Question

Question #ba207

Answer 1

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.

`"25.24% S"` => `"25.24 g S"`
`"74.76% F"` => `"74.76 g F"`

Convert the mass of each element to moles using the molar mass of each element. The molar mass of sulfur = `"32.065g/mol"`. The molar mass of fluorine = `"18.9984032g/mol"`.

`"25.24 g S"` x `"1 mol S"/"32.065g S"` = `"0.787 mol S"`

`"74.76 g F"` x `"1 mol F"/"19 g F"` = `"3.937 mol F"`

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

`"S"` =>`"0.787mol"/"0.787mol"` = `"1"`

`"F"` => `"3.937mol"/"0.787mol"` = `"5"`

The empirical formula is `"SF"_5"`.

The empirical formula is `"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 (`"SF"_5"`)2

which is `S_2` `F_10`