Abbreviated Report Week 1

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Georgia Institute Of Technology *

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1211K

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Chemistry

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Apr 3, 2024

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Precision, Accuracy, and Precipitation Reactions 29 March 2024 CHEM 1211K Laboratory Data and Results Unknown letter: A Unknown + sodium sulfate Unknown + lead(II) nitrate Clear after unknown added and no precipitate formed Clear after unknown added and no precipitate formed Unknown + sodium hydroxide Unknown + sodium chloride Turned light blue in color when unknown added and precipitate formed Clear after unknown added and no precipitate formed Unknown + barium nitrate Clear after unknown added and no precipitate formed Identity of unknown solute: Copper(II) Nitrate Table 1. Determination of the identity of an unknown solution of an ionic salt via precipitation reactions. Masses of water delivered (g) Instrument Trial 1 Trial 2 Trial 3 Mean St. dev. Serological pipet 10.144 9.76 10.118 10.007 0.215 Volumetric pipet 10.147 10.185 10.275 10.202 0.066 Graduated cylinder 9.566 10.118 9.83 9.838 0.276 Most accurate: Serological Pipet Most precise: Volumetric Pipet Table 2. Determination of precision and accuracy of laboratory glassware designed to deliver 10 mL of liquid.

0 100 200 300 400 500 600 20 30 40 50 60 70 80 90 100 f(x) = 0.11 x + 33.18 R² = 0.95 Time (s) Temperature (°C) Figure 1. Graph depicting the heating of 50mL of water over 600 seconds.

Discussion Experiment A involved determining the chemical formula of an unknown salt solution, my group chose Unknown A, by creating precipitates with other know solutions. Through the mixing of Unknown A with various solutions and the formation of a precipitate, I have found Unknown A to be the compound copper(II) nitrate. It was observed that sodium hydroxide formed a precipitate with Unknown A and was the only known solution to do so. Cu ¿ When given the possible cations of Cu 2+ , Ba 2+ , Ag + , or Na + along with the possible anions of Cl – , NO 3 – , or SO 4 2– ; a total of 12 different salt solutions can be made. Using solubility rules however helps narrow down the selection to just a single salt solution. Any salt solutions containing sodium were eliminated because salts of group 1 cations are soluble and would therefore not create a precipitate with sodium hydroxide. Salt solutions containing silver could be ruled out since a sulfate becomes insoluble when combined with silver, thus a precipitate would have been observed when mixing Unknown A with sodium sulfate which was not the case. The same explanation can be made with the use of barium as well, leaving only the copper salt solutions. Copper(II) chloride would have formed a precipitate with lead(II) nitrate and copper(II) sulfate would have formed a precipitate with barium nitrate, both of which were not observed in actual testing. Simple process of elimination leads to the only possible salt solution of copper(II) nitrate to be the Unknown A solution in Table 1. Table 2 gives data on experiment 2 where we were tasked to measure out 10 mL of water with three different pieces of glassware. The glassware used was a graduated cylinder, a volumetric pipet, and a serological pipet in order to test their accuracy and precision. After collecting our data we observed that the accuracy of the glassware from most accurate being the serological pipet, then the graduated cylinder, and the least accurate being the volumetric pipet. The order of accuracy was determined by the percent error, which shows how close each tool actually got to the 10 mL of water. Percent error can be calculated using the formula: PE = x meas − x true x true × 100 Calculations provide the serological pipet had a 1.18 percent error in trial 3, the graduated cylinder had a 1.18 percent error in trial 2, and the volumetric pipet had a 1.47 percent error in trial 1. Although the serological pipet and the graduated cylinder have equal percent errors, the next closest in accuracy was the serological pipet with a percent error of 1.47 in trial 1, while the graduated cylinder only had a percent error of 1.70 in trial 3 making the serological pipet more accurate. However when it comes to precision, the volumetric pipet was the most precise, then the serological pipet, and the least precise was the graduated cylinder. Precision shows how consistent the glassware

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is at yielding the same volumes of water and unlike accuracy is measured using standard deviation. Precision can be calculated using the formula: σ = √ 1 N ∑ i N ¿¿¿ Calculations provide that the volumetric pipet had a standard deviation of 0.066, the serological pipet had a standard deviation of 0.215, and the graduated cylinder had a standard deviation of 0.276 thus confirming the order of precision listed above. Experiment C is depicted in Figure 1 as 50 mL of water was heated and the temperature was measured every 5 seconds for a period of 600 seconds. Given that we need to calculate the average heating rate of the hot plate in joules per minute, the first thing that needs to be solved for is the time it takes to raise the temperature of the water by one degree Celsius using the line of best fit. The equation is set up where the y is replaced by one plus the y-intercept (1 + 33.175) at x=0, therefore y would be 34.175 and thus reflecting the increase by one degree Celsius. Then solve for x, representing the time it takes to raise the temperature of the water by one degree Celsius. y = 0.1142 x + 33.175 34.175 = 0.1142 x + 33.175 1 = 0.1142 x x = 8.757 seconds This can be saved for later as 1°C/8.757 seconds. Next we need to know the specific heat of water, the energy required to raise the temperature of 1 gram of water by 1°C, which is stated to be 4.184 J/g°C. Additionally we will use the density of water which is 1g/mL. From here a series of dimensional analysist can be performed to get the final answer into the correct units using the equation: mL× g mL × J g°C × ° C s × s min = J min 50 mL× 1 g 1 mL × 4.184 J 1 g° C × 1 ° C 8.757 s × 60 s 1 min = 1433 Joules per minute By plugging in the values solved beforehand to their respective positions the numerical answer can be calculated. Going from left to right, first there is the volume of water (mL), 50mL, in correspondence to the amount of water put in the beaker. Next is the density of water (g/mL) stated to be 1g/1mL, followed by the specific heat of water (J/g°C) stated to be 4.184 J/g°C. Then comes the time it takes to raise the temperature of the water by one degree Celsius (°C/s), coming from the first calculation of 1°C/8.757

seconds. Last is the conversion of seconds to minutes (s/min) known to be 60s/min. We can now determine that the average heating rate of the hot plate to be 1433 J/min.