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A compound on analysis gave the following percentage composition: Na=14.31% S = 9.97%, H = 6.22%, O = 69.5%, calcualte the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation. Molecular mass of the compound is 322 [Na = 23, S = 32, H = 1, 0 = 16].

Answer» <html><body><p></p>Solution :<img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/FM_CHE_XI_V01_C01_E01_043_S01.png" width="80%"/> <br/>All H combines with 10 oxygen <a href="https://interviewquestions.tuteehub.com/tag/atoms-887421" style="font-weight:bold;" target="_blank" title="Click to know more about ATOMS">ATOMS</a> to form as `10H_(2)O` <br/>So the empirical formula is `Na_(2)SO_(4).10H_(2)O` <br/>Empirical formula mass = `(23xx2) + (<a href="https://interviewquestions.tuteehub.com/tag/32-307188" style="font-weight:bold;" target="_blank" title="Click to know more about 32">32</a> <a href="https://interviewquestions.tuteehub.com/tag/xx-747671" style="font-weight:bold;" target="_blank" title="Click to know more about XX">XX</a> <a href="https://interviewquestions.tuteehub.com/tag/1-256655" style="font-weight:bold;" target="_blank" title="Click to know more about 1">1</a>) + (16xx4) + (10xx 18)` <br/>= 46 + 32 + <a href="https://interviewquestions.tuteehub.com/tag/64-330663" style="font-weight:bold;" target="_blank" title="Click to know more about 64">64</a> + 180 = 322 <br/>`n=("Molecular mass")/("Emperical formula mass")=322/322=1` <br/>Molecular formula=`Na_(2)SO_(4).10H_(2)O`</body></html>


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