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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.
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