Question

Determine the number of moles of (a) \(\mathrm{N}_{2} \mathrm{O}_{4}\) in a \(115 \mathrm{g}\) sample (b) \(\mathrm{N}\) atoms in \(43.5 \mathrm{g}\) of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{N}\) atoms in a sample of \(\mathrm{C}_{7} \mathrm{H}_{5}\left(\mathrm{NO}_{2}\right)_{3}\) that has the same number of \(\mathrm{O}\) atoms as \(12.4 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

Short answer

The number of moles in a 115 g sample of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is approximately 1.25 moles. The number of moles of \(\mathrm{N}\) atoms in 43.5 g of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is approximately 0.588 moles. The number of moles of \(\mathrm{N}\) atoms in a sample of \(\mathrm{C}_{7} \mathrm{H}_{5}\left(\mathrm{NO}_{2}\right)_{3}\) that has the same number of \(\mathrm{O}\) atoms as 12.4 g of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is approximately 0.207 moles.

Step-by-step solution

Calculate moles of \(\mathrm{N}_{2} \mathrm{O}_{4}\)

Firstly, find the molar mass of \(\mathrm{N}_{2} \mathrm{O}_{4}\). According to the periodic table, the molar mass of nitrogen (\(N\)) is approximately 14 g/mol and that of oxygen (\(O\)) is 16 g/mol. Therefore, the molar mass of \(\mathrm{N}_{2} \mathrm{O}_{4}\) is \((2 \times 14) + (4 \times 16) = 92 \mathrm{g}/\mathrm{mol}\). Then, using the formula \(n = \frac{m}{M}\), plug in the given mass (115 g) and the calculated molar mass to find the number of moles: \(n = \frac{115 \, \mathrm{g}}{92 \, \mathrm{g}/\mathrm{mol}} = 1.25 \, \mathrm{mol}\).

Calculate moles of \(\mathrm{N}\) atoms in \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\)

Firstly, calculate the molar mass of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\). The molar mass of magnesium (\(Mg\)) is approximately 24 g/mol, nitrogen is 14 g/mol and oxygen is 16 g/mol. Therefore, the molar mass of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is \(24 + 2[(1 \times 14) + (3 \times 16)] = 148 \mathrm{g}/\mathrm{mol}\). Next, calculate the number of moles in 43.5 g of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) using the formula: \(n = \frac{43.5 \, \mathrm{g}}{148 \, \mathrm{g}/\mathrm{mol}} \approx 0.294 \, \mathrm{mol}\). As there are two \(\mathrm{N}\) atoms for each molecule of \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\), the number of moles of \(\mathrm{N}\) atoms is \(2 \times 0.294 \, \mathrm{mol} = 0.588 \, \mathrm{mol}\).

Calculate moles of \(\mathrm{N}\) atoms in a sample of \(\mathrm{C}_{7} \mathrm{H}_{5}\left(\mathrm{NO}_{2}\right)_{3}\)

First determine how many moles of \(O\) are present in 12.4 g of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\). The molar mass of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is \(6 \times 12 + 12 \times 1 + 6 \times 16 = 180 \mathrm{g}/\mathrm{mol}\). The number of moles of \(O\) in 12.4 g of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) is \( \frac{12.4 \, \mathrm{g}}{180 \, \mathrm{g}/\mathrm{mol}} = 0.069 \, \mathrm{mol}\). Given that each molecule of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) contains 6 oxygens, the mole quantity of \(O\) atoms is \(0.069 \, \mathrm{mol} \times 6 = 0.414 \, \mathrm{mol}\). Now the same number of \(O\) atoms are said to be present in \(\mathrm{C}_{7} \mathrm{H}_{5}\left(\mathrm{NO}_{2}\right)_{3}\). Each molecule of \(\mathrm{C}_{7} \mathrm{H}_{5}\left(\mathrm{NO}_{2}\right)_{3}\) has 6 oxygen atoms and 3 nitrogen atoms. Therefore, the mole quantity of \(N\) atoms will be \( \frac{3}{6} \times 0.414 \, \mathrm{mol} = 0.207 \, \mathrm{mol}\).

