Problem 111 - Chemistry and Chemical Reactivity

Question

A piece of nickel foil, \(0.550 \mathrm{mm}\) thick and \(1.25 \mathrm{cm}\) square, is allowed to react with fluorine, \(\mathrm{F}_{2},\) to give a nickel fluoride. (a) How many moles of nickel foil were used? (The density of nickel is \(8.902 \mathrm{g} / \mathrm{cm}^{3} .\) ) (b) If you isolate 1.261 g of the nickel fluoride, what is its formula? (c) What is its complete name?

Step-by-Step Solution

Answer

(a) 0.01304 moles of Ni. (b) Formula is NiF2. (c) Nickel(II) fluoride.

Step 1: Calculate Volume of Nickel Foil

The nickel foil is a 1.25 cm x 1.25 cm square with a thickness of 0.550 mm. Convert the thickness from millimeters to centimeters for uniformity: 0.550 mm = 0.055 cm. Calculate the volume of the foil:\[ \text{Volume} = \text{length} \times \text{width} \times \text{thickness} = 1.25 \times 1.25 \times 0.055 \ \text{cm}^3 = 0.0859375 \ \text{cm}^3\]

Step 2: Determine Mass of Nickel Foil

Using the volume calculated, find the mass of the nickel foil. Use the density formula \(\rho = \frac{m}{V}\), where \(m\) is mass and \(V\) is volume.\[ m = \rho \times V = 8.902 \ \text{g/cm}^3 \times 0.0859375 \ \text{cm}^3 = 0.7653796875 \ \text{g} \approx 0.765 \ \text{g} \]

Step 3: Convert Mass to Moles of Nickel

Calculate the number of moles of nickel using its molar mass (58.69 g/mol):\[ \text{Moles of Ni} = \frac{0.765 \ \text{g}}{58.69 \ \text{g/mol}} \approx 0.01304 \ \text{moles}\]

Step 4: Determine Formula of Nickel Fluoride

Assume the formula of nickel fluoride is \(\text{NiF}_x\). Calculate moles of nickel in the fluoride product using its mass:\[ \text{Total mass of product} = 1.261 \ \text{g}\]Calculate mass of fluorine in the compound:\[ \text{Mass of Ni in product} = 0.765 \ \text{g} \rightarrow \text{Mass of F} = 1.261 \ \text{g} - 0.765 \ \text{g} = 0.496 \ \text{g} \]Convert fluorine mass to moles (F atomic mass = 19.00 g/mol):\[ \text{Moles of F} = \frac{0.496 \ \text{g}}{19.00 \ \text{g/mol}} \approx 0.0261 \ \text{moles}\]From this, the ratio is approximately 2:1 (F:Ni), suggesting the formula is \(\text{NiF}_2\).

Step 5: Name the Nickel Fluoride

The nickel fluoride is \(\text{NiF}_2\). Since nickel fluoride is ionic and nickel can have multiple oxidation states, check for charge balancing. \(\text{F}^- ext{ ions} = 2 \text{F}^- ext{ per } \text{Ni}^{2+}\), this indicates the compound is nickel(II) fluoride.

Key Concepts

Chemical Reactions

Chemical reactions are processes where reactants transform into products, often involving rearrangement of atoms and breaking/forming bonds. In this exercise, a chemical reaction occurs between nickel foil and fluorine gas, leading to the formation of nickel fluoride. This reaction is an example of a combination reaction, where multiple reactants combine to form a single product.

Here's what happens in this specific reaction:

• The nickel, initially metallic, reacts with fluorine gas.
• This process involves the transfer of electrons from nickel to fluorine, creating ionic bonds and forming nickel fluoride compound.
• The reaction can be represented by the equation: \( \text{Ni (s)} + \text{F}_2 (g) \rightarrow \text{NiF}_2 (s) \).

The result is a solid compound, nickel fluoride, indicating a successful chemical transformation from elemental to compound form.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in a chemical reaction. It is central to predicting the outcomes of chemical reactions and is crucial in scaling reactions for practical use.

In this exercise, stoichiometry helps us:

• Understand how much nickel reacts with fluorine.
• Predict the mass and moles of products formed.

Using the known density of nickel and its dimensions, we determine the mass of nickel. Then, by converting mass to moles using the molar mass of nickel, we calculate the moles of nickel reacting. The stoichiometric ratio derived—between nickels and fluorides—is crucial for identifying the chemical formula of the new compound as seen in our conclusion of NiF₂ formation.

Molecular Formula Determination

Determining the molecular formula of a compound involves finding out the ratio of different elements present in it. This is achieved by mass measurements and molar conversions.

To find the molecular formula for nickel fluoride:

• We first isolate the mass of nickel present in our compound through prior calculations.
• The next step involves subtracting the mass of nickel from the total mass of nickel fluoride, which reveals the mass of fluorine present.
• Converting the mass of fluorine to moles allows us to deduce the molar ratio between nickel and fluorine.
  For example, the mass of fluorine divided by its atomic mass (\(19.00 \ \text{g/mol}\)) gives us the moles of fluorine.

This molar ratio helps us derive the empirical formula, in this case, NiF₂. Hence, the formula reflects the simplest whole number molar ratio aligning with our experimental results, leading to the conclusion that nickel(II) fluoride is the compound formed, as confirmed by the stoichiometric and reactional analysis.