A gaseous compound which contains only C, H and S is burnt with oxygen under such conditions that individual volumes of the reactant and product can be measured at the same temperature and pressure. It is found that 3 volumes of the compound react with oxygen to yield 3 volumes of CO_(2), 3 volumes of SO_(2) and 6 volumes of water vapour. What volume of oxygen is required for teh combustion? What is the formula of the compound ? Is this an empirical formula or molecular formula ?

A gaseous compound which contains only C, H and S is burnt with oxygen

1.	A gaseous compound which contains only C, H and S is burnt with oxygen under such conditions that individual volumes of the reactant and product can be measured at the same temperature and pressure. It is found that 3 volumes of the compound react with oxygen to yield 3 volumes of CO_(2), 3 volumes of SO_(2) and 6 volumes of water vapour. What volume of oxygen is required for teh combustion? What is the formula of the compound ? Is this an empirical formula or molecular formula ?
Answer» Solution :`{:("COMPOUND",+,O_(2),rarr,CO_(2),+,SO_(2),+,H_(2)O),(3 "vol",,v"vol(say)",,3 "vol",,3"vol",,6"vol"),(or " 3 moles",,v"moles",,3"moles",,3"moles",,6"moles"):}` Applying POAC for O atoms, moles of O in `O_(2)` =moles of O in `CO_(2) +` moles of O in `SO_(2)`+ moles of O in `H_(2)O 2xx` moles of `O_(2)` `=2 xx` moles of `CO_(2)+2 xx` moles of `SO_(2) +1 xx` moles of `H_(2)O` `2v=2 xx 3+2 xx3 +1 xx 6= 18, v= 9` `therefore` volume of `O_(2)` required for combustion is 9. Again, moles of C in `CO_(2)=1 xx` moles of `CO_(2)=3` moles of S in `SO_(2)=1XX` moles of `SO_(2)=3` moles of H in `H_(2)O=2 xx` moles of `H_(2)O=12` `therefore` formula is `C_(3)S_(3)H_(12) or CSH_(4)` Now we see 3 moles of the compound contain 3 moles of C atoms in `C_(3)S_(3)H_(12)`. Hence i mole of the compound should contain 1 mole of C atoms and therefore, the molecular formula is `CSH_(4)`.