Astudent performed an analysis of a sample for its calcium content and obtained the following results:
14.92 g, 14.91 g, 14.88 g, 14.92 g
the actual amount of calcium in the sample is 20.90 g. for this lab, the instructor asked students to obtain results within 5% error and rsd and to redo the experiment if that was not achieved.
(a) report the student’s results for the average and the standard deviation of the measured masses. show your calculations below. do the calculations by hand, and if you have a calculator or excel you can use it to check your answer.
. ± .
(average) (standard deviation)
b) calculate the percent error, using the average mass, for this student’s experiment.
(c) calculate the relative standard deviation for this student’s experiment.
(d) briefly comment on the accuracy vs. the precision obtained by the student. does the student need to redo the experiment?

14.9075 g, 28.67%, 0.11%

Explanation:

The mean concentration of calcium = summation x / frequency

= ( 14.92 + 1491 + 14.88 + 14.92 ) /4 = 14.9075 g

Standard deviation = √(summation (x - μ)² /n) = √ ( ((14.92 - 14.9075)² +(14.91 - 14.9075)² + (14.88 - 14.9075)² + ( 14.92 - 14.9075)²) / 4) = 0.0164

b)  percent error = abs(14.9075 - 20.90) / 20.90 × 100 = 28.67%

c) relative standard deviation = standard deviation / mean × 100 = 0.0164 /  14.9075 × 100 = 0.11%

d) The accuracy of the measure is the measurement compared to the actual which according to the standard set by the instructor (5%error) is not very accurate because the percent error is high (28.67%) while the relative standard deviation is quite low ( 0.11%) which means the measurement precision is very high.

The student will have to redo the experiment because the experiment was not too accurate since the percent error is way higher than the set value (5%) although the precision was high.

fe 55.85 g/mol and h2o 18.02 g/mol.