ipl-logo

Pipetting Techniques Lab Report

419 Words2 Pages

Pipetting accuracy not only depends on the conditions of the lab pipettes, but also on our pipetting technique. The way to know whether our pipetting is accurate is repeat the steps for many times. Then, the standard deviation is calculated and this is our pipetting errors. In this practical, since our standard deviations are zero, this means our pipetting was accurate.
The accuracy can also be determined by comparing the expected mass with the mean mass that gained in practical. For the volume of 738.5 μL (0.7385ml), since the density should be one, by deriving the formula of density = mass/volume, the expected mass is 0.7385g. By comparing the expected mass (0.7385g) with the mean mass (0.7100) that gained from the practical, the difference of the values is small (0.0285g), so …show more content…

By comparing the expected mass (0.0785g) with the mean mass (0.0700) that gained from the practical, the difference of the values is small (0.0085g), so this means that the accuracy is high.
For the volume of 7.8 μL (0.0078ml), since the density should be one, by deriving the formula of density = mass/volume, the expected mass is 0.0078g. By comparing the expected mass (0.0078g) with the mean mass (0.0100) that gained from the practical, the difference of the values is small (0.0022g), so this means that the accuracy is high.
When 790μL of absolute ethanol is pipetted, the pipette drips. This may because absolute ethanol is the liquid with high vapour pressure force air cushion to expand. Besides, ethanol has lower density. This will increase the size of air cushion. Techniques such as reverse pipetting and pre-wetting of the tip can be used to prevent this condition, so that may be why the pipette does not drip for the second time during the practical. However, these techniques can only be used to obtain better accuracy, dripping still

Show More
Related
631 Words | 3 Pages

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) /

Read More
1513 Words | 7 Pages

Another possible source of error could come from the complete filtration of the precipitate (CuCO3). This would lead to lower measurement of the mass of CuCO3 and filter paper and a lower overall percent yield. In future experiments, to eliminate errors in the percent yield, the experiment should include multiple trials for more reliable data and a close actual percent yield to 100 percent. The experiment could also be performed with a higher grade filter paper to ensure most precipitates do not pass through to the filtered solution, leading to a more precise measurement of the mass of CuCO3 and filter paper and an overall more accurate calculation of the actual percent yield. The measured percent yield may have been affected by the errors,

Read More
425 Words | 2 Pages

To calculate the experimental mass the substance of each bag and the bag its self was measured using a balance. After gathering the mass subtract the mass of the empty bag to the mass of the unknown substance, in order to just have the mass of the substance. Afterward the mass of the unknown substance was divided by the number of moles recorded on the bag of the substance. The measurements are displayed on the table

Read More
1067 Words | 5 Pages

The experimental molar mass was calculated to be 87.5005 grams. Some source of error could be that there was not enough time to let the solution warm up to room temperature, letting the solid formed in the 11 dram dial to melt, causing the temperature probe to possibly read incorrectly. Also, while using a plastic pipet, the drops added were not consistent and varied in size. This could have led to different masses between the first and second set of 20 glycerol drops used in trial 2 and 3 of the experiment. These mistakes and errors can account for the percent error of 4.98%.

Read More
1222 Words | 5 Pages

The hypothesis made, the density calculated in the experiment will stay the same because the density of the unidentified object will never change, was supported. The results support the hypothesis because in every trial the density always came out to 9g/mL. In trial one the mass was 71.16g, the volume was 8mL, and the density was 8.895g/mL, but when rounded to the proper sig fig came out to 9g/mL. In trial two the mass was 71.12g, the volume was 8mL, and the density was 8.89g/mL, but when rounded to the proper sig fig came out to 9g/mL. In trial three the mass was 71.14g, the volume was 8mL, and the density was 8.8925g/mL, but when rounded to the proper sig fig came out to 9g/mL. When averaged the mass was 71.14g, the volume was 8mL, and the density was 9g/mL. Errors that could have occurred are, not calculating the density correctly, not completely submerging the unidentified object with water in the graduated cylinder to get the volume, not rounding the sig figs correctly when finding the density, not measuring the unidentified object’s mass in grams, not measuring the unidentified object’s volume in milliliters, and not writing the correct units with the proper number or not the correct unit at all.

