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3 years ago

Calculate the empirical formula for a compound containing 67.6% Hg 10.8% S 21.6% O

Chemistry

1 answer:

pashok25 [27]3 years ago

3 0

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

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Explanation:

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pH is calculated by taking negative logarithm of hydrogen ion concentration.

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According to stoichiometry,

1 mole of $Ba(OH)_2$ gives 2 mole of $OH^-$

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Putting in the values:

$pOH=-\log[0.0596]=2.82$

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$pH=14-2.82$

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