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Question: 10 g of hydrofluoric acid gas occupies 5.6 liters of volume at STP. If the empirical formula of the ...

10 g of hydrofluoric acid gas occupies 5.6 liters of volume at STP. If the empirical formula of the gas is HF, then its molecular formula in the gaseous state will be:
A. HFHF
B. H2F2{H_2}{F_2}
C. H3F3{H_3}{F_3}
D. H4F4{H_4}{F_4}

Explanation

Solution

Empirical formula of a chemical compound is defined as the simplest positive integer ratio of atoms present in a compound. It does not provide any information about the arrangement or number of atoms. Also if we do the component analysis for any compound, we get the empirical formula.

Complete step by step answer:
As per the statement of the question, the empirical formula of the gas is given as HFHF. And by the definition of empirical formula, the empirical mass of HFHF is equal to the sum of the mass of hydrogen (HH) and mass of Fluorine (FF). The mass of the hydrogen atom is known to be 1g and the mass of fluorine is given as 19g. The empirical mass of HFHF is calculated as:
Empirical Mass = Mass of Hydrogen + Mass of Fluorine
1+19=20g\Rightarrow 1 + 19 = 20g
At standard temperature and pressure, we know the volume of one mole of any gas 22.4 litres. Weight by mass ratio can be provided as:
wM=5.622.4\dfrac{w}{M} = \dfrac{{5.6}}{{22.4}}
10M=14\Rightarrow \dfrac{{10}}{M} = \dfrac{1}{4}
M=40g\Rightarrow M = 40g
As we know the empirical mass as 20g, The valency of the element (nn)is calculated as:
n=Molecular MassEmpirical Massn = \dfrac{{Molecular{\text{ }}Mass}}{{Empirical{\text{ }}Mass}}
4020=2\Rightarrow \dfrac{{40}}{{20}} = 2
\therefore Molecular Formula = n ×\times Empirical Formula
2×HF=H2F2\Rightarrow 2 \times HF = {H_2}{F_2}

So, the correct answer is Option B.

Note: Empirical formula is the chemical formula that we obtain using composition analysis. The molecular formula of a compound represents the actual number of each element that occurs in the smallest existing unit. To determine the molecular formula of a compound, we need its composition and molecular weight of the compound. We can also use empirical formulas in the case where all data is not available.