Analytical Lab 1 Paper
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Chemistry
Date
Oct 30, 2023
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Robert Billings Lab Report 1 Calculations / Table of Data Calculations 1-5 1.
From the 5 times the same beaker was weighed we received the following data points (30.4515g, 30.4517g, 30.4514g, 30.4519g, 30.4516g) To calculate the mean mass we will add all 5 and divide by 5. (
30.4515 + 30.4517 + 30.4514 + 30.4519 + 30.4516)/5
= 30.4516g. The range is 5E-4 due to subtracting the smallest value from the largest value. The standard deviation is 0.00019. The %RSD is 0.00019(100)/30.4516
= 0.006% 2.
10 mL:
(mass of 4 weights) (9.8973g, 9.8517g, 9.9342g, 9.8504g) (avg = 9.8834g) V (10mL) : ࠵?/࠵?
(
!.#!$%&
’.(()!&/+,
, !.#-’$&
’.(()!&/+,
,
!.!%.)&
’.(()!&/+,
,
!.#-(.&
’.(()!&/+,
) Avg Volume (
!.##%.&
’.(()!&/+,
) (9.867ml, 9.823ml, 9.905ml, 9.822ml) (9.855ml) Standard deviation = 0.0398 mL %RSD = (.(%!#+/(’(()
!.#--+/
= 0.403% %Error =
. 145࠵?࠵?(100)/10࠵?࠵?
= -1.45% 5mL (mass of 4 weights) (5.0416g, 4.9393g, 4.9259g, 4.9583g) (avg = 4.966g) V(5mL) : ࠵?/࠵?
(
-.(.’2&
’.(()!&/+,
, ..!%!%&
’.(()!&/+,
,
..!)-!&
’.(()!&/+,
,
..!-#%&
’.(()!&/+,
) Avg Volume (
..!22&
’.(()!&/+,
) (5.027ml, 4.925ml, 4.912ml, 4.944ml) (4.952ml)
Standard deviation: 0.0517mL %RSD = 0.0517࠵?࠵?(100)/4.952࠵?࠵?
= 1.04% %Error = 0.48mL(100)/50mL = 0.96% 3.
1000uL:
(mass of 4 weights) (0.9733g, 0.9875g, 0.9681g, 0.9830g, 0.9959g) (avg = 0.98156g) V (1000uL) : ࠵?/࠵?
(
(.!$%%&
’.(()!&/+,
, (.!#$-&
’.(()!&/+,
,
(.!2#’&
’.(()!&/+,
,
(.!#%(&
’.(()!&/+,
, (.!!-!&
’.(()!&/+,
)
Avg Volume (
(.!#’-2&
’.(()!&/+,
) = (0.9787ml) Volumes : (0.970ml, 0.985ml, 0.9653ml, 0.980ml, 0.993ml) Standard deviation = 0.0111 mL %RSD = (.(’’’+/(’(()
(.!$#$&/+/
= 1.1% %Error =
21࠵?࠵?(100)/1000࠵?࠵?
= -2.1% 500uL:
(mass of 4 weights) (0.3993g, 0.4368g, 0.4940g, 0.4815g, 0.4448g) (avg = 0.45128g) V (500uL) : +
3
(
(.%!!%&
!.##$%&
’(
, (..%2#&
!.##$%&
’(
,
(..!.(&
!.##$%&
’(
,
(..#’-&
!.##$%&
’(
, (
....
#&
!.##$%&
’(
)
Avg Volume: (
(..-’)#&
’.(()!&/+,
) = (0.449ml) Volumes: (0.398ml, 0.4355ml, 0.4926ml, 0.480ml, 0.4435ml) Standard deviation = 0.03765 mL %RSD = (.(%$2-+/(’(()
(...!&/+/
= 8.3% %Error =
51࠵?࠵?(100)/500࠵?࠵?
= -10.2%
100uL:
(mass of 4 weights) (0.0985g, 0.0954g, 0.0987g, 0.0961g, 0.0995g) (avg = 0.09764g) V (100uL) : ࠵?/࠵?
