Question

Cyclohexene, a hydrocarbon, has a molar mass of 82.1 g/mole. if the combustion of 0.8300 g cyclohexene produces 0.9102 g h2o and 2.668 g co2, what is the molecular formula of this compound?

Answer 1

Answer is:  the molecular formula is C₆H₁₀.

Chemical reaction: C₆Hₓ + yO₂ → 6CO₂ + x/2H₂O.

m(H₂O) = 0.9102; mass of the water.

n(H₂O) = m(H₂O) ÷ m(H₂O)  

n(H₂O) = 0,9102 g ÷ 18 g/mol

n(H₂O) = 0,0505 mol.

From chemical reaction: n(H) = 2 · n(H₂O).

n(H) = 0.101 mol.

m(CO₂) = 2.668 g; mass of carbon dioxide.

n(CO₂) = 2.668 g ÷ 44 g/mol.

n(CO₂) = 0.0606 mol; amount of the substance.

n(CO₂) = n(C) = 0.0606 mol.

n(C) : n(H) = 0.606 mol : 0.101 mol.

n(C) : n(H) = 6 : 10; proportion of carbon and hydrogen atoms in cyclohexene.

Answer 2

Answer:

• The molecular formula will be C6H10.

Explanation;

Given Data:

• mass of compound =   0.8300 g
• mass of H2O          =   0.9102 g
• mass of CO2          =   2.668 g

Solution:

% of C = (mass of CO2/mass of compound)× (12/44)×100

% of C = (2.66÷0.8300)×(12/44)× 100 = 87.40 %

% of H = (mass of H2O/mass of compound)× (2/18)×100

% of H =  (0.910÷0.8300)× (2/18)×100 = 12.18 %

Niiw we calculate moles of C and H from % as follow

moles of C = 87.40/12 = 7.2 mol

moles of H = 12.18/1 = 12.18 mol

In we take atomic ratio

                 C     :     H

         12.18/7.2 :   7.2/7.2

           1.69       :     1

multiply atomic ratio by 3 to get whole no.

  C : H = 3 (1.69:1) = 5 : 3

So empirical formula is C3H5.

Calculations of molecular formula:

empirical formula mass of C3H5 = 41

  n = molar mass of compound /empirical formula mass

 n = 82.1/41   = 2

Molecular mass =  n × empirical formula = 2 × C3H5 = C6H10