Chemistry 145 Experiment 1 Report

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Jan 9, 2024

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Name: Elias Douramanis Partners: Austin Cincotta Date: 09/07/2023 EXPERIMENT 1 TITLE: Non-Additive Volumes OBJECTIVES: 1. To investigate the effects of combining equal amounts of 95% ethanol and deionized water. 2. To investigate the precision of different types of glassware. PROCEDURE REFERENCE: Suny Broome Community College Department of Chemistry MODIFICATIONS: 1. Only Case E will be tested during this experiment. 2. One test of Case E will be done.

DATA/RESULTS: TABLE 1: TABLE 2: glassware dry mass of glassware using the balance (g) volume of water added as read from the glassware (mL) mass of glassware + water (g) mass of water by difference volume of water using col 5 and a density of 0.9982 g/mL (mL) % error in the volume measurement (mL) 50 mL beaker 29.132 g 20 mL 48.942 g 19.810 g 19.846 mL .77% 150 mL beaker 77.319 g 100 mL 163.278 g 85.959 g 86.114 mL 16.1% 10 mL cylinder 36.415 g 7 mL 43.507 g 7.092 g 7.105 g 1% 100 mL cylinder 137.725 g 50 mL 187.196 g 49.471 g 49.560 g .89% 25.00 mL volumetric pipette 29.132 g 25.00 mL 54.113 g 24.981 g 25.026 g 0.1039% volume of ethanol (mL) volume of DI water (mL) mixture volume (mL) average of three mixture volumes (mL) % difference between mixture & nominal values average density of mixture (g/mL) Case E trial; 100 mL nominal volume 50 mL 50 mL 97.6 mL 97.6 mL 2.4% 0.926 g/mL

CALCULATIONS: Table 1 1. Mass of water by difference: Calculated by subtracting the mass of water plus glassware from the dry mass of the glassware. 50 mL beaker: 48.942 g – 29.132 g = 19.810 g 2. Volume of water: Calculated using the calculated mass of water and a density of 0.9982 g/mL. 50 mL beaker: 𝐷 = 𝑀 𝑉 0.9982 = 19.810 𝑉 .9982𝑉 = 19.810 𝑉 = 19 .846 3. % error in the volume measurement: Calculated using the formula for percent error 50 mL beaker: % error = 𝐴???𝑎𝑙 𝑉𝑎𝑙??−𝐸𝑥𝑝????? 𝑉𝑎𝑙?? 𝐸𝑥𝑝????? 𝑉𝑎𝑙?? × 100% % ?𝑟𝑟𝑜𝑟 = 19.846 𝑚𝐿 − 20 𝑚𝐿 20 𝑚𝐿 × 100% % ?𝑟𝑟𝑜𝑟 = .77% All values recorded in Table 1 above Table 2 1. % difference between mixture & nominal values: Calculated using % difference equation. Case E trial: % ?𝑖?? = |?𝑖???𝑟?𝑛??| 𝐴??𝑟𝑎?? × 100% %?𝑖?? = 2.4 𝑚𝐿 98.8 𝑚𝐿 × 100% %?𝑖?? = 2 .4% 2. Average density of mixture: Calculated using density equation. Case E trial: 𝐷 = 𝑀 𝑉 𝐷 = (50 𝑚𝐿 × 0.9982?/𝑚𝐿 + 50𝑚𝐿 × 0.8101?/𝑚𝐿) 97.6 𝑚𝐿 𝐷 = 0 .926? /𝑚𝐿

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All values recorded in Table 2 above RESULTS AND CONCLUSION: When combining 50 mL of both 95% ethanol and deionized water, the resulting mixture ends up with less mass than the masses of the original liquids. The resulting mass of the mixture was 97.6 g meaning that the loss was about 2.4% of the total mass of both liquids before mixture. These liquids seemingly did not follow the law of conservation of mass. Several different types of glassware were also tested during this experiment, and most of them were accurate within 1% error, except for the 150 mL beaker which had a percent error of 16.1%. QUESTIONS: 1. Explain why it is important to use the same balance when making the determination of mass by difference (section VII, parts 1 - 3). Using the same balance ensures that all factors remain constant when measuring the mass of different things. A different balance may be slightly inaccurate, or the air flow in a different part of the room could change which could skew the results. 2. Whenever liquids are transferred from one container to another, incomplete mass transfer will occur. Describe how you might make sure that the non-additive volumes were not simply due to mass transfer losses. By making sure that all containers are washed with the liquids that are being transferred within them beforehand, and tapping the glass to make sure that as much liquid as possible has transferred, you minimize any losses or miscalculations. If these precautions are made, the loss in mass from the mixture would have been too great for it to be from simply mass transfer losses.