The empirical formula of the compound obtained from the question given is NaBrO₃
<h3>Data obtained from the question </h3>
- Sodium (Na) = 15.24%
- Bromine (Br) = 52.95%
- Oxygen (O) = 31.81%
<h3>How to determine the empirical formula </h3>
The empirical formula of the compound can be obtained as illustrated below:
Divide by their molar mass
Na = 15.24 / 22.99 = 0.663
Br = 52.95 / 79.90 = 0.663
O = 31.81 / 16 = 1.988
Divide by the smallest
Na = 0.663 / 0.663 = 1
Br = 0.663 / 0.663 = 1
O = 1.988 / 0.663 = 3
Thus, the empirical formula of the compound is NaBrO₃
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brainly.com/question/24297883
It would be A) Cl, K, Ar
Because Cl=35,K=39,Ar=40
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.
