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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Solution

Empirical formula:- a chemical formula showing the simplest ratio of elements in a compound rather than the total number of atoms in the molecule CH₂O is the empirical formula for glucose.
Potassium(k) containing = $47.9 %$
beryllium(Be) containing= $5.5 %$
flurine(F) containing= $46.6 %$
Element %age Atomic mass $\frac{% a g e}{A t o m i c m a s s}$ Simplest atomic ratio Simplest whole number ratio
Potassium(k) 47.9 39 $\frac{47.9}{39} = 1.22$ $\frac{1.22}{0.61} = 2$ 2
beryllium(Be) 5.5 9 $\frac{5.5}{9} = 0.61$ $\frac{0.61}{0.61} = 1$ 1
flurine(F) 46.6 19 $\frac{46.6}{19} = 2.45$ $\frac{2.45}{0.61} = 4$ 4
Empirical formula= $K_{2} B e_{1} F_{4} = K_{2} B e F_{4}$

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Similar questions

(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).

Q. 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]

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.

Determine the empirical formula of a compound containing 43.3% sodium, 11.3% Carbon, 45.28% Oxygen by mass

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Element	%age	Atomic mass	$\frac{% a g e}{A t o m i c m a s s}$	Simplest atomic ratio	Simplest whole number ratio
Potassium(k)	47.9	39	$\frac{47.9}{39} = 1.22$	$\frac{1.22}{0.61} = 2$	2
beryllium(Be)	5.5	9	$\frac{5.5}{9} = 0.61$	$\frac{0.61}{0.61} = 1$	1
flurine(F)	46.6	19	$\frac{46.6}{19} = 2.45$	$\frac{2.45}{0.61} = 4$	4

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)

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 = $39$ )