Determine the empirical formula of a compound containing 47.9% potassium, 5.5% beryllium, and 46.6% fluorine by mass. (Atomic weight of Be=9;F=19; K=39) Work to one decimal place.

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Solution

Determination of Empirical Formula:
The empirical formula can be determined from the percentage composition of the compound as follows:
1. The percentage weight of oxygen, carbon, or hydrogen (necessary elements in every organic compound) is sometimes not given. Hence, it is first determined by subtracting the sum of other elemental percentages from 100.
2. The percentage weight of each element is divided by its atomic weight. This gives the ratio of the number of atoms in a molecule of the compound.
3. To get a whole number ratio of atoms of different elements and water molecules, the figures obtained in step 2 are divided by the smallest number.
4. The empirical formula is now derived by writing the symbols of various elements side by side with the number of atoms of each one as the subscript to the lower right of its symbol.
Element Percentage weight Atomic mass Relative no. of moles Simple ratio of atoms Whole number ratio
K 47.9 39 47.9/39 = 1.2 1.2/0.6=2 2
Be 5.5 9 5.5/9 = 0.6 0.6/0.6=1 1
F 46.6 19 46.6/19 = 2.4 2.4/0.6=4 4
Result:
Hence, the empirical formula of the compound is K₂BeF₄.

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Determine the empirical formula of a compound containing $47.9 %$ potassium, $5.5 %$ beryllium and $46.6 %$ fluorine by mass.
(Atomic weight of Be = $9$ ; F = $19$ ; K = $39$ )

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Element	Percentage weight	Atomic mass	Relative no. of moles	Simple ratio of atoms	Whole number ratio
K	47.9	39	47.9/39 = 1.2	1.2/0.6=2	2
Be	5.5	9	5.5/9 = 0.6	0.6/0.6=1	1
F	46.6	19	46.6/19 = 2.4	2.4/0.6=4	4