The following results were obtained for the determination of calcium in a NIST limestone sample: % CaO = 50.32, 50.22, 50.40, 50.21, and 50.44. Five gross samples were then obtained for a carload of limestone. The average percent CaO values for the gross samples were found to be 49.47, 50.10, 49.58, 49.76, and 50.49. Calculate the relative standard deviation associated with the sampling step. Relative standard deviation =

The following results were obtained for the determination of calcium in a NIST limestone sample: % CaO = 50.32, 50.22, 50.40, 50.21, and 50.44. Five gross samples were then obtained for a carload of limestone. The average percent CaO values for the gross samples were found to be 49.47, 50.10, 49.58, 49.76, and 50.49. Calculate the relative standard deviation associated with the sampling step. Relative standard deviation =

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter11: Atomic Mass Spectrometry

Section: Chapter Questions

Problem 11.10QAP

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The following results were obtained for the determination of calcium in a NIST limestone sample: %CaO = 50.33, 50.22, 50.36, 50.21, and 50.44. Five gross samples were then obtained for a carload of limestone. The average %CaO values for the gross samples were found to be 49.53, 50.12, 49.60, 49.87, and 50.49. Calculate the relative standard deviation associated with the sampling step
Analysis of several plant-food preparations for potassium ion yielded the following data: Sample Percent K+ 1 3.90, 3.96, 4.16, 3.96 2 4.48, 4.65, 4.68, 4.42 Find the mean ; standard deviation variance and RSD for each sample.
Calculate the standard deviation for an element whose % in a sample were found to be20.8, 21.6, 22.1, 22.0, 23.3, 21.9 and 22.8%
Titration of an oxalate sample gave the following percentages: 15.07 % , 15.58 % , and 15.42 % . Calculate the average. Calculate the standard deviation.
An atomic absorption method for the determination of the amount of iron present in used jet engine oil was found from pooling 30 triplicate analyses to have a standard deviation s = 3.6 mg Fe/mL. If s is a good estimate of σ, calculate the 95 and 99% confidence intervals for the result 18.5 mg Fe/mL if it was based on (a) a single analysis, (b) the mean of two analyses, and (c) the mean of four analyses.
A young researcher was evaluating a standard method for determining the methylmercury content in blue fin tuna using high performance liquid chromatography (HPLC). She determined the standard deviation (?)(s) for the method to be 0.440.44 ppb and assumed that ?s was a good approximation of ?σ . As a test, she used this method to evaluate the methylmercury content in a National Institutes of Standards and Technology (NIST) standard and determined the unknown amount to be within 0.210.21 ppb (?)(μ) of the known mean with 99% probability. How many replicate measurements of the NIST standard did the researcher perform?
An atomic absorption method for the determination of the amount of iron present in used jet engine oil was found from pooling 30 triplicate analyses to have a standard deviation s = 3.1 µg Fe/mL. If s is a good estimate of o, calculate the 95 and 99% confidence intervals for the result 21.7 µg Fe/mL if it was based on the following criteria. (a) a single analysis 95% confidence interval= 21.7 + 99% confidence interval = 21.7 + (b) the mean of two analyses 95% confidence interval = 21.7 + 99% confidence interval= (c) the mean of six analyses 21.7 ± 95% confidence interval = 21.7 + 99% confidence interval= 21.7 + μg Fe/mL μg Fe/mL μg Fe/mL μg Fe/mL μg Fe/mL μg Fe/mL
To test the quality of the work of a commercial laboratory, duplicate analyses of apurified benzoic acid (68.8% C, 4.953% H) sample were requested. It is assumed that therelative standard deviation of the method is sr →  = 4 ppt for carbon and 6 ppt forhydrogen. The means of the reported results are 68.5% C and 4.882% H. At the 95%confidence level, is there any indication of systematic error in either analysis?
You are developing a procedure for determining traces of copper in biological materials using a wet digestion followed by measurements by atomic absorption spectrophotometry. In order to test the validity of the method, you obtain a NIST orchard leaves standard reference material and analyze this material with your procedure five times and obtain a mean of 12.27 ppm with a standard deviation of 0.097 ppm. The NIST sample is listed as 11.7 ppm. Does your method give a statistically valid value at the 95% confidence level?
A volume of 250 ml of a 0.05 M solution of a reagent of formula weight (relative molecular mass) 40 was made up, the weighing being done by difference. The standard deviation of each weighing was 0.0001 g. 1 (a) What were the standard deviation and relative standard deviation of the weight of reagent used?
One method for the quantitative determination of the concentration of constituents in a sample analyzed by gas chromatography is area normalization. Here, complete elution of all the sample constituents is necessary. The area of each peak is then measured and corrected for differences in detector response to the different eluates. This correction involves dividing the area by an empirically determined correction factor. The concentration of the analyte is found from the ratio of its corrected area to the total corrected area of all peaks. For a chromatogram containing three peaks, the relative areas were found to be 16.4, 45.2 and 30.2, in order of increasing retention time. Calculate the percentage of each compound if the relative detector responses were 0.60, 0.78 and 0.88, respectively.
In a volumetric determination of an analyte A, the data obtained and their standard deviations are as follows: Initial buret reading 0.19 mL 0.02 mL Final buret reading 7.86 mL 0.03 mL Sample mass 63.0 mg 0.2 mg From the data, find the coefficient of variation of the final result for the %A that is obtained by using the equation that follows and assuming there is no uncertainty in the equivalent mass. %A= titrant volume x equivalent mass x 100 %/ sample mass Coefficient of variation = Submit Answer % Try Another Version 10 item attempts remaining