Question

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

Answer 1

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

Answer 2

Answer: The empirical formula for the given compound is $Al_2S_3$

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:

• Step 1: Converting the given masses of the elements into moles.

To calculate moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of aluminium = $\frac{35.94g}{27g/mol}=1.33mol$

Moles of sulfur = $\frac{64.06g}{32g/mol}=2mol$

• Step 2: Now, dividing each calculated moles by the smallest number of moles, which is 1.33 moles, to calculate the mole ratio of elements.

Mole ratio of aluminium = $\frac{1.33}{1.33}=1$

Mole ratio of sulfur = $\frac{2}{1.33}=1.5$

• Step 3: Converting the mole ratio into whole number ratio.

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

Mole ratio of aluminium = $1\times 2=2$

Mole ratio of sulfur = $1.5\times 2=3$

• Step 4: Writing each mole ratio as the subscripts of the elements to make the empirical formula.

Empirical formula for the given compound is $Al_2S_3$