Question

What is the mass of sulfur in a 18.7 g sample of aluminum sulfate?

Answer 1

Q171944

What is the mass of sulfur in an 18.7 g sample of aluminum sulfate?

Solution:

The formula of aluminum sulfate is Al₂(SO₄)₃.

There is 2 Al, 3 S, and 12 O atoms in each unit formula of Al₂(SO₄)₃.

molar mass of Al₂(SO₄)₃. = 2 * atomic mass of Al + 3 * atomic mass of S + 12 * atomic mass of O

= 2 * 26.982 g/mol + 3 * 32.065 g/mol + 12 * 15.999 g/mol

= 53.964 g/mol ++ 96.195 g/mol + 191.988 g/mol

= 342.147 g/mol

The mass of Al₂(SO₄)₃. given to us is 18.7 grams.

Covert this to moles by using the molar mass of Al₂(SO₄)₃.

Step 1: Convert 18.7 g of Al₂(SO₄)₃ to moles.

$moles \space of \space Al_2(SO_4)_3 = 18.7 g \space of \space Al_2(SO_4)_3 * \frac{ 1 mol \space of \space Al_2(SO_4)_3}{342.147 g \space of \space Al_2(SO_4)_3}$

= 0.05465 mol of Al₂(SO₄)₃.

Step 2: Find the moles of the Sulphur atom.

We know that each unit of Al₂(SO₄)₃. contains 3 atoms of S.

So, 3 mol of S ≡ 1 mol of Al₂(SO₄)₃.

$moles \space of\space S = 0.05465 \space mol \space of\space Al_2(SO_4)_3. * \frac{3 \space mol \space of \space S }{ 1 \space mol \space of\space Al_2(SO_4)_3 }$

= 0.1640 mol of S.

Step 3: Convert 0.1640 mol of S to grams by using the atomic mass of S.

Atomic mass of S = 32.065 g/mol

$grams\space of\space S = 0.1640 \space mol\space of\space S * \frac{32.065\space g\space of\space S }{ 1\space mol\space of\space S }$

= 5.26 grams of S.

In the question, we are given the mass of aluminum sulfate in 3 significant figures, so our final answer must also be in 3 significant figures.

Hence there is 5.26 grams of S in the given grams of aluminum sulfate.