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A dibasic organic acid gave the following results: C = 34.62%, H = 3.84%, 0.1075g of this acid consumes 20mL of 0.1N NaOH for complete neutralisation. Find out the molecular formula of the acid.

Answer» <html><body><p></p>Solution :Calculation of empirical formula: <br/> `{:("Element","Percentage","At.mass","Relative number of atoms","Simplest ratio of atoms"),("Carbon",34.62,<a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a>,(34.62)/(12)=2.88,(2.88)/(2.88)=1xx3=3),("Hydrogen",3.84,1,(3.84)/(1)=3.84,(3.84)/(2.88)=1.33xx3 =4),("Oxygen",overset(61.54)("by difference"),16,(61.54)/(16)=3.84,(3.84)/(2.88)=1.33 xx 3 =4):}` <br/> Empirical formula of the acid `= C_(3)H_(4)O_(4)` <br/> Empirical formula mass `= (3 xx 12) +(4 xx1) +(4 xx 16) = <a href="https://interviewquestions.tuteehub.com/tag/104-266430" style="font-weight:bold;" target="_blank" title="Click to know more about 104">104</a>`<br/> Calculation of molecular mass: <br/> `20mL 0.1N NaOH -= 0.1075g` acid <br/> `20 xx 0.1 <a href="https://interviewquestions.tuteehub.com/tag/ml-548251" style="font-weight:bold;" target="_blank" title="Click to know more about ML">ML</a> <a href="https://interviewquestions.tuteehub.com/tag/n-568463" style="font-weight:bold;" target="_blank" title="Click to know more about N">N</a> NaOH -= 0.1075g` acid <br/> So, `1000 mL 1N NaOH -= (0.1075)/(20 xx 0.1) xx 1000g` acid <br/> `-= 53.75g` acid <br/> Eq. mass of the acid `= 53.75` <br/> <a href="https://interviewquestions.tuteehub.com/tag/mol-1100196" style="font-weight:bold;" target="_blank" title="Click to know more about MOL">MOL</a>. mass of the acid = Eq. mass `xx` Basicity <br/> `= 53.75 xx 2 = 107.50` <br/> `n = ("Mol. mass")/("Emp. mass") = (107.50)/(104.0) ~~1` <br/> Molecular formula `= C_(3)H_(4)O_(4)`</body></html>


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