Sodium Bicarbonate And Calorimetry Lab Report

conclusion: The results show that when sodium bicarbonate is heated, it decomposes to form sodium carbonate (1.51g), and carbon dioxide gas. This is most likely because the percent yield was only 3.1% off. Sodium carbonate is known to be a solid crystal substance which agrees with the crystal structure substance formed on the crucible furthering the idea that sodium carbonate actually formed. We could have also tested this by dropping water onto the substances produced to see if it dissolves or not since sodium carbonate dissolves in water. In this experiment NaHCO3 decomposes into Na2CO3, CO2 and H2O when heated.

The last goal was to determine the percent yield of a product formed during a reaction with the unknown compound. Experimental Design The first day of lab consisted of various preliminary tests that helped identify the unknown compound.

Lab 27. Stoichiometry and Chemical Reactions Report In our lab we were asked Which Balanced Chemical Equation Best Represents the Thermal Decomposition of Sodium Bicarbonate. Sodium Bicarbonate is a chemical compound with the formula NaHCO3, also known as baking soda. In the process to answer our guiding question we have to determine how atoms are rearranged during a chemical reaction.

Procedure The various solids and liquids were used for this lab. The first experiment had effects of the temperature towards the chemical reaction. The first experiment had two test tubes filled with same amount of substances to react, but one was at 10 ºC while the other one was at 50 ºC. The second experiment involved different molar concentration of substances reacting. Individual test tube was filled with 1.0,

After the reaction finishes, the amount of CO2 that was released is calculated and then using the molemap, it is possible to solve for the amount of the unknown substance. The grams of the unknown powder are then divided by the moles of the substance to get the molar mass. This molar mass is then compared to the molar mass of the known substances. Unknown substance A is lithium carbonate since the molar mass produced in the experiment is 66.68 g/mol and the molar mass of lithium carbonate is 73.89 g/mol. Unknown substance B has a molar mass of 146.70 g/mol which is inconclusive since it is too different from the other molar masses.

Using this number as well as the theoretical yield, the percentage yield was calculated and can be seen below in Figure 5. % Yield: (0.1805 g)/(0.3064 g) x 100 = 58.92% Figure 5: The calculations of the percent yield. The percentage yield was calculated to be 58.92% and this represents that a portion of the reactants were lost during the experiment due to a small spill or improper collection of the product during the purification process. Next, the spectrum of the product was obtained.

Then, using two coffee cups, observe the temperature increase from the first thermometer in the calorimeter. After it has stopped increasing, record the final temperature. Next, weight a small portion of your unknown sample and record the mass and place it in the water bath for ten minutes and record the temperature. Pour the unknown element into the calorimeter. Monitor the temperature and record it in the data table.

These equations are as follows: NaHCO3 (s) → NaOH (s) + CO2 (g) 2NaHCO3 (s) → Na2CO3 (s) + CO2 (g) + H2O (g) 2NaHCO3 → Na2O (s) + 2CO2 (s) + H2O (s) NaHCO3 → NaH (s) + CO (g) + O2 (g) This lab will determine the correct

The needed mass for reactant A was 3.52 grams and the needed mass for reactant B was 2.12 grams. The moles of product C (CaCO3) were also required to have been calculated using mass- to- mole conversion, but the mass was already established, so its mass was not calculated. Results Calculations used to determine the percent error 2.3 g CaCO3 - 2.00 g CaCO3 2.00 g CaCO3 x 100 = 15 % Reactants A and B’s masses were not similar, but the numbers were also not drastically different. Obviously, product C’s solution is different because moles were calculated as opposed to a mass.

In this experiment, groups were given two different unidentified substances with the goal of being able to figure out what chemical each substance was. Unidentified substance “2” was proven to be Tin (II) Chloride and the Unidentified substance “4” turned out to be Sodium Chloride. The conclusion that substance “2” was Tin (II) chloride could be proven due to the fact that it reacted with Zinc and Sodium Hydroxide. When Tin (II) chloride reacted with Zinc, it began bubbling. This occurred because there was a single replacement reaction between Tin (II) Chloride and Zinc.

Purpose The purpose of the lab was to determine the mass percent of carbon in sodium bicarbonate, also known as baking soda, by the means of measuring the mass of carbon dioxide gas produced in the chemical reaction. Expected Results The expected, or theoretical result was 52.39 percent composition of carbon dioxide in the equation. To find this value, the molar mass of carbon dioxide was calculated, being 44.01g, which was utilized in the formula to find percent composition, %CE

Results The data obtained from the experiment had undergone statistical analysis using t-tests and the results were recorded in Figure 1.0 and Figure 1.1 above. According to the data obtained in Figure 1.0, the p-value is less than 0.05 in all 5 treatment solutions. It is also shown intensity Figure 1.0, the calculated t-value of each concentration of NaHCO3 in each treatment is greater than the critical t-value.

The actual yield of the reaction was 4.411 grams of copper and was obtained through the experiment

Investigation of the effect of NaCl concentration on the evaporation rate of water. Chemistry HL Internal Assessment Vitaebella Tsang Ao Ling Contents page Contents page 2 Introduction 3 Design 4 Research question 4 Variables 4 Method 5 Results 6 Discussion 9 Evaluation 9 Bibliography 10 Introduction Many recipes call for salt to be added to the water when cooking pasta to add flavor, but there has been common belief or misunderstanding that adding salt will make the water boil faster. However, it is now known that that is not the fact, and that adding salt will do the opposite and make the water boil more slowly instead.

However, the percent yield of this experiment was 109.96% (Fig. 5). Ultimately, the product mass collected after the reaction was a greater amount than the initial mass used in the original