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Lera25 [3.4K]
3 years ago
13

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

Chemistry
2 answers:
bija089 [108]3 years ago
8 0

<u>Answer:</u> The empirical formula for the given compound is Al_2S_3

<u>Explanation:</u>

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:

  • <u>Step 1:</u> 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

  • <u>Step 2:</u> 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

  • <u>Step 3:</u> 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

  • <u>Step 4:</u> 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

Marat540 [252]3 years ago
6 0

<u>Given:</u>

% Al = 35.94

% S = 64.06

<u>To determine:</u>

Empirical formula of a compound with the above composition

<u>Explanation:</u>

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

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