Question

A sample of the compound prepared in this experiment, weighing 0.1021 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 $5.16\times {10}^{-3}\mathrm{M}$a. How many moles of Cu were in the original sample, which had been effectively diluted to a volume of $200\mathrm{mL}?$ moles______. 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 formula of the copper chloride compound?

Short answer

a. 0.0001032 moles of Cu were in the original sample. b. 0.00656 g of Cu were in the sample. c. 0.09554 g and 0.00270 moles of Cl were in the sample. d. The formula of the copper chloride compound is ${\mathrm{CuCl}}_{26}$.

Step-by-step solution

Calculate the concentration of Cu2+ ions in the 200 mL sample

We are given the concentration of the $\left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4^{2+}\right]$ complex in the mixture containing 10 mL aliquot and 10 mL of 6 M NH3 as $5.16\times {10}^{-3}\mathrm{M}$. The total volume of this solution is 20 mL. Considering a simple dilution process, we can calculate the concentration of Cu2+ ions in the 200 mL sample using the following equation: $$C_1{V}_1=C_2{V}_2$$ Here, $C_1$ is the concentration in the 200 mL sample, $V_1$ is the volume of the 200 mL sample (200 mL), $C_2$ is the concentration in the 20 mL mixture ($5.16\times {10}^{-3}\mathrm{M}$), and $V_2$ is the volume of the 20 mL mixture (20 mL).

Solving for the concentration of Cu2+ ions in the 200 mL sample

$$C_1=\frac{C_2{V}_2}{V_1}=\frac{(5.16\times {10}^{-3}\mathrm{M})(20\mathrm{mL})}{200\mathrm{mL}}$$ $$C_1=0.000516\mathrm{M}$$ Now that we have the concentration of Cu2+ ions in the 200 mL sample, we can find the moles of Cu in the original sample.

Calculate the moles of Cu in the original sample

$$MolesofCu=C_1\times V_1$$ $$MolesofCu=(0.000516\mathrm{M})\times (0.200\mathrm{L})$$ $$MolesofCu=0.0001032moles$$ Answer a: 0.0001032 moles of Cu were in the original sample.

Calculate the grams of Cu in the sample

To find the grams of Cu in the sample, we multiply the moles of Cu by the molar mass of Cu. $$GramsofCu=MolesofCu\times MolarmassofCu$$ $$GramsofCu=0.0001032moles\times 63.55\frac{g}{mol}$$ $$GramsofCu=0.00656g$$ Answer b: 0.00656 g of Cu were in the sample.

Calculate the grams and moles of Cl in the sample

We know the total mass of the compound sample (0.1021 g) and the grams of Cu (0.00656 g). From this, we can find the grams of Cl in the sample. $$GramsofCl=Totalmassofthesample-GramsofCu$$ $$GramsofCl=0.1021g-0.00656g$$ $$GramsofCl=0.09554g$$ Now, we can calculate the moles of Cl in the sample using the molar mass of Cl. $$MolesofCl=\frac{GramsofCl}{MolarmassofCl}$$ $$MolesofCl=\frac{0.09554g}{35.45\frac{g}{mol}}$$ $$MolesofCl=0.00270moles$$ Answer c: 0.09554 g and 0.00270 moles of Cl were in the sample.

Determine the formula of the copper chloride compound

Now that we have the moles of Cu (0.0001032 moles) and Cl (0.00270 moles), we can find the ratio between Cu and Cl to determine the formula of the compound. $$Ratio=\frac{MolesofCl}{MolesofCu}$$ $$Ratio=\frac{0.00270moles}{0.0001032moles}$$ Now, we round the ratio to the nearest whole number to obtain the empirical formula. $$Ratio\approx 26$$ Therefore, the formula of the copper chloride compound is ${\mathrm{CuCl}}_{26}$. Answer d: The formula of the copper chloride compound is ${\mathrm{CuCl}}_{26}$.

Key concepts

Dilution Process

When working in a chemistry lab, you might need to dilute a solution. This means making a solution weaker by adding more solvent, like water, without changing the amount of solute. A common formula used to calculate dilution is: $$C_1{V}_1=C_2{V}_2$$

• Here, $C_1$ is the initial concentration of the solution, and $V_1$ is its initial volume.
• $C_2$ is the final concentration you want, and $V_2$ is the final volume of the solution.

For example, if we know that the concentration of copper ions in a small, concentrated solution is $5.16\times {10}^{-3}\text{M}$, and we want to know what it would be when diluted to a larger volume, we use the same equation. In this way, we find that our concentration of copper ions decreases as we increase the volume, but the total amount of copper stays the same. By understanding this relationship, chemists can easily adjust the concentration of their solutions to suit their needs.

Molar Mass Calculation

To calculate chemical quantities, we often need to know the molar mass of a substance. The molar mass is the weight of one mole of a chemical element or compound, usually measured in grams per mole. Here's how to calculate it:

• Find the atomic masses of all elements in the compound from the periodic table.
• Multiply each element's atomic mass by the number of atoms of that element in the compound.
• Sum all these values for a total molar mass.

For example, copper has a molar mass of $63.55\text{ g/mol}$. When you know the moles of copper, you can find the mass in grams using the relationship: $$extMass=extMoles\times extMolarMass$$In the exercise, by multiplying the moles of copper found in the experiment by copper's molar mass, we determine the exact grams of copper present. This practice is crucial for converting between measures in chemical reactions and analyses.

Empirical Formula Determination

The empirical formula of a compound tells us the simplest whole-number ratio of elements within it. To find it, there's a simple process:

• Determine the moles of each element present in the compound.
• Divide all the mole values by the smallest number of moles calculated.
• Round the resulting numbers to the nearest whole number to get the ratios.

In the example above, we found the moles of copper and chlorine and compared them to find the ratio. However, unusual outcomes can occur, such as unexpectedly high ratios, which should prompt a check for calculation or interpretation errors. Finding the correct empirical formula offers valuable insights into the composition of unknown substances, supporting further chemical analysis and understanding.