Problem 5 - Chemical Principles in the Laboratory

Question

A sample of the compound prepared in this experiment, weighing 0.0998 g, is dissolved in HNO, and diluted to a volume of \(100 \mathrm{mL}\). A \(10-\mathrm{mL}\) aliquot of that solution is mixed with \(10 \mathrm{mL} 6 \mathrm{M} \mathrm{NH}_{3}\). The \(\left[\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\right]\) in the resulting solution is found to be \(5.12 \times 10^{-3} \mathrm{M}.\) a. How many moles of Cu were in the original sample, which had been effectively diluted to a volume of \(200 \mathrm{mL} ?\) ________________ g b. How many grams of Cu were in the sample? ____________ g c. How many grams of Cl were in the sample? How many moles? ____________ g ___________ moles d. What is the empirical formula of the copper chloride compound?

Step-by-Step Solution

Answer

a. 1.024e-3 moles Cu; b. 0.06506 g Cu; c. 0.03474 g Cl, 0.00098 moles; d. CuCl

Step 1: Determine Moles of Cu

The concentration of the copper ammonium complex in the resulting solution is given as \(5.12 \times 10^{-3} \mathrm{M}\). Since this is for a \(10 \text{ mL}\) (or \(0.01 \text{ L}\)) aliquot, the total moles of \([\mathrm{Cu(NH}_3\mathrm{)}_4]^{2+}\) in this aliquot can be calculated as:\[\text{moles of Cu} = 5.12 \times 10^{-3} \times 0.01 = 5.12 \times 10^{-5} \text{ moles}\]

Step 2: Scale Moles of Cu to Original Solution

Since the \(10 \text{ mL}\) aliquot corresponds to a \(100 \text{ mL}\) initial dilution, we scale the moles of copper to find the total in the \(100 \text{ mL}\):\[\text{total moles of Cu in } 100 \text{ mL} = 5.12 \times 10^{-5} \times 10 = 5.12 \times 10^{-4} \text{ moles}\]

Step 3: Determine Moles of Cu in Original 200 mL Solution

The solution was further diluted when 10 mL was mixed with 10 mL of 6 M \(\mathrm{NH}_{3}\) (making a total of \(20 \text{ mL}\)). So, we relate this to the original \(200 \text{ mL}\):\[\text{moles of Cu in } 200 \text{ mL} = 5.12 \times 10^{-4} \times 2 = 1.024 \times 10^{-3} \text{ moles}\]

Step 4: Convert Moles of Cu to Grams

The molar mass of copper (Cu) is approximately \(63.55 \text{ g/mol}\). The mass of copper in the sample can be found using:\[\text{mass of Cu} = 1.024 \times 10^{-3} \text{ moles} \times 63.55 \text{ g/mol} = 0.06506 \text{ g}\]

Step 5: Calculate Total Mass of the Sample for Cl

The total mass of the initial sample is \(0.0998 \text{ g}\). To find the mass of chlorine, subtract the mass of copper:\[\text{mass of Cl} = 0.0998 \text{ g} - 0.06506 \text{ g} = 0.03474 \text{ g}\]

Step 6: Convert Mass of Cl to Moles

The molar mass of chlorine (Cl) is approximately \(35.45 \text{ g/mol}\). Calculate the moles of chlorine in the sample:\[\text{moles of Cl} = \frac{0.03474 \text{ g}}{35.45 \text{ g/mol}} = 0.00098 \text{ moles}\]

Step 7: Determine Empirical Formula

Determine the molar ratio of Cu to Cl. The moles of Cu are \(1.024 \times 10^{-3}\) and the moles of Cl are \(0.00098\). The ratio is found by dividing both values by \(0.00098\):\[\text{Cu:Cl} \approx \frac{1.024 \times 10^{-3}}{0.00098} : \frac{0.00098}{0.00098} \approx 1.05 : 1\]The empirical formula closely matches CuCl.

Key Concepts

Molar Mass Calculation

Understanding molar mass calculation is crucial when determining the amount of a substance within a sample. The molar mass of an element is the mass of one mole of its atoms. To calculate the molar mass, you use the atomic mass from the periodic table, measured in grams per mole (g/mol). For instance, copper (Cu) has a molar mass of approximately 63.55 g/mol, meaning one mole of copper atoms weighs 63.55 grams.
To calculate the mass of a particular element in a sample, you multiply the number of moles by the molar mass. For example, if you know you have 1.024 x 10^-3 moles of copper, you multiply by its molar mass, resulting in approximately 0.06506 grams of copper. This step in calculations helps determine how much of a particular element exists in a given sample, aiding in empirical formula determination.

Stoichiometry

Stoichiometry is the part of chemistry that deals with the quantities and ratios of reactants and products in chemical reactions. It is essential for understanding how substances interact in chemical formulas and reactions. In determining the empirical formula from a given data set, stoichiometry helps us figure out the relative number of moles of each element in the compound.
Consider determining the empirical formula for a copper chloride compound. By calculating the moles of copper and chlorine, the stoichiometric ratios can be derived. In this example, dividing the number of moles of each element by the smallest number of moles gives us the simplest whole number ratio of Cu:Cl. For instance, with 0.001024 moles of Cu and 0.00098 moles of Cl, dividing both by 0.00098 yields a ratio close to 1.05:1, suggesting an empirical formula of CuCl. Stoichiometry is, therefore, a foundational concept for understanding chemical composition and reactions.

Chemical Solution Concentration

The concentration of a chemical solution tells us how much solute, like a dissolved substance, is present in a given volume of solvent. It is often expressed in molarity (M), which means moles of solute per liter of solution. Understanding concentration is vital in calculations related to chemical reactions and solution preparation.
In experimental setups, such as when measuring \([\text{Cu(NH}_3\text{)}_4]^{2+}\) concentration, knowing the initial solution concentration is critical. In our exercise, a 10-mL solution with a concentration of 5.12 x 10^-3 M indicated the amount of Cu ion present. Multiplying this concentration by the volume (in liters) gives the total moles of Cu ions. Scaling these results to larger volumes helps determine the concentration after dilution, ensuring accurate empirical results. Therefore, understanding chemical solution concentrations allows precision in both laboratory and practical applications.