Answer to Question #256719 in Organic Chemistry for Ore

Question

20cm³ of a gaseous hydrocarbon were mixed with 80cm³ of oxygen in excess and exploded. the final gaseous mixture contained 40cm³ of carbon iv oxide and 30cm' of unreacted oxygen. all gaseous volume were measured under the same conditions. determine the formula of the hydrocarbon

Answer 1

The general unbalanced equation is:

?C_xH_y + ?O₂ --> ?CO₂ + ?H₂O

The volume of oxygen that has been involved in the reaction is 80 - 30 = 50 cm³.

According to the Avogadro's law, under the same conditions volumes and number of molecules (moles) of gases are directly proportional regardless of the nature of the gas. Therefore, knowing that 20 cm³ of C_xH_y and 50 cm³ of O₂ reacted to produce 40 cm³ of CO₂ and water (still unknown amount), the stoichiometric coefficients before each species should be proportional to their volumes:

2C_xH_y + 5O₂ --> 4CO₂ + ?H₂O

At this step, the coefficient before water can also be determined by balancing oxygen atoms: it is known that there are 10 oxygens on the left side, so the coefficient "2" is required for water to balance oxygens:

2C_xH_y + 5O₂ --> 4CO₂ + 2H₂O

Now the overall equation is balanced, so the subscripts x and y should give 4 carbon and 4 hydrogen atoms in total (the same as on the product side). Obviously, both x and y = 2, and the formula of a hydrocarbon is C₂H₂.

Answer: C₂H₂