A13.00 g sample of a compound contains 4.15 g potassium (k), 3.76 g chlorine (cl), and oxygen (o). calculate the empirical formula.

To solve this problem, let us all convert the mass of each element into number of moles using the formula:

moles = mass / molar mass

Where,

molar mass K = 39.10 g / mol

molar mass Cl = 35.45 g / mol

molar mass O = 16 g / mol

and mass O = 13 g – 4.15 g – 3.76 g  = 5.09 g

 

moles K = 4.15 g / (39.10 g / mol) = 0.106 mol

moles Cl = 3.76 g / (35.45 g / mol) = 0.106 mol

moles O = 5.09 g / (16 g / mol) = 0.318 mol

 

The ratio becomes:

0.106 K: 0.106 Cl: 0.318 O

We divide all numbers with the smallest number, in this case 0.106. This becomes:

K: Cl: 3O

 

Therefore the empirical formula is:

Bthe s the

The activity series is shown in the attached picture. if the element is below the other, then you can displace the other. following the activity series as a guide, the reactions are as follows: nickel and aluminum chloride3 ni + 2 alcl₃  → 3 nicl₂ + 2 albarium and copper bromideba + cubr  →  no reaction because cu is less oxidizable than bapotassium nitrate and aluminum3 kno₃ + al  → al(no₃)₃ + 3 ksodium phosphate and zincnaso₄ + zn  → znso₄ + na