Question

A metal, M, was converted to the sulfate, \(\mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). Then a solution of the sulfate was treated with barium chloride to give barium sulfate crystals, which were filtered off. \(\mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}(a q)+3 \mathrm{BaCl}_{2}(a q) \longrightarrow\) $$ 2 \mathrm{MCl}_{3}(a q)+3 \mathrm{BaSO}_{4}(s) $$ If \(1.200 \mathrm{~g}\) of the metal gave \(6.026 \mathrm{~g}\) of barium sulfate, what is the atomic weight of the metal? What is the metal?

Short answer

The atomic weight of the metal is approximately 139.53 g/mol, suggesting it is Lanthanum (La).

Step-by-step solution

Write the Reaction Equation

The chemical reaction is given as: \[ \mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}(aq) + 3 \mathrm{BaCl}_{2}(aq) \rightarrow 2 \mathrm{MCl}_{3}(aq) + 3 \mathrm{BaSO}_{4}(s) \] This signifies that each mole of metal sulfate \(\mathrm{M}_{2}(\mathrm{SO}_{4})_{3}\) produces 3 moles of \(\mathrm{BaSO}_{4}\).

Calculate Moles of Barium Sulfate Formed

We start by calculating the moles of \(\mathrm{BaSO}_{4}\). The molar mass of \(\mathrm{BaSO}_{4}\) is 233.39 g/mol. Therefore, the moles of \(\mathrm{BaSO}_{4}\) formed is calculated as follows: \[moles\ of\ BaSO_4 = \frac{6.026\ g}{233.39\ g/mol} = 0.0258\ mol\ \]

Relate Moles of Metal Sulfate to Barium Sulfate

From the balanced equation: 1 mole of \(\mathrm{M}_{2}(\mathrm{SO}_{4})_{3}\) results in 3 moles of \(\mathrm{BaSO}_{4}\), hence: \[ moles\ of\ \mathrm{M}_{2}(\mathrm{SO}_{4})_{3} = \frac{0.0258\ mol\ of\ BaSO_4}{3} = 0.0086\ mol \] This means 0.0086 moles of \(\mathrm{M}_{2}(\mathrm{SO}_{4})_{3}\) were originally present.

Calculate Molar Mass of Metal Sulfate

Since 1.200 g of the metal was converted into its sulfate: \[Molar\ mass\ of\ \mathrm{M_{2}(SO_{4})_{3}} = \frac{1.200\ g}{0.0086\ mol} = 139.53\ g/mol\ \]

Calculate Atomic Mass of the Metal

The formula \(\mathrm{M}_{2}(\mathrm{SO}_{4})_{3}\) contains:3 moles of \(\mathrm{SO}_{4}^{2-}\), which is 96.07 g/mol each. Hence, \[Molar\ mass\ of\ (SO_4)_3 = 3 \times 96.07 = 288.21\ g/mol\] Thus, the molar mass of 2 atoms of metal M in \(\mathrm{M}_{2}(\mathrm{SO}_{4})_{3}\) is: \[2M = 139.53 - 288.21 = 279.06\ g/mol\]Calculating for one atom of metal M:\[M = \frac{279.06}{2} = 139.53\ g/mol\] The atomic weight of \(\mathrm{M}\) closely resembles the atomic mass of Lanthanum (La), 138.91 g/mol.

Key concepts

Chemical Reactions

Chemical reactions are processes where substances, called reactants, transform into different substances known as products. In this case, the metal M is initially converted into its sulfate form, \(\mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}\). When this compound reacts with barium chloride \(\mathrm{BaCl}_{2}\), an exchange occurs forming metal chloride \(2 \mathrm{MCl}_{3}\) and barium sulfate \(3\mathrm{BaSO}_{4}\), the latter being a solid that can be filtered out.

This type of reaction is a double displacement reaction, where the anions and cations of the reactants trade places. Understanding this transformation helps to deduce how different compounds can be isolated and analyzed in laboratory environments.

Stoichiometry

Stoichiometry involves calculating the quantities of reactants and products in a chemical reaction. In this exercise, stoichiometry helps us relate the amount of metal sulfate initially present to the amount of barium sulfate produced.

By examining the balanced chemical equation, we can see that 1 mole of \(\mathrm{M}_{2}( \mathrm{SO}_{4})_{3}\) yields 3 moles of \(\mathrm{BaSO}_{4}\).

• This ratio (1:3) allows us to determine the quantity of the metal sulfate based on the barium sulfate formed.
• We start with 0.0258 moles of \(\mathrm{BaSO}_{4}\), therefore dividing by 3 gives 0.0086 moles of \(\mathrm{M}_{2}\left(\mathrm{SO}_{4}\right)_{3}\).

This logical flow shows how stoichiometry is critical for predicting quantities and confirming reaction outcomes.

Molar Mass Calculation

Calculating molar mass is an essential skill in chemistry, used to understand the mass of substances involved in reactions. Here, we need to find the molar mass of the metal sulfate \(\mathrm{M}_{2}(\mathrm{SO}_{4})_{3}\) since we know the mass of the reactant metal is \(1.200\ \mathrm{g}\).

Using the moles calculated (0.0086), the molar mass of the metal sulfate is found through the equation:\[ Molar\ mass = \frac{1.200\ \mathrm{g}}{0.0086\ \mathrm{mol}} = 139.53\ \mathrm{g/mol}\]To derive the atomic mass of the metal, we subtract from this the known mass of the sulfate groups:

• Each sulfate ion, \(\mathrm{SO}_{4}^{2-}\), is about 96.07 g/mol, and with three of them, it totals 288.21 g/mol.
• Subtracting this from the overall molar mass yields the molar mass for two metal atoms.

This meticulous calculation can pinpoint the atomic weight of the element involved.

Metal Identification

Identifying a metal in a chemical compound often relies on calculating its atomic weight, once the formula of the compound and its molar mass are known.

After determining that the molar mass for two atoms of metal is 279.06 g/mol, we divide this value by 2 to get the atomic weight of the individual metal atom, which is 139.53 g/mol.

• This closely matches the atomic weight of Lanthanum (La), which is 138.91 g/mol.

Metal identification combines theoretical computation with experimental data to match with known elements' properties, validating the methodology used in chemical analysis.