Introduction Baking soda is a white crystalline powder (NaHCO3) also known as sodium bicarbonate, which is formed with an acid and a base. It can react mildly with other compounds. Vinegar is a compound that contains acetic acid (CH3COOH) and water (H20) it is the most easily available mild acid currently, it is not only used for cooking but for medical and chemical purposes as well.Even though vinegar is used for everyday things such as cooking and cleaning it is still a very dangerous and toxic acid. An estimated or calculated amount of a substance is the theoretical amount. The exact amount that is actually calculated during the end of the experiment is the actual yield, to calculate the percentage yield you must take the the actual yield …show more content…
The percentage yield was a yield of 110% of the sodium acetate. A source of error that was made during this experiment was the transferring of the baking soda to the flask some of the dissolved baking soda might have still been in the beaker after it was pouring into the flask. A solution to this error would be to have better skills or to have tried to get out a much of it as we could by getting a tool to try and scrape some of it out. Another source of error that was made during this experiment was the measuring of the baking soda, so it could have been more exact. A solution to this error would have been to use more precise scales like scales that measure to 3 decimal places rather than to 1 decimal place to get a more exact value instead of a less accurate measurement. A final source of error that was made during the experiment is the concentration of the products used were low so there might have been other thing in the vinegar to add mass to the final solution. A solution to this error could have been to use a more concentrated acetic acid because the vinegar could have other products that could ruin your calculations in the end if it did not dissolve with the …show more content…
3. 233.4 g - 228.9 g=4.5 g ∴ The actual yield was 4.5 g. 4. 4.5 g x 100% = 110% 4.102g 5. I believe that it is not common to obtain an exact 100% yield because that would mean that the chemists doing the experiment would have to do everything perfectly without any flaws. Like measuring off by a slight amount or leaving behind some of the substance if transferring it over to another flask or beaker. A person could not do something so perfectly because humans always do errors while doing things, it would be very unlikely for a chemists to get a 100% yield after for a chemical reaction. 6. The percentage yield was not 100%. Some factors that could have affected the experiments end result could have been through measuring the all the reactants and equipment their might have been an error somewhere there. Another factor that could have been is that the vinegar was a lower concentrated acetic acid and could have other components in it as well which could have ruined the end measurement of the final product by getting a heavier measurement that actually
To find the mass percent of acetic acid in vinegar, the molar mass of acetic acid is 60.05 g/mole, and 1.00 g/mol of density, then 0.96 mol×(60.05 g/1 mol) = 57.65 g 57.65 g/1000= 0.0576 ×100 = 5.76% The average mass % of acetic acid in vinegar = 5.53% The average % of acetic acid was 5.53%, which is close to the acidity of the vinegar that was taken in the lab which was 5%, for the different percent we had it could be because of errors in calculations or errors in collecting data. 3.
Calculation: Initial Mass(g)-Final Mass (g)=Change in Mass (g) Trial 1 74.5-62.0=12.5(g) Trial 2 272.7-271.5=1.2(g) Percent Error: 272.7-271.5 x 100 272.7 =0.440% Percent Change: 74.5-62.0 x 100 74.5 (Trial 1) =16.778% 272.7-271.5 x 100 272.7 (Trial 2) =0.440%
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.
I. Purpose: To experimentally determine the mass and the mole content of a measured sample. II. Materials: The materials used in this experiment a 50-mL beaker, 12 samples, a balance and paper towels. III.
As seen in table 1, the theoretical yield was .712 g of C_17 H_19 NO_3. The % yield of this experiment was 7.51 % of C_17 H_19 NO_3. . This low yield can be explained from a poor recrystallization technique combined with potential contamination. Throughout the experiment, the mixture changed color from green, orange, to yellowish lime, and eventually clear.
In this lab, human error could have possibly been that the salt wasn’t fully dissolved or even the Kool-Aid wasn’t fully dissolved. To fix this next time, both mixtures can be stirred a little longer. A third human error could have been when putting 20 drops into the test tubes, some drops were bigger than others causing there to be more than mL of mixture in the test tube. At the end of the lab, a red and yellow M&M were used to do a home material test. I dissolved the color off the shell with warm water and placed a dot of each color onto a strip of chromatograph and placed them in two individual
3. In this experiment, the percent yield was 90%. This number implies that there was little error in this experiment. However, this result could have been caused by certain external factors.
The percent recovery of the copper was calculated using the equation, percent recovery = (the mass of the copper recovered after all the chemical reactions/the initial mass of the copper) x 100. The amount of copper that was recovered was 0.32 grams and the initial mass of the copper was 0.46 grams. Using the equation, (0.32 grams/0.46 grams) x 100 equaled 69.56%. The amount of copper recovered was slightly over two-thirds of the initial amount.
(150.22g/mol)(3.5 x 10^-3 mol of nucleophile) = 0.525 g Actual yield = 0.441 g, Percent Yield = (0.441g/0.525g) x 100% = 84% 10. Percent recovery from recrystallization = (0.172g/0.441g) x 100% = 38% 11.
It was impossible to accurately measure the volume of liquid at any given moment, as the meniscus was moving side to side. Secondly, the distillation was ended while there was still liquid in to round bottom flask. The composition and volume of this liquid were unaccounted for in the calculated
The final product weight for percent yield was only the solid E product, which missed one half of the final product produce. If both products were weight, the percent yield would have been larger that it was. Instead of 22.33%, it could have been 44.66%. To prove that both products were obtained, but only one of the two products was analyze, a TLC plate of the DCM layer, that contains both products, and of the final product, was obtain.
The actual data is the result on our experiment vs theoretical, which is based on the calculations above. I have also learned to pay more attention to draining out all of the product completely before continuing to test the experiment, as any small drop of contaminant can veer our results into a different
For example, in the response experiment, a yeast solution was prepared without sugar mistakenly and thus had to be prepared again. This suggests that other errors in preparation and measurement could have been encountered. For the future, careful measurements using clean uncontaminated flasks would eliminate possibilities of such error. A source of error for the metabolism experiment involves the yeast’s yellow hue. It is possible that the color of the yeast caused the solution to look more
Ali Atwi : Internal assesment – calculating of the concentration of ethanoic acid in vinegar AIM : To calculate the concentration of ethanoic acid CH3COOH in vinegar using stoichiometric equations, ( Yamaha brand ) Introduction : I personally like to add a little bit of vinegar on my food because it makes it taste better, yet I know that vinegar contains acid, and I also know the consequences of highly concentrated acid intake, like severe itching and stomach ache, vomiting. Venigar contains a small percentage of ethanoic acid Ch3COOH. This practical aims to find out the concentration of the of the vinegar against a standard solution of sodium hydroxide soloution of concentration 0.1 mol dm3 through acid-base titration, the label on the bottle says 6%.
That caused a new initial reading of NaOH on the burette (see Table1 & 2). The drops were caused because the burette was not tightened enough at the bottom to avoid it from being hard to release the basic solution for titrating the acid. The volume of the acid used for each titration was 25ml. The volume of the solution was then calculated by subtracting the initial volume from the final volume. We then calculated the average volume at each temperature.