Question #ba207

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Question #ba207

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Answer 1 · 2014-11-26T17:43:10

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.

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

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

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

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

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

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

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

The empirical formula is ${SF}_{5}$ $"SF"_5"$ .

The empirical formula is ${SF}_{5}$ $"SF"_5"$ .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 ( ${SF}_{5}$ $"SF"_5"$ )2

which is $S_{2}$ $S_2$ $F_{10}$ $F_10$

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