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Question: An experiment showed that in a lead chloride solution,6.21 g of lead combined with 4.26 g of chlorin...

An experiment showed that in a lead chloride solution,6.21 g of lead combined with 4.26 g of chlorine. The empirical formula of this chloride is:
(a) PbCl2PbC{{l}_{2}}
(b) Pb2Cl2P{{b}_{2}}C{{l}_{2}}
(c) PbClPbCl
(d) PbCl4PbC{{l}_{4}}

Explanation

Solution

In order to find out the empirical formula, first we have to find out the moles of the atoms using their molar masses and then by that you can easily predict the molecular formula of that very compound.

Complete step by step solution:
First of all, we should know what an empirical formula is. By the empirical formula we mean the simple whole number ratio of each atom in a comp[found. It can be calculated as; suppose that you have the 100g of the compound and the percentage of each atom is considered to be taken as; x g of the atom in 100g of the compound.
Now, the next step is to convert the given masses into the moles using the molar mass of the given atoms and then there, molar ratio gives the empirical formula of the compound.
Now, let’s calculate the empirical formula of the given compound according to above given points:
6.21g of Pb means 6.21 g of Pb in 100g of the lead chloride. if;
207 g of lead atom contains = 1 mole of Pb
1 g of the carbon atom contains=1207  \dfrac{1}{\begin{aligned} & 207 \\\ & \\\ \end{aligned}} mole of Pb
Similarly,
6.21 g of the carbon atom contains=1207  ×6.21\dfrac{1}{\begin{aligned} & 207 \\\ & \\\ \end{aligned}}\times 6.21 mole of Pb
=0.03= 0.03 molesmoles ofof PbPb
Thus, for Cl atom,
4.26g of Cl means 4.26 g of the Cl in 100g of the lead chloride. if;
35.5 g of hydrogen atom contains = 1 mole of Cl
1 g of Cl atom contains=135.5\dfrac{1}{35.5} mole of Cl
4.26 g of the Cl atom contains= 135.5×4.26\dfrac{1}{35.5}\times 4.26mole of Cl
=0.12= 0.12 molesmoles ofof ClCl
Thus, the mole ratio of lead Pb and chloride Cl is as;
    0.03:0.12\implies 0.03:0.12
    1:4\implies 1:4
So, the empirical formula=PbCl4PbC{{l}_{4}}

Hence, the empirical formula of this chloride is =PbCl4PbC{{l}_{4}} which is option (D).

Note: Don’t get confused in the molecular and the empirical formulas . they both are different terms. Molecular formula shows the number of each type of atom which is present in a molecule whereas the empirical formula is the simplest whole ratio of the atoms in the compound.