Question

What is the chemical formula of \[{C_4}{H_9}\]with molar mass of $114.26g/mol?$

Answer 1

Hint :The empirical formula of a compound gives the simplest ratio of the number of different atoms present, whereas the molecular formula gives the actual number of each different atom present in a molecule.

Complete Step By Step Answer:
$ - $ The empirical formula is the simplest whole-number ratio of atoms in the compound.
$ - $ A chemical formula that gives the total number of atoms of each element in each molecule of a substance.
$ - $The molecular formula is always a multiple of the empirical formula; and the multiple might be $1$.
$ - $The empirical formula mass of a compound refers to the sum of the atomic masses of the elements present in the empirical formula.
$ - $The molecular mass of a compound is a multiple of the empirical formula mass.
${M_r} = n \times empirical formula mass$
Where ${M_r} = $molecular mass
$n = $A whole number.
Alternatively, you can relate the molar mass of the molecular formula and the empirical formula in the same way:
Molar mass of the molecular formula $ = $ $n \times $ molar mass of empirical formula.
$ - $Empirical formula can be calculated from the percentage composition of a compound.
So, ${[empiricalformula]_n} = [molecularformula]$
Now, we have to find the multiple $n$ for the empirical formula ${C_4}{H_6}$.
$(4 \times 12.011 + 1.00794)gmo{l^{ - 1}} \times n$.
$ = 114.26gmo{l^{ - 1}}$
Here we use the atomic masses of carbon and hydrogen.
The multiple is$n = 2$
So, molecular formula
$ = {[EmpiricalFormula]_n} = 2 \times ({C_4}{H_9})$
$ = {C_8}{H_{18}}$.

Note :
When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition