Problem 53 - Chemistry: The Central Science

Question

Magnesium can be used as a "getter" in evacuated enclosures, to react with the last traces of oxygen. (The magnesium is usually heated by passing an electric current through a wire or ribbon of the metal.) If an enclosure of \(0.382 \mathrm{~L}\) has a partial pressure of \(\mathrm{O}_{2}\) of \(3.5 \times 10^{-6}\) torr at \(27{ }^{\circ} \mathrm{C}\), what mass of magnesium will react according to the following equation? $$ 2 \mathrm{Mg}(s)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$

Step-by-Step Solution

Answer

The mass of magnesium required to react with the given amount of oxygen in the enclosure is approximately \(3.67 \times 10^{-9}\, \mathrm{g}\).

Step 1: Find the number of moles of oxygen gas using Ideal Gas Law

The ideal gas law equation is given by \(PV = nRT\), where \(P\) represents pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the ideal gas constant, and \(T\) is the temperature in Kelvin. With the given values, using the ideal gas law, we can find the number of moles of \(\mathrm{O}_{2}\). Given volume, \(V = 0.382 \mathrm{~L}\), partial pressure, \(P = 3.5 \times 10^{-6}\, \mathrm{torr}\), and temperature \(T = 27{ }^{\circ} \mathrm{C}\). First, convert the temperature from Celsius to Kelvin: $$ T(K) = 27 + 273.15 = 300.15 \mathrm{K} $$ Next, convert the pressure from torr to atm, as we will be using the ideal gas constant in atm: $$ P(\mathrm{atm}) = 3.5 \times 10^{-6}\, \mathrm{torr} \times \frac{1\, \mathrm{atm}}{760\, \mathrm{torr}} = 4.61 \times 10^{-9}\, \mathrm{atm} $$ Now, use the ideal gas law to find the number of moles of \(\mathrm{O}_{2}\): $$ n(\mathrm{O}_{2}) = \frac{PV}{RT} = \frac{(4.61 \times 10^{-9}\, \mathrm{atm})(0.382\, \mathrm{L})}{(0.0821\, \mathrm{L\, atm/mol\, K})(300.15\, \mathrm{K})} = 7.55 \times 10^{-11}\, \mathrm{mol} $$

Step 2: Use stoichiometry to find the number of moles of magnesium required

From the given balanced chemical equation, the mole ratio of magnesium to oxygen is \(2:1\): $$ 2 \mathrm{Mg}(s) + \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{MgO}(s) $$ Using this mole ratio, calculate the amount of \(\mathrm{Mg}\) required to react with the given amount of \(\mathrm{O}_{2}\): $$ n(\mathrm{Mg}) = 2 \times 7.55 \times 10^{-11}\, \mathrm{mol} = 1.51 \times 10^{-10}\, \mathrm{mol} $$

Step 3: Convert the number of moles of magnesium to mass

To find the mass of magnesium required, multiply the number of moles of magnesium by the molar mass of magnesium: $$ \mathrm{Mass\, of\, Mg} = n(\mathrm{Mg}) \times \mathrm{Molar\, mass\, of\, Mg} $$ Given the molar mass of magnesium: \(24.31\, \mathrm{g/mol}\) $$ \mathrm{Mass\, of\, Mg} = (1.51 \times 10^{-10}\, \mathrm{mol}) (24.31\, \mathrm{g/mol}) = 3.67 \times 10^{-9}\, \mathrm{g} $$ Therefore, the mass of magnesium required to react with the given amount of oxygen in the enclosure is approximately \(3.67 \times 10^{-9}\, \mathrm{g}\).

Key Concepts

Ideal Gas Law

The ideal gas law is a fundamental equation in chemistry that describes the behavior of gases. It connects four key properties of a gas: pressure (P), volume (V), the number of moles (n), and temperature (T). The equation is expressed as:
\[ PV = nRT \]where:

• P is the pressure of the gas in units like ATM or torr,
• V is the volume in liters,
• n is the number of moles, and
• R is the ideal gas constant (0.0821 L atm/mol K).

In the provided problem, the ideal gas law helps determine the number of moles of oxygen gas (\text{O}_2) present in the enclosure. Temperature is first converted from Celsius to Kelvin. Conversion is essential because gas laws use Kelvin. Pressure is also converted from torr to atm to align with common units used with the ideal gas constant. By organizing units correctly, we ensure precision and appropriate conversions when solving the equation.

Chemical Reactions

Chemical reactions are processes where reactants are transformed into products. In the context of this problem, magnesium (\text{Mg}) reacts with oxygen gas (\text{O}_2) to form magnesium oxide (\text{MgO}). The balanced equation given is:
\[ 2 \text{Mg} (s) + \text{O}_2 (g) \rightarrow 2 \text{MgO} (s) \]
The stoichiometry of a chemical reaction provides vital insights. It shows the proportion of reactants needed to form products. In this case, two moles of magnesium react with one mole of oxygen to make two moles of magnesium oxide. Understanding stoichiometry helps us use molecular proportions to predict amounts of materials required. This prediction is crucial for processes like the one described where we calculate the mass of magnesium needed to fully react with a specified amount of oxygen.

Molar Mass

Molar mass is a critical concept when converting between the number of moles and the mass of a substance. It is the mass of one mole of a given element or compound, usually reported in grams per mole (\text{g/mol}). For magnesium, this value is 24.31 \text{g/mol}.
Knowing the molar mass allows us to link macroscopic measurements (in grams) with microscopic amounts (in moles). For instance, once the number of moles of magnesium required is determined through stoichiometry, it can be easily converted into mass using its molar mass:
\[\text{Mass of Mg} = n(\text{Mg}) \times \text{Molar mass of Mg} \].
In practice, molar mass is vital for any laboratory work involving precise quantities of chemicals, making it a cornerstone of stoichiometric calculations.

Gas Pressure Conversions

Gas pressure can be measured in several units, such as atmospheres (ATM), torr, or pascals (Pa). Understanding these units and being adept at converting between them is crucial for solving problems involving gas pressure.
In this exercise, oxygen's partial pressure was initially given in torr. However, for calculations with the ideal gas law, it was converted into atmospheres:
\[ P(\text{atm}) = P(\text{torr}) \times \frac{1\, \text{atm}}{760\, \text{torr}} \]The conversion is necessary because the ideal gas law constant (R) is typically used with atm units. Ensuring all units match across measurements is crucial not only for accuracy but also for maintaining the consistency and integrity of mathematical calculations in chemistry.