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A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it in oxygen gives 3.38 g carbon dioxide,0.690 g of water and no other products. A volume of 10.0 L ( measured at STP) of this weldinggas is found to weight 11.6g. Calculate : (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

Answer» <html><body><p> <br/> <br/> </p>Solution :`44g CO_(2) = 12 g` carbon <br/> `3.38 g CO_(2) = (?)` <br/> `=(12)/(44) xx 3.38g = 0.9218` g carbon<br/> `18 g H_(2)O = 2g H_(2)` <br/> `0.69 g H_(2)O = (2)/(18) xx 0.69` <br/> `= 0.0767` g <a href="https://interviewquestions.tuteehub.com/tag/hydrogen-22331" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROGEN">HYDROGEN</a> <br/> Total mass of <a href="https://interviewquestions.tuteehub.com/tag/compound-926863" style="font-weight:bold;" target="_blank" title="Click to know more about COMPOUND">COMPOUND</a> `= 0.9218+0.0767` <br/> `= 0.9985g` <br/> Masspercent of `C = (0.9218)/(0.9985) xx 100 = 92.32%` <br/> Mass <a href="https://interviewquestions.tuteehub.com/tag/percent-1150333" style="font-weight:bold;" target="_blank" title="Click to know more about PERCENT">PERCENT</a> of `H=(0.0767)/(0.9985) xx 100 = 7.68%` <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/KPK_AIO_CHE_XI_P1_C01_E01_078_S01.png" width="80%"/> <br/> Empirical formula `= <a href="https://interviewquestions.tuteehub.com/tag/ch-913588" style="font-weight:bold;" target="_blank" title="Click to know more about CH">CH</a>` <br/> mass of 10 L gas at STP `= 11.6g` <br/> mass of 22.4 L at STP `=(11.6xx22.4)/(10) = 25.984` <br/> Molar mass `=25.984 g/mol ~=<a href="https://interviewquestions.tuteehub.com/tag/26-298265" style="font-weight:bold;" target="_blank" title="Click to know more about 26">26</a> g/mol` <br/> Empirical formula mass `= 12 +1= 13` <br/> `:.n = ("Molar mass")/("Empirical formula mass")` <br/> `=(26)/(13)=2` <br/> Molecular formula `=n xx` Empirical formula <br/> `=2xxCH=C_(2)H_(2)`</body></html>


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