Question

An ionic compound ${\mathrm{MX}}_3$ is prepared according to the following unbalanced chemical equation. $$\mathrm{M}+{\mathrm{X}}_2⟶{\mathrm{MX}}_3$$ A 0.105 -g sample of ${\mathrm{X}}_2$ contains $8.92\times {10}^{20}$ molecules. The compound ${\mathrm{MX}}_3$ consists of 54.47$\mathrm{\%}\mathrm{X}\mathrm{by}$ 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 ${\mathrm{MX}}_3\mathrm{can}$ be prepared?

Short answer

The compound MX3 is Yttrium trichloride (YCl3). The mass of YCl3 that can be prepared from 1.00 g each of Y and Cl2 is approximately 0.609 g.

Step-by-step solution

Calculate the molar mass of X from Avogadro's number

We know the mass of X2 (0.105 g) and are given the number of molecules in the sample ($8.92\times {10}^{20}$). To calculate the molar mass of X, we use Avogadro's number, which represents the number of entities (atoms or molecules) in one mole of a substance. Number of moles = ${\displaystyle \frac{Numberofmolecules}{Avogadr{o}^{\prime }snumber}}$ $$MolesofX2=\frac{8.92\times {10}^{20}}{6.022\times {10}^{23}}\approx 1.482\times {10}^{-3}moles$$ Now, we will determine the molar mass of X2: Molar mass of X2 = ${\displaystyle \frac{Mass}{Moles}}$ $$MolarmassofX2=\frac{0.105g}{1.482\times {10}^{-3}moles}\approx 70.8g/mol$$ Since X2 is diatomic, the molar mass of X will be half of the molar mass of X2: Molar mass of X = ${\displaystyle \frac{70.8g/mol}{2}}=35.4g/mol$ Using the periodic table, we see that X matches the element chlorine (Cl) with a molar mass of 35.4 g/mol.

Calculate the molar mass of M and find its identity

We know that MX3 consists of 54.47% X by mass. Therefore, the mass percentage of M can be given as: Percent M = 100% - 54.47% = 45.53% Let's assume 100 g of MX3. In this case, the mass of M in MX3 will be 45.53 g and the mass of 3 Cl (3X) atoms will be 54.47 g. As we know the molar mass of Cl (35.4 g/mol), we can calculate the moles of Cl in 54.47 g: Moles of Cl = ${\displaystyle \frac{54.47g}{35.4g/mol}}=1.538moles$ Since there are three times the number of moles of Cl atoms as M atoms in MX3: Moles of M = ${\displaystyle \frac{1.538moles}{3}}=0.513moles$ Now, to find the molar mass of M, we can use its mass and moles: Molar mass of M = ${\displaystyle \frac{Mass}{Moles}}$ $$MolarmassofM=\frac{45.53g}{0.513moles}\approx 88.8g/mol$$ Using the periodic table, we see that M matches the element yttrium (Y) with a molar mass of 88.8 g/mol. So, the compound MX3 is Yttrium trichloride (YCl3).

Determine the mass of YCl3 that can be prepared from the given masses of Y and Cl2

We are given 1.00 g each of Y and Cl2. We will first calculate the moles of Y and Cl2: Moles of Y = ${\displaystyle \frac{1.00g}{88.8g/mol}}=0.0113moles$ Moles of Cl2 = ${\displaystyle \frac{1.00g}{70.8g/mol}}=0.0141moles$ Now, we will write the balanced chemical equation for the formation of YCl3: $$2Y+3Cl2⟶2YCl3$$ From the balanced equation, we can see that 3 moles of Cl2 (X2) are needed to react with 2 moles of Y to form 2 moles of YCl3. However, we do not have an equal ratio of moles as per the balanced equation. To identify the limiting reactant, we will divide the moles of Y and Cl2 by their respective stoichiometric coefficients: For Y: ${\displaystyle \frac{0.0113}{2}}=0.00565$ For Cl2: ${\displaystyle \frac{0.0141}{3}}=0.0047$ The smaller value (0.0047) corresponds to the limiting reactant, which in this case is Cl2. Now, we can use the moles of the limiting reactant to calculate the moles of YCl3 formed: Moles of YCl3 = $0.0047\times {\displaystyle \frac{2molesYCl3}{3molesCl2}}=0.00313moles$ Finally, we can calculate the mass of YCl3 formed using its molar mass (Y: 88.8 g/mol, 3Cl: 3 x 35.4 g/mol = 106.2 g/mol): Mass of YCl3 = $0.00313moles\times (88.8g/mol+106.2g/mol)\approx 0.609g$ So, the mass of YCl3 that can be prepared from 1.00 g each of Y and Cl2 is 0.609 g.

