A. Calculate the empirical formula of a molecule with percent compositions: 55.3% potassium (K), 14.6% phosphorus (P), and 30.1% oxygen (O). Complete the table below to help with the calculations. (5 points)

Element
Mass Percent
Mass of 100g of the compound
Number of Moles in 100 G

K
55.3%

P
14.6%

O
30.1%

empirical formula => K₃PO₄

Explanation:

For empirical formula problems, if not given in percent per 100 weight convert given data to percent per 100 weight and then follow the following sequence:

% => grams/100wt => moles => mole ratio => reduce mole ratio to smallest whole no. ratio => empirical ratio

K => 55.3% => 55.3 grams/100 wt => 55.3 g/ 39.0 g·mol⁻¹ = 1.418 mol

P => 14.6% => 14.6 grams/100 wt => 14.6 g/ 30.9 g·mol⁻¹ = 0.4725 mol

O => 30.1% => 30.1 grams/100 wt => 30.1 g/16.0 g·mol⁻¹ = 1.8813 mol

K : P : O => 1.418 : 0.4725 : 1.8813

Mole ratio => K : P : O => 1.418 : 0.4725 : 1.8813  

Reduce mole ratio by dividing all values by smallest mole value ...  

K : P : O => 1.418/0.4725 : 0.4725/0.4725 : 1.8813/0.4725 => 3 : 1 : 4  

Multiply reduced ratio by whole number multiple that will give a set of whole number values => empirical ratio. For this problem, whole no. multiple = 1 => empirical ratio => 3 : 1 : 4

empirical formula => K₃PO₄

moles per 100wt.

formula weight = 3K + 1P + 4O = [3(39) + 1(30.9) + 4(16)] g/mol = 211.9 g/mol

moles per 100wt = 100 g / 211.9 g·mol⁻¹ = 0.472 mol. K₃PO₄

false

because it does not have the potential to be entirely used up