Question

Styrene has the empirical formula $ CH $ . It has a molar mass of $ 104{\text{ }}g $ . What is the multiplier to get the molecular formula?

Answer 1

Hint: Styrene is a derivative of benzene. It is a colorless oily liquid. The empirical formula and molar mass are provided. Find out the molecular formula by using the $ Molar{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}empirical{\text{ }}formula \times n = molar{\text{ }}mass{\text{ }}of{\text{ }}the{\text{ }}compound $ , where n is any whole number.

Complete answer:
Styrene is a synthetic chemical compound. Another name is vinyl benzene and ethenylbenzene. It is a derivative of benzene. It is a colorless oily liquid that evaporates very easily. It has a sweet smell.
Styrene has several uses such as it is an important material for the production of polystyrene, its products are found in multiple essential products we use every day. The plastic made of styrene is a thermoplastic polymer which has high formability.
In the question we are given the empirical formula of styrene that is $ CH $ and its molar mass that is $ 104g $ , we need to calculate the molecular formula. We know that the molecular formula is always a whole number multiple of the empirical formula.
Thus $ empirical{\text{ }}formula \times n = molecular{\text{ }}formula $
This can also be written as:
Molar mass of the empirical formula $ \times {\text{ }}n = $ molar mass of the compound
 $ \Rightarrow CH \times n = 104g $
 $ \Rightarrow (12 + 1) \times n = 104g $
 $ \Rightarrow 13 \times n = 104g $
 $ \Rightarrow n = \dfrac{{104g}}{{13}} $
 $ \Rightarrow n = 8 $
Hence the multiplier to get the molecular formula is $ 8 $ .
Thus the molecular formula $ = {(CH)_8} $
 $ = {C_8}{H_8} $
Therefore the molecular formula of styrene is $ {C_8}{H_8} $ .

Note:
The empirical formula is the simplest whole-number combination in that compound and the molecular formula is the actual whole-number ratio between the elements in that compound. These two topics are an important part of the mole concept in chemistry.