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uzer
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when K is burnt in excess O2, a compound is produced that contains 54.9% K.

Calculate the percentage of oxygen present & hence calculate the empirical formula of this compound
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Nirgilis
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Moved to chemistry help
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Dominic1905
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(Original post by uzer)
when K is burnt in excess O2, a compound is produced that contains 54.9% K.

Calculate the percentage of oxygen present & hence calculate the empirical formula of this compound

The key information in this question is that the K is being burnt in an excess of O2; this is important because it means that O2 isn't a limiting reagent, there's more than enough of it around to react fully with the K, so we can assume that the rest of the compound is oxygen.

Therfore % Oxygen = 100% - K%

100%-54.9%=45.1%

Now we know the percentage composition of each component of the compound we can divide the relevant percentage by the molar mass of that element and work out the empirical formula:

K
54.9%/39=1.4

O
45.1%/16=2.8

Dividing by the lowest number give the ratio in its simplest form:

1.4/1.4=1 therefore we have 1 K atom present

2.8/1.4=2 therefore we have 2 O atoms present

This means the empirical formula is KO2
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