Key concepts

Molar Mass

Molar mass is a fundamental concept in chemistry used to determine the mass of one mole of a given substance. It is derived from the atomic masses found on the periodic table. Understanding molar mass is essential for performing mole calculations and is typically expressed as grams per mole (g/mol).
To find the molar mass of a compound, you simply sum the molar masses of all the atoms within its chemical formula. For instance, if you are asked to find the molar mass of dinitrogen tetroxide \(\mathrm{N}_2 \mathrm{O}_4\), you would calculate as follows:
- Two nitrogen atoms: \(2 \times 14 \text{ g/mol} = 28 \text{ g/mol}\)
- Four oxygen atoms: \(4 \times 16 \text{ g/mol} = 64 \text{ g/mol}\)
So, the molar mass of \(\mathrm{N}_2 \mathrm{O}_4\) is \(28 + 64 = 92 \text{ g/mol}\). Knowing the molar mass allows you to convert between the mass of a sample and the number of moles it contains, using the relationship \(n = \frac{m}{M}\), where \(n\) is the number of moles, \(m\) is the mass, and \(M\) is the molar mass.

Stoichiometry

Stoichiometry is the aspect of chemistry that involves calculating the relationships between the amounts of reactants and products in chemical reactions. It heavily relies on the understanding of mole calculations and molar ratios.
In stoichiometric calculations, the balanced chemical equation provides a "road map" to solve problems. For example, the number of moles in reactants might dictate the amount of products formed. This requires converting grams of a substance to moles using its molar mass and leveraging the coefficients from the balanced equation to find desired quantities.
For example, when calculating the number of moles of \(\mathrm{N}\) atoms in \(\mathrm{Mg}(\mathrm{NO}_3)_2\), we found that each molecule contains 2 nitrogen atoms. Knowing the moles of \(\mathrm{Mg}(\mathrm{NO}_3)_2\) allows multiplying by this stoichiometric factor (2:1) to find the moles of nitrogen atoms. This crucial technique enables precise predictions about reactants and products.

Chemical Formula

The chemical formula provides a concise way of expressing how many and which types of atoms are present in a molecule. Through symbolic representation, chemical formulas play a crucial role in determining the properties and behavior of compounds.
Consider \(\mathrm{N}_2\mathrm{O}_4\) - the formula immediately tells us that each molecule consists of 2 nitrogen atoms and 4 oxygen atoms. This direct insight into its molecular composition enables us to calculate its molar mass effectively.
Chemical formulas also help identify relationships within complex reactions. For example, looking at \(\mathrm{Mg}(\mathrm{NO}_3)_2\), knowing how many of each atom exists allows us to perform stoichiometry and mole calculations, establishing how various species interact in a chemical reaction.
Understanding chemical formulas is foundational for navigating chemistry as they depict the very elements that are pivotal for calculations like those in stoichiometry and determining molar masses.

Nitrogen Atoms

Nitrogen atoms, represented as \(N\), are frequently encountered in diverse chemical compounds. Understanding their quantity within a compound's formula is integral to mole calculations involving nitrogen atoms.
In molecules such as \(\mathrm{N}_2\mathrm{O}_4\), nitrogen atoms are straightforward to calculate, but in more complex compounds like \(\mathrm{Mg}(\mathrm{NO}_3)_2\) or \(\mathrm{C}_7 \mathrm{H}_5\left(\mathrm{NO}_2\right)_3\), careful attention to the formula is necessary. \(\mathrm{Mg}(\mathrm{NO}_3)_2\) contains two nitrate ions, each with one nitrogen, which shows the compound has two nitrogen atoms per molecule.
This understanding allows you to calculate exactly how many moles of nitrogen atoms are present when you know the moles of the compound, applying stoichiometric principles. Overall, understanding the presence and role of nitrogen atoms aids significantly in chemical calculations and equation balancing.

Oxygen Atoms

Oxygen atoms, often represented as \(O\), are key players in various chemical reactions and compounds. They are essential to many atmospheric, biological, and synthetic processes.
In the sample problem provided, oxygen atoms are found within several compound formulas, including \(\mathrm{N}_2\mathrm{O}_4\) and \(\mathrm{C}_6 \mathrm{H}_{12}\mathrm{O}_6\). Recognizing the quantity of oxygen atoms in these formulas can assist in understanding reactions' requirements and products.
Specifically, \(\mathrm{C}_6 \mathrm{H}_{12}\mathrm{O}_6\) contains 6 oxygen atoms per molecule, which is vital in calculations where the number of oxygen atoms in other compounds must equate to those in \(\mathrm{C}_6 \mathrm{H}_{12}\mathrm{O}_6\). This concept underscores calculations in problems involving comparisons or constraints based on the number of oxygen atoms.
In chemistry, oxygen is ubiquitous, and accurately understanding its quantity in chemical formulas facilitates more accurate stoichiometric and mole-based calculations.