Problem 155 - 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 \(1.261 \mathrm{g}\) 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) Formula is NiF2; (c) Name: Nickel(II) fluoride.

Step 1: Convert Thickness to cm

The thickness of the nickel foil is given as 0.550 mm, which we convert to cm for consistency with the given density units: \(0.550 \text{ mm} = 0.055 \text{ cm}\).

Step 2: Calculate the Volume of Nickel Foil

Since the foil is square-shaped with an area of \(1.25 \text{ cm} \times 1.25 \text{ cm}\), the volume \(V\) is calculated as: \[V = ext{Area} \times ext{Thickness} = 1.25^2 \times 0.055 = 0.0859375 \text{ cm}^3\]

Step 3: Calculate the Mass of Nickel Foil

Using the density \(\rho\) of nickel, the mass \(m\) is found by: \[m = \rho \times V = 8.902 \text{ g/cm}^3 \times 0.0859375 \text{ cm}^3 = 0.765 \text{ g}\]

Step 4: Calculate Moles of Nickel Used

The atomic mass of Ni is approximately 58.69 g/mol. Calculate the moles \(n\) of nickel:\[n = \frac{0.765 \text{ g}}{58.69 \text{ g/mol}} = 0.01303 \text{ mol}\]

Step 5: Determine Moles of Nickel in Fluoride Before

For 0.01303 mol of Ni reacting with \(\text{F}_2\), use the mass of the product \(1.261 \text{ g}\) to decide the formula. Assume NiF or NiF2 and use its mass data to calculate the moles assuming a various stoichiometry of the compound, finding the matching mole ratio for fluoride through comparison.

Step 6: Calculate Formula Mass and Identify Formula

If NiF (104.69 g/mol): - 1 atom each Ni and F, wouldn't match expected mass with the moles calculated in Step 4.If NiF2 (116.69 g/mol): - 1 Ni and 2 F leads to calculation:\[\text{Moles of NiF}_2 = \frac{1.261 \text{ g}}{116.69 \text{ g/mol}} = 0.0108 \text{ mol}\] Results show approximate reaction formula NiF2 after balancing.

Step 7: Name the Compound

Since the name refers to a nickel ion nickel(II) and two fluoride ions, the complete name is: Nickel(II) fluoride.

Key Concepts

Moles of a Metal

In the world of chemistry, moles are essential for quantifying substances at the atomic level. When determining how many moles of nickel were used in this exercise, we first need to find the mass of the nickel foil. By knowing the density of nickel, which is 8.902 g/cm³, we calculated the mass from the volume of the foil. This mass is then used to find the number of moles. The formula to calculate moles is:- \[ n = \frac{\text{mass (g)}}{\text{atomic mass (g/mol)}} \]For nickel, the atomic mass is approximately 58.69 g/mol. Hence, with the mass of the nickel calculated to be 0.765 g, the moles of nickel is found to be 0.01303 mol. This conversion to moles is crucial because it allows for direct comparison with moles of another substance involved in the chemical reaction.

Density Calculation

Density is a measure of how much mass is contained in a given volume. It is vital when converting between mass and volume, especially in problems like this one involving nickel foil. The density formula is given by:- \[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]In our exercise, the density of nickel was provided as 8.902 g/cm³. With this information and the calculated volume of the nickel foil, we could accurately determine the mass of the foil. This step is critical when needing the exact mass to find the moles of a substance, as in our piece of nickel. Understanding how these calculations are interlinked helps make sense of the way physical properties like density influence chemical calculations.

Chemical Formulas

Chemical formulas are a symbolic representation of the elements in a compound and the ratio of these elements. When we consider the chemical reaction between nickel and fluorine to form a nickel fluoride, the challenge is to determine the correct chemical formula of the product. - For instance, potential chemical formulas are NiF and NiF$_2$. - Through trial and comparable analysis using the mass of the product (1.261 g) and possible stoichiometries, we determined that NiF$_2$ is likely the correct formula. The molar mass of NiF$_2$ is around 116.69 g/mol. Working backwards through the reaction's stoichiometry, the expected mass of the resultant compound can be matched, verifying that two fluorine atoms pair with one nickel atom. Properly understanding chemical formulas lays the foundation for predicting reaction products and their properties.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. In this exercise, stoichiometry plays a key role in determining the formula and name of the resulting nickel fluoride. - By examining the moles used and produced, you can determine the product's chemical formula. - The exercise assumed two potential products, NiF and NiF$_2$, illustrating stoichiometry's role in analyzing each potential outcome through calculated molecular masses. For the correct stoichiometric balance, a Ni:F ratio of 1:2 was established, leading us to the formula NiF$_2$. This corresponds to the product being nickel(II) fluoride, where each nickel atom bonds with two fluorine atoms. Stoichiometry is thus the tool for expressing exact chemical relationships and guiding understanding in chemical processes.