What is the empirical formula of a compound with 35.94% aluminum and 64.06% sulfur?

Given:

% Al = 35.94

% S = 64.06

To determine:

Empirical formula of a compound with the above composition

Explanation:

Atomic wt of Al = 27 g/mol

Atomic wt of S = 32 g/mol

Based on the given data, for 100 g of the compound: Mass of Al = 35.94 g and mass of S = 64.06 g

# moles of Al = 35.94/27 = 1.331

# moles of S = 64.06/32 = 2.002

Divide by the smallest # moles:

Al = 1.331/1.331 = 1

S = 2.002/1,331 = 1.5 ≅ 2

Empirical formula = AlS₂

The empirical formula for the given compound is

Explanation:

Let us assume that the mass of the compound be 100 grams.

So, the percentage of the elements will be equal to their respective masses.

Mass of aluminium = 35.94 g

Mass of sulfur = 64.06 g

To find the empirical formula for a compound, we follow some steps:

To calculate moles, we use the equation:

Moles of aluminium =

Moles of sulfur =

Mole ratio of aluminium =

Mole ratio of sulfur =

Multiplying both the ratios by 2, to make both the mole ratios to be a whole number.

Mole ratio of aluminium =

Mole ratio of sulfur =

Empirical formula for the given compound is

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