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 \(12.4 \mathrm{g} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\)

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: \(n = \frac{115 \, \mathrm{g}}{92 \, \mathrm{g}/\mathrm{mol}} = 1.25 \, \mathrm{mol}\).

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 \(2 \times 0.294 \, \mathrm{mol} = 0.588 \, \mathrm{mol}\).

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 \( \frac{3}{6} \times 0.414 \, \mathrm{mol} = 0.207 \, \mathrm{mol}\).