Question

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?

Answer 1

@ganeshie8 @Rohangrr

Answer 2

@Vincent-Lyon.Fr

Answer 3

302 is thwe diameter

Answer 4

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

Answer 5

lol

Answer 6

Any Idea @Vincent-Lyon.Fr

Answer 7

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 \(C_4H_7\)

Answer 8

Thxxx...)

Answer 9

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 !

Answer 10

Ok....)