JEE Challenger
More from Some Basic Concepts of Chemistry
ChemMedAdv 2022 • Paper 2

Empirical Formula and Composition Analysis of Binary Compounds

To check the principle of multiple proportions, a series of pure binary compounds represented as PmQn\text{P}_m\text{Q}_n were analyzed, with their mass percentages tabulated below. Select the correct option(s):

CompoundWeight % of PWeight % of Q15050244.455.634060\begin{array}{|c|c|c|} \hline \text{Compound} & \text{Weight \% of P} & \text{Weight \% of Q} \\ \hline \mathbf{1} & 50 & 50 \\ \hline \mathbf{2} & 44.4 & 55.6 \\ \hline \mathbf{3} & 40 & 60 \\ \hline \end{array}

Options

A

If compound 3 has an empirical formula of P3Q4\text{P}_3\text{Q}_4, then compound 2 possesses an empirical formula of P3Q5\text{P}_3\text{Q}_5.

B

Assuming the empirical formula of compound 3 is P3Q2\text{P}_3\text{Q}_2 and element P has an atomic mass of 2020, then Q's atomic mass is 4545.

Correct
C

If compound 2 has an empirical formula given by PQ\text{PQ}, then the empirical formula of compound 1 becomes P5Q4\text{P}_5\text{Q}_4.

Correct
D

When the atomic weights of P and Q are 7070 and 3535 respectively, compound 1 has an empirical formula of P2Q\text{P}_2\text{Q}.

Topics & Concepts

Some Basic Concepts of ChemistryMole ConceptStoichiometry

Step-by-Step Solution

An analysis of the given mass compositions for the three binary compounds PmQn\text{P}_m\text{Q}_n provides the following mass ratios:

  • Compound 1: Mass ratio Mass of PMass of Q=5050=1\text{Mass ratio } \frac{\text{Mass of P}}{\text{Mass of Q}} = \frac{50}{50} = 1

  • Compound 2: Mass ratio Mass of PMass of Q=44.455.6=45\text{Mass ratio } \frac{\text{Mass of P}}{\text{Mass of Q}} = \frac{44.4}{55.6} = \frac{4}{5}

  • Compound 3: Mass ratio Mass of PMass of Q=4060=23\text{Mass ratio } \frac{\text{Mass of P}}{\text{Mass of Q}} = \frac{40}{60} = \frac{2}{3}

Let MPM_P and MQM_Q be the atomic masses of elements P\text{P} and Q\text{Q}, respectively.

Evaluation of Options:

Option (A):

If the empirical formula of Compound 3 is P3Q4\text{P}_3\text{Q}_4: 3MP4MQ=23  ⟹  MPMQ=89\frac{3 M_P}{4 M_Q} = \frac{2}{3} \implies \frac{M_P}{M_Q} = \frac{8}{9}

For Compound 2: nPnQ=Mass of P/MPMass of Q/MQ=(44.455.6)×MQMP=45×98=910\frac{n_P}{n_Q} = \frac{\text{Mass of P} / M_P}{\text{Mass of Q} / M_Q} = \left(\frac{44.4}{55.6}\right) \times \frac{M_Q}{M_P} = \frac{4}{5} \times \frac{9}{8} = \frac{9}{10}

Thus, the empirical formula of Compound 2 should be P9Q10\text{P}_9\text{Q}_{10}, not P3Q5\text{P}_3\text{Q}_5.
Option (A) is incorrect.

Option (B):

If the empirical formula of Compound 3 is P3Q2\text{P}_3\text{Q}_2 and MP=20M_P = 20: 3MP2MQ=23\frac{3 M_P}{2 M_Q} = \frac{2}{3}

Substitute MP=20M_P = 20: 3×202MQ=23  ⟹  602MQ=23  ⟹  4MQ=180  ⟹  MQ=45\frac{3 \times 20}{2 M_Q} = \frac{2}{3} \implies \frac{60}{2 M_Q} = \frac{2}{3} \implies 4 M_Q = 180 \implies M_Q = 45

Option (B) is correct.

Option (C):

If the empirical formula of Compound 2 is PQ\text{PQ}: 1⋅MP1⋅MQ=44.455.6=45  ⟹  MPMQ=45\frac{1 \cdot M_P}{1 \cdot M_Q} = \frac{44.4}{55.6} = \frac{4}{5} \implies \frac{M_P}{M_Q} = \frac{4}{5}

For Compound 1: nPnQ=(5050)×MQMP=1×54=54\frac{n_P}{n_Q} = \left(\frac{50}{50}\right) \times \frac{M_Q}{M_P} = 1 \times \frac{5}{4} = \frac{5}{4}

Thus, the empirical formula of Compound 1 is P5Q4\text{P}_5\text{Q}_4.
Option (C) is correct.

Option (D):

Given MP=70M_P = 70 and MQ=35M_Q = 35, for Compound 1: Moles of P=5070=57\text{Moles of P} = \frac{50}{70} = \frac{5}{7} Moles of Q=5035=107\text{Moles of Q} = \frac{50}{35} = \frac{10}{7}

The mole ratio is: nP:nQ=57:107=1:2n_P : n_Q = \frac{5}{7} : \frac{10}{7} = 1 : 2

Thus, the empirical formula of Compound 1 is PQ2\text{PQ}_2, not P2Q\text{P}_2\text{Q}.
Option (D) is incorrect.

Conclusion:

The correct options are (B) and (C).

PreviousNext
Empirical Formula and Composition Analysis of Binary Compounds | Chemistry PYQ Solution - JEE Challenger