Problem 151 - 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, \(F_{2},\) 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 \(1.261 \mathrm{g}\) of the nickel fluoride, what is its formula? (c) What is its complete name?

Step-by-Step Solution

Answer

(a) 0.01303 mol Ni (b) NiF₂ (c) Nickel(II) fluoride.

Step 1: Calculate the Volume of Nickel Foil

First, convert the thickness of the nickel foil from mm to cm: \[ 0.550 \text{ mm} = 0.055 \text{ cm} \]The foil is a square, so the surface area is:\[ 1.25 \text{ cm} \times 1.25 \text{ cm} = 1.5625 \text{ cm}^2 \]The volume of the foil is therefore:\[ \text{Volume} = \text{Area} \times \text{Thickness} = 1.5625 \text{ cm}^2 \times 0.055 \text{ cm} = 0.0859375 \text{ cm}^3 \]

Step 2: Calculate the Mass of Nickel Foil

Using the density of nickel, calculate the mass of the foil:\[ \text{Mass} = \text{Density} \times \text{Volume} = 8.902 \text{ g/cm}^3 \times 0.0859375 \text{ cm}^3 = 0.765 \text{ g} \]

Step 3: Convert Mass of Nickel Foil to Moles

Calculate the moles of nickel using its molar mass:\[ \text{Moles of Ni} = \frac{\text{mass}}{\text{molar mass}} = \frac{0.765 \text{ g}}{58.69 \text{ g/mol}} = 0.01303 \text{ mol} \]

Step 4: Determine the Formula of Nickel Fluoride

Let the nickel fluoride formula be \(\text{NiF}_x\). The moles of isolated \(\text{NiF}_x\) can be calculated as \(\frac{1.261 \text{ g}}{\text{the molar mass of NiF}_x}\). Assuming you have all nickel reacted:\[ 0.01303 \text{ mol NiF}_x = \frac{1.261 \text{ g}}{58.69 + 18.998x \text{ g/mol NiF}_x} \]Solve for \(x\) by trial or algebraic simplification. It shows \(x = 2\), matching known nickel fluoride compounds.

Step 5: Identify the Complete Name of Nickel Fluoride

With the formula \(\text{NiF}_2\), based on known valences, Nickel(II) fluoride is the complete name because nickel is in a +2 oxidation state in this compound.

Key Concepts

Chemical Reactions

Chemical reactions involve the transformation of substances through breaking and forming bonds between atoms. In this exercise, nickel foil reacts with fluorine gas to produce nickel fluoride, a classic example of a chemical reaction.
Naturally, reactions involve changes in chemical bonds and the rearrangement of atoms. The key players in this reaction are the elemental nickel and fluorine, represented as Ni and F₂, respectively.
Fluorine is highly reactive, especially with metals, making it an ideal reactant to form compounds like nickel fluoride.
In chemical reactions, the reactants, in this case, nickel and fluorine, undergo a transformation to form the product, nickel fluoride. Keeping track of the moles of reactants and products is crucial to understanding the stoichiometry of the reaction, which tells us the proportions in which the substances react and form products.
This stoichiometric relationship is helpful when determining the amounts of each substance in a complete reaction. In our example problem, we calculated the moles of nickel used, a fundamental step for determining the final product's formula and name.

Molar Mass

Molar mass is a measure of the mass of a given substance per mole of that substance. It plays a crucial role in chemical calculations, especially when converting between grams and moles. Knowing the molar mass allows chemists to proportionally use amounts of chemicals, analyze reactions, and predict yields.
The molar mass of an element is essentially its atomic weight expressed in grams per mole. For nickel (Ni), this number is approximately 58.69 g/mol. So, if we have the mass of nickel, we can easily convert it into moles by dividing the mass by the molar mass.
In the exercise, this conversion was necessary to calculate the moles of nickel used for the reaction. This conversion is crucial because moles express the amount of substance in classical chemistry, relating directly to the number of atoms or molecules involved in a reaction.

Nickel Compounds

Nickel compounds, including nickel fluoride, have significant industrial and chemical roles. Nickel itself, a transition metal, has various oxidation states and forms several compounds depending on the elements it combines with.
In this exercise, nickel reacts with fluorine to form nickel fluoride, specifically ext{NiF}_2. This compound is known as nickel(II) fluoride, where the roman numeral indicates nickel's +2 oxidation state in the compound.
Nickel fluoride has various applications, including in metal surface treatments and as catalysts. In chemical terms, understanding the formula and the oxidation state is vital for predicting the behavior and potential uses of the compound.
For students, recognizing the link between reaction conditions and compound formation is crucial in grasping how compounds like nickel fluoride come into existence. The structure of nickel(II) fluoride correlates to its production and application, providing insight into the broader field of nickel chemistry.

Density Calculations

Density is the mass of a substance per unit volume, expressed typically in g/cm³ or kg/m³. It’s a vital property in materials science and chemistry for understanding how substances are compacted in space.
In the exercise, we calculated the mass of nickel from the known density and volume of the nickel foil, using the formula: \[ \text{Mass} = \text{Density} \times \text{Volume} \]Understanding how to use density in calculations enables the determination of unknown masses when given a material's dimensions.
This is especially useful when working with solid materials or when the precise mass cannot be directly measured. Knowing both the density and volume allows us to determine the mass, which is crucial for further calculations, like converting to moles.
Density calculations are not only fundamental in laboratory work but also in real-world applications where material properties are of utmost importance. Recognizing how volume, mass, and density interrelate is key to grasping concepts in chemistry and physics, enabling precise experimental and industrial processes.