I have a question regarding problem #11 from Ch.2 of Principles of Modern Chemis

Question

I have a question regarding problem #11 from Ch.2 of Principles of Modern Chemistry 7th Ed. by Oxtoby. The problem statement is as follows:

Aluminum oxide (Al2O3) occurs in nature as a mineral called corundum, which is noted for its hardness and resistance to attack by acids. Its density is 3.97 g cm-3. Calculate the number of atoms of aluminum in 15.0 cm3 corundum.

I begin by calculating the mass using mass = volume * density. 15 cm3 * 3.97 g cm-3 = 59.55 g

Then I calculate the mole ratio. Knowing that Al2O3 molar mass is 101.9612 g, I divide 59.55 g by 101.9612 g and get 0.58405 mol.

Up to this point, my work is correct compared to the Chegg Solution. I take Avogadro's number (6.0221x1023) and multiply it by 0.58405 mol to get 3.5172x1023. This is incorrect compared to the solution which says I should multiply by 2 mol Al (my understanding is from Al2) which gives 7.03x1023 atoms as the solution.

Why is it multiplied by 2 mol Al? If one mole of Al2O3 is 6.0221x1023, how can the solution be a value higher than Avogadro's number when our given volume results in 0.58 moles. Quoting the text, "One mole of a substance is the amount that contains Avogadro

Explanation / Answer

You understand well the concept, and there is ambiguity as you said. Here I put some examples for you:

1 mol of O2 have 6.02x1023molecules of O2, But if you want to know moles of atoms of O, in each mol there is 2x6.02x1023 moles of atoms of O. Note that the difference is if you talk about moleculs or atoms.

Another example Mg3(PO4)2 . One mol of  of Mg3(PO4)2 has 6.02x1023 "molecules" of Mg3(PO4)2

But it has 3 x 6.02x1023 atoms of Mg

2 x 6.02x1023 atoms of P

8 x   6.02x1023 atoms of O

2 x   6.02x1023   PO43- ions.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.