Question #2612f

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Question #2612f

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Answer 1 · 2017-07-31T17:11:12

I think you have $C H_{2}$ $CH_2$ .....

Explanation:

$2.65 \cdot g$ $2.65*g$ of a hydrocarbon, $C_{x} H_{y}$ $C_xH_y$ was burnt in air to give $8.33 \cdot g$ $8.33*g$ $C O_{2}$ $CO_2$ and $3.41 \cdot g$ $3.41*g$ water.........

$Moles of carbon dioxide = \frac{8.33 \cdot g}{44.01 \cdot g \cdot m o l^{- 1}} = 0.189 \cdot m o l$ $"Moles of carbon dioxide"=(8.33*g)/(44.01*g*mol^-1)=0.189*mol$

$Moles of water = \frac{3.41 \cdot g}{18.01 \cdot g \cdot m o l^{- 1}} = 0.189 \cdot m o l$ $"Moles of water"=(3.41*g)/(18.01*g*mol^-1)=0.189*mol$ , i.e. $0.189 \cdot m o l \times 2 H$ $0.189*molxx2H$

The empirical formula is thus $C H_{2}$ $CH_2$ ......

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

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Question #2612f

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