(
(.(!#-&
’.(()!&/+,
, (.(!-.&
’.(()!&/+,
,
(.(!#$&
’.(()!&/+,
,
(.(!2’&
’.(()!&/+,
, (.(!!-&
’.(()!&/+,
)
Avg Volume : (
(.(!$2.&
’.(()!&/+,
) = (0.0974ml) Volumes: (0.0982ml, 0.0951ml, 0.0984ml, 0.0958ml, 0.0992ml) Standard deviation = 0.00178 mL %RSD = (.((’$#+/(’(()
(.(!$.&/+/
= 1.8% %Error =
2.64࠵?࠵?(100)/100࠵?࠵?
= -2.64% 50uL:
(mass of 4 weights) (0.0468g, 0.04644g, 0.0488g, 0.0490g, 0.0467g) (avg = 0.04754g) V (50uL) : ࠵?/࠵?
(
(.(.2#&
’.(()!&/+,
, (.(.2..&
’.(()!&/+,
,
(.(.##&
’.(()!&/+,
,
(.(.!(&
’.(()!&/+,
, (.(.2$&
’.(()!&/+,
)
Avg Volume: (
(.(.$-.&
’.(()!&/+,
) = (0.0474ml) Volumes: (0.0467ml, 0.0463ml, 0.0487ml, 0.0489ml, 0.0466ml) Standard deviation = 0.00125 mL %RSD = (.((’)-+/(’(()
(.(.$.&/+/
= 2.6%
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%Error =
2.6࠵?࠵?(100)/50࠵?࠵?
= -5.2% 4.
As expected, my error is higher than what is traditionally expected from the factory certification pipette. The following is the error expected vs the error I received respectively. (0.01mL,0.48mL) (0.02mL,0.145mL) (6uL,21uL) (6uL,51uL) (0.5uL,2.64uL) (0.25uL,2.60uL) This is clearly due to human error, to prevent the aforementioned. More practice with the instruments should be put into place as well as caution with ensuring that a new cap is used with every measurement when using the micropipette. Furthermore, ensuring that both the zeroing of the scale and the measurement of the water is taken with the glass screens closed on the balance. 5.
N/A TABLE
Questions 1.
Random error is unpredictable and usually occur due to uncontrollable factors. This can be due to limitations or the fault of measurement equipment, environment, or human error. A specific example is if the analytical balance was influenced by air currents or noise. Systematic error is error usually caused by flaws or certain bias either by the human or the experiment. These errors are consistent and repeatable. For example, the thermometer may read one degree due to high but due to a calibration error. This isn’t something that is random but something that can and should be fixed. 2.
The number one scenario in which you should use a median instead of mean is during instances of an outlier. A median is much less sensitive to outliers and therefore should be used if an outlier is present as to not skew the data set. A mean considers all parts of the data set to calculate average and therefore can present a mis representation of the totality of the data set. 3.
TC glassware (to contain) is designed to “contain” a specific amount of volume of liquid accurately and precisely. This glassware is used when it is pivotal that an exact volume is measured properly. TD glassware is designed to deliver a specific volume of liquid accurately when the liquid is dispensed. TD glassware considers the volume remaining in the glassware after dispensing, making it a great choice for when someone wants to accurately dispense liquid. TC glassware would be (Erlenmeyer flask, Beaker, test tubes, volumetric pipette) TD glassware would be (Buret, Pipette of most kinds, Syringe) 4.
To calibrate my volumetric flask, I would proceed with the following steps: 1. Ensure my flask is clean and dry 2. Grab a certified solution or certified weight as a comparison 3. Use a funnel to carefully transfer the certified weight to the flask 4. Add DI water to calibration mark on flask 5. Re weigh the flask with contents and compare to expected weight of substances known weight 6. Record all measurements and calculate weight of flask. 5.
