Question

Question #2612f

Answer 1

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.