Question

An ionic compound \(\mathrm{MX}_{3}\) is prepared according to the following unbalanced chemical equation. $$ \mathrm{M}+\mathrm{X}_{2} \longrightarrow \mathrm{MX}_{3} $$ A \(0.105-\mathrm{g}\) sample of \(\mathrm{X}_{2}\) contains \(8.92 \times 10^{20}\) molecules. The compound \(\mathrm{MX}_{3}\) consists of \(54.47 \% \mathrm{X}\) by mass. What are the identities of \(\mathrm{M}\) and \(\mathrm{X}\), and what is the correct name for \(\mathrm{MX}_{3}\) ? Starting with \(1.00 \mathrm{~g}\) each of \(\mathrm{M}\) and \(\mathrm{X}_{2}\), what mass of \(\mathrm{MX}_{3}\) can be prepared?

Short answer

The identities of M and X are Iodine (I) and Chlorine (Cl), respectively. The compound is named Iodine trichloride (ICl3). The correct balanced equation is \( \text{I} + 3\text{Cl}_2 \longrightarrow \text{ICl}_3 \). When starting with 1.00 g each of Iodine and chlorine gas, the mass of ICl3 that can be prepared is 1.836 g.

Step-by-step solution

Determine molar mass of X2

First, we have the mass and the number of molecules of X2. We can use Avogadro's number to determine the number of moles of X2 and then calculate its molar mass. The number of moles of X2 can be calculated using the following equation: $$ \text{moles of X2} = \frac{\text{number of molecules}}{\text{Avogadro's number}} $$ Plug in the values given: $$ \text{moles of X2} = \frac{8.92 \times 10^{20}}{6.022\times 10^{23} \text{molecules/mol}} $$ $$ \text{moles of X2} = 1.482 \times 10^{-3} \text{mol} $$ Now, we can find the molar mass of X2: $$ \text{molar mass of X2} = \frac{\text{mass of X2}}{\text{moles of X2}} $$ $$ \text{molar mass of X2} = \frac{0.105\text{g}}{1.482 \times 10^{-3}\text{mol}} $$ $$ \text{molar mass of X2} = 70.85 \text{g/mol} $$

Identify the element X and its molar mass

Since X2 has a molar mass of 70.85 g/mol, it means the molar mass of X must be half of that. $$ \text{molar mass of X} = \frac{\text{molar mass of X2}}{2} $$ $$ \text{molar mass of X} = \frac{70.85 \text{g/mol}}{2} $$ $$ \text{molar mass of X} = 35.43 \text{g/mol} $$ Looking at the periodic table, the element with a molar mass closest to 35.43 g/mol is chlorine (Cl), with a molar mass of 35.45 g/mol.

Calculate molar mass of M

The compound MX3 consists of 54.47% X by mass. Since we now know that X is chlorine, we can determine the molar mass of M by subtracting the mass contribution of chlorine from the total molar mass. $$ \text{mass \% of M in MX3} = 100 - 54.47 = 45.53\% $$ Let x be the molar mass of MX3. The equation to find the molar mass of M is: $$ 0.4553x = (3 \times 35.45) $$ Solve for x: $$ x = \frac{(3 \times 35.45)}{0.4553} = 233.1 \text{g/mol} $$ Subtract the molar mass of chlorine (3 x 35.45) to find the molar mass of M: $$ \text{molar mass of M} = 233.1 - 3 \times 35.45 = 126.8 \text{g/mol} $$

Identify the element M

Checking the periodic table again, the element with a molar mass close to 126.8 g/mol is Iodine (I), with a molar mass of 126.90447 g/mol.

