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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 litre (Measured at STP) of this welding gas is found weigh `11.6 g`. Calculate (i) empirical formula, (ii) molar mass of the gas, and (iii) molecular formula. |
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Answer» Amount of carbon in `3.38 g CO_(2)` `=12/44xx3.38 g=0.9218 g` Amount of hydrogen in `0.690 g H_(2)O` `=2/18xx0.690 g=0.0767 g` As the compound contains only `C` and `H`, therefore, total mass of the compound `=0.9218+0.0767 g=0.9985 g` `%` of `C` in the compound`=0.9218/0.9985xx100=92.32` `%` of `H` in the compound`=0.0767/0.9985xx100=7.68` `|{:("Element",% "by mass","Atomic mass","Moles of the element","Simplest molar ratio","Simplest whole number molar rato"),(C,92.32,12,92.32/12=7.69,1,1),(H,7.68,1,7.68/1=7.68 1,1,):}|` `:.` Empirical formula `=CH` `10.0 L` of the gas at `STP` weigh`=11.6 g` `:. 22.4 L` of the gas at `STP` will weigh `=11.6/10.0xx22.4` `=25.984 g ~~26 g` :. Molar mass `=26 g mol^(-1)` |
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