Question

Suppose you have a cube of pure manganese metal measuring 5.25 \(\mathrm{cm}\) on each side. You find that the mass of the cube is 1076.6 \(\mathrm{g}\) . What is the molar concentration of manganese in the cube?

Short answer

The molar concentration of manganese in the cube is approximately \(0.1354\,\mathrm{mol/cm^3}\).

Step-by-step solution

Calculate the volume of the cube

Given, the cube has a side length of 5.25 cm. To find the volume of the cube, use the formula for the volume of a cube: \[V = a^3\] where \(V\) is the volume, and \(a\) is the side length. Plug in the given side length and solve for the volume: \[V = (5.25\,\mathrm{cm})^3 = 144.703125\,\mathrm{cm^3}\]

Determine the number of moles of manganese

We are given the mass of the cube, which is 1076.6 g. To find the number of moles of manganese in the cube, divide the mass of the cube by the molar mass of manganese. Manganese has a molar mass of 54.94 g/mol. Therefore, the number of moles of manganese, \(n\), can be found using: \[n = \frac{mass}{molar\,mass}\] \[n = \frac{1076.6\,\mathrm{g}}{54.94\,\mathrm{g/mol}} = 19.594793755\,\mathrm{mol}\]

Calculate the molar concentration of manganese in the cube

Now that we have the volume of the cube and the number of moles of manganese, we can find the molar concentration, \(C\), by dividing the number of moles by the volume. To find the concentration in \(\mathrm{mol/cm^3}\), make sure the volume is in \(\mathrm{cm^3}\): \[C = \frac{n}{V}\] \[C = \frac{19.594793755\,\mathrm{mol}}{144.703125\,\mathrm{cm^3}} = 0.135362627\,\mathrm{mol/cm^3}\] The molar concentration of manganese in the cube is approximately \(0.1354\,\mathrm{mol/cm^3}\).

Key concepts

Manganese

Manganese is a critical transition metal with the chemical symbol Mn. It is widely used in industrial applications due to its strength and durability. Manganese is essential in the manufacture of steel to improve hardness and wear resistance. It appears as a hard, brittle gray metal and is often found in compounds such as manganese dioxide in nature.
Manganese also has significant biological importance. It acts as a cofactor for many enzymes in the human body, playing a vital role in metabolism, bone formation, and the functioning of antioxidative enzymes.
In chemical experiments like the one in the exercise, manganese is considered in its pure metal form, making it simpler to calculate metrics such as molar concentration.

Chemical Stoichiometry

Chemical stoichiometry is the branch of chemistry that deals with the quantitative relationships of reactants and products in a chemical reaction. It is foundational in understanding how elements and compounds interact to form new substances. Stoichiometry uses balanced chemical equations to help predict the amounts of reactants needed or products formed in a reaction.
In relation to the exercise, stoichiometry helps us translate between mass, moles, and volume of manganese. It forms the basis for all calculations related to the conversion from grams of a substance to moles, leveraging the concept of molar mass as a conversion factor.

Molar Mass

Molar mass is defined as the mass of one mole of a substance, expressed in grams per mole ( ext{g/mol}). It is a vital constant in chemistry used to convert between mass and moles of a substance. For example, in this exercise, the molar mass of manganese is given as 54.94 ext{g/mol}.
Understanding molar mass is crucial in stoichiometry as it helps determine the quantity of atoms or molecules in a given mass of a substance. This allows chemists to precisely calculate the number of moles from a known mass, which can then be used to determine the molar concentration in a mixture. Ensuring accurate usage of molar mass is essential for reliable laboratory calculations and to facilitate scaling reactions from a laboratory setting to industrial processes.

Volume Calculation

Volume calculation is important in solving problems related to the space an object occupies. For regular shapes like cubes, the calculation becomes straightforward. The volume of a cube is determined using the formula \[V = a^3\], where \(V\) is the volume and \(a\) is the length of one side.
In this exercise, with each side of the manganese cube measuring 5.25 cm, its volume is computed as \(5.25^3 = 144.703125 \, \mathrm{cm^3}\). This step is critical for further calculations – like finding molar concentration – because it provides the necessary volume to correlate with moles of the substance. A precise understanding and calculation of volume ensure accurate determination of properties like density and concentration in scientific experiments.