CHEM Experiment 7 - Chem 106 lab report PDF

Preview

Summary

Chem 106 lab report...

Description

Rida Ahmed Salma Benitez Nadeesha Wijerathne 11 March 2020 Experiment 7: How much is a mole? – “Avogadro’s Number Dilemna”

Introduction: A chemical reaction involves the rearrangement of billions of atoms or molecules. Since it is irrational to count or even visualize all of these tiny molecules, scientists developed a way to refer to an entire quantity of atoms that could be measured and observed. The solution to this dilemma is Avogadro’s number and the mole. The mole is a simple SI unit that allows scientists to calculate the number of atoms or molecules in a certain mass of a given substance. It is defined as the amount of a substance that contains as many particles as there are atoms in 12 grams of carbon-12. Thus, 1 mol contains 6.022×1023 particles of a substance, or Avogadro’s number. This number is fundamental to understanding both the chemical makeup of molecules as well as their interactions with other molecules. In this experiment, we attempted to estimate Avogadro’s number by forming a monolayer of stearic acid (C18H36O2) molecules on the surface of water. We calculated Avogadro’s number based on the given cross-sectional area of a stearic acid molecule and the amount of stearic acid required to cover the water’s surface, and later compared these results to the rest of the class.

Materials: 

1

50 mL-beaker



1 mL acetone



3 mL hexane



10 mL-graduated cylinder



Pasteur pipet (and pipet bulb)



Watch glass



3 mL stearic acid/hexane solution



Ruler



Calculator



Laboratory safety goggles



Laptop or device with access to Microsoft Excel



General Chemistry Laboratory (2019) by G. Smeureanu and S. Geggier



Carbon Copy Laboratory Notebook

Observations and Experimental: Part A – Calculations:

Part 1: Determine the volume of one drop of hexane Trial 1: volume of 1 drop =

3 mL ¿ of drops

=

3 mL 211 drops

= 0.01422 mL

Trial 2: volume of 1 drop =

3 mL ¿ of drops

=

3 mL 209 drops

= 0.01435 mL

Trial 3: volume of 1 drop =

3 mL ¿ of drops

=

3 mL 210 drops

= 0.01429 mL

Average drop volume =

2

0.01422mL+ 0.01435 mL+0.01429 mL 3

= 0.01429 mL

Part 2: Determine the number of drops required to form a monolayer Diameter of watch glass = 14.8 cm Trial 1: 25 drops Trial 2: 27 drops Trial 3: 24 drops surface area of water =

π r 2=π

area of SA molecule = 0.21

(

)

2

14.8 cm =172.0336 c m2 2

(

n m2

)

2

1 cm =2.1 ×10−15 c m 2 7 10 nm 2

number of SA molecules in monolayer =

average volume of solution added =

172.0336 c m = 8.192 −15 2 2.1 ×10 c m

16

×10 molecules

( 25+273 +24 ) 0.01429 mL=0.362mL ¿ 3.62

moles of SA =

3.62

−4

×10 L

Avogadro’s number (NA) =

(

1 mole =1.272 (0.10 gL )( 284.5 g)

8.192 × 10 16 molecules ) −7 1.272 ×10 mol

−4

×10 L −7

×10 mol

¿ 6.438 ×1023 molecules / mol

Part A – Observations: 23

¿ 6.022 × 10 ∨¿ ∙100 %=7 .3 % 23 23 ¿ 6.438 ×10 −6.022 × 10 ∨ ¿ ¿ % error=¿ Group Initials MS TWRS WE 3

Avogadro’s Number 9.74 ×10 23 23 6.438 ×10 23 7.12 × 10

% Error 61.8 7.3 18.24

RJ BJ MJ MMSN LC DS Class Mean Class Standard Deviation

1.23 × 1024 23 6.3537 ×10 24 1.288 × 10 23 9.07 ×10 2.02 × 1024 8.5 ×1023

104.3 5.5 113.88 50.66 235.6 41.15 9.261 ×1023 23 5.725 × 10

Part B – Calculations:

