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In the "Determining the Length of a Molecule and the Diameter of a Carbon Atom" lab, the purpose was to determine the thickness of a mono layer of oleic acid molecules floating perpendicularly on the surface of water, which also would conclude to finding the diameter of a single carbon atom. The oleic acid contains a fatty acid with 18 carbon atoms bound together with one carbon atom at the head and the other 17 as a tail. Once the oleic acid is dropped on the surface of the water it creates a circular thin film called a mono layer. The oleic acid contains a density of 0.895 grams per milliliters allowing it to float on the surface of water. This was all shown by first having to determine the volume of one droplet of oleic acid-ethanol*…show more content…*

Firstly we measured the volume of one droplet of oleic acid. This was done by setting up a dimensional analysis chart of one droplets of oleic acid divided by our average number of droplets needed to reach 1.00 ml, resulting in .0313 ml. Secondly was to find the volume of just the oleic acid contained in one droplet. This was carried out by multiplying the volume of one droplet of oleic acid and 2% because each droplet is only 2% oleic acid and 98% ethanol. Concluding the calculation, the volume came to be .000626 ml. Thirdly we were to find the area of the mono layer film with the equation pie times the average diameter from our data divided by two squared. The area cam to be 801.7 centimeters squared. Next was to find the height of the circular, mono layer film by dividing the volume of just the oleic acid in one droplet by the area of the mono layer film, resulting to .000000781 centimeters squared as the height of the circular film. The percent error of the actual thickness of an oleic acid molecule came out to be 51.8% from out calculations, measured by the true value of 1.62 times 10 to the power of negative sixth(true value) and subtracting that by our thickness result divided by the true value and multiplied by 100. Next came the diameter of a single carbon atom. We measured this by taking the mono layer thickness and dividing it by the number of carbon atoms in the tail(17), which concluded to .0000000459 centimeters. Finally, we took the percent error of an outside source saying the diameter of a carbon atom was 1.54 times ten to the negative tenth power. Concluding this calculation, the result came out to being 29,800% of percent error from our measured

Firstly we measured the volume of one droplet of oleic acid. This was done by setting up a dimensional analysis chart of one droplets of oleic acid divided by our average number of droplets needed to reach 1.00 ml, resulting in .0313 ml. Secondly was to find the volume of just the oleic acid contained in one droplet. This was carried out by multiplying the volume of one droplet of oleic acid and 2% because each droplet is only 2% oleic acid and 98% ethanol. Concluding the calculation, the volume came to be .000626 ml. Thirdly we were to find the area of the mono layer film with the equation pie times the average diameter from our data divided by two squared. The area cam to be 801.7 centimeters squared. Next was to find the height of the circular, mono layer film by dividing the volume of just the oleic acid in one droplet by the area of the mono layer film, resulting to .000000781 centimeters squared as the height of the circular film. The percent error of the actual thickness of an oleic acid molecule came out to be 51.8% from out calculations, measured by the true value of 1.62 times 10 to the power of negative sixth(true value) and subtracting that by our thickness result divided by the true value and multiplied by 100. Next came the diameter of a single carbon atom. We measured this by taking the mono layer thickness and dividing it by the number of carbon atoms in the tail(17), which concluded to .0000000459 centimeters. Finally, we took the percent error of an outside source saying the diameter of a carbon atom was 1.54 times ten to the negative tenth power. Concluding this calculation, the result came out to being 29,800% of percent error from our measured

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