Problem 1 - Chemical Principles in the Laboratory

Question

A student prepared \(7.2 \mathrm{g}\) of \(\mathrm{Co}_{\mathrm{x}}\left(\mathrm{NH}_{3}\right)_{y} \mathrm{Cl}_{z}\). She then analyzed the compound by the procedure in this experiment. A. In the gravimetric determination of chloride, she weighed out 0.2988 g of the compound. The following data were obtained: Mass of crucible plus AgCl 19.0020 \mathrm{g} Mass of crucible 18.4628 \mathrm{g} ______________ g MM AgCl = 143.323 g Mass of AgCl ______________ g MM Cl\(^{-}\) = 35.453 g Mass of \(\mathrm{Cl}^{-}\) in \(\mathrm{AgCl}\) ______________ Moles \(\mathrm{Cl}^{-}\) in sample ______________ Moles \(\mathrm{Cl}^{-}\) in 100 -g sample ______________ B. In the colorimetric determination of cobalt, she used a sample weighing 0.5120 g. The molarity of cobalt ion in the solution from the volumetric flask was 0.085 M. Moles cobalt ion in 25 mL solution = moles cobalt in sample ______________ Moles cobalt per 100-g sample ______________ C. In the volumetric determination of ammonia, the sample weighed \(0.9985 \mathrm{g}\). In the titration, \(0.1000 \mathrm{M} \mathrm{HCl}\) was used. She found that 40.00 mL of the \(\mathrm{NH}_{3}\) solution required \(39.62 \mathrm{mL}\) of the HCl to reach the end point. No. moles HCl used ______________ = no. moles \(\mathrm{NH}_{3}\) in \(40 \mathrm{mL} \mathrm{NH}_{3}\) solution No. moles \(\mathrm{NH}_{3}\) in \(250 \mathrm{mL} \mathrm{NH}_{3}\) ______________ = no. moles \(\mathrm{NH}_{3}\) in sample No. moles \(\mathrm{NH}_{3}\) per \(100-\mathrm{g}\) sample ______________ D. Calculation of the formula of \(\operatorname{Co}_{x}\left(\mathrm{NH}_{3}\right)_{y} \mathrm{Cl}_{z}\): In 100 g of sample, no. moles Co ion \(=\) ______________ no. moles \(\mathrm{NH}_{3}=\) ______________ no. moles \(\mathrm{Cl}^{-}=\) ______________ Dividing by smallest number, no. moles Co ion ______________ no. moles \(\mathrm{NH}_{3}\) ______________ no. moles \(\mathrm{Cl}^{-}\) ______________ Formula of compound ______________ (This may or may not be the formula of the compound you will be making!)

Step-by-Step Solution

Answer

The empirical formula is Co(NH₃)Cl₃.

Step 1: Determine Mass of AgCl

Given the mass of the crucible with AgCl is 19.0020 g and the mass of the crucible is 18.4628 g. To find the mass of AgCl, subtract the mass of the crucible from the total mass: \[ \text{Mass of AgCl} = 19.0020 \, \text{g} - 18.4628 \, \text{g} = 0.5392 \, \text{g} \]

Step 2: Calculate Mass of Cl⁻ in AgCl

The molar mass of AgCl is 143.323 g/mol, and the molar mass of Cl⁻ is 35.453 g/mol. The mass percentage of Cl⁻ in AgCl is calculated as: \[ \text{Mass percentage of Cl}^- = \frac{35.453}{143.323} \times 100\% = 24.74\% \]Then, the mass of Cl⁻ in the AgCl sampled is: \[ \text{Mass of Cl}^- = 0.5392 \, \text{g} \times \frac{35.453 \, \text{g/mol}}{143.323 \, \text{g/mol}} = 0.1334 \, \text{g} \]

Step 3: Find Moles of Cl⁻ in Sample

Use the mass of Cl⁻ to find the number of moles. \[ \text{Moles of Cl}^- = \frac{0.1334 \, \text{g}}{35.453 \, \text{g/mol}} = 0.00376 \, \text{mol} \]

Step 4: Calculate Moles of Cl⁻ per 100 g Sample

Given 0.2988 g of compound analyzed, scale the moles from the sample to a 100 g basis: \[ \text{Moles of Cl}^- \text{ per 100 g sample} = \frac{0.00376 \, \text{mol}}{0.2988 \, \text{g}} \times 100 \, \text{g} = 1.258 \, \text{mol} \]

Step 5: Determine Moles of Cobalt Ion in Solution

From the second procedure, use the molarity of cobalt: \[ \text{Moles of Co ion in 25 mL} = 0.085 \, \text{M} \times 0.025 \, \text{L} = 0.002125 \, \text{mol} \]

