Question #2612f

1 Answer
anor277
Jul 31, 2017

I think you have CH_2.....

Explanation:

2.65*g of a hydrocarbon, C_xH_y was burnt in air to give 8.33*g CO_2 and 3.41*g water.........

"Moles of carbon dioxide"=(8.33*g)/(44.01*g*mol^-1)=0.189*mol

"Moles of water"=(3.41*g)/(18.01*g*mol^-1)=0.189*mol, i.e. 0.189*molxx2H

The empirical formula is thus CH_2......

We need a molecular weight determination in order to assess the molecular formula.

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