91Ó°ÊÓ

Grignard reagents, like \(\mathrm{CH}_{3} \mathrm{MgBr}\) in our exercise, are key players in organic chemistry. They are formed by reacting an alkyl or aryl halide with magnesium metal, creating a compound that is highly reactive. This reactivity is harnessed to introduce new carbon-carbon bonds in synthetic organic chemistry.

When a Grignard reagent reacts with a compound containing active hydrogens—such as alcohols, acids, or water—it forms a hydrocarbon. In our exercise, compound \(\mathrm{Z}\) reacts with \(\mathrm{CH}_{3} \mathrm{MgBr}\) to release methane (\(\mathrm{CH}_{4}\)). This indicates that each active hydrogen in compound \(\mathrm{Z}\) has reacted to form \(\mathrm{CH}_{4}\).

• The Grignard reagent acts as a nucleophile, attacking electrophilic centers.
• Each molecule of methane produced corresponds to an active hydrogen in the sample.

Knowing the amount of evolved gas helps in deducing the active hydrogens.

The molar volume is a crucial concept in understanding gas reactions. At normal temperature and pressure (NTP), one mole of any ideal gas occupies 22.4 liters or 22,400 milliliters. This standard condition simplifies calculations between volume and moles.

In our exercise, knowing that 224 milliliters of \(\mathrm{CH}_{4}\) gas was evolved allows us to convert this volume into moles. By using the relationship:\[\text{Moles of gas} = \frac{\text{Volume in ml}}{22,400 \text{ ml/mol}}\]we can find the moles of methane formed and thus determine the number of active hydrogens in compound \(\mathrm{Z}\).

• This volume-to-moles conversion is crucial for stoichiometric calculations.
• It helps relate gas volume to the number of particles involved in the reaction.

Understanding molar volumes allows for accurate predictions and analysis in gas-producing reactions.

Mole calculations are a core pillar of chemical quantitative analysis. The mole is a fundamental unit in chemistry, representing \(6.022 \times 10^{23}\) particles of a substance. For gas reactions, this concept ties volume measurements to the quantity of substance through molar relationships.

In the given problem, we compute moles of both \(\mathrm{CH}_{4}\) formed and compound \(\mathrm{Z}\) used. These calculations help us connect the quantities of reactants and products:

• The moles of \(\mathrm{CH}_{4}\) are derived from its volume at NTP, which we calculated to be 0.01 mol.
• The moles of \(\mathrm{Z}\) are determined by dividing its mass by its molar mass, giving 0.005 mol.

This relationship between moles of reactants and products is essential for identifying stoichiometric ratios in reactions. Such calculations form the basis of most quantitative work in chemistry.

Determining the molar mass of a compound allows chemists to relate a substance's mass to the number of moles, facilitating stoichiometric calculations. The molar mass is calculated by summing the atomic masses of all atoms in a molecule.

For compound \(\mathrm{Z}\), we know its molar mass is 90 g/mol. This value allows us to determine the number of moles of \(\mathrm{Z}\) from a given mass. The calculation:\[\text{Moles of } \mathrm{Z} = \frac{0.450 \text{ g}}{90 \text{ g/mol}} = 0.005 \text{ mol}\]

• The given molar mass is integral for converting between mass and moles.
• This conversion is critical for stoichiometric calculations that predict product formation and reactant usage.

Understanding how molar mass interrelates with other chemical quantities enhances problem-solving in both experimental and theoretical chemistry.