Question

A silicious rock contains the mineral \(\mathrm{ZnS}\). To analyze for \(\mathrm{Zn}\), a sample of the rock is pulverized and treated with \(\mathrm{HCl}\) to dissolve the ZnS (silicious matter is insoluable). Zinc is precipitated from solution by the addition of potassium ferrocyanide \(\mathrm{K}_{4} \mathrm{Fe}(\mathrm{CN})_{6}\). After filtering, the precipitate is dried and weighed. The reactions which occur are $$ \begin{gathered} \mathrm{ZnS}+2 \mathrm{HCl} \rightarrow \mathrm{ZnCl}_{2}+\mathrm{H}_{2} \mathrm{~S} \\ 2 \mathrm{ZnCl}_{2}+\mathrm{K}_{4} \mathrm{Fe}(\mathrm{CN})_{6} \rightarrow \mathrm{Zn}_{2} \mathrm{Fe}(\mathrm{CN})_{6}+4 \mathrm{KCl} \end{gathered} $$ If a 2 gram sample of rock yields \(0.969\) gram of \(\mathrm{Zn}_{2} \mathrm{Fe}(\mathrm{CN})_{6}\), what is the percentage of \(\mathrm{Zn}\) in the sample? Atomic weight \(\mathrm{Zn}=65.4\), molecular weight \(\mathrm{Zn}_{2} \mathrm{Fe}(\mathrm{CN})_{6}=342.6\).

Short answer

The percentage of Zn in the rock sample is 18.5%.

Step-by-step solution

Determine the stoichiometric ratios of the chemical reactions

Using the balanced chemical equations provided, we can observe the stoichiometric ratios for the reactions involved: 1 ZnS reacts with 2 HCl to produce 1 ZnCl2 and 1 H2S. 2 ZnCl2 reacts with 1 K4Fe(CN)6 to produce 1 Zn2Fe(CN)6 and 4 KCl.

Calculate the moles of Zn2Fe(CN)6 in the precipitate

We are given the weight of the precipitate (Zn2Fe(CN)6) as 0.969 grams. To find the moles, we can use the molecular weight of Zn2Fe(CN)6, which is given as 342.6 g/mol. Moles of Zn2Fe(CN)6 = (0.969 g) / (342.6 g/mol) = 0.00283 mol

Calculate the moles of Zn in the precipitate

By examining the stoichiometry of the balanced equation involving Zn2Fe(CN)6, we can see that 1 mol of Zn2Fe(CN)6 contains 2 mols of Zn. So, to find the moles of Zn in the precipitate, we simply multiply the moles of Zn2Fe(CN)6 by 2. Moles of Zn = (0.00283 mol Zn2Fe(CN)6) × (2 mol Zn / 1 mol Zn2Fe(CN)6) = 0.00566 mol Zn

Calculate the weight of Zn in the precipitate

To find the weight of Zn in the precipitate, we can multiply the moles of Zn by its atomic weight, which is given as 65.4 g/mol. Weight of Zn = (0.00566 mol Zn) × (65.4 g/mol Zn) = 0.3699 g Zn

Calculate the percentage of Zn in the rock sample

Finally, to find the percentage of Zn in the original 2 gram rock sample, we can divide the weight of Zn in the precipitate by the weight of the rock sample and multiply by 100. Percentage of Zn = (0.3699 g Zn / 2 g rock sample) × 100 = 18.5% Thus, the percentage of Zn in the rock sample is 18.5%.

Key concepts

Understanding ZnS Analysis in a Silicious Rock

Zinc sulfide (ZnS) analysis in geological samples involves a series of steps to isolate and quantify the zinc content present in a mineral form within rocks. The process starts with pulverizing the rock to release the ZnS particles. When reacting these particles with hydrochloric acid (HCl), ZnS is transformed into soluble zinc chloride (ZnCl₂) while producing hydrogen sulfide (H₂S) gas, as shown in the first chemical reaction.

This solubilization is essential because zinc must be in solution to react with potassium ferrocyanide (K₄Fe(CN)₆). The next stage is precipitating zinc as zinc ferrocyanide (Zn₂Fe(CN)₆), which allows for its quantification upon drying and weighing. The accurate measurement of its weight reflects the zinc content originally present, thereby facilitating a precise analysis of its percentage in the rock.

Deciphering Stoichiometric Ratios in Chemical Reactions

Stoichiometric ratios play a critical role in quantitative chemistry. They are the proportional relationships between the reactants and products in a balanced chemical equation. These ratios are obtained by inspecting the coefficients of each substance in the equation, which tell you how many moles of each substance are involved in the reaction.

For example, the equation for the reaction between ZnS and HCl indicates that for every mole of ZnS, two moles of HCl are required to yield one mole of ZnCl₂ and one mole of H₂S. Similarly, the reaction with K₄Fe(CN)₆ shows that two moles of ZnCl₂ will react with one mole of K₄Fe(CN)₆ to produce one mole of Zn₂Fe(CN)₆. Understanding these ratios is essential for converting between moles of different substances in the reaction and ultimately determining the quantity of Zn in the sample.

Calculating Moles for Quantitative Analysis

Mole calculation serves as the bridge between the microscopic world of atoms and molecules and the macroscopic world that can be measured in a laboratory. The mole concept allows chemists to count particles by weighing them. In the given exercise, mole calculation is used to convert the mass of the precipitated Zn₂Fe(CN)₆ to the number of moles, using its molecular weight as the conversion factor.

The equation for this conversion is simply moles = mass (in grams) / molar mass (in g/mol). Once you have the moles of Zn₂Fe(CN)₆, you can use stoichiometry to determine the moles of elemental zinc, and from there, calculate the mass of zinc by using the atomic weight. Divide this by the original rock sample mass and multiply by 100 to find the percentage of Zn, which is the final objective of the analysis. This demonstration of mole calculation is a fundamental technique in chemistry to quantitative measurements and is crucial for any type of stoichiometric analysis.