1.

A crystalline compound when heated become anhydrous by losing 51.2 % of the mass. On analysis, the anhydrous compound gave the following percentage composition : Mg = 20.0 % , S = 26.66 % and O = 53.33 %. Calculate the molecular formula of the anhydrous compound and crystalline compound. The molecular mass of anhydride compound is 120 u.

Answer» <html><body><p></p>Solution :Step I. <a href="https://interviewquestions.tuteehub.com/tag/determination-949893" style="font-weight:bold;" target="_blank" title="Click to know more about DETERMINATION">DETERMINATION</a> of empirical formula of the anhydrous compound <br/> `{:("Element","Percentage","Atomic mass","Gram atoms (Moles)","Atomic ratio (Molar ratio)","Simplest whole no. ratio"),("<a href="https://interviewquestions.tuteehub.com/tag/mg-1095425" style="font-weight:bold;" target="_blank" title="Click to know more about MG">MG</a>",20.0,24,(24.0)/(24)=0.83,(0.83)/(0.83)=1,1),("S",26.66,32,(26.66)/(32)=0.83,(0.83)/(0.83)=1,1),("O",53.33,16,(53.33)/(16)=3.33,(3.33)/(0.83)=4.01,4):}` <br/> `:.` The empirical formula of anhydrous compound is `MgSO_(4)` <br/> Step II. Determination of the molecular formula of the anhydrous compound <br/> Empirical formula mass `= 24+32 +4 <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> 16 = 24 + 32 + 64 = <a href="https://interviewquestions.tuteehub.com/tag/120-270396" style="font-weight:bold;" target="_blank" title="Click to know more about 120">120</a>` u <br/> Molecular mass = 120 u (Given) , `n=("Molecular mass")/("Empirical formula mass")=(120)/(120)=1` <br/> `:.` Molecular formula `=nxx` Empirical formula `=1 xx MgSO_(4)=MgSO_(4)` <br/> `:.` Molecular formula of anhydrous compound `= MgSO_(4)` <br/> Step III. Determination of molecular formula of the crystalline compound <br/> Let us first calculate the number of molecules of water of crystallisation in the compound. For that, <br/> Let the weight of hydrated compound = 100 u <br/> Weight of water lost = 51.2 u <br/> `:.` Weight of anhydrous compound `= 100-51.2 = 48.9` u <br/> Now, if the weight of anhydrous compound is 48.8 u, then that of water 51.2 u <br/> If the weight of anhydrous compound is 120 u, then that of water `= (51.2)/(48.8)xx120 = 125.9` u <br/> Molecular mass of water `(H_(2)O)=2xx1+16 = 18` u <br/> `:.` No. of `H_(2)O` molecules `= ("Weight of water")/("Molecular mass of water")=(125.9)/(18)=7` (approx) <br/> Thus, the formula of crystalline compound `= MgSO_(4).7H_(2)O`.</body></html>


Discussion

No Comment Found

Related InterviewSolutions