Problem 85 - General Chemistry

Question

Copper has compounds with copper(I) ion or copper(II) ion. A compound of copper and chlorine was treated with a solution of silver nitrate, \(\mathrm{AgNO}_{3}\), to convert the chloride ion in the compound to a precipitate of AgCl. A \(59.40-\mathrm{mg}\) sample of the copper compound gave \(86.00 \mathrm{mg} \mathrm{AgCl}\). a. Calculate the percentage of chlorine in the copper compound. b. Decide whether the formula of the compound is \(\mathrm{CuCl}\) or \(\mathrm{CuCl}_{2}\)

Step-by-Step Solution

Answer

The compound is likely CuCl, with chlorine at 35.79%.

Step 1: Determine the mass of chlorine in AgCl

The first step is to calculate the mass of chlorine in the silver chloride (\(\mathrm{AgCl}\)) precipitate. The formula for silver chloride is AgCl, which consists of silver (Ag) and chlorine (Cl). Using the molar masses, calculate the fraction of chlorine in AgCl. The molar mass of Ag is approximately 107.87 g/mol and that of Cl is about 35.45 g/mol. Thus, the total molar mass of AgCl is 143.32 g/mol. The chlorine percentage in AgCl is given by:\[\text{Fraction of Cl in AgCl} = \frac{35.45}{143.32} \on\approx 0.2473\]

Step 2: Calculate the mass of chlorine in the sample

Using the fraction obtained in Step 1, calculate the mass of chlorine present in the AgCl precipitate:\[\text{Mass of Cl} = 86.00\, \text{mg} \times 0.2473 = 21.26\, \text{mg}\]This is the amount of chlorine in the sample derived from the conversion of chloride ions to AgCl.

Step 3: Calculate percentage of chlorine in copper compound

To find the percentage of chlorine in the original copper compound sample, divide the mass of chlorine by the mass of the original sample and multiply by 100:\[\text{Percentage of Cl} = \left(\frac{21.26\, \text{mg}}{59.40\, \text{mg}}\right) \times 100 \approx 35.79\%\]

Step 4: Determine possible molecular formulas

Now we need to compare this chlorine percentage against the known chlorine percentages in CuCl and CuCl2. Calculate the chlorine percentage for both:For \(\mathrm{CuCl}\) (molar mass = 63.55 + 35.45 = 98 g/mol):\[\%\,\text{Cl in CuCl} = \left(\frac{35.45}{98.00}\right) \times 100 \approx 36.18\%\]For \(\mathrm{CuCl}_2\) (molar mass = 63.55 + 2\times 35.45 = 134.45 g/mol):\[\%\,\text{Cl in CuCl}_2 = \left(\frac{70.90}{134.45}\right) \times 100 \approx 52.74\%\]

Step 5: Compare empirical data with theoretical values

The empirical chlorine percentage in the compound (35.79%) is much closer to the theoretical chlorine percentage in \(\mathrm{CuCl}\) (36.18%) than that in \(\mathrm{CuCl}_2\) (52.74%). Therefore, the formula of the compound is likelier to be \(\mathrm{CuCl}\).

Key Concepts

Copper(I) Ion

The Copper(I) ion, frequently denoted as Cu⁺, plays a crucial role in the formation of many copper compounds. In these compounds, copper forms a single positive charge, which significantly influences the compound's properties and reactions.

It's common to find Copper(I) in coordination with chloride ions to form simple compounds like Copper(I) chloride, \(\mathrm{CuCl}\). In \(\mathrm{CuCl}\), the Copper(I) ion is balanced by one chloride ion.

Copper(I) ions are less stable in solution than Copper(II) ions, often leading to oxidation into a higher oxidation state.

• Cu⁺ ions can potentially provide crucial clues about the compound's formula by looking at the related weights and amounts.
• The Copper(I) oxidation state is typically more straightforward, often forming binary compounds.

Knowledge about Copper(I) is essential for identifying the correct chemical formula when analyzing copper and chlorine compounds.

Copper(II) Ion

Copper(II) ion, represented as Cu²⁺, is another vital form of copper found in many compounds, alongside Copper(I) ion. This ion carries a +2 charge, which means it commonly pairs with more than one negatively charged ion, like in \(\mathrm{CuCl}_2\).

Due to this charge, Copper(II) tends to form complexes and associates with more complex structures compared to the simpler Copper(I) compounds. For example, when Copper(II) ions bond with two chloride ions, the compound \(\mathrm{CuCl}_2\) is formed.

Because of its +2 charge, Copper(II) ion is more stable and appears frequently in aqueous solutions:

• Its high stability makes it prevalent in nature and many industrial applications.
• Understanding Copper(II) ion is essential for explaining differences in compound structures and proportions of elements.

Recognizing the oxidation state is key to determining chemical formulas in laboratory settings or chemical reactions.

Chemical Formulas

Chemical formulas represent the types and numbers of atoms in a compound, providing insight into its composition. Understanding and deriving these formulas is vital in chemistry.

In our example, copper forms compounds either with the formula \(\mathrm{CuCl}\) or \(\mathrm{CuCl}_2\). These formulas indicate:

• The number of copper and chlorine atoms present.
• The stoichiometry or the proportion between the elements in the compound.

Chemical formulas help predict how compounds will behave during chemical reactions and how they interact with other substances. For instance, by knowing a compound's formula, chemists can predict:

• Its reaction with silver nitrate to form a precipitate of \(\mathrm{AgCl}\).
• Its molar mass, which is essential for various calculations.

Calculating the percentage of elements in these formulas helps decide whether \(\mathrm{CuCl}\) or \(\mathrm{CuCl}_2\) is present in a compound.

Molar Mass Calculation

Calculating molar mass is foundational in determining the proportions of elements within compounds. When you calculate molar mass, you essentially weigh the elements involved in one mole of a substance.

For example, the molar mass of Copper(I) chloride (\(\mathrm{CuCl}\)) is determined by adding the mass of one mole of copper (about 63.55 g/mol) to the mass of one mole of chlorine (about 35.45 g/mol). This results in a molar mass of 98 g/mol.

For Copper(II) chloride (\(\mathrm{CuCl}_2\)), you add two moles of chlorine since there are two chlorine atoms, which results in a molar mass of 134.45 g/mol:

• This calculation helps in determining the percentage composition of each element in the compound.
• It is crucial for converting between grams and moles, which is often necessary in practical laboratory settings.

Thus, knowing how to accurately calculate molar mass is essential for various chemical determinations, including empirical formula derivation.

Percentage Composition

Percentage composition is used to understand what fraction of a compound's mass comes from a particular element. It helps chemists make comparisons and draw conclusions about the compound.

In our example, we calculate the percentage composition of chlorine in both \(\mathrm{CuCl}\) and \(\mathrm{CuCl}_2\) to decide which compound fits our empirical data. This process involves finding the chlorine's contribution to the total weight of the compound:

• Evaluating the percentage of chlorine helps in comparing it with the empirical data collected.
• The closer the empirical data is to the theoretical data, the more likely it confirms the compound's formula.

Given \(\mathrm{CuCl}\) and \(\mathrm{CuCl}_2\) have distinct percentage compositions, this information becomes critical in deducing which compound we have in our experiment. Thus, percentage composition acts as a decisive tool in confirming the compound's identity.