Question

If a molecule has an empirical formula of C2H2O and a molecular mass of 84.0 g/mol, what is the molecular formula?

Answer 1

Answer:

$C_2 H_2 O \times 2=C_4 H_4 O_2$ is the molecular formula

Explanation:

Empirical formula is the simplest form of a Molecular formula.

For example  

Molecular formula of glucose is $C_6 H_{12} O_6$

Its empirical formula is $C_1 H_2 O_1$

The subscript numbers of the molecular formula is divided by a common number n=6 here.

Another example:

Molecular formula of Octane is $C_8 H_{18}$ and Its empirical formula will be $C_4 H_9$  

Subscripts divided by n=2.

To find n we make use of the formula  

$\frac {(molecular mass )}{(empirical formula mass)}=n$

$= \frac {84.0g}{(C_2H_2O mass)}\\\\=\frac {84.0g}{(24+2+16)g}=\frac {84.0g}{42g}=2$

So, we see n=2 here

Empirical formula × n = molecular formula

$C_2 H_2 O \times 2=C_4 H_4 O_2$ is the molecular formula

(Answer)

Please note:

Molar mass is the mass of 1 mole of the substance and its unit is g/mol .

n is the number of moles by which the empirical formula is multiplied to get the molecular formula of the compound.

Answer 2

Answer:

=C₄H₄O₂

Explanation:

Given the empirical formula of a molecule, the he the quotient of the molecular mas and and the empirical mass=constant.

84.0 g/mol/mass of(C₂H₂O)=constant

=84/(12×2+1×2×16)

=84/42

=2

Therefore, the molecular formula is (C₂H₂O)₂=C₄H₄O₂