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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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3 years ago

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Read 2 more answers

A chemistry student weighs out 0.027 kg of an unknown solid compound X and adds it to 550. mL of distilled water at 30.° C. Afte

QveST [7]

Answer:

1. Yes

2.The solubility of X is 34.55g/L

Explanation:

Solubility of solute refers to how readily a solute will dissolve in a solvent at a particular temperature. Its the amount of moles or grams required to saturate 1dm $^{3}$ or 1 Litre of water.

From the problem, when the liquid was drained off and amount of X which didn't dissolve was measured, it weighed 0.008kg, this means out of 0.027kg, 0.027-0.008 actually dissolved

= 0.019kg*1000 = 19g.

if 19g is required to saturate 550mL at 30°C,

then $\frac{19*1000}{550}$ will saturate 1L

= 34.545g will saturate 1Litre

The solubility thus is 34.55g/L

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3 years ago