mL of Acetic Acid 45.17% of NaHCO3 reacted

?? 100%

100% x 35mL/ 45.17% = 77.48 almost 78 mL of Acetic acid needed to react the entire amount of NaHCO3

Mass % of NaHCO3 VS. Acetic Acid

y Mass % of NaHCO3 0 5 10 15 20 25 30 35 26.92 29.58 57.24 42.77 38.71 40.93 39.54 45.17
Vinegar mL