# What is the molecular formula for a "164.2 g" sample of a compound that contains "46.92 g K", "38.52 g S" and oxygen? The molecular weight of the compound is "270.4 g/mol".

Jun 15, 2015

${K}_{2}$${S}_{2}$${O}_{8}$

#### Explanation:

Let's start calculating the weight percentage of the elements in the compound:
potassium is 46.92/162.24 = 28.92 %
sulfur is 38.52/162.24=23.74 %
by difference oxygen is 100-28.92-23.74 = 47.34 %

Now divide the percentage mass by the atomic mass and obtain the atomic ratio
$K = \frac{28.92}{39.1} = 0.74$
$S = \frac{23.74}{32.1} = 0.74$
$O = \frac{47.34}{16} = 2.96$
The ratio among elemnts is $0.74 : 0.74 : 2.96$ i.e. $1 : 1 : 4$,
therefore the minimal molecular formula is $K$$S$${O}_{4}$,
with molecular mass of $135.2$
Molar mass is $270.4$, the double of molecular mass, and the minimal formula has to be doubled : ${K}_{2}$${S}_{2}$${O}_{8}$ (potassium peroxydisulphate)

Jun 15, 2015

The molecular formula for the compound is ${\text{K"_2"S"_2"O}}_{8}$.

#### Explanation:

1. Determine the empirical formula.
2. Determine the empirical formula weight.
3. Divide the molecular formula weight by the empirical formula weight.
4. Multiply the empirical formula weight times the result.

EMPIRICAL FORMULA
Determine the mass of each element.

$\text{K"=46.92"g}$
$\text{S"=38.52"g}$
$\text{O"=162.4-(46.92"g"+38.52"g")=76.80"g}$

Determine the molar mass of each element, which is its atomic weight in g/mol.

$\text{K"=39.0983"g/mol}$
$\text{S"=32.06"g/mol}$
$\text{O"=15.999"g/mol}$

Determine the moles of each element by multiplying the given masses by inverse of the molar mass: $\text{1 mol"/"mass}$ .

$\text{K": 46.92cancel"g K"xx("1 mol K")/(39.0983cancel"g K")="1.200 mol K}$

$\text{S": 38.52cancel"g S"xx("1 mol S")/(32.06 cancel"g S")="1.202 mol S}$

$\text{O": 76.80cancel"g O"xx("1 mol O")/(15.999 cancel"g O")="4.8 mol O}$

Determine the mole ratios for the empirical formula by dividing the number of moles of each element by the smallest number.

"K": ("1.200 mol")/("1.200 mol")=1.000

$\text{S": "1.202 mol"/"1.200 mol} = 1.002$

$\text{O": "4.800 mol"/"1.200 mol} = 4.000$

The empirical formula is $\text{K"_1"S"_1"O"_4"=KSO"_4}$ .

EMPIRICAL FORMULA WEIGHT
Multiply the subscripts times the molar mass for each element and add the results.

("1 x 39.0983 g/mol")+("1 x 32.06 g/mol")+("4 x 15.999g/mol")="135.15 g/mol"

MOLECULAR FORMULA
Divide the molecular weight $\text{270.4 g/mol}$ by the empirical formula weight $\text{135.15 g/mol}$.

$\text{270.4 g/mol"/"135.15 g/mol"="2.000}$

Multiply the empirical formula $\text{KSO"_4}$ by $2$ to get the molecular formula $\text{K"_2"S"_2"O"_8}$ .