## Yahoo! Group Title Complete combustion of 5.90 g of a hydrocarbon produced 18.8 g of CO2 and 6.75 g of H2O. What is the empirical formula for the hydrocarbon? one year ago one year ago

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@ganeshie8 @Rohangrr

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@Vincent-Lyon.Fr

3. Rohangrr Group Title

302 is thwe diameter

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Lol...@Rohangrr Wat Do u Mean..)

5. Rohangrr Group Title

lol

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Any Idea @Vincent-Lyon.Fr

7. Vincent-Lyon.Fr Group Title

It is very easy, and you do not even need the information about the 5.9 g of hydrocarbon. It is just a matter of proportionality. Look how you solve this kind of problem. 1. Write general chemical equation: $$\large C_nH_m+xO_2\rightarrow nCO_2+\frac m2 H_2O$$ 2. Work out proportionality of masses using molar masses 44 g/mol for carbon dioxide and 18 g/mol for water: Mass of $$CO_2$$ produced is 44n Mass of $$H_2O$$ produced is 18m/2 = 9m Hence: $$\Large \frac{44n}{9m} = \frac{18.8}{6.75}$$ => $$\Large \frac mn = \frac 74$$ 3. After rearranging for a simple fraction 7/4 : Empirical formula is $$C_4H_7$$

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Thxxx...)

9. Vincent-Lyon.Fr Group Title

Note: if mass of oxygen used is mentioned, you might have to write the stoichiometric coefficient for oxygen as n + m/4 I was just lazy because is was not useful in your example !

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Ok....)