In this experiment, you will determine the mass percent of carbon produced in the decomposition of sodium bicarbonate produced from the following chemical equation:

Reacting sodium bicarbonate (sodium hydrogen carbonate) with sulfuric acid will produce a salt, water, and carbon dioxide gas. According to the Law of Definite Composition, the mass percentage of carbon produced by the bicarbonate should be constant no matter how much sodium bicarbonate is being decomposed. The experiment should be performed two times and the results averaged.


Wear safety goggles in the lab at all times. Wash your hands immediately upon contact with chemicals.


1. Place an empty, dry 100 mL beaker on the balance pan. Zero the balance.

2. Add approximately 0.70g of NaHCO3 to the beaker and record the exact mass of NaHCO3.

3. Remove the beaker and zero balance.

4. Place the dropper of 1M H2SO4 and the beaker with the NaHCO3 on the balance. Record the mass as the initial total mass.

5. At your lab station SLOWLY add the H2SO4 to the beaker one drop at a time. Swirl the contents to ensure complete mixing. Continue releasing the acid drops in this manner until effervescence ceases.

6. Place the beaker and dropper on the balance. Record as the final total mass.

7. Discard the contents of the beaker in the sink, rinse and dry the beaker.

8. Repeat procedure for trial two.

Table 1: Masses of Reactants and Products Through Trials 1 and 2

 Trial 1Trial 2
Mass of NaHCO30.71 g0.70 g
Initial Total Mass89.12 g85.26 g
Final Total Mass88.68 g84.87 g
Mass CO20.44 g0.39 g


1.) Calculate the mass of CO2 lost in each trial. (initial total mass – final total mass)

Trial 1: 89.12 – 88.68 = 0.44 g           

Trial 2: 85.26 – 84.87 = 0.39 g           

2.) Calculate the theoretical % C in CO2. Remember to round all % to two digits after the decimal.

Atomic Mass of C: 12.01 g/mol

Atomic Mass of O: 16.00 g/mol → 2 M of Oxygen = 32.00 g/mol

Mass of Compound:   1C + 2O

                                    = 12.01 + 32.00 g/mol

= 44.01 g/mol

Mass % of C = 12.01/44.01×100=27.29%

3.) Calculate the actual mass C for each trial. (%C as decimal X mass of CO2)

Trial 1:

Mass % of C: 27.29%

Mass of CO2: 0.44 g

Mass of C = 0.2729 x 0.44 = 0.12 g

Trial 2:

Mass % of C: 27.29%

Mass of  CO2: 0.39 g

Mass of C = 0.2729 x 0.39 = 0.11 g

4.) Calculate the %C for each trial. (mass C/mass NaHCO3). Remember to round all % to two digits after the decimal.

%C for Trial 1: 12.00/0.71 = 16.90 %

%C for Trial 2: 11.00/0.7 = 15.71 %

5.) Calculate the average % C.

16.90 + 15.71 = 32.61

32.61/2 = 16.31

Average Percent of Carbon = 16.31%

6.) Calculate the theoretical %C in NaHCO3. Remember to round all % to two digits after the decimal.

%C = 12.01(100)/(Na + H + C + 3O)

%C = 12.01(100)/(22.99 + 1 + 12.01 + 3(16))

%C = 12.01(100)/84

%C = 14.30%

7.) Calculate your % error. [(|actual-theoretical|/theoretical) x 100]

 [(|actual-theoretical|/theoretical) x 100]

= [(|16.31-14.30|/14.30) x 100]

= 14.06 %


1.) What effects would there be in your results if you did not decompose all of the sodium hydrogen carbonate in each of your trials? Would the percentage of C that you determine be lower or higher than it should have been?

Our results would not be as accurate or completely correct if we did not decompose all of the sodium hydrogen carbonate in our trials. This is because with the Law of Definite Composition, there is a specific amount of each element found in a chemical compound where the percentage composition is constant. Since a specific amount is required with the formation of a compound, in this case CO2, this law also applies with the decomposition of a compound. If a specific compound is decomposed, the precise amounts of each element within the compound should be produced. If the compound (Sodium Hydrogen Carbonate) was not fully decomposed (dissolved), then there may be the possibility that all precise amounts of the element are not completely separated then the expected amount will not be produced. The percentage of Carbon would be lower than it should have been if the decomposition process is not thoroughly completed.

2.) If we used this experiment to determine the % of C in various brands of sodas, would you expect the results to illustrate the Law of Definite Composition? Why or Why not?

It would be expected that if this experiment was conducted to determine the % of C in various brands of soda, it would illustrate the Law of Definite Composition. This is because as the Law of Definite Composition states that regardless of the amount, a chemical compound will always contain the same elements in the same proportions by mass. The element of Carbon remains the same, no matter what. Therefore, putting it together with different types of sodas that may have different masses of CO2 altogether, the percentage composition of Carbon will still remain the same based on the total mass of CO2.

author avatar
William Anderson (Schoolworkhelper Editorial Team)
William completed his Bachelor of Science and Master of Arts in 2013. He current serves as a lecturer, tutor and freelance writer. In his spare time, he enjoys reading, walking his dog and parasailing. Article last reviewed: 2022 | St. Rosemary Institution © 2010-2024 | Creative Commons 4.0

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