Problem 96 - Chemistry: An Atoms-Focused Approach

Question

Some catalytic converters in automobiles contain two manganese oxides: \(\mathrm{Mn}_{2} \mathrm{O}_{3}\) and \(\mathrm{MnO}_{2}.\) a. What are the names of these compounds? b. What is the manganese content of each (expressed as a percent by mass)? c. Explain how \(\mathrm{Mn}_{2} \mathrm{O}_{3}\) and \(\mathrm{MnO}_{2}\) are consistent with the law of multiple proportions.

Step-by-Step Solution

Answer

Answer: Mn2O3 is named Manganese (III) oxide, and MnO2 is named Manganese (IV) oxide. Mn2O3 has a manganese content of 69.6% by mass, while MnO2 has a manganese content of 63.2% by mass. Mn2O3 and MnO2 are consistent with the law of multiple proportions since the ratio between the amounts of oxygen in each compound (3/4) is a simple whole number.

Step 1: Identify the elements

The given compounds are manganese oxides, which means that they consist of manganese (Mn) and oxygen (O).

Step 2: Name the compounds using IUPAC rules

Using IUPAC nomenclature, we name the compounds as follows: 1. Mn2O3: Manganese (III) oxide 2. MnO2: Manganese (IV) oxide #b. Calculating the manganese content of each compound#

Step 1: Calculate the molar mass of each compound

Using the molar masses of Mn and O, we can calculate the molar mass of Mn2O3 and MnO2: 1. Mn2O3: \((2 \times 54.94 \ \text{g/mol}) + (3 \times 16.00 \ \text{g/mol}) = 157.88 \ \text{g/mol}\) 2. MnO2: \((1 \times 54.94 \ \text{g/mol}) + (2 \times 16.00 \ \text{g/mol}) = 86.94 \ \text{g/mol}\)

Step 2: Calculate the manganese content as a percent by mass

To find the manganese content as a percent by mass, divide the mass of manganese by the total mass of the compound, and multiply it by 100. 1. Mn2O3: \(\frac{2 \times 54.94 \ \text{g/mol}}{157.88 \ \text{g/mol}} \times 100 = 69.6 \%\) 2. MnO2: \(\frac{1 \times 54.94 \ \text{g/mol}}{86.94 \ \text{g/mol}} \times 100 = 63.2 \%\) #c. Explaining how Mn2O3 and MnO2 are consistent with the law of multiple proportions#

Step 1: Recall the law of multiple proportions

The law of multiple proportions states that when two elements form more than one compound, the ratios of their masses in the compounds are simple whole numbers.

Step 2: Apply the law to Mn2O3 and MnO2

In Mn2O3, there are 2 moles of Mn and 3 moles of O. In MnO2, there are 1 mole of Mn and 2 moles of O. The ratio between moles of Mn and O in Mn2O3 is \(\frac{2}{3}\) and that in MnO2 is \(\frac{1}{2}\). To find the simplest whole number ratio between the amounts of oxygen in each compound, we can take their ratio and multiply by the least common denominator: \(\frac{\frac{1}{2}}{\frac{2}{3}} = \frac{3}{4}\) This shows that the amount of oxygen in MnO2 is 3/4 times the amount of oxygen in Mn2O3, which is a simple whole number ratio, thereby demonstrating that Mn2O3 and MnO2 are consistent with the law of multiple proportions.

Key Concepts

IUPAC nomenclature

When it comes to naming chemical compounds, the International Union of Pure and Applied Chemistry (IUPAC) provides clear guidelines. The goal is to give each chemical compound a unique and systematic name that unambiguously describes its structure and composition. For manganese oxides, we need to consider the oxidation state of manganese present in the compound.

• For the compound with the formula \( \mathrm{Mn}_{2}\mathrm{O}_{3} \), the oxidation state of manganese is +3, because the two Mn atoms together total +6, and the three O atoms total -6. This compound is named **Manganese (III) oxide**.
• For \( \mathrm{MnO}_{2} \), manganese is in the +4 oxidation state, as one Mn atom is +4 and two O atoms are -4. This is named **Manganese (IV) oxide**.

Hence, by following the IUPAC rules, we ensure clarity and avoid confusion about what the compound's formula and structure are. This consistency is key in scientific communication across different fields.

molar mass calculation

Calculating the molar mass of a compound involves summing the product of each element's atomic mass and the number of times each element appears in the formula. Let's break down the molar mass calculation for the manganese oxides mentioned.For **Manganese (III) oxide** \( \mathrm{Mn}_{2}\mathrm{O}_{3} \):

• Manganese (Mn) has an atomic mass of 54.94 g/mol and there are two Mn atoms.
• Oxygen (O) has an atomic mass of 16.00 g/mol and there are three O atoms.
• Therefore, the molar mass is \( (2 \times 54.94) + (3 \times 16.00) = 157.88 \ \text{g/mol} \).

For **Manganese (IV) oxide** \( \mathrm{MnO}_{2} \):

• Again, Mn: 54.94 g/mol. This time only one Mn atom.
• Oxygen: Still 16.00 g/mol, but two O atoms.
• The calculation is \( (1 \times 54.94) + (2 \times 16.00) = 86.94 \ \text{g/mol} \).

By understanding molar mass calculation, you gain insight into molecular weight, which is crucial for converting between moles and grams.

percent composition

Percent composition can help understand how much of a compound is made up of a certain element. It is the ratio of the total mass of a specific element to the total mass of the compound, multiplied by 100.For **Manganese (III) oxide** \( \mathrm{Mn}_{2}\mathrm{O}_{3} \):

• The molar mass is 157.88 g/mol, and the manganese mass totals \( 2 \times 54.94 = 109.88 \ \text{g/mol} \).
• Percent manganese in \( \mathrm{Mn}_{2}\mathrm{O}_{3} \) is:\[ \frac{109.88}{157.88} \times 100 \approx 69.6\% \]

For **Manganese (IV) oxide** \( \mathrm{MnO}_{2} \):

• The compound has a molar mass of 86.94 g/mol with manganese contributing 54.94 g/mol.
• The calculation is:\[ \frac{54.94}{86.94} \times 100 \approx 63.2\% \]

Calculating percent composition helps in determining the proportion of each element in a compound, which is valuable for understanding and predicting chemical behaviors.

law of multiple proportions

The law of multiple proportions is a fundamental concept in chemistry. It states that if two elements form more than one compound, the masses of one element that combine with a fixed mass of the other are in ratios of small whole numbers.Applying this to manganese oxides \( \mathrm{Mn}_{2}\mathrm{O}_{3} \) and \( \mathrm{MnO}_{2} \):

• In \( \mathrm{Mn}_{2}\mathrm{O}_{3} \), we have 2 moles of Mn for every 3 moles of O.
• In \( \mathrm{MnO}_{2} \), it's 1 mole of Mn for every 2 moles of O.

Let's analyze the oxygen atom ratios:

• The ratio from Mn \( \mathrm{Mn}_{2}\mathrm{O}_{3} \) to \( \mathrm{MnO}_{2} \) is \( \frac{3}{2} \div \frac{2}{1} \), simplifying to \( \frac{3}{4} \).

This simple whole number ratio confirms the compounds observe the law of multiple proportions. This illustrates that even though their composition differs, the ratios of their elements remain constant and whole, proving a fundamental feature of chemical compounds.