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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 welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula.

Answer» <html><body><p></p>Solution :Amount of carbon in 3.38 g `CO_(<a href="https://interviewquestions.tuteehub.com/tag/2-283658" style="font-weight:bold;" target="_blank" title="Click to know more about 2">2</a>) = 12/44 xx 3.38 = 0.9218 g` <br/> Amount of <a href="https://interviewquestions.tuteehub.com/tag/hydrogen-22331" style="font-weight:bold;" target="_blank" title="Click to know more about HYDROGEN">HYDROGEN</a> in 0.690 g `H_(2)O =2/18 xx 0.690 = 0.0767 g` <br/> As the fuel gas contains only carbon and hydrogen, total mass of the gas burnt =` 0.9218 + 0.0767 = 0.9985` g <br/> `therefore` of C in the fuel gas = `(0.9218)/0.9985 xx 100 = 92.318` <br/> % of H in the fuel gas = `0.0767/0.9985 xx 100 = 7.682`<br/> (i) Calculation of empirical formula: <br/> <img src="https://doubtnut-static.s.llnwi.net/static/physics_images/NTN_HCS_ISC_CHE_XI_P1_C01_E10_034_S01.png" width="80%"/> <br/> `therefore` Empirical formula = CH <br/> (ii) Calculation of molar mass: <br/> `therefore 10.0 <a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>` of the gas at <a href="https://interviewquestions.tuteehub.com/tag/stp-633132" style="font-weight:bold;" target="_blank" title="Click to know more about STP">STP</a> weigh = 11.6 g<br/> `therefore 22.4 L` of the gas at STP will weigh `=11.6/10.0 xx 22.4 = 25.984 g = 26 g` <br/> Molar mass of the gas `=26 g "mol"^(-1)` <br/> (<a href="https://interviewquestions.tuteehub.com/tag/iii-497983" style="font-weight:bold;" target="_blank" title="Click to know more about III">III</a>) Calculation of molecular formula : <br/> `n=("Molar mass")/("Empirical formula mass") = 26/(12 + 1) = 26/13 = 2` <br/> `therefore` Molecular formula of the gas `=2 xx (CH) = C_(2)H_(2)`</body></html>


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