An unknown compound has a percent composition of 52.10% potassium, 15.8% carbon, and 32.1% oxygen. the molar mass of the compound is 150.22 g/mol. what is the empirical formula and the molecular formula of this compound?

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.

K2C2O3.

The correct answer should be "electrons orbit the nucleus in circular motions exactly as shown on the classic bohr model."

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