Question

Convert 6.35 grams of aluminum sulfate to moles​

Answer 1

Answer:

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

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 $\rm {SO_4}^{2-}$ will give a charge of -6.
• Two aluminum ions $\rm Al^{3+}$ will give a charge of +6.

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

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

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 $\rm Al_2(SO_4)_3$.

Hence the formula mass of $\rm Al_2(SO_4)_3$:

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

How many moles of formula units in 6.35 grams of $\rm Al_2(SO_4)_3$?

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