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# Does anyone know how to do this?

a 2.314 g sample of a particular compound was found to contain 0.8735g of potassium, 0.7931 g of chlorine with the balance being oxygen. calculate the simpliest formula of the compound

### 2 Answers

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• (0.8735 g K) / (39.0983 g K/mol) = 0.0223411 mol K

(0.7931 g Cl) / (35.4532 g Cl/mol) = 0.0223703 mol Cl

((2.314 g total) - (0.8735 g K) - (0.7931 g Cl)) / (15.99943 g O/mol) =

0.0404639 mol O

Divide by the smallest number of moles:

(0.0223411 mol K) / 0.0223411 mol = 1.000

(0.0223703 mol Cl) / 0.0223411 mol = 1.0013

(0.0404639 mol O) / 0.0223411 mol = 1.811

If you round off now, you get an empirical formula of KClO2.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

If you try to get closer to integers by first multiplying by 5, you get K5Cl5O9, which is unreasonable in inorganic chemistry.

But the moles of oxygen in the last calculation line above is fishy.  It should be much closer to an integer than 1.811 .

If you started with a sample mass of 2.3814 g (instead of "2.314 g") and do the calculations again as above, you get very close to 2 for the coefficient for O:

KClO2.

So it looks to me like the given sample mass got an "8" deleted somehow from its value.

• Calculate the moles of each element

moles K = 0.8735/39 = 0.0224

moles Cl = 0.7931/35.4 = 0.0224

g O =2.314 - 0.8735 - 0.7931 = 0.6474

moles O = 0.6474/16 = 0.0405

So the ration of K:Cl:O is approximately 1:1:2

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