A volumetric pipette is designed to be highly accurate and precise when measurements of fixed volumes are required. These are usually used when conducting quantitative experiments that require precise measurements. A Mohr pipette is used when variable volumes must be transferred. Maybe precise volume measurements aren’t at the up most importance rather the transfer is important. You would use a Mohr pipette in that case. Mohr pipette don’t have a single volume mark like a volumetric pipette and for that reason are used for variable volume.
6.
Most of the times this method is used to account the mass of a substance that can’t be easily transferred or highly reactive. This technique is done by weighing the container, adding a substance, reweighing the container, and then subtracting the total from the empty container. This seems simple but it is very useful when one wants to avoid random error, avoid reagent loss, and avoid contamination. 7.
I would ensure my analytical balance is calibrated properly to ensure accuracy. Grab a clean and dry container and place it on the balance which I will proceed to tare. I will carefully add the sample to the container ensuring none of the sample is lost. On top of that I will ensure that the conditions are the exact same as when I zeroed the balance except for the addition of the sample. I will record the mass and add a (+-) value and then the uncertainty value depending on the type of analytical scale and glassware that was used for the weight. This should be done 4 more times to account for uncertainty. 8.
Average = 0.1278 + 0.1282 + 0.1275 + 0.1296/4
= 0.128275 Standard Deviation = (0.1278 − 0.128275)
)
+ (0.1282 − 0.128275)
)
+
(0.1275 − 0.128275)
)
+
((.’)!24(.’)#)$-)
$
/3
= 2.935࠵?
42
Standard Deviation = ࠵?࠵?࠵?࠵?(2.935࠵?
42
) = 0.00171
•
F
࠵?࠵? 0.1278࠵?:
࠵? =∣ 0.1278 − 0.128275/∣ 0.00171 = 0.278
•
࠵?࠵?࠵? 0.1282࠵?:
࠵? =∣ 0.1282 − 0.128275/∣ 0.00171 = 0.044
•
࠵?࠵?࠵? 0.1275࠵?:
࠵? =∣ 0.1275 − 0.128275/∣ 0.00171 = 0.453
•
࠵?࠵?࠵? 0.1296࠵?:
࠵? =∣ 0.1296 − 0.128275/∣ 0.00171 = 0.775
None exceeds critical value and therefore none should be an outlier. 9.
࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?2࠵?3 = 325.811 ࠵?/࠵?࠵?࠵?
࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?2࠵?3 =
0.1968࠵?
325.811࠵?/࠵?࠵?࠵?
= 6.04࠵? − 4 ࠵?࠵?࠵?࠵?࠵?
࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?: ࠵?࠵?2࠵?3 + 6࠵?࠵?࠵?3 > 2࠵?࠵?(࠵?࠵?3)3 + 3࠵?2࠵?
࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?(࠵?࠵?3)3 ∶ (6.04࠵?
4.
࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?2࠵?3) (2 ࠵?࠵?࠵? ࠵?࠵?(࠵?࠵?3)3)
= 1.208࠵?^ − 3 ࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?(࠵?࠵?3)3
1.208࠵?^ − 3 ࠵?࠵?࠵?࠵?࠵? / 1 ࠵?࠵?࠵?࠵?࠵? = 1.208࠵?^ − 3࠵?
1.208࠵?^ − 3 ࠵?࠵?࠵?࠵?࠵? (324.92 ࠵?/࠵?࠵?࠵?) = 0.3925࠵? ࠵?࠵?(࠵?࠵?3)3
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392.5 ࠵?࠵?/1 ࠵?࠵?࠵?࠵?࠵? = 392.5࠵?࠵?࠵?
10.
࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? = 2࠵?࠵?࠵?࠵?3 + ࠵?࠵?2࠵?࠵?3 > ࠵?࠵?2࠵?࠵?3 + 2࠵?࠵?࠵?࠵?3
࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?2࠵?࠵?3 = 0.3572࠵?/105.98࠵? = 3.37࠵?^ − 3 ࠵?࠵?࠵?࠵?࠵?