Determine the name and balanced equation for the compound MX3

Now that we know M is Iodine (I) and X is Chlorine (Cl), the compound MX3 is ICl3 (Iodine trichloride). The balanced chemical equation is: $$ \text{I} + 3\text{Cl}_2 \longrightarrow \text{ICl}_3 $$

Calculate the mass of ICl3 prepared

When starting with 1.00 g each of I and Cl2, we can use stoichiometry to calculate the mass of ICl3 produced. First, we need to convert grams to moles: $$ \text{moles of I} = \frac{1.000 \text{g}}{126.90447 \text{g/mol}} = 0.007878 \text{mol} $$ $$ \text{moles of Cl2} = \frac{1.000\text{g}}{70.9\text{g/mol}} = 0.01411 \text{mol} $$ Now, using the balanced chemical equation, we can find the limiting reactant and calculate the moles of ICl3 produced using the stoichiometry of the reaction: $$ \text{I} + 3\text{Cl}_2 \longrightarrow \text{ICl}_3 $$ Iodine, in this case, is our limiting reactant since for each mole of iodine, we need 3 moles of Cl2. Convert the moles of ICl3 produced back to grams: $$ \text{mass of ICl3} = \text{moles of I produced} \times \text{molar mass of ICl3} $$ $$ \text{mass of ICl3} = 0.007878 \text{mol} \times 233.1 \text{g/mol} $$ $$ \text{mass of ICl3} = 1.836 \text{g} $$ When starting with 1.00 g each of Iodine and chlorine gas, the mass of ICl3 that can be prepared is 1.836 g.

Key concepts

Molar Mass Calculation

Molar mass calculation is a fundamental concept in stoichiometry that involves determining the mass of one mole of a substance. A mole is a standard unit of measurement in chemistry that represents a huge number of particles, specifically Avogadro's number, which is approximately \(6.022 \times 10^{23}\) particles. This allows chemists to work with a manageable amount of substance when performing experiments.

For instance, in the given problem, calculating the molar mass of \(\mathrm{X}_{2}\) was the first step. The mass of the sample was divided by the number of moles to find the molar mass. If you have the number of molecules and the mass of a substance, you can use the formula:

\[ \text{molar mass} = \frac{\text{mass of the sample}}{\text{moles of substance}} \]

Remember, the molar mass is expressed in grams per mole (g/mol). This calculation is crucial because it lays the groundwork for later steps, such as identifying elements based on the calculated molar mass and further stoichiometric calculations.

Limiting Reactant

The concept of the limiting reactant is essential in predicting the amount of product that can be formed in a chemical reaction. It refers to the reactant that will be completely consumed first in a chemical reaction, limiting the extent of the reaction and, consequently, the amount of product made.

Considering the provided example, where \(\mathrm{I}\) and \(\mathrm{Cl}_{2}\) react to form \(\mathrm{ICl}_{3}\), identifying the limiting reactant was done after determining the moles of each reactant from their respective masses. Then, the stoichiometry of the reaction was examined:

\[ \mathrm{I} + 3\mathrm{Cl}_{2} \longrightarrow \mathrm{ICl}_{3} \]

Given that iodine reacts with chlorine gas in a 1:3 mole ratio, one could pinpoint the limiting reactant by comparing the mole ratio of the reactants to the mole ratio required by the balanced equation. When you establish which reactant is limiting, you can calculate the maximum amount of product generated, not just the theoretical yield under ideal conditions.

Chemical Equation Balancing

Balancing chemical equations is a critical step in stoichiometry as it ensures the law of conservation of mass is upheld. This means that in a chemical reaction, the number of atoms of each element in the reactants side must equal the number of atoms of that element in the products side. An equation is balanced when there are equal numbers of atoms for each element involved in the reaction on both sides of the equation.

To balance a chemical equation, one must change the coefficients, which are the numbers placed before reactants and products. The coefficients indicate the ratios of the amounts of substances that react or are produced. An example from our exercise is the balancing of the formation of \(\mathrm{ICl}_{3}\):

\[ \mathrm{I} + 3\mathrm{Cl}_{2} \longrightarrow \mathrm{ICl}_{3} \]

Here, the coefficient '3' before \(\mathrm{Cl}_{2}\) signifies that three moles of chlorine gas react with one mole of iodine to produce one mole of \(\mathrm{ICl}_{3}\). Correctly balancing the equation is vital for correctly applying stoichiometry to predict reactant consumption and product formation in chemical reactions.