Question

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%

Answer 1

The way you calculate the empirical formula is to firstly assume 100g. To find each elements moles you take each elements percentage listed, times it by one mole and divide it by its atomic mass. (ex: moles of K =55.3g x 1 mole/39.1g, therefore there is 1.41432225 moles of Potassium) Once you’ve completed this for every element you list each elements symbol beside it’s number of moles and divide by the smallest number because it can only go into its self once. After you’ve done this, you’ve found your empirical formula, which is the simplest whole number ratio of atoms in a compound. I’ve added an example of a empirical question I completed last semester :)

Answer 2

Answer:

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

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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