The empirical formula of a gaseous fluorocarbon is cf2. at a certain temperature and pressure, a 1-l volume holds 8.93 g of this fluorocarbon, whereas under the same conditions, the 1-l volume holds only 1.70 g gaseous fluorine (f2). determine the molecular formula of this compound.

C₄F₈

Explanation:

Using their mole ratio to compute their mass

molar mass of carbon = 12.0107 g/mol

molar mass of fluorine gas = 37.99681

let x = mass of carbon

given mass of fluorine = 1.70 g

x / 12.01067 = 1.70 / 37.99687

cross multiply

x = ( 1.70 × 12) / 37.99687 = 20.4 / 37.99687 = 0.53688 g

mass of one mole of CF₂ = 0.53688 + 1.70 = 2.23688 g

number of mole of CF₂ = 8.93 g / 2.23688 = 3.992 approx 4

molecular formula of CF₂ = 4 (CF₂) = C₄F₈

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :

b is the correct answer

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