# Question be093

Aug 26, 2017

Empirical Formula => ${K}_{2} S {O}_{4}$

#### Explanation:

Given Compound with the following composition by weight
Potassium (K) = 7.8 grams
Sulfur (S) = 3.2 grams
Oxygen (O) = 6.4 grams
Wt(K) + Wt(S) + Wt(O) = 7.8g + 3.2g + 6.4g = 17.4g
-Convert composition to Percent per 100-wt.
%K = (7.8/17.4)100% = 44.83% w/w

%S = (3.2/17.4)100% = 17.98% w/w

%O = (6.4/17.4)100% = 37.19%# w/w

From % per 100 wt. values the following sequence is effective. Only if fractions of 0.25, 0.75 or 0.50 follow redution of mole ratios does one multiply by 4 if 0.25 or 0.75, and 2 if 0.50 in order to obtain the small whole number ratio as empirical ratio.

Calculation sequence is therefore ...
%/100 wt. => grams/100 wt. => moles => mole ratio => reduce mole ratio => Empirical Ratio.

If the molecular weight is given, then the molecular formula may be determined from the following expression...

Molecular Weight = Empirical Formula Weight x Whole No. Multiple(n) => ${\left(A B C\right)}_{n}$ => Molecular Formula => ${A}_{n} {B}_{n} {C}_{n}$

For the posted problem ...
%K = 44.83% => 44.83g => (44.83g)/(39g/mol) = 1.1495 mole K
%S = 17.98% => 17.98g => (17.98g)/(32g/mol) = 0.5619 mole S
%O = 13.19% => 37.19g => (37.19g)/(16g/mol) = 2.3244 mole O

Set up mole ratio and reduce by dividing each mole value by the smallest mole value of the ratio ...

$m o l \left(K\right) : m o l \left(S\right) : m o l \left(O\right)$ => $\left(\frac{1.1495}{0.5619}\right)$:$\left(\frac{0.5619}{0.5619}\right)$:$\left(\frac{2.3244}{0.5619}\right)$

=> K:S:O Empirical Ratio => 2:1:4 => Empirical Formula => ${K}_{2} S {O}_{4}$