Question

How many oxygen atoms are in 6.90 g of Al2(SO4)3? Express your answer using scientific notation with two decimal places.

Answer 1

Answer:

There are 1.44 x 10^23 atoms of oxigen in 6.90 grs of Al2(SO4)3.

Explanation:

To solve this problem you have to find:

1) The molecular mass of Al2(SO4)3;

2) Estimates the percentage of oxigen and express it in terms of its moles.

3) Calculate the number of atoms using Avogadro´s Number.

1) Molecular mass Al2(SO4)3= 2x26,98 + 3x31.06 + 12x16 =339.14 grs/mol.

2) If in the molecular mass of the compound there are 192 grs/mol (12x16) of oxigen, in 6.90 g will be:

192 grs/mol Oxigen ------------- 339.14 grs/mol Al2(SO4)3

x= 3.90 grs/mol--------------------  6.90 grs/mol  Al2(SO4)3

Then if one mol of oxigen has a mass of 16 grs, 3.90 grs represents....

1 mol oxigen ------------- 16 grs of oxigen

x= 0.24 moles------------ 3.90 grs of oxigen

3) There are 6,022 x 10^23 (Avogrado´s number) atoms of oxigen in a mol of oxigen, so in 0.24 moles will be

1 mol oxigen ---------------- 6,022 x 10^23 atoms of oxigen

0.24 moles oxigen ------- x = 1.44 x 10^23 atoms of oxigen.

Summarizing there are 1.44 x 10^23 atoms of oxigen in 6.90 grs of Al2(SO4)3.

Answer 2

First, you need to know the molar mass of the compound. The compound Aluminum Sulfate has a molar mass of 342.15 g/mol. From the subscript of the compound, there are 4*3 = 12 moles of oxygen in 1 mole of the compound. The solution is as follows:

6.90 g Al₂(SO₄)₃*(1 mol/342.15 g)*(12 mol O/1 mol Al₂(SO₄)₃)*(6.022×10²³ atoms/mole) = 1.457×10²³ atoms of oxgyen