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Standard IX

Chemistry

Empirical & Molecular Formula

The empirical...

Question

The empirical formula of a compound containing 47.9 % potassium, 5.5 % beryllium and 46.6 % fluorine by mass is
[At. weight of Be = 9; F= 19; K=39]

K_{2} B e F_{2}

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K B e F_{4}

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K_{2} B e_{2} F_{4}

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K_{2} B e F_{4}

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Solution

The correct option is C $K_{2} B e F_{4}$
Element Percentage composition Atomic weight Relative No. of atoms Simplest ratio
Potassium 47.9 39 47.9/39 = 1.2 1.2/0.6 = 2
Beryllium 5.5 9 5.5/9 = 0.6 0.6/0.6 = 1
Fluorine 46.6 19 46.6/19 = 2.4 2.4/0.6 = 4
Hence, empirical formula is $K_{2} B e F_{4}$ .

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Similar questions

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

(a) 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.
(b) Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120 g of copper oxide? The equation for the reaction is:
3CuO + 2NH₃ $\overset{}{\to}$ 3Cu + 3H₂O + N₂
(Volume occupied by 1 mole of gas at STP is 22.4 litres).

(i) 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).
(ii) Given that the relative molecular mass of copper oxide is 80, what volume of ammonia (measured at STP) is required to completely reduce 120 g of copper oxide?
The equation for the reaction is:
3CuO + 2NH₃ $\overset{}{\to}$ 3Cu + 3H₂O + N₂
(Volume occupied by 1 mole of gas at STP is 22.4 litres).

Water is added to

3.52

g of

U F_{6}

. The products formed are

3.08

g of a solid (containing only

U, O

and

F

) and only

0.8

g of gas. The gas (containing fluorine and hydrogen only) contains

95

percent by mass fluorine. Assume that the empirical formula is the same as the molecular formula. (

U = 238, F = 19

The empirical formula of the gas is:

Q. The empirical formula weight of a compound containing carbon and hydrogen is

13

. The molecule of the compound is

39

times heavier than a molecule of hydrogen. The molecular formula of the compound is:

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Element	Percentage composition	Atomic weight	Relative No. of atoms	Simplest ratio
Potassium	47.9	39	47.9/39 = 1.2	1.2/0.6 = 2
Beryllium	5.5	9	5.5/9 = 0.6	0.6/0.6 = 1
Fluorine	46.6	19	46.6/19 = 2.4	2.4/0.6 = 4

The empirical formula of a compound containing 47.9 % potassium, 5.5 % beryllium and 46.6 % fluorine by mass is [At. weight of Be = 9; F= 19; K=39]

The correct option is C K2BeF4Element Percentage composition Atomic weight Relative No. of atoms Simplest ratioPotassium 47.9 39 47.9/39 = 1.21.2/0.6 = 2Beryllium 5.5 9 5.5/9 = 0.60.6/0.6 = 1Fluorine 46.6 19 46.6/19 = 2.42.4/0.6 = 4Hence, empirical formula is K2BeF4.

The empirical formula of a compound containing 47.9 % potassium, 5.5 % beryllium and 46.6 % fluorine by mass is
[At. weight of Be = 9; F= 19; K=39]

The correct option is C $K_{2} B e F_{4}$
Element Percentage composition Atomic weight Relative No. of atoms Simplest ratio
Potassium 47.9 39 47.9/39 = 1.2 1.2/0.6 = 2
Beryllium 5.5 9 5.5/9 = 0.6 0.6/0.6 = 1
Fluorine 46.6 19 46.6/19 = 2.4 2.4/0.6 = 4
Hence, empirical formula is $K_{2} B e F_{4}$ .