Step 6: Calculate Moles of Cobalt per 100 g Sample

The weighed sample is 0.5120 g for Co determination: \[ \text{Moles of Co ion per 100 g sample} = \frac{0.002125 \, \text{mol}}{0.5120 \, \text{g}} \times 100 \, \text{g} = 0.4150 \, \text{mol} \]

Step 7: Find Moles of NH₃ in Sample

In 40 mL of NH₃ solution, HCl used was 39.62 mL at 0.1000 M, therefore: \[ \text{No. moles HCl} = 39.62 \, \text{mL} \times 0.1000 \, \text{M} = 0.003962 \, \text{mol} \]

Step 8: Moles of NH₃ in Sample

Assuming equivalence, moles of NH₃ = moles of HCl in 40 mL. The sample is 0.9985 g, which provides: \[ \text{Moles of NH}_3 \text{ per 100 g sample} = \frac{0.003962 \, \text{mol}}{0.9985 \, \text{g}} \times 100 \, \text{g} = 0.3966 \, \text{mol} \]

Step 9: Calculate the Formula of the Compound

Using the numbers derived: \- Moles of Co ion = 0.4150 \- Moles of NH₃ = 0.3966 \- Moles of Cl⁻ = 1.258 \Divide each by the smallest amount to find the simplest ratio: \[ \text{Ratio Co} : \text{NH}_3 : \text{Cl}^- = \frac{0.4150}{0.3966} : \frac{0.3966}{0.3966} : \frac{1.258}{0.3966} \approx 1 : 1 : 3 \] Thus, the empirical formula is approximately \( \text{Co(NH}_3\text{)}\text{Cl}_3 \).

Key Concepts

Moles Calculation

Understanding moles is crucial in chemistry as they serve as a bridge between the microscopic world of atoms and the macroscopic world we observe. Moles allow chemists to count particles by weighing them. Each mole of a substance contains Avogadro's number of particles, which is approximately \(6.022 \times 10^{23}\) particles. This large number makes it feasible to perform macroscopic experiments.
For instance, calculating moles from a given mass involves using the formula: \[\text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}}\] In our exercise, the calculations of moles for \(\text{Cl}^-\), cobalt, and \(\text{NH}_3\) were conducted using this relationship. Such calculations help determine the amounts of reactants and products involved in chemical reactions.

Empirical Formula Determination

Empirical formula determination is an important method used to find the simplest whole-number ratio of atoms in a compound. This is especially important in chemical analysis because it provides foundational information about a compound's composition.
Once the moles of each component (like \(\text{Cl}^-\), \(\text{Co}\), and \(\text{NH}_3\)) are calculated, these values are used to identify ratios. By dividing each mole value by the smallest number of moles among the components, you can find an empirical formula.
For our Co-containing compound, the results led us to the empirical formula \(\text{Co(NH}_3\text{)}\text{Cl}_3\), indicating the compound contains one cobalt atom for every ammonia molecule and three chloride ions.

Colorimetric Analysis

Colorimetric analysis is a technique used to determine the concentration of colored compounds in a solution. It is based on beer-lambert law, which relates the absorption of light to the properties of the material through which the light is traveling.
In the exercise, this method was used to determine the concentration of cobalt ions. The known absorbance and molarity helped calculate the moles of cobalt ions present. By comparing the sample's absorbance to a standard curve or predefined data, one can determine the concentration of the solution.
This technique is powerful in identifying concentrations of substances like transition metals with distinct colorful ions, making it an invaluable tool in chemistry labs.

Volumetric Analysis

Volumetric analysis is a technique often seen in titration processes, where a solution of known concentration is used to determine the concentration of an unknown solution. In the exercise, this was employed during the titration of ammonia with hydrochloric acid.
By carefully measuring the volume of the titrant (HCl in this case) needed to react completely with the analyte (\(\text{NH}_3\)), the exact concentration of ammonia can be determined, using the relation: \[\text{Moles of HCl} = \text{Volume (L)} \times \text{Molarity (mol/L)}\] Since moles of \(\text{NH}_3\) and HCl are equal at equivalence, this calculation provides a way to find the moles of ammonia in the sample.

Stoichiometry

Stoichiometry fundamentally deals with the quantitative relationships of reactants and products in a chemical reaction. It relies heavily on the concept of the mole and is vital for calculating the amounts needed or produced in a chemical reaction.
In stoichiometry, each substance in a reaction is represented by its formula, which incorporates its molar mass, allowing conversion between mass, moles, and ultimately the count of atoms or molecules.
Our exercise example, involving the determination of a compound's formula using experimental data, is an excellent example of stoichiometry in action. By understanding these relationships, one can predict accurately how much of a substance is needed or how much will be produced in given reactions. This principle is applicable in many fields, including pharmaceuticals, materials science, and environmental science.