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A 2.6540 -g sample of an iron ore, which is \(53.51 \% \mathrm{w} / \mathrm{w} \mathrm{Fe}\), is dissolved in a small portion of concentrated \(\mathrm{HCl}\) and diluted to volume in a \(250-\mathrm{mL}\) volumetric flask. A spectrophotometric determination of the concentration of Fe in this solution yields results of 5840,5770,5650 , and \(5660 \mathrm{ppm} .\) Determine whether there is a significant difference between the experimental mean and the expected value at \(\alpha=0.05\).

Short Answer

Expert verified
No significant difference between experimental and expected Fe ppm at \(\alpha=0.05\).

Step by step solution

01

Calculate the expected ppm of Fe

First, let's determine the amount of Fe (in grams) present in the original sample. The sample is 53.51% Fe by weight, so the mass of Fe is \[\text{mass of Fe} = \frac{53.51}{100} \times 2.6540 \,\text{g} = 1.4201 \,\text{g} \] Now, calculate the expected concentration in ppm. The solution volume is 250 mL or 0.250 L. Therefore, the ppm is calculated as \[\text{expected ppm} = \frac{1.4201 \,\text{g Fe}}{0.250 \,\text{L}} \times 10^6 = 5680.4 \,\text{ppm} \]
02

Calculate the experimental mean ppm of Fe

To find the mean ppm of Fe from the experimental data, add the given ppm values and divide by the number of observations.\[\text{mean ppm} = \frac{5840 + 5770 + 5650 + 5660}{4} = 5730 \]
03

Calculate the standard deviation of the data

First, find the variance by determining the squared differences of each data point from the mean, averaging these, and then taking the square root. \[s^2 = \frac{(5840 - 5730)^2 + (5770 - 5730)^2 + (5650 - 5730)^2 + (5660 - 5730)^2}{4 - 1} \]\[s^2 = \frac{110^2 + 40^2 + (-80)^2 + (-70)^2}{3} = \frac{12100 + 1600 + 6400 + 4900}{3} = \frac{25000}{3} \approx 8333.33 \]Standard deviation, \(s\), is \[s = \sqrt{8333.33} \approx 91.27 \]
04

Perform the t-test

With the standard deviation and mean, calculate the t-value to test if there's a significant difference from the expected ppm. Use the t-test formula \[t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]where \(\bar{x} = 5730\), \(\mu = 5680.4\), \(s = 91.27\), and \(n = 4\).\[t = \frac{5730 - 5680.4}{91.27/\sqrt{4}} = \frac{49.6}{45.635} \approx 1.087 \]
05

Determine significance at \(\alpha = 0.05\)

For \(n - 1 = 3\) degrees of freedom and \(\alpha = 0.05\), the critical t-value from the t-table (two-tailed) is approximately \(3.182\). Since our calculated \(|t| = 1.087\) is less than \(3.182\), we fail to reject the null hypothesis. Thus, there is no significant difference between the experimental mean and the expected value.

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Key Concepts

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

t-test
A t-test is a statistical analysis used to compare the means of two groups to understand if they are significantly different from each other. In our scenario, we're determining if the actual observed mean iron concentration deviates substantially from the expected concentration. The null hypothesis assumes there is no difference between the observed and expected means.

The formula for the t-test is:
  • \( t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \)
Where:
  • \( \bar{x} \) = sample mean
  • \( \mu \) = population mean
  • \( s \) = standard deviation
  • \( n \) = sample size
Once you compute the t-value, it is compared to a critical value from the t-distribution table, based on your chosen significance level (\( \alpha \)) and degrees of freedom. If the calculated t-value is larger than the critical value, you reject the null hypothesis.
Spectrophotometry
Spectrophotometry is a technique used to measure how much light a chemical substance absorbs by measuring the intensity of light at different wavelengths. In iron ore analysis, spectrophotometry is used to determine the concentration of iron in solutions.

The basic principle is that different compounds absorb light at different wavelengths. A spectrophotometer passes light through your iron ore solution and measures how much light is absorbed (related to concentration) and transmitted. By comparing to a calibration curve or standard, we can quantify the iron concentration in parts per million (ppm).

Through spectrophotometry, we obtain quantitative analysis efficiently and accurately, making it a crucial component in chemical analysis and quality assurance in industrial applications.
Standard Deviation
Standard deviation is a measure that quantifies the amount of variation or dispersion in a set of values. A low standard deviation indicates that the values tend to be close to the mean, whereas a high standard deviation indicates a wider range of values. In the context of iron ore analysis, understanding the standard deviation helps in assessing how consistent the measured concentrations of iron are.

Here's how it's calculated:
  • Compute the mean (average) of the sample data.
  • Subtract the mean from each observed value, and square the result.
  • Sum these squared differences.
  • Divide by the number of observations minus one, giving you variance.
  • Take the square root of the variance to get the standard deviation.
This process provides a sense of how much the measurements deviate from the expected value.
Iron Ore Analysis
Iron ore analysis involves determining the iron concentration in ore samples. This is significant for evaluating ore quality and economic viability. Typically, analyses focus on quantifying the iron content using methods such as spectrophotometry.

The process starts with preparing the sample, often involving dissolving a known weight of the ore in an acid solution to ensure the iron is in a measurable form. In our case, the concentration of iron in this diluted solution is compared against expected values, checking consistency and accuracy using statistical methods like t-tests.

Analysing iron ore samples helps companies decide whether to extract the mineral economically. Accurate analysis prevents financial loss by ensuring that only ore with proper concentration levels goes through further processing.

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Most popular questions from this chapter

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