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The seller of a mining claim took a random ore sample that weighed approximately \(5 \mathrm{lb}\) and had an average particle diameter of \(5.0 \mathrm{~mm}\). Inspection revealed that about \(1 \%\) of the sample was argentite (see Problem 8-9), and the remainder had a density of about \(2.6 \mathrm{~g} / \mathrm{cm}^{3}\) and contained no silver. The prospective buyer insisted on knowing the silver content of the claim with a relative error no greater than \(5 \%\). Determine whether the seller provided a sufficiently large sample to permit such an evaluation. Give the details of your analysis.

Short Answer

Expert verified
Yes, the sample size is sufficient to meet the 5% relative error requirement.

Step by step solution

01

Convert Ore Sample Weight

First, convert the weight of the ore sample from pounds to grams:We know that 1 lb is approximately 453.592 grams.Therefore, the weight of the sample in grams is: \[ 5 \text{ lb} \times 453.592 \frac{\text{g}}{\text{lb}} = 2267.96 \text{ g} \]
02

Calculate Weight of Argentite in Sample

Given the sample is 1% argentite, calculate its mass:\[ \text{Weight of Argentite} = 0.01 \times 2267.96 \text{ g} = 22.6796 \text{ g} \]
03

Estimate Silver Content in Argentite

Argentite is Ag2S, with a molar mass of 247.8 g/mol (Ag: 107.87 each, S: 32.06).The silver part with a molar mass of \(107.87 \text{ g/mol} \times 2 = 215.74 \text{ g/mol}\).Thus, the silver content proportion in argentite:\[ \frac{215.74}{247.8} \approx 0.87 \]Thus, the mass of silver in the sample is:\[ 22.6796 \times 0.87 = 19.933252 \text{ g} \approx 19.93 \text{ g} \]
04

Determine Relative Error Requirement

The buyer requires a relative error of no more than 5%, meaning the error in estimating silver content should be less than 5% of the average measurement of silver.For the given sample:\[ \text{Acceptable Error} = 0.05 \times 19.93 \text{ g} = 0.9965 \text{ g} \]
05

Check Sufficient Sample Size

Given that the entire estimated silver is 19.93g with an acceptable error of 0.9965g: The sample is sufficiently large for the required relative error as it allows an evaluation within the permissible margin of error. Thus, a sample of 5 lb providing approximately 19.93 g of silver supports a relative error requirement of 5%.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Silver Content Analysis
Silver content analysis is a method used to determine the amount of silver in a sample. This is crucial in mining and refining processes, where precise knowledge of valuable metal content can impact both evaluation and pricing decisions. In analytical chemistry, silver analysis often involves converting the sample from its raw form into a more measurable form, using the known composition of chemical compounds.

For example, in this exercise, the ore sample consists of argentite, which is a compound known as silver sulfide ( Ag_2S ), making it essential to understand its silver composition. The molar mass of argentite is 247.8 g/mol, derived from two silver atoms and one sulfur atom. By knowing the molar masses, we can calculate the percentage of silver in argentite, which simplifies the determination of actual silver content.
  • The silver to argentite ratio is determined by dividing the molar mass of silver in the compound by the total molar mass of argentite.
  • In this case, silver's molar mass is 215.74 g/mol, and it's divided by 247.8 g/mol, giving a silver proportion of about 87%.
  • This proportion helps in calculating actual silver mass in any argentite sample.
Understanding the silver content accurately is vital for financial and operational decisions in mining ventures.
Relative Error Calculation
Relative error calculation is a means to assess the accuracy of a measurement compared to its true value. It's expressed as a percentage of the true or accepted value, indicating the size of the error in relation to the actual value measured. This measurement is key in ensuring the reliability of experimental data.

In the context of the exercise, the prospective buyer desired a relative error of no more than 5%, meaning that the allowable discrepancy between the measured and true silver content should be minimal. To determine this:
  • Compute the acceptable error margin by multiplying the total estimated silver mass by the desired relative error ( 5% in this case).
  • The buyer's target was that this error should not surpass 0.9965 grams, derived from 5% of the estimated 19.93 grams of silver.
  • A small relative error confirms accurate and trustworthy measurements, important for buyer decisions on resource investments.
Relative error is particularly critical in industries such as metals and mining, where an erroneous assessment can lead to significant financial loss.
Argentite Sample Analysis
Analyzing an argentite sample involves quantifying the presence of silver in the mineral form of silver sulfide ( Ag_2S ). Argentite is often processed in mining operations to extract elemental silver. Hence, understanding its composition is vital.

