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Millikan Oil Drop Experiment

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Determining Smallest Possible Charge on Droplet Using Millikan Oil Drop Experiment Simulation
Abstract: The purpose of this lab was to determine the value of the smallest charge using a Millikan oil drop simulation on the TI-83+ graphing calculator. This was done by using a program on the TI-83+ graphing calculator, where an oil droplet was placed on the screen and cursor keys were used to adjust the voltage until the droplet is suspended. The droplets radius, voltage, plate separation, and charge are then stored into the system. The simulation was repeated for 40 droplets. After the mass and charge of the oil was calculated, then rearranged depending on step size to find smallest charge, which was 1.67E-19. The smallest …show more content…

All the values of electric charge seemed to be a multiple of a smaller value, which was 1.67*10-19. This smallest charge had a %error of 4.2% from the original elementary charge of

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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.

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An error that could have been present during the lab includes not letting the zinc react completely with the chloride ions by removing the penny too early from the solution. For instance, the percent error of this lab was 45.6%, which was determined by the subtraction of the theoretical percent of Cu 2.5% and the experimental percent of Cu 3.64% and dividing by the theoretical percent of Cu 2.5%. This experiment showed how reactants react with one another in a solution to drive a chemical reaction and the products that result from the

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In this lab there were five different stations. For the first station we had to determine an unknown mass and the percent difference. To find the unknown mass we set up the equation Fleft*dleft = Fright*dright. We then substituted in the values (26.05 N * 41cm = 34cm * x N) and solved for Fright to get (320.5g). To determine the percent difference we used the formula Abs[((Value 1 - Value 2) / average of 1 & 2) * 100], substituted the values (Abs[((320.5 - 315.8) /

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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%

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1294 Words | 6 Pages

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

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563 Words | 3 Pages

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.

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How does the type of dissolvent in the water affect the number of drops that can fit on a penny? We will attempt to find the answer to this question using the hypothesis “If we use salt water solution, then there will be more drops on the penny. ” We will use the materials salt, sugar, lemonade mix, flour, a beaker, a pipette, paper towels, a stirring rod, a graduated cylinder, and some tap

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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

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But the difference was no bigger than 0.08, and after the values were rounded the same empirical formula was deduced. So the experiment can be concluded as successful. Evaluation: The method used was simple and easy to follow; however, it did not include how much oxygen was needed to react completely. Also it didn 't mention what magnesium oxide looked like after it finished reacting, so it was a guesswork of determining whether the reaction was finished or not.

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While reading this, Phil pointed out that there were many small

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What is the effect of surface area to volume ratio on the rate of diffusion of the colour from the agar jelly cube? INTRODUCTION: Diffusion is the movement of spreading particles from high concentration to low concentration in an environment such as a cell. This major procedure is used in cells to source them with nutrients, water, oxygen, and to transport unwanted wastes such as carbon dioxide out of the cell or to different cellular organelles.

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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.

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The mass of vinegar used during the experiment was 4.108 grams. It was determined that there were .003129 moles of CH3COOH in the vinegar sample. Using this information and the molar mass of CH3COOH, which was 60.05 g/mol, the mass of acetic acid in the vinegar was calculated: 4.Vinegar is a 5% aqueous solution of acetic acid. Since the mass of acetic acid within the vinegar was calculated as .18789 g in step 3, the percent of CH3COOH was calculated using the following equation: To calculate the percent error, the experimental value of 4.5% acetic acid in vinegar was subtracted by the theoretical value of 5% and divided by 5% to yield a percent error of 8.54%. The following is a copy of the calculations done using decimals: 5.The equivalence point of the titration curve measured in step 1 was 25.25 mL of NaOH.

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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.

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