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Tasya [4]
3 years ago
14

Calculate the empirical formula of a compound that has a composition of 5.9% (by mass) hydrogen and 94.1% (by mass) oxygen.​

Chemistry
2 answers:
adelina 88 [10]3 years ago
8 0

Answer:

The empirical formula is the simplest form;

Given:

Oxygen O at 94.1% and

H at 5.9%

Assume 100grams.

94% = 0.941 x 100gm. = 94.1 gm x 1mole/16gm. = 5.88 moles of O

5.9% = 0.059 x 100gm. = 5.9gm. X 1moleH/1.002gm. = 5.88 moles of H

There is one mole of O for each mole of H so the empirical formula is O_1H_1

and written as OH.

ra1l [238]3 years ago
4 0

Explanation:

given \: that \: oxygen \: by \: 94.1\% (.941)\\ hydrogen \: by5.9\%(.059) \\ in \: 100gram \:  \\ oxygen = 100 g\times .941 = 94.1 \times  \frac{1(mol)}{16g}  \\  = 5.88moles \: of \: oxygen \\ in \: hydrogen \:  = .059\times 100 = 5.9 \times  \frac{1mol}{1.002gram} \\   = 5.88mole \: of \: hydrogen \\  \: so \: here \:  \: both \: oxygen \: andhydrogen = 5.88 \\their \: ratio = 1 \: 1  \\ so \: emparical \: formula = oh \\ thank \: you

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