Problem 147 - General Chemistry

Question

A metal, \(\mathrm{M}\), 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?

Step-by-Step Solution

Answer

The metal is gallium with an atomic weight of approximately 69.77 g/mol.

Step 1: Determine Moles of Barium Sulfate

First, we need to calculate the moles of barium sulfate (\( \text{BaSO}_4 \)). The molar mass of \( \text{BaSO}_4 \) is calculated by adding the atomic masses of its elements:\[ \text{Molar mass of BaSO}_4 = 137.33 \, \text{(Ba)} + 32.07 \, \text{(S)} + 4 \times 16.00 \, \text{(O)} = 233.39 \, \text{g/mol} \]Now, calculate the moles of \( \text{BaSO}_4 \) using the given mass:\[ \text{Moles of BaSO}_4 = \frac{6.026 \, \text{g}}{233.39 \, \text{g/mol}} = 0.0258 \, \text{mol} \]

Step 2: Relate Moles to Moles of Metal Sulfate

From the balanced chemical equation, we see that 3 moles of \( \text{BaSO}_4 \) are produced for every mole of \( \text{M}_2(\text{SO}_4)_3 \). Thus, the moles of \( \text{M}_2(\text{SO}_4)_3 \) can be calculated:\[ \text{Moles of } \text{M}_2(\text{SO}_4)_3 = \frac{0.0258 \, \text{mol}}{3} = 0.0086 \, \text{mol} \]

Step 3: Calculate Moles of Metal

In \( \text{M}_2(\text{SO}_4)_3 \), there are 2 moles of metal M per mole of sulfate. Therefore, multiply the moles of sulfate by 2 to find the moles of the metal M:\[ \text{Moles of M} = 2 \times 0.0086 \, \text{mol} = 0.0172 \, \text{mol} \]

Step 4: Determine the Atomic Weight of Metal

The atomic weight of metal M can now be calculated by dividing the mass of the metal by the moles calculated:\[ \text{Atomic weight of M} = \frac{1.200 \, \text{g}}{0.0172 \, \text{mol}} = 69.77 \, \text{g/mol} \]

Step 5: Identify the Metal

Compare the calculated atomic weight to known atomic weights. The atomic weight closest to 69.77 g/mol corresponds to gallium (Ga), which has an atomic weight of approximately 69.72 g/mol.

Key Concepts

Chemical Reactions

A chemical reaction involves the transformation of substances through the breaking and forming of chemical bonds. In this exercise, a metal \( \mathrm{M} \) undergoes a reaction to form a sulfate compound \( \mathrm{M}_2(\mathrm{SO}_4)_3 \). This compound is further reacted with barium chloride \( \mathrm{BaCl}_2 \) to produce barium sulfate \( \mathrm{BaSO}_4 \) as a precipitation. These types of reactions are often used to discern the characteristics and components of a compound.

The equation \[ \text{M}_2(\text{SO}_4)_3 + 3\text{BaCl}_2 \longrightarrow 2\text{MCl}_3 + 3\text{BaSO}_4 \] illustrates a double displacement reaction where the compounds swap parts, forming new compounds. It's critical to balance such equations to ensure that the same number of each type of atom appears on both sides of the equation, adhering to the law of conservation of mass.

Molar Mass

Molar mass is a central concept in stoichiometry, referring to the mass of one mole of a chemical substance. It is calculated by summing the atomic masses of all atoms present in a molecule. For instance, in barium sulfate \( \mathrm{BaSO}_4 \), the molar mass is determined by adding the atomic masses of barium, sulfur, and oxygen: \( 137.33 \, \text{g/mol (Ba)} + 32.07 \, \text{g/mol (S)} + 4 \times 16.00 \, \text{g/mol (O)} = 233.39 \, \text{g/mol} \).

Knowing molar mass allows us to convert between grams and moles, a critical step in translating laboratory measurements into meaningful chemical quantities. In the example provided, given a certain mass of barium sulfate, we use its molar mass to find moles, bridging the gap between mass and chemical equations.

Barium Sulfate Precipitation

Precipitation is a process where a solid forms from a solution during a chemical reaction. In the given exercise, barium sulfate \( \mathrm{BaSO}_4 \) is the precipitate formed. The reaction between metal sulfate and barium chloride yields this white, insoluble solid.

Precipitation reactions are useful in analytical chemistry for determining the presence and quantity of particular ions in a solution. Here, the quantity of barium sulfate precipitated allows us to indirectly calculate the amount of the original metal, aiding in determining the metal's identity. The stoichiometric relation between the reactants and products guides the calculation of the precipitated compound, which is essentially "counting molecules" through their mass relationships.

Atomic Weight Calculation

Calculating atomic weight involves determining how much one atom of an element weighs, typically expressed as grams per mole. In this problem, after obtaining the mass of metal \( \mathrm{M} \) and calculating the moles through the stoichiometry of the reaction, we find the atomic weight. It's given by dividing the mass by the number of moles resulting in \( \text{Atomic weight of M} = \frac{1.200 \, \text{g}}{0.0172 \, \text{mol}} = 69.77 \, \text{g/mol} \).

This calculated weight is compared against known atomic weights from a periodic table, identifying the element as gallium, based on its proximity to the calculated weight. This method of determination is common in stoichiometric evaluations and emphasizes understanding molecular conversions from compound masses.