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

Composition:

5.9% Hydrogen and 94.1% Oxygen (by mass)

in a 100 gram sample:

5.9 gram Hydrogen

94.1 gram Oxygen

Finding the number of moles:

Moles of Hydrogen:

moles = given mass/ molar mass

moles = 5.9 / 1 [molar mass of Hydrogen = 1g/mol]

moles of hydrogen = 5.9 moles

Moles of Oxygen:

number of moles = given mass / molar mass

number of moles = 94.1 / 16 [molar mass of Oxygen = 16g/mol]

moles of Oxygen = 5.88 ≈ 5.9 moles

Empirical Formula:

Moles of Hydrogen : Moles of Oxygen

5.9 : 5.9

Moles of Hydrogen : Moles of Oxygen = 1:1

Empirical Formula = OH

Answer 2

`oxygen \: at \: 94.1\% \: hydrogen \: at \: 5.9\% \\ at \: 100gram \: oxygen \: \\ \: 94.1\% = .941 * 100g \\ = 94.1 * (1mol)/(16g) = 5.88g \\ hydrogen \: \\ 5.9\% = .059 * 100 = 5.9 * (1mol)/(1.002g) = 5.88g \\ here \: oxygen = hydrogen \: so \: ratio \\ = 1 \: \: \: 1 \: \\ emperical \: formula \: = oh\\ thank \: you`