(3.37࠵?^ − 3 ࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?2࠵?࠵?3) (2 ࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?࠵?࠵?3)
= 6.74࠵?^ − 3 ࠵?࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?࠵?࠵?3
࠵?ℎ࠵?࠵? ࠵?࠵?࠵?࠵?࠵? ࠵?࠵? 6.74 ࠵?࠵?࠵?࠵?/0.07216࠵? = 93.4࠵?࠵?
࠵?࠵?࠵?࠵?࠵?/࠵?࠵?࠵?࠵?࠵?࠵? = ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?
࠵?࠵?࠵?࠵?࠵?/࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? = ࠵?࠵?࠵?࠵?࠵?࠵?
11.
(0.38࠵?/࠵? ࠵?2࠵?࠵?࠵?4 ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?) (1.37 ࠵?/࠵?࠵?) = 0.5206 ࠵?/࠵?࠵?
0.5206࠵?/࠵?࠵? / 194.19࠵? ࠵?࠵? ࠵?2࠵?࠵?࠵?4 = 0.00268 ࠵?࠵?
0.00268࠵?࠵? (1࠵?࠵?࠵? ࠵?࠵?࠵?4/࠵?2࠵?࠵?࠵?4) (1000࠵?࠵?) = 2.68 ࠵?
12.
࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵? = ࠵?࠵?࠵?࠵?3 + ࠵?࠵?3࠵?࠵?࠵?4 > ࠵?࠵?3࠵?࠵?࠵?4 + 3࠵?࠵?࠵?࠵?3
࠵?࠵?࠵?࠵?3 = 0.177࠵?, ࠵?࠵?࠵?࠵?࠵?࠵? = 1.5࠵?
࠵?࠵?࠵?࠵?࠵? = (0.177࠵?)(1.5࠵?)
= 0.2655 ࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵?3
࠵?࠵?3࠵?࠵?࠵?4 = (0.122࠵?)(0.5࠵?) = 0.061 ࠵?࠵?࠵?࠵?࠵? ࠵?࠵?3࠵?࠵?04 = 0.061 ࠵?࠵?࠵?࠵?࠵? ࠵?࠵?3࠵?࠵?࠵?4
࠵?࠵?࠵?࠵?࠵? ࠵?࠵?࠵?࠵? ࠵?࠵?3࠵?࠵?࠵?4 = 462.53 ࠵?/࠵?࠵?࠵?
࠵?࠵?࠵?࠵? ࠵?࠵? ࠵?࠵?3࠵?࠵?࠵?4 = (462.53࠵?/࠵?࠵?࠵?)(0.061࠵?࠵?࠵?࠵?࠵?) = 28.21 ࠵?࠵?࠵?࠵?࠵?
Notebook (Attached as separate document)
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DELL
2.
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(9.75 x 103)(8.4 x 10-6)
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Using the following mass standard data, calculate the Rx value.
mass of
mass of
area of
area of
isopentyl
acetate (g)
isopentyl
alcohol (g)
isopentyl
acetate peak
isopentyl
alcohol peak
80.0
1.10
1.28
189
Please report your answer to 3 decimal places. (No units required)
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GRAVITY AND ACCELERATION (II)
Name
ihe distance covered by a freely falling body is calculated by the following formula,
d = at?
12
Example 1: How far will an object fall in 5 seconds?
Answer:
d = 9.8 m/s) (5s)2
= 122.5 meters
where d =distance
a = acceleration
t= time
Example 2: What is the average velocity of a ball
that attains a velocity of 39.2 m/s after 4 seconds?
(2
Answer:
Va
= V,- V, =
39.2 - 0 =
19.6 m/s
Solve the following problems.
11 How far will a rubber ball fall in 10 seconds?
Answer:
2 How far will a rubber ball fall in 20 seconds?
Answer:
3. How long will it take an object dropped from a window to fall a distance of
78.4 meters?
Answer:
4 Calculate the final velocity of the ball in Problem 1.
Answer:
5. What is the average velocity of the ball in Problem 1?
Answer:
6. An airplane is traveling at an altitude of 31,360 meters. A box of supplies is dropped
from its cargo hold. How long will it take to reach the ground?
Answer:
7. At what yelocity will the box in Problem…
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