Key concepts

Molar Mass Calculation

When you want to know how heavy one mole of a chemical substance is, you're talking about molar mass. Molar mass allows chemists to convert between the mass of a substance and the amount of particles that make it up. Let's break down how we find it.

In the exercise, we calculated the molar mass of a diatomic molecule, $X_2$. We needed to know the mass of a specific number of these molecules, which was given as $0.105$ grams for $8.92\times {10}^{20}$ molecules.

Here's how we find it:

• Use Avogadro's number $(6.022\times {10}^{23})$ to determine how many moles are in the sample: $$\text{Moles of }{X}_2=\frac{8.92\times {10}^{20}}{6.022\times {10}^{23}}\approx 1.482\times {10}^{-3}$$
• Then, calculate the molar mass: $$\text{Molar mass of }{X}_2=\frac{0.105\text{ g}}{1.482\times {10}^{-3}\text{ moles}}\approx 70.8\text{ g/mol}$$
• Because $X_2$ is diatomic, divide by 2 to find the molar mass of $X$: $$35.4\text{ g/mol}$$

Simple! This gives us the identity of $X$ as chlorine $(Cl)$, matching the periodic table.

Limiting Reactant

Understanding the limiting reactant helps determine the maximum amount of product that can be formed during a chemical reaction.

In our exercise, we mixed equal masses of $Y$ and $C{l}_2$ to make $YC{l}_3$. But which one runs out first and stops the reaction?

To figure this out:

• Calculate the moles of each reactant: $$\text{Moles of }Y=\frac{1.00\text{ g}}{88.8\text{ g/mol}}=0.0113\text{ moles}$$$$\text{Moles of }C{l}_2=\frac{1.00\text{ g}}{70.8\text{ g/mol}}=0.0141\text{ moles}$$
• Use the balanced equation to find the ratio of reactants needed $(2Y+3C{l}_2\to 2YC{l}_3)$.
• Divide the moles by their coefficients:$$\text{For }Y:\frac{0.0113}{2}=0.00565$$ $$\text{For }C{l}_2:\frac{0.0141}{3}=0.0047$$
• The smaller value, $0.0047$, means $C{l}_2$ is the limiting reactant, controlling how much $YC{l}_3$ can be made.

This concept is crucial for determining yields in chemistry.

Chemical Equation Balancing

Balancing a chemical equation ensures that the same number of each atom appears on both sides of the equation, abiding by the law of conservation of mass.

For the synthesis of $YC{l}_3$, the unbalanced equation given was: $$\text{M}+{\text{X}}_2\to {\text{MX}}_3$$

After identifying $M$ as yttrium $(Y)$ and $X$ as chlorine $(Cl)$, the balanced chemical equation becomes:$$2Y+3C{l}_2\to 2YC{l}_3$$

Here’s why it's balanced:

• 2 yttrium atoms react with 3 chlorine diatomic molecules.
• This process forms 2 $YC{l}_3$ molecules, with each product consisting of one yttrium and three chlorine atoms.

Balancing ensures that all reactant molecules are accounted for and that mass is conserved. It's a fundamental skill in chemistry allowing accurate predictions and measurements in reactions.