Unknown Solid Acid

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.

* Answers to questions above... 1) 4.078 + 4.056 + 4.095 + 4.014 = 16.24/4 = 4.061 ±0.008 grams 2.253 + 2.256 + 2.261 + 2.249 = 9.019/4 = 2.255 ±0.012 grams 2) Mass of water = 4.061 – 2.255 = 1.806 ±0.020 grams Then convert to moles  =1.806 ±0.020 / 18 g = 0.100 ±0.020 mol H2O

Using the equation m = ΔTf/Kf , the molality of the unknown solution was found. Then, moles of unknown were calculated, which was used to calculate the average molar mass of unknown. Theory: After the experiment was completed, the data

.1-amEr (c) Gram molecular mass of HNO3 = Mass of 1-molecule of HNO3x NA = 63 amu x NA = 63 gm X NA = 63 gram NA Solved Example-4:Find out the mass of carbon -12 that would contain 1.0 x1019carbon-12 atoms. Solution : Mass of 6.022 x1023 carbon-12 atoms = 12 g Mass of 1.0 x1019 carbon-12 atoms = 12x1x10 19 g 6.022 x1023 = 1.99 x10-4g Solved Example-5: How many molecules are present in 100 g sample of NH3?

Lastly, add the molar masses of each element in the compound. 3. Explain

CONCLUSION As can be seen from the graph above, the mass percentage difference decreases as molarity increases. In other words, the molarity and the mass percentage difference are inversely proportional. According to figure 2, the mass of the trial 1 potato which will be submerged in the sugar solution with a molarity of 0.0, is 0.49 grams. The same potato piece after being soaked for 24 hours in the sugar solution of molarity 0.0, is 0.69 grams.

3. Upon adding 20 drops of NaOH, a white precipitate was formed signifying acidic impurity. In the second NaOH mixture, about 20 drops were administered and no precipitate formed indicating that the ample is more pure than before. Data: Weight of flask = 75.10 grams Weight of the flask with solids =

Weighed 1 gram of NaC2H3O2 and mixed it with ionized water. Boiled 12 mL of 1.0M Acetic Acid added into a beaker containing the sodium carbonate on a hot plate until all the liquid is evaporated

The unknown solution contained 20+ -0.05 mL of the unknown in a 40 mL beaker. A 10 mL graduated cylinder was used to accurately measure. The pH value of the unknown was recorded, and then the probe was removed again and cleaned. Last, the pH probe was placed into potassium nitrate. The potassium nitrate was contained in a 40 mL beaker, .520

Each group was assigned a different percent of sucrose solution out of the four variables; 0% , 5%, 10%, and 15%. After we filled the beaker we then got two potato cores. Once we had the cores we cut the skin off the ends. Following this we then cut the two potato cores into four 2.00 cm potato cores. After they were cut into 2.00 cm each we found the mass.

In step 4 NaOH, a strong base, was added drop by drop to NaCl, a salt of a strong acid-strong base. The addition of NaOH to NaCl resulted in the pH dramatically going up which is the opposite of what happened in step 1. The initial pH of NaCl was 4.71 and when just one drop of NaOH was added the pH went up to 5.14. After all three drops of NaOH were added to NaCl the pH was tested to be 8.21, the highest pH so far. So the NaCl most certainty did not buffer the pH changes.

The stopcock was opened while the KHP solution was gently stirred. The solution turned faint pink, the stopcock was closed, and the final volume was recorded. The titration was

The equation of the reaction between sodium hydroxide and ethanoic acid is as follows: CH3COOH + NaOH → CH3COONa + H2O We can measure the end point of titration process and we can also measure the amount of reactants. The concentration of ethanoic acid in the vinegar can be determined through stoichiometric calculations, Using the values obtained from the titration, and also the chemical equation as a reference. Phenolphthalein indicator is used in this acid-base titration Equipment and materials:

Procedure A. Preparation of NaOH solution The molarity of a solution is the ratio of the number of solutes dissolved in a liter of solution. To figure out the needed mass (in grams) of NaOH pellets to be dissolved in a 0.25 L of water, remember that a mole is equivalent to the quotient of mass over the molar mass of the substance. This was used to rearrange the base formula and to derive the mathematical equation of mass in terms of molarity. mass (g) =

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.