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A compound contains 4.07% hydrogen, 24.27% carbon and 71.65% chlorine. Its molar mass is 98.96g. What are its empirical and molecular formulas ?

Answer» <html><body><p> <br/> <br/> </p>Solution :Step-1 : Conversion of mass per cent to grams : <br/> Since we are having mass percent. It is convenient to use 100 g of the compound as the starting material. Thus, in the 100 g sample of the above compound, 4.07g hydrogen is present, 24.27 g carbon is present and 71.65 g chlorine is present. <br/> Step-2 : Convert into number moles of each <a href="https://interviewquestions.tuteehub.com/tag/element-969236" style="font-weight:bold;" target="_blank" title="Click to know more about ELEMENT">ELEMENT</a> : <br/> Divide the <a href="https://interviewquestions.tuteehub.com/tag/masses-1088822" style="font-weight:bold;" target="_blank" title="Click to know more about MASSES">MASSES</a> obtained above by respective atomic masses of various elements . <br/> Moles of Hydrogen `=(4.07g)/(1.008g) = 4.04` <br/> Moles of Carbon `=(24.27g)/(12.01g) = 2.021` <br/> Moles of Chlorine `=(71.65 g)/(35.453g)=2.021` <br/> Step-3 : Divide the mole value obtained above by the smallest number : <br/> Since 2.021 is the smallest value, division by it gives a ratio of `2:1:1` for `H:C:Cl` <br/> In case the ratios are not whole number, then, they may be converted into whole number by multiplying by the suitable coessicient. <br/> Step-4 : <a href="https://interviewquestions.tuteehub.com/tag/write-746491" style="font-weight:bold;" target="_blank" title="Click to know more about WRITE">WRITE</a> empirical <a href="https://interviewquestions.tuteehub.com/tag/formula-464310" style="font-weight:bold;" target="_blank" title="Click to know more about FORMULA">FORMULA</a> by mentioning the numbers after writing the symbols of respective elements : <br/> `CH_(2)Cl` is the empirical formula of the above compound. <br/> Step-5 : Writing molecular formula : <br/> (a) Determine empirical formula mass <br/> Add the atomic masses of various <a href="https://interviewquestions.tuteehub.com/tag/atoms-887421" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMS">ATOMS</a> present in the empirical formula . <br/> For, `CH_(2)Cl` empirical formula mass is<br/> `12.01 + 2 xx 1.008 + 35.453=49.48g` <br/> (b) Divide molar mass by empirical formula mass : `("Molar mass")/("Empirical formula mass")=(98.96g)/(49.48g) =2 =(n)` <br/> (c) Multiply empirical formula by n obtained above to get the molecular formula. <br/> Empirical formula `=CH_(2)Cl, n=2` <br/> Hence molecular formula is `=C_(2)H_(4)Cl_(2)`</body></html>


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