Respuesta :
1) You will need to use the atomics masses of the elements. So list them:
K: 39 g/mol
C: 12 g/mol
O: 16 g/mol
2) Your base is 100 grams, so you have:
K: 52.1 g
C: 15.8 g
O: 32.1 g
3) Convert grams to moles (dividing each element by its atomic mass)
K: 52.10g / 39 g/mol = 1.336 mol
C: 15.8 g / 12 g/mol = 1.317 mol
O: 32.1 g / 16 g/mol = 2.006 mol
4) Divide all the mol numbers by the least mol number:
K: 1.336 / 1.317 = 1.01
C: 1.317 / 1.317 = 1.00
O: 2.006 / 1.316 = 1.52
5) To have only whole numbers multiply by 2 and round to the nearest whole number
K = 2
C = 2
O = 3
So the empirical formula is K2C2O3
6) Find the molar mass of the empirical formula:
2*39 + 2*12 + 3*16 = 150.
7) Find the number of times that the molecular formula containd the mass of the empirical formula:
150 / 150.22 = 1
Then the molecular formula is the same empirical formula.
Answer: K2C2O3.
K: 39 g/mol
C: 12 g/mol
O: 16 g/mol
2) Your base is 100 grams, so you have:
K: 52.1 g
C: 15.8 g
O: 32.1 g
3) Convert grams to moles (dividing each element by its atomic mass)
K: 52.10g / 39 g/mol = 1.336 mol
C: 15.8 g / 12 g/mol = 1.317 mol
O: 32.1 g / 16 g/mol = 2.006 mol
4) Divide all the mol numbers by the least mol number:
K: 1.336 / 1.317 = 1.01
C: 1.317 / 1.317 = 1.00
O: 2.006 / 1.316 = 1.52
5) To have only whole numbers multiply by 2 and round to the nearest whole number
K = 2
C = 2
O = 3
So the empirical formula is K2C2O3
6) Find the molar mass of the empirical formula:
2*39 + 2*12 + 3*16 = 150.
7) Find the number of times that the molecular formula containd the mass of the empirical formula:
150 / 150.22 = 1
Then the molecular formula is the same empirical formula.
Answer: K2C2O3.