24,8% of potassium ,34,77% of Mn,40,5% of oxygen how do i find the emperical formular thanks in advance
From the periodic table, the atomic masses of each are:
K = 39
Mn = 55
O = 16
Divide the percentage to the respective atomic mass:
K: 24.8 / 39 = 0.636
Mn: 34.77 / 55 = 0.632
O: 40.5 / 16 = 2.531
Divide each by the smallest number. The smallest number here is 0.632:
K: 0.636 / 0.632 = 1.006
Mn: 0.632 / 0.632 = 1
O: 2.531 / 0.632 4.005
Approximate each number, so that it becomes the number of atoms of that particular element in the formula. Therefore, the formula is KMnO4.
Thank you
To find the empirical formula, we need to determine the simplest whole number ratio of different elements in a compound.
1. Start by converting the percentage composition of each element into grams.
Given percentages:
Potassium (K) = 24.8%
Manganese (Mn) = 34.77%
Oxygen (O) = 40.5%
Assuming we have 100 grams of the compound, we can calculate the masses of each element:
K = (24.8/100) * 100 g = 24.8 g
Mn = (34.77/100) * 100 g = 34.77 g
O = (40.5/100) * 100 g = 40.5 g
2. Convert the mass of each element into moles by dividing it by its molar mass.
The molar mass can be found by looking up the atomic masses of the elements from the periodic table:
K (potassium) = 39.1 g/mol
Mn (manganese) = 54.9 g/mol
O (oxygen) = 16.0 g/mol
Moles of each element:
moles of K = 24.8 g / 39.1 g/mol ≈ 0.63 mol
moles of Mn = 34.77 g / 54.9 g/mol ≈ 0.63 mol
moles of O = 40.5 g / 16.0 g/mol ≈ 2.53 mol
3. Determine the simplest whole number ratio of moles by dividing the number of moles of each element by the smallest number of moles obtained.
In this case, the smallest number of moles is 0.63 mol, so divide the moles of each element by 0.63:
moles of K = 0.63 mol / 0.63 mol ≈ 1
moles of Mn = 0.63 mol / 0.63 mol ≈ 1
moles of O = 2.53 mol / 0.63 mol ≈ 4
4. Finally, write the empirical formula using the ratio of whole numbers obtained.
The empirical formula for the compound is K1Mn1O4, which can be simplified as KMnO4.