Problem 14 - General Chemistry: Principles and Modern Applications

Question

A 3.104 g sample of an oxide of manganese contains 1.142 grams of oxygen. Write a balanced chemical equation for the reaction that produces the compound from \(\mathrm{Mn}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g})\)

Step-by-Step Solution

Answer

The balanced chemical reaction is \(\mathrm{2Mn(s)} + \mathrm{O_{2}(g)} \rightarrow \mathrm{2MnO(s)}\)

Step 1: Calculation of moles

The first step is to convert the given masses of Mn and O into moles. The molar mass of manganese (Mn) is approximately 54.94 g/mol, and for oxygen (O), it is 16.00 g/mol. To calculate the moles, divide the given mass by the molar mass. Hence, the moles of Mn = 3.104g/54.94 g/mol = 0.056 moles (approx) and the moles of O = 1.142g/16.00 g/mol = 0.071 moles (approx).

Step 2: Determining the empirical formula

To ascertain the empirical formula of manganese oxide, find the simplest whole number ratio of moles of Mn to moles of O. This is done by dividing each by the smallest quantity of moles. So, Mn:O = 0.056/0.056 : 0.071/0.056 = 1:1.27. Rounding to the nearest whole number gives Mn:O = 1:1 (Approximately). Hence, the empirical formula of the oxide is MnO.

Step 3: Writing the chemical equation

Considering the empirical formula obtained in Step 2, the balanced chemical equation for the reaction that produces manganese oxide from \(\mathrm{Mn}(\mathrm{s})\) and \(\mathrm{O}_{2}(\mathrm{g})\) is:\[\mathrm{2Mn(s)} + \mathrm{O_{2}(g)} \rightarrow \mathrm{2MnO(s)}\]

Key Concepts

Manganese Oxide

Manganese oxide is a compound formed from the reaction between manganese (Mn) and oxygen (O extsubscript{2}). In chemistry, when we talk about oxides of elements like manganese, we're typically discussing a combination of that element with oxygen. Manganese can form a variety of oxide compounds, and each may have different properties. Here, we focus on one particular oxide that has the empirical formula MnO.

This oxide is created through a direct reaction between solid manganese and gaseous oxygen. Understanding this compound involves examining the component elements and how they combine at a molecular level.

• Manganese is a transition metal that is part of group 7 on the periodic table.
• Oxygen is a life-sustaining element which is essential in many chemical reactions including oxidation processes.

When manganese reacts with oxygen, they bond in a set ratio, in this case, one atom of manganese for every atom of oxygen, forming manganese(II) oxide, or MnO. This stoichiometry is determined through mole calculations and empirical formula derivation.

Empirical Formula

The empirical formula represents the simplest ratio of atoms present in a compound. It provides insight into the basic makeup of the compound without specifying the actual number of atoms. The task of determining an empirical formula begins with finding the amount of each element present as calculated by moles.

In our example with manganese oxide, we determine the empirical formula by comparing moles of manganese to moles of oxygen:

• Calculate moles from mass using the formula: moles = mass (g) / molar mass (g/mol).
• Determine the ratio of these moles by dividing each by the smallest quantity.
• The result is the simplest whole number ratio of the components in the compound.

For manganese oxide, the empirical formula comes out as MnO, meaning one manganese atom combines with one oxygen atom. This formula ensures a neutral and stable compound. In reality, variations in oxidation states of manganese can create different types of manganese oxides, each with unique properties and applications.

Mole Calculation

Mole calculation is a foundational concept in chemistry involving the conversion of mass into moles – a unit used to express amounts of a chemical substance. The mole is central to understanding chemical reactions and stoichiometry.

To convert between mass and moles, one uses the molar mass of an element, which is the mass of one mole of atoms of that element. Each atomic species has its molar mass, e.g., Manganese has a molar mass of approximately 54.94 g/mol, and oxygen 16.00 g/mol. Calculation of moles allows us to determine how many chemical units, such as atoms or molecules, we have:

• Use the formula: moles = mass / molar mass.
• Division of the mass of each element by its respective molar mass provides the mole quantity.
• This calculation is crucial for establishing ratios in empirical formula determination.

In our manganese oxide example, converting the given sample masses into moles allows us to properly derive the empirical formula and write balanced chemical reactions.