# Question #ba207

Nov 26, 2014

The empirical formula represents the lowest whole-number ratio of elements in a compound.

Let us assume that we started with 100 g of compound. On analysis it gives 25.24 % or 25.24 g of Sulfur, and 74.76% or 74.76g of fluorine.

$\text{25.24% S}$ => $\text{25.24 g S}$
$\text{74.76% F}$ => $\text{74.76 g F}$

Convert the mass of each element to moles using the molar mass of each element. The molar mass of sulfur = $\text{32.065g/mol}$. The molar mass of fluorine = $\text{18.9984032g/mol}$.

$\text{25.24 g S}$ x $\text{1 mol S"/"32.065g S}$ = $\text{0.787 mol S}$

$\text{74.76 g F}$ x $\text{1 mol F"/"19 g F}$ = $\text{3.937 mol F}$

Divide the number of moles by the lowest number of moles to find the lowest whole-number ratio.

$\text{S}$ =>$\text{0.787mol"/"0.787mol}$ = $\text{1}$

$\text{F}$ => $\text{3.937mol"/"0.787mol}$ = $\text{5}$

The empirical formula is $\text{SF"_5}$.

The empirical formula is $\text{SF"_5}$ .The substance has empirical formula mass: One mole of S has mass 32.06 g /mol and 5 moles of F has mass 5x 19 g/mol = 95 g/mol

Empirical formula mass is 32.06 g /mol + 95 g /mol = 127.06 g/mol
Molecular formula mass is given as 254.1 g /mol.

Calculating n = Molecular formula mass / Empirical formula mass

n =254.1 g/mol / 127.06 g/mol

n = 2

Molecular formula is ($\text{SF"_5}$)2

which is ${S}_{2}$ ${F}_{10}$