Q84E - Chemistry (OpenStax)

Question

The principal component of mothballs is naphthalene, a compound with a molecular mass of about 130 amu, containing only carbon and hydrogen. A 3.000-mg sample of naphthalene burns to give 10.3 mg of CO2. Determine its empirical and molecular formulas.

Step-by-Step Solution

Answer

Empirical formula is C₅H₄.

 

Molecular Formula is C₁₀H₈.

Data given

Assume the combustion reaction as follows

 

     C _xH _y   +   O₂     \( \to \)           CO₂   +  H₂O

  3.00 mg                       10.3 mg

 

 

3.00 mg   =  0.003 g of C _xH _y   

10.3 mg   =   0.0103 g of CO2 

Determine the required formula

0.0103 g of CO₂ x 1 mol CO₂   x   1 mol C   x 12 g of C

                             44 g of CO₂      1 mol CO₂   1 mol C

                      =  0.00281 g C

C _xH _y      =  0.003 g

Then as we know carbon amount, we can calculate H by subtraction.

H   =    0.003 - 0.00281 =    1.9 x 10^-4 g of H

We will use the formula

          Grams   \( \to \)    mols    \( \to \)    mol  ratio     \( \to \)   empirical formula   \( \to \)       molecular formula

 

0.00281 g of C x   1 mol C =    2.3397   x 10 ^-4  mol C

                             12 g C

1.9 x 10^-4 g of H  x   1 mol H   =    1.8849  x 10 ^-4 mol H

                               1 g H

Mole ratio 

1.25 mol C has to be whole number . So multiply it to get whole number. 

2.3397  x 10 ^-4  mol C  =    1.25 mol C  x  4   =  5 mol C

1.8849  x 10 ^-4 mol

 

1.8849  x 10 ^-4 mol H =  1 mol H  x 4  = 4 mol H

1..8849   x 10 ^-4 mol

So empirical formula is C₅H₄.

Molecular mass/ Empirical mass =   130/ 64.082 = 2 

Molecular formula =  C₁₀ H₈