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jacob193

Answer:

There are 0.0186 moles of formula units in 6.35 grams of aluminum sulfate [tex]\rm Al_2(SO_4)_3[/tex].

Explanation:

What's the empirical formula of aluminum sulfate?

Sulfate is an anion with a charge of -2 per ion. When sulfate ions are bonded to metals, the compound is likely ionic.

Aluminum is a group III metal. Its ions tend to carry a charge of +3 per ion.

The empirical formula of an ionic compound shall balance the charge on ions with as few ions as possible.

The least common multiple of 2 and 3 is 6. That is:

  • Three sulfate ions [tex]\rm {SO_4}^{2-}[/tex] will give a charge of -6.
  • Two aluminum ions [tex]\rm Al^{3+}[/tex] will give a charge of +6.

Pairing three [tex]\rm {SO_4}^{2-}[/tex] ions with two [tex]\rm Al^{3+}[/tex] will balance the charge. Hence the empirical formula: [tex]\rm Al_2(SO_4)_3[/tex].

What's the mass of one mole of aluminum sulfate? In other words, what's the formula mass of [tex]\rm Al_2(SO_4)_3[/tex]?

Refer to a modern periodic table for relative atomic mass data:

  • Al: 26.982;
  • S: 32.06;
  • O: 15.999.

There are

  • two Al,
  • three S, and
  • twelve O

in one formula unit of [tex]\rm Al_2(SO_4)_3[/tex].

Hence the formula mass of [tex]\rm Al_2(SO_4)_3[/tex]:

[tex]\underbrace{2\times 26.982}_{\rm Al} + \underbrace{3\times 32.06}_{\rm S} + \underbrace{12\times 15.999}_{\rm O} = \rm 342.132\;g\cdot mol^{-1}[/tex].

How many moles of formula units in 6.35 grams of [tex]\rm Al_2(SO_4)_3[/tex]?

[tex]\displaystyle n = \frac{m}{M} = \rm \frac{6.35\;g}{342.132\;g\cdot mol^{-1}} = 0.0186\;mol[/tex].

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