Answer:
HgSO₄
Explanation:
% => g => moles => ratio => reduce => empirical ratio
%Hg = 67.6% => 67.6g/201g/mol = 0.34mol
%S = 10.8% => 10.8g/32g/mol = 0.34mol
%O = 21.6% => 21.6g/16g/mol = 1.35mol
Hg:S:O => 0.34:0.34:1.35
Reduce to whole number ratio by dividing by the smaller mole value...
Hg:S:O => 0.34/.34:0.34/.34:1.35/.34 => Empirical Ratio = 1:1:4
∴ Empirical Formula is HgSO₄
Empirical formula of the compound is HgSO4.
The empirical formula yields the compound with the lowest whole number ratio of atoms. Empirical formula specifies the relative number of atoms in each element in the compound. An empirical formula, in other words, provides us the relative ratios of different atoms in a compound.
A compound is a material formed by the interaction of more than one distinct chemical element.
Calculation of empirical formula is shown as below:
Now,67.6% of Hg = 67.6 g of Hg.
10.8% of S = 10.8 g of S
21.6% of O = 21.6 g of O
Molar mass of each component which exist in compound, Hg = 200.59 g/mol, O = 15.9994 g/mol, S = 32.065 g/mol
Gram to mole conversion:
[tex]Hg\;=\;67.6\;.\;\frac{1\; mol}{200.59\; g}\;=\;0.337\;mol\;Hg\\S\;\;=\;10.8\;.\;\frac{1\; mol}{32.065\; g}\;=\;0.337\;mol\;S\\O\;\;=\;21.6\;.\;\frac{1\; mol}{15.9994\; g}\;=\;1.35\;mol\;O[/tex]
Calculation of mole of elements.
[tex]{Mole\;of\;S}\;=\;\frac{0.337\;mol}{0.337\;mol} =\frac{1}{1} \\\frac{Mole\;of\;O}{Mole\;of\;S}\;=\;\frac{1.35\;mol}{0.337\;mol} =\frac{4}{1}[/tex]
Hence, the empirical formula will be HgSO4.
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