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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?

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@Vincent-Lyon.Fr
302 is thwe diameter

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Other answers:

Lol...@Rohangrr Wat Do u Mean..)
lol
Any Idea @Vincent-Lyon.Fr
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 \(C_4H_7\)
Thxxx...)
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 !
Ok....)

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