Question

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

Answer 1

Step-by-step explanation:

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

Answer 2

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.