Colorimetric Analysis Lab

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Concept: The purpose of this lab is to determine the density of a solid and an unknown liquid in order to determine the unknowns from a list of substances provided in the lab instruction. A method to identify the substance is to figure out the density (d=m/v) where d is the density equals to the mass divided by the volume of the substance. When measuring the mass, reset the balance to zero to obtain only the mass of the object in grams (g) and not anything else. When measuring the volume, read at the meniscus for an accurate measurement. Procedure: We obtained an unknown metal with an identification number (6) from the instructor. Make sure not to get fingerprints on the sample otherwise it will interfere with identifying the density of …show more content…

We used a crucible tongs to place a dry, empty graduated cylinder on a zero out balance. We recorded the mass of it which was 26.30(g). After we pour 3.14-mL of the unknown liquid into the graduated cylinder; this is sample 1. We used the same procedure to read the graduated cylinder of the unknown metal by reading it at the meniscus with an index card from behind. We recorded the volume. We then weigh the first sample and recorded the mass to be 28.64. Do not remove the first sample from the graduated cylinder. Next we added an additional 3-4-mL of the unknown liquid to the first sample. The combine number is 7.35-mL to the nearest .01-mL for sample 2. We then recorded the mass of the unknown sample, 32.14(g), in grams on a balance. Again, do not pour out the unknown liquid from the graduated cylinder. We added more of the unknown liquid until it was nearly but not over 10-mL. We recorded the volume for sample 3 to be 9.39-mL. Then weigh the sample to be 33.78(g). Pour the liquid back into the original container; do not pour the liquid down the …show more content…

For sample 1, we subtracted 28.64(g) from 26.30(g) and got a mass of 2.34(g). We did the same process for the rest of the samples and got a mass of 5.84(g) (32.14(g) – 26.30(g)) for sample 2 and 7.49(g) (33.78(g) – 26.30(g)) for sample 3. We divided the mass of the each samples by their own volumes. For sample 1 we divided 2.34(g) by 3.14(mL) and got a density of .745(g/mL). We did the same process for the other samples and got .795(g/mL) (5.84(g)/7.35(mL)) for sample 2 and .797(g/mL) (7.49(g)/9.39(mL)) for sample 3. We then add up all the densities of the sample and then divided by 3 to get the average density, .779(g/mL). The identity of the unknown liquid (5) was cyclohexane which has a density of .792(g/mL). We had to calculate the percent experimental error just like the unknown metal. We took .779(g/mL) minus .792(g/mL) and then divided it by .792(g/mL). Then multiply the product by 100 and got -1.64% error. After calculating all the values, we put it in a graph to compare the density, mass, and volume by using

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