Question

a certain quantity of a gasious hydrocarbon burned with a certain quantity of oxygen gass.the volume of all reactants is is 600ml.after the burning the volume of CO2 and H20 is found 700ml in the same position.what is the molecular formula of compound?

Answer 1

Answer on Question #68930, Physics / Other

A certain quantity of a gaseous hydrocarbon burned with a certain quantity of oxygen gass. The volume of all reactants is 600ml. After the burning the volume of CO2 and H2O is found 700ml in the same position. What is the molecular formula of compound?

Solution:

Hydrocarbons burns in excess oxygen based on the following equation:

$$C _ {x} H _ {y} (g) + \left(x + \frac {y}{4}\right) O _ {2} (g) \rightarrow x C O _ {2} (g) + \frac {y}{2} H _ {2} O (l)$$

Hence, if $1\mathrm{cm}^3$ of is completely burned in oxygen,

Volume of oxygen used $= (x + y / 4)\mathrm{cm}^3$

Volume of carbon dioxide produced $= x\mathrm{cm}^3$

Volume of water produced (as liquid) $= y / 2\mathrm{cm}^3$

Hence, we can directly take out the ratio of moles that reacted:

$$\left\{ \begin{array}{c} 1 + x + \frac {y}{4} = 6 \\ x + \frac {y}{2} = 7 \end{array} \right.$$ $$\left\{ \begin{array}{l} x + \frac {y}{4} = 5 \\ x + \frac {y}{2} = 7 \end{array} \right.$$

From first equation

$$x = 5 - \frac {y}{4}$$

Substituting to second equation

$$5 - \frac {y}{4} + \frac {y}{2} = 7$$ $$\frac {y}{4} = 2$$ $$y = 8$$ $$x = 5 - \frac {y}{4} = 5 - 2 = 3$$

Hence, the molecular formula of the hydrocarbon is $C_3H_8$.

Answer: $C_3H_8$

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