How to Calculate Standard Deviation
Essay by Kashfia Farah • March 17, 2016 • Lab Report • 358 Words (2 Pages) • 1,157 Views
Results
Table 1: volume of stock and water (ml), mass of phosphorus (µg), absorbance at 660nm
Mass of phosphorus (µg) | Volume of stock and water (5mL) | Absorbance at 660 nm | ||
1 | 2 | average | ||
10 | 0.1 + 4.9 | 0.034 | 0.037 | 0.036 |
20 | 0.2 + 4.8 | 0.057 | 0.082 | 0.069 |
30 | 0.3 + 4.7 | 0.097 | 0.086 | 0.091 |
40 | 0.4 + 4.6 | 0.111 | 0.143 | 0.127 |
50 | 0.5 + 4.5 | 0.140 | 0.166 | 0.153 |
Unknown x | 5 | 0.038 | 0.035 | 0.036 |
Unknown y | 5 | 0.108 | 0.104 | 0.106 |
Sample calculation to show the volume of stock:
There is 100 (µg) of phosphorus in 1 mL of phosphate solution
Therefore, 10 (µg) of phosphorus will be = = 0.1 mL of phosphate solution[pic 1]
The calculation is repeated to find the stock solution for 20, 30, 40 and 50 (µg) of phosphorus respectively.
[pic 2]
Figure 1: the graph representing the relationship between absorbance at 660nm and the mass of phosphorus (µg).
To find the mass of phosphorus from the standard curve,
For unknown X
Given, y = 0.003x
0.036 = 0.003x
Therefore, x = 12 (µg)
Concentration of unknown x = 12 / 5 = 2.4(µg/mL)
For unknown Y
Given, y = 0.003x
0.106 = 0.003x
Therefore, x = 35.3 (µg)
Concentration of unknown y = 35.3 / 5 = 7.06 (µg/mL)
Calculation to find the standard deviation for unknown X and Y
Formula for standard deviation,
S = [pic 3]
Unknown x1,
From graph, y= 0.003x
0.038 = 0.003x
Therefore, x = 12.7 (µg)
Concentration for x1, 12.7 / 5 = 2.5(µg/mL)
Unknown x2,
From graph, y= 0.003x
0.035 = 0.003x
Therefore, x = 11.7 (µg)
Concentration for x2, 11.7 /5 = 2.3(µg/mL)
Hence, the standard deviation for the average concentration
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