Question

What is the molar mass of diacidic organic Lewis base \((B)\), if \(12 \mathrm{~g}\) of chloroplatinate salt \(\left(B \mathrm{H}_{2} \mathrm{PtCl}_{6}\right)\) on ignition produced \(5 \mathrm{gm}\) residue of Pt? (a) 52 (b) 58 (c) 88 (d) None of these

Short answer

58 g/mol

Step-by-step solution

Understand the problem

The problem involves determining the molar mass of a diacidic organic Lewis base (B). The mass of the chloroplatinate salt is given, along with the mass of platinum that remains after ignition. The chloroplatinate salt is of the form \(B \mathrm{H}_{2} \mathrm{PtCl}_{6}\), and upon ignition, it leaves a residue of platinum \(Pt\).

Calculate the molar mass of the residue (Pt)

Using the periodic table, look up the atomic mass of platinum (Pt) to find that the molar mass of platinum is approximately 195.08 g/mol.

Determine the grams of Pt from the given mass of residue

The problem states that the residue after ignition is 5 g of \(Pt\). This means that there were 5 grams of \(Pt\) in the original 12 g of chloroplatinate salt.

Compute the moles of Pt

To find the moles of Pt in the residue, use the molar mass of Pt: \[ \text{moles Pt} = \frac{\text{mass of Pt residue}}{\text{molar mass of Pt}} = \frac{5 \text{ g}}{195.08 \text{ g/mol}} \]

Relate moles of Pt to moles of \(B \mathrm{H}_{2} \mathrm{PtCl}_{6}\)

In the compound \(B \mathrm{H}_{2} \mathrm{PtCl}_{6}\), there is one mole of Pt for every mole of the compound. Therefore, the moles of Pt are equal to the moles of \(B \mathrm{H}_{2} \mathrm{PtCl}_{6}\).

Calculate the molar mass of \(B \mathrm{H}_{2} \mathrm{PtCl}_{6}\)

The molar mass of the compound \(B \mathrm{H}_{2} \mathrm{PtCl}_{6}\) is the mass of the original sample (12 grams) divided by the number of moles calculated in the previous step.

Determine the molar mass of \(B\)

To find the molar mass of \(B\), subtract the molar mass of \(\mathrm{H}_{2} \mathrm{PtCl}_{6}\) from the molar mass of \(B \mathrm{H}_{2} \mathrm{PtCl}_{6}\) using their known atomic weights. \(\text{Molar mass of} \mathrm{H}_{2} \mathrm{PtCl}_{6} = 2\times1.008 + 195.08 + 6\times35.45\). Finally, subtract this from the molar mass of the entire compound to find the molar mass of \(B\).

Key concepts

Diacidic Organic Lewis Bases

A diacidic organic Lewis base is a type of compound that can donate two pairs of electrons to an acid, often forming complexes with transition metals. In general, Lewis bases are species that have at least one lone pair of electrons which can be shared with a Lewis acid to form a coordinate covalent bond.

In the context of chloroplatinate salts, the organic Lewis base acts as a ligand that provides the electron pairs needed to bond with the metal center, platinum in this instance. Understanding the behavior of diacidic organic Lewis bases is essential for solving problems related to complexation and salt formation which is often encountered in coordination chemistry.

Chloroplatinate Salts

Chloroplatinate salts, such as the one described in the given exercise \(B \mathrm{H}_{2} \mathrm{PtCl}_{6}\), consist of a central platinum ion surrounded by chloride ions and possibly other ligands, such as the diacidic organic Lewis base mentioned. The platinum in chloroplatinate salts will typically be in the +2 or +4 oxidation state and these salts are formed through reactions between platinum compounds and chlorides.

When heated, or 'ignited', chloroplatinate salts undergo decomposition, where the chloride and any organic ligands are removed, leaving behind a residue of metallic platinum. This reaction is pertinent to the field of analytical chemistry for determining the composition and purity of chloroplatinate complexes.

Stoichiometry

Stoichiometry is the branch of chemistry that deals with the quantitative relationships of the reactants and products in a chemical reaction. It enables chemists to predict the amounts of substances consumed and produced in a given reaction.

In our problem, we are utilizing stoichiometry to deduce the molar mass of an organic Lewis base from the molar mass of its chloroplatinate salt and the mass of platinum residue after ignition. The steps include determining the molar mass of platinum (from the periodic table), calculating the number of moles of platinum in the residue, and then relating these moles to the original chloroplatinate salt to eventually find the molar mass of the organic Lewis base. Through stoichiometry, the quantitative relationship between the reactant (chloroplatinate salt) and the residue (platinum) is established, illustrating the practical importance of stoichiometry in solving real-world chemical problems.