Question

12) The complete combustion of 0.645 g of a compo und that contains only carbon and hydrogen produces 1008 ml of CO2(g), measured at STP, and 0.945 g of H20. The 0.645 g gas sample occupies 351 ml at 50 c and 0.567 atm. What is the molecular formula of the compound? Setup:

Answer 1

Let the compound be of the molecular formula C_yH_x

So, the molecular mass of the compound m = molar mass of C*y + molar mass of H*x = 12y+x = m

Given that the combustion of C_yH_x is taking place so we have the following reaction -

C_yH_x + O2-> CO2 + H2O

It produces 1008 mL of CO2 at STP so, moles of CO2 produced = vol produced in L/ 22.4 (At STP all ideal gases have a molar volume of 22.4 L) = 1.008/22.4 = 0.045 moles

also, moles of CO2 = moles of C atoms in it = 0.045 moles

So, all the C atoms of CO2 must've come from C_yH_x (from the chemical equation)

moles of C atom in C_yH_x = moles of C_yH_x*y

moles of C_yH_x = 0.645/m (moles = weight/molar mass)

so, moles of C atom = (0.645/m)*y = 0.045 moles ----- Eqn1

Also, for C_yH_x at Pressure P = 0.567 atm, temperature T = 50+273 = 323 K, Volume V = 351 mL = 351/1000 = 0.351 L , so, by applying ideal gas equation PV=nRT where R = 0.0821 and n is moles of gas used

we know that n = weight/molar mass = 0.645g/m

so, we have 0.567*0.351 = (0.645/m)*0.0821*323 so, m = 0.645*0.0821*323/(0.567*0.351) = 86

So, molar mass of C_yH_x = 86 g/mol

12y + x = 86 --- Eqn2

From eqn1 we have (0.645/m)*y = 0.045 and now as we have m = 86 we get

(0.645/86)*y = 0.045 so, y = 6

and putting value of y in Eqn2 we get 12*6 + x = 86 so, x = 86 - 72 = 14

So, molecular formula of the compound C_yH_x is C₆H₁₄