In the given problem, the argentite forms only 1% of the ore, yet holds the key to estimating available silver. The analytical method includes steps like:
  • Estimating the bulk ore sample's weight (converted into grams).
  • Calculating the argentite weight as a percentage of this total weight.
  • Utilizing the fixed silver proportion in argentite to determine actual silver content.
Such detailed sample analysis provides essential data for evaluating a mining claim's potential profitability. The sample's analysis methodically accounts for every step, ensuring precise results.

Handling and evaluating argentite are standard practices in mineral exploration, where comprehensive calculations assist in decision-making about claim worthiness and investment potential.

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Most popular questions from this chapter

The method of standard additions was used to determine nitrite in a soil sample. A \(1.00-\mathrm{mL}\) portion of the sample was mixed with \(24.00 \mathrm{~mL}\) of a colorimetric reagent, and the nitrite was converted to a colored product that produced a blank-corrected absorbance of \(0.300\). To \(50.00 \mathrm{~mL}\) of the original sample, \(1.00 \mathrm{~mL}\) of a standard solution of \(1.00 \times 10^{-3} \mathrm{M}\) nitrite was added. The same color-forming procedure was followed, and the new \(\mathrm{ab}-\) sorbance was \(0.530\). What was the concentration of nitrite in the original undiluted sample?

A study was made to determine the activation energy \(E_{\mathrm{A}}\) for a chemical reaction. The rate constant \(k\) was determined as a function of temperature \(T\), and the data in the table below obtained. $$ \begin{array}{ll} T, \mathrm{~K} & {k, \mathrm{~s}^{-1}} \\ \hline 599 & 0.00054 \\ 629 & 0.0025 \\ 647 & 0.0052 \\ 666 & 0.014 \\ 683 & 0.025 \\ 700 & 0.064 \\ \hline \end{array} $$ The data should fit a linear model of the form \(\log k=\) \(\log A-E_{\mathrm{A}} /(2.303 R T)\), where \(A\) is the preexponential factor, and \(R\) is the gas constant. (a) Fit the data to a straight line of the form \(\log k=\) \(a-1000 b / T\). *(b) Find the slope, intercept, and standard error of the estimate. *(c) Noting that \(E_{\mathrm{A}}=-6 \times 2.303 R \times 1000\), find the activation energy and its standard deviation (Use \(R=1.987 \mathrm{cal} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\) ). *(d) A theoretical prediction gave \(E_{\mathrm{A}}=41.00 \mathrm{kcal}\) \(\mathrm{mol}^{-1} \mathrm{~K}^{-1}\). Test the null hypothesis that \(E \mathrm{~A}\) is this value at the \(95 \%\) confidence level.

The following table gives the sample means and standard deviations for six measurements each day of the purity of a polymer in a process. The purity is monitored for 24 days. Determine the overall mean and standard deviation of the measurements and construct a control chart with upper and lower control limits. Do any of the means indicate a loss of statistical control? $$ \begin{array}{c|c|c||c|c|c} \text { Day } & \text { Mean } & \text { SD } & \text { Day } & \text { Mean } & \text { SD } \\ \hline 1 & 96.50 & 0.80 & 13 & 96.64 & 1.59 \\ 2 & 97.38 & 0.88 & 14 & 96.87 & 1.52 \\ 3 & 96.85 & 1.43 & 15 & 95.52 & 1.27 \\ 4 & 96.64 & 1.59 & 16 & 96.08 & 1.16 \\ 5 & 96.87 & 1.52 & 17 & 96.48 & 0.79 \\ 6 & 95.52 & 1.27 & 18 & 96.63 & 1.48 \\ 7 & 96.08 & 1.16 & 19 & 95.47 & 1.30 \\ 8 & 96.48 & 0.79 & 20 & 96.43 & 0.75 \\ 9 & 96.63 & 1.48 & 21 & 97.06 & 1.34 \\ 10 & 95.47 & 1.30 & 22 & 98.34 & 1.60 \\ 11 & 97.38 & 0.88 & 23 & 96.42 & 1.22 \\ 12 & 96.85 & 1.43 & 24 & 95.99 & 1.18 \\ \hline \end{array} $$

What factors determine the mass of a gross sample?

The mishandling of a shipping container loaded with 750 cases of wine caused some of the bottles to break. An insurance adjuster proposed to settle the claim at \(20.8 \%\) of the value of the shipment, based on a random 250-bottle sample in which 52 were cracked or broken. Calculate (a) the relative standard deviation of the adjuster's evaluation. (b) the absolute standard deviation for the 750 cases (12 bottles per case). (c) the \(90 \%\) confidence interval for the total number of bottles. (d) the size of a random sampling needed for a relative standard deviation of \(5.0 \%\), assuming a breakage rate of about \(21 \%\).

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