Answer is: an oxybromate compound is KBrO₄ (x = 4).
ω(Br) = 43.66% ÷ 100%.
ω(Br) = 0.4366; mass percentage of bromine.
If we take 100 grams of compound:
m(Br) = ω(Br) · 100 g.
m(Br) = 0.4366 · 100 g.
m(Br) = 43.66 g; mass of bromine.
n(Br) = m(Br) ÷ M(Br).
n(Br) = 43.66 g ÷ 79.9 g/mol,
n(Br) = 0.55 mol; amoun of bromine.
From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).
m(K) = 0.55 mol · 39.1 g/mol.
m(K) = 21.365 g; mass of potassium in the compound.
m(O) = 100 g - 21.365 g - 43.66 g.
m(O) =34.97 g; mass of oxygen.
n(O) = 34.97 g ÷ 16 g/mol.
n(O) = 2.185 mol.
n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.
n(K) : n(Br) : n(O) = 1 : 1 : 4.
The mass of the compound is the sum of the mass of all the elements in the compound. The value of X in the oxybromate compound is 4.
The given oxybromate compound has 43.66 % bromine. The molar mass of the compound is the sum of all the elements in the compound.
The molar mass will be:
[tex]\rm KBrO_X=K+Br+O_X\\KBrO_X=39+80+16X\\[/tex]
The percent of elements in the compound is given as:
[tex]\rm Percent=\dfrac{Mass}{Total\;mass}\;\times\;100[/tex]
The mass of bromine in the compound is 80 grams.
Substituting the values for X as:
[tex]\rm 43.66=\dfrac{80}{39+80+16X}\;\times\;100\\\\43.66(119+16X)=80\times 100\\X=4[/tex]
The value of X for the oxobromate compound, [tex]\rm \bold{KBrO_X}[/tex] is 4.
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