Oxygen Atoms In N2O3 Calculation And Explanation

In the fascinating realm of chemistry, understanding the composition of molecules is paramount. This article will guide you through the process of calculating the number of oxygen atoms present in a given mass of dinitrogen trioxide ($N_2O_3$). We will meticulously break down the steps involved, ensuring a clear and comprehensive understanding of the underlying principles. This exploration is not just about arriving at the correct answer; it's about grasping the fundamental concepts of stoichiometry and the mole concept, which are cornerstones of chemical calculations. Mastering these concepts will equip you with the tools to tackle a wide array of chemical problems, from simple molar mass calculations to complex reaction stoichiometry.

To begin, we must first define what the question is asking. We are given 160 grams of $N_2O_3$ and asked to determine the number of oxygen atoms present within this sample. This necessitates a journey through several key concepts: molar mass, moles, and Avogadro's number. Each of these concepts plays a crucial role in bridging the gap between macroscopic measurements (grams) and the microscopic world of atoms and molecules. The journey may seem complex at first, but with careful attention to each step, the solution will unfold logically and elegantly.

The molar mass of a compound is the mass of one mole of that compound, expressed in grams per mole (g/mol). It serves as a conversion factor between the mass of a substance and the amount of substance in moles. To calculate the molar mass of $N_2O_3$, we must sum the atomic masses of all the atoms present in the molecule. The atomic masses of nitrogen (N) and oxygen (O) can be found on the periodic table. Nitrogen has an atomic mass of approximately 14.01 atomic mass units (amu), and oxygen has an atomic mass of approximately 16.00 amu. In one molecule of $N_2O_3$, there are two nitrogen atoms and three oxygen atoms. Therefore, the molar mass of $N_2O_3$ can be calculated as follows:

Molar mass of $N_2O_3$ = (2 × Atomic mass of N) + (3 × Atomic mass of O)

= (2 × 14.01 g/mol) + (3 × 16.00 g/mol)

= 28.02 g/mol + 48.00 g/mol

= 76.02 g/mol

This calculation tells us that one mole of $N_2O_3$ has a mass of 76.02 grams. This value is crucial for converting the given mass of $N_2O_3$ (160 grams) into moles, which is the next step in our calculation. The molar mass acts as a bridge, allowing us to move from the tangible world of grams to the abstract but incredibly useful world of moles. Without the molar mass, we would be unable to relate the macroscopic measurement of mass to the microscopic count of molecules and atoms. This foundational step is essential for all stoichiometric calculations, making it a vital concept to master.

Having determined the molar mass of $N_2O_3$ (76.02 g/mol), we can now convert the given mass of 160 grams into moles. The mole is the SI unit of amount of substance and represents a fixed number of entities (atoms, molecules, ions, etc.), specifically Avogadro's number ($6.022 imes 10^{23}$). To convert grams to moles, we use the following formula:

Moles = Mass / Molar mass

In this case:

Moles of $N_2O_3$ = 160 g / 76.02 g/mol

≈ 2.105 moles

This calculation reveals that 160 grams of $N_2O_3$ corresponds to approximately 2.105 moles. This conversion is a pivotal step, as it allows us to work with the number of molecules rather than their mass. The mole concept provides a convenient way to count vast numbers of atoms and molecules, which are far too numerous to count individually. By converting grams to moles, we are essentially translating a macroscopic measurement into a microscopic count. This is a fundamental principle in chemistry, enabling us to understand and predict the outcomes of chemical reactions.

Now that we know the number of moles of $N_2O_3$, we can determine the number of moles of oxygen atoms present. From the chemical formula $N_2O_3$, we see that each molecule of $N_2O_3$ contains three oxygen atoms. Therefore, one mole of $N_2O_3$ contains three moles of oxygen atoms. To find the total moles of oxygen atoms in 2.105 moles of $N_2O_3$, we multiply by the stoichiometric ratio:

Moles of O atoms = Moles of $N_2O_3$ × (Moles of O atoms / Moles of $N_2O_3$)

Moles of O atoms = 2.105 moles × (3 moles O / 1 mole $N_2O_3$)

Moles of O atoms ≈ 6.315 moles

This calculation shows that there are approximately 6.315 moles of oxygen atoms in 160 grams of $N_2O_3$. This step is crucial because it directly addresses the question of how many oxygen atoms are present. We have now successfully linked the mass of the compound to the amount of a specific element within it. This highlights the importance of understanding chemical formulas and their role in determining the composition of compounds. The stoichiometric ratio derived from the chemical formula acts as a bridge, allowing us to relate the amount of the compound to the amount of its constituent elements.

The final step is to convert moles of oxygen atoms to the number of individual oxygen atoms. To do this, we use Avogadro's number ($6.022 imes 10^{23}$ atoms/mol), which represents the number of atoms, molecules, or other entities in one mole. The formula for this conversion is:

Number of atoms = Moles × Avogadro's number

In this case:

Number of O atoms = 6.315 moles × $6.022 imes 10^{23}$ atoms/mol

Number of O atoms ≈ $3.80 imes 10^{24}$ atoms

Therefore, there are approximately $3.80 imes 10^{24}$ oxygen atoms in 160 grams of $N_2O_3$. This final calculation provides the answer to the original question, expressing the number of oxygen atoms in a scientifically meaningful way. Avogadro's number serves as the ultimate bridge, connecting the macroscopic world of moles to the microscopic world of individual atoms. This constant is a cornerstone of chemistry, allowing us to count the incredibly small entities that make up matter. This step completes the journey, providing a concrete answer to the initial question.

In summary, we have successfully calculated the number of oxygen atoms in 160 grams of $N_2O_3$ by following a step-by-step approach:

1. Calculate the molar mass of $N_2O_3$ (76.02 g/mol). This step established the crucial link between mass and moles.
2. Convert grams of $N_2O_3$ to moles (2.105 moles). This allowed us to work with the amount of substance rather than its mass.
3. Determine moles of oxygen atoms (6.315 moles). This step utilized the chemical formula to relate the amount of the compound to the amount of a specific element.
4. Calculate the number of oxygen atoms ($3.80 imes 10^{24}$ atoms). This final step converted moles to the number of individual atoms, providing the answer to the original question.

Therefore, the correct answer is C. $3.80 imes 10^{24}$.

This exercise demonstrates the power of stoichiometry and the mole concept in solving chemical problems. By understanding these fundamental principles, we can confidently navigate the world of chemical calculations and gain a deeper appreciation for the composition of matter. The ability to convert between mass, moles, and number of atoms is a cornerstone of chemical understanding, enabling us to make predictions and interpret experimental results. This journey through the calculation of oxygen atoms in $N_2O_3$ serves as a valuable example of how these concepts are applied in practice, reinforcing their importance in the study of chemistry.