DIAMETER CONCENTRATION AVG DROP VOL TRIAL 1 VOL TRIAL 2 VOL TRIAL 3 VOL 14.3 0.1 1.21E-02 23 20 27 14.6 0.1 1.54E-02 15 17 16 14.2 0.1 1.45E-02 21 24 23 14.5 0.1 1.13E-02 15 21 18 14.9 0.1 1.44E+00 25 23 24 14.5 0.1 1.36E-02 22 23 21 14.2 0.1 1.15E-02 23 24 20 14.7 0.1 1.63E-02 28 24 26 15 0.1 1.43E-02 26 27 24 14.6 0.1 1.27E-02 25 27 21 SURFACE AREA SA MOLECULES IN LAYER AVG VOL ADDED MOLES OF SA AVOGADRO'S NUMBER ERROR 160.6013338 7.64768E+16 2.82E-04 9.92E-08 7.71E+23 2.80E+01 167.410535 7.97193E+16 2.46E-04 8.66E-08 9.20E+23 5.28E+01 158.363015 7.5411E+16 3.29E-04 1.16E-07 6.53E+23 8.40E+00 165.1250938 7.8631E+16 2.03E-04 7.15E-08 1.10E+24 8.26E+01 174.3611038 8.30291E+16 3.46E-02 1.21E-05 6.84E+21 9.89E+01 165.1250938 7.8631E+16 2.99E-04 1.05E-07 7.48E+23 2.42E+01 158.363015 7.5411E+16 2.57E-04 9.03E-08 8.35E+23 3.87E+01 169.7116838 8.08151E+16 4.24E-04 1.49E-07 5.43E+23 9.91E+00 176.709375 AVG 8.41473E+16 3.67E-04 1.29E-07 6.52E+23 8.31E+00 7.64E+23 167.410535 SD 7.97193E+16 3.09E-04 1.09E-07 7.34E+23 2.19E+01 1.58E+23 OVERALL ERROR 2.68E+01 Discussion and Conclusion: 4

In this experiment, we performed a series of tests in order to estimate Avogadro’s number. After we were able to observe the formation of a monolayer of stearic acid over water, we were able to measure Avogadro's number to be approximately

23 6.438 ×10 molecules / mol , which

we later compared to the rest of the class’s data. We then calculated the class mean, standard deviation, and overall percent error to estimate and compare the precision and accuracy of our data values. According to our measurements, our experimental value appears to be accurate because it is close to the actual or true value of Avogadro’s number, which is 6.022×1023 molecules/mol. This is supported by the calculation of a low percent error, which was 7.3%. The smaller the percent error, the more accurate the data. However, our experimental value was not precise when compared to the rest of the class data. This can be demonstrated by the significantly high standard deviation value, 5.725×1023, which indicates that the data values are spread out over a wider range and are not as precise to one another. Although our percent error is relatively low, it is not 0%, which would indicate that our approximated Avogadro’s number is identical to the true value. We can account for our percent error through numerous limitations or experimental errors that may have occurred while conducting the various tests during this experiment. One such experimental error may have been in our technique or method for delivering drops of hexane using a Pasteur pipet in order to determine the volume of one drop of hexane in Part 1. While conducting the trials for this procedure, there were a few times when the pipet dispensed an unequal number of drops due to the difficulty in controlling the pressure exerted on the pipet bulb. It is also possible that we may have miscounted the number of drops due to this unequal number. Since this volume measurement is used in the calculations to estimate Avogadro’s number, a slight error in the volume can cause a change in our estimation. Another experimental error that may have occurred