Read More
467 Words | 2 Pages

Error Analysis We may have made a few errors along the way. We could have had a blunder because we may not have waited long enough for the penny to stay on the scale. We could have made a human error because we could have had a little more or less sodium hydroxide in our graduated cylinder and eyeballed it wrong. Another error you could have made was an instrumental limitations because read the cylinder wrong on the side by

Read More
848 Words | 4 Pages

These small errors may be arised as a result of temperature. Because the accurate mesuring process took much time and during this time the temperature of water was decreasing. Among two methods the density bottle gives more accurate measurement of the volume, resulting in more accurate determination of density. One of disadvantages of hydrometer is that hydrometer has Operator dependent readings, therefore has limited accuracy. Morever, for measuring the density using hydrometer large sample volume is required.

Read More
511 Words | 3 Pages

This method, not only allows to calculate the average, but also assures that the data is close to the actual value. After these sets of procedures were conducted, the unknown metal sample was identified. The unknown substance had an average density of 7.64 g/ml, which was close to the density of Tin with 7.3 g/ml as its density. The possible explanation for the differences in their values can be due to experimental errors. One of the main experimental errors that contribute to this would be, that the pennies might not be completely

Read More
478 Words | 2 Pages

The water percent is determined to be 42.06%. To find a percent error, a theoretical percent water must be used. To find the theoretical percent error divide the mass of water by the mass of magnesium sulfate heptahydrate and multiply by 100 to get a percent. The theoretical percent water is

Read More
1237 Words | 5 Pages

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

Read More
506 Words | 3 Pages

By conducting the Collecting Gas lab, it was determined that we had an actual yield of 0.0031 moles of H2 compared to a theoretical yield of 0.0030 moles of H2. By plugging in the actual and theoretical yield into the percent error formula, it was determined that there was a +3.33% error. In order to come to the conclusion of the +3.333% error, students had to complete a series of calculations. These calculations included: Converting the recorded room temperature from Celsius to Kelvin in order to plug the temperature value into the the ideal gas equation (PV = nRT). Converting the recorded room pressure from inches Hg to millimeters Hg in order to find the total pressure to plug into Pressure Conversion and Adjustment equation, which finds the pressure of H2.

Read More
217 Words | 1 Pages

The first step in this lab was to weigh the mass of each pipe using the scale. We then wrote down the weight (kg) of each pipe on our data sheet and measured the radius ( in terms of meter) of pipe. Next, we hypothesized which of the pipes would roll down the first and reasoned why. After finishing up with our hypothesis, we calculate each pipes I value with the formulas I = ½(mr2) for solid cylinder & I = mr2 for ring. Now we can set up for the experiment by creating a ramp by stacking the textbooks and leaning the wooden board against the textbooks.

Read More
762 Words | 4 Pages

Our procedure for part two of the lab affected the mass much less than part one because we had found an adequate way to collect most of the gas in the flask. Unlike part one where some of it went

Read More
362 Words | 2 Pages

In this lab, we were trying to figure out the density of unknown metals. We determined the density at the end by first measuring the mass of the metal on a scale, then using water displacement, finding the volume of the metal. First, we decided to use metal B. We took out one piece of the metal and measured the mass by placing it on the scale. Then, we filled up the graduated cylinder with exactly 20 mL of water.

Read More
864 Words | 4 Pages

Due to the unaccountability of the inconsistency in droplet size, many of the numbers may be varied because in one trial a huge droplet may count as one, but in another trial, I may have counted a small droplet as one, which causes results to possibly be

Read More

More about Pipetting Techniques Lab Report

Related Topics

Open Document