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Question: In a chemical determination of the atomic weight of vanadium , \( 2.8934g \) of pure \( VOC{l_3} \) ...

In a chemical determination of the atomic weight of vanadium , 2.8934g2.8934g of pure VOCl3VOC{l_3} was allowed to undergo a set of reactions as a result of which all the chlorine contained in this compound was converted to AgClAgCl which weighed 7.1801g.7.1801g. Calculate the atomic weight of vanadium. (Ag=107.868,Cl=35.453)  \left( {Ag = 107.868,Cl = 35.453} \right)\;

Explanation

Solution

Hint : To calculate the atomic weight of vanadium, we need to consider the number of moles of chlorine in each compound. It is said that pure VOCl3,VOC{l_3}, which is vanadium oxytrichloride, undergoes several chemical reactions before its full chlorine content is converted to AgCl,AgCl, which is silver chloride. The number of moles of chlorine in vanadium oxytrichloride is the same as the number of moles of chlorine in silver chloride. We can use the equation:
Number of moles of a substance = Mass of the substance in the compoundMolar mass of the substance{\text{Number of moles of a substance = }}\dfrac{{{\text{Mass of the substance in the compound}}}}{{{\text{Molar mass of the substance}}}} .

Complete Step By Step Answer:
First, let us calculate the number of moles of chlorine in VOCl3VOC{l_3} and AgCl.AgCl. We know that:
Number of moles of chlorine in VOCl3=Number of moles of chlorine in AgCl{\text{Number of moles of chlorine in }}VOC{l_3} = {\text{Number of moles of chlorine in }}AgCl
3 moles of chlorine in VOCl3=1 mole of chlorine in AgCl\Rightarrow 3{\text{ moles of chlorine in }}VOC{l_3} = 1{\text{ mole of chlorine in }}AgCl
In VOCl3,VOC{l_3}, 3 moles of chlorine in VOCl3=3×2.8934x+16+(3×35.453)3{\text{ moles of chlorine in }}VOC{l_3} = \dfrac{{3 \times 2.8934}}{{x + 16 + \left( {3 \times 35.453} \right)}}
=8.6802x+122.359(1)= \dfrac{{8.6802}}{{x + 122.359}} \to (1)
Where 'x' is the atomic weight of vanadium. The atomic weight of oxygen is 1616 . Unit has been ignored as the final expression is in moles.
In AgCl,AgCl, 1 mole of chlorine in AgCl=1×7.1801107.868+35.4531{\text{ mole of chlorine in }}AgCl = \dfrac{{1 \times 7.1801}}{{107.868 + 35.453}}
=7.1801143.321= \dfrac{{7.1801}}{{143.321}}
=0.05009moles(2)= 0.05009\,moles \to (2)
Equating (1) and (2),{\text{(}}1{\text{) and (}}2{\text{),}} we get:
8.6802x+122.359=0.05009\dfrac{{8.6802}}{{x + 122.359}} = 0.05009
Rearranging,
x=(8.68020.05009)122.359x = \left( {\dfrac{{8.6802}}{{0.05009}}} \right) - 122.359
x=173.292122.359x = 173.292 - 122.359
Hence,
x=50.933gx = 50.933g
The atomic weight of vanadium is 50.933g.50.933g.

Note :
The principle used in this question is POAC or the Principle of atom conservation. In terms of moles, this principle states that the number of moles of an element in a compound are conserved throughout the reaction undergone by the compound. According to the question, the principle has been applied to chlorine, as the number of moles of chlorine in vanadium oxytrichloride has been conserved as the compound undergoes change into silver chloride.