5

could have been during the procedure in Part 2, when we had to keep adding stearic acid/hexane solution dropwise to the water in the watch glass. An error may have been that we were unable to accurately recognize the first instance when the drops stopped disappearing into the surface of the water, possibly due to adding the drops one after the other too quickly. A failure to accurately determine the number of these drops would directly affect the calculation for the average volume of solution added, which would then also affect our estimation of Avogadro’s number. Another experimental error involving the watch glass may be that we did not completely fill the water to its brim around the entire circumference of the glass due to inattention or unawareness. This would also affect the number of drops that we would observe disappearing in the water. Another source of error may include the handling of lab materials, such as the watch glass in between the three trials of Part 2. In our first trial, we were not able to see the stearic acid/hexane drop disappear within a second in the watch glass, which can be attributed to improper cleaning of the watch glass prior to the trial. This required us to restart the first trial. In conclusion, experimental errors are intrinsic to any lab procedure. While they can be minimized via multiple trials, they cannot be completely eliminated. Several experimental errors that may have occurred during this lab include delivering an unequal number of drops, a failure to count the number of drops accurately, negligence in filling the water up to the brim, and improper cleaning of lab equipment. Although we cannot eliminate experimental errors, we can learn to become more aware of them in future lab experiments.

Focus Questions: Part A: 1. What is Avogadro’s number based on your measurements?

6

Based on our measurements in this lab, Avogadro' s number is approximately 23

6.438 ×10 molecules /mol .

2. How does your experimental value compare to the actual value and the class average? Our experimental value appears to be accurate because it is close to the actual accepted value for Avogadro’s number. This is demonstrated by our low percent error, 7.3%. However, our experimental value was not precise when compared to the rest of the class data. This can be demonstrated by the high standard deviation value, 5.725×1023, which indicates a wide spread in the data values from the class mean.

Part B: 1. How accurate is the data from your peers from last year? The data from last year appears to be accurate due to a low percent error. We calculated the overall percent error to be approximately 26.8%, which is relatively low. This demonstrates that the experimental values are close to the actual value of Avogadro’s number, 6.022 ×1023 . 2. How precise is it? The data from last year appears to be precise due to a low standard deviation. We calculated the standard deviation to be approximately 1.58 ×1023 , which is relatively low. This small standard deviation demonstrates that there is a low discrepancy among all the experimental values, which indicates less dispersion within the data values from the mean.

References:

7



“Avogadro’s Number and the Mole”: https://courses.lumenlearning.com/introchem/chapter/avogadros-number-and-the-mole/



“How Was Avogadro’s Number Determined?”: https://www.scientificamerican.com/article/how-was-avogadros-number/

Post-Lab Assessment Questions: 1. How many grams of ammonia are necessary to weigh out 3.4 moles? 3.4 moles NH 3 ×

17.03052 g =57.904 g NH 3 1 mole NH 3

2. How many molecules of ammonia are there in 5 moles of NH3? 5 moles NH 3 ×

6.022 ×1023 molecules =3.011 ×1024 molecules 1mole NH 3

NH 3

3. How many moles of H are there in 6.0 g of H2O? 1 mole H 2 O 2 moles H =0.666 moles H × 18.01528 g 1 mole H 2 O

6.0 g H 2 O×

4. How many molecules of CH4 are there in 24.3 g of methane? How many moles of H atoms are there? 24.3 g CH 4 ×

1 mole CH 4 6.022 ×1023 molecules 23 =9.122 ×10 molecules CH 4 × 1 mole CH 4 16.04276 g

24.3 g CH 4 ×

1 mole CH 4 4 moles H × =6.059 moles H 16.04276 g 1 mole CH 4

5. If a gumdrop is 1.5 cm3 in volume, what would the volume of a dozen gumdrops be? 12 gumdrops ×

1.5 cm3 =18 cm3 1 gumdrop

6. If a raindrop is 0.07 cm3 in volume, what would the volume of an Avogadro number of raindrops be? 8

6.022 × 1023 raindrops ×

0.07 cm 3 =4.215 ×10 22 cm3 1raindrop

7. Lauric acid is used to make the detergents found in shampoo and toothpaste. Lauric acid is similar to stearic acid, but only has 12 carbons. Show the Lewis structure for lauric acid.

9...