Help on a chemistry homework question

I am trying to help to my kid with a chemistry question but I cannot seem to get the correct answer.
Question is -- 4.27x10E25 atoms of aluminum exist in a sample of Aluminum Oxide,how many milligrams of Aluminum oxide is contained in this sample?
Can someone tell me how to solve this? Thanks

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Re: Help on a chemistry homework question

[quote="Born_N_Belize"]I am trying to help to my kid with a chemistry question but I cannot seem to get the correct answer.
Question is -- 4.27x10E25 atoms of aluminum exist in a sample of Aluminum Oxide,how many milligrams of Aluminum oxide is contained in this sample?
Can someone tell me how to solve this? Thanks[/quote]
Aluminium Oxide is Al2O3; For every two aluminiums in the sample, there are three Oxygens.

The MASS of an Aluminium ATOM is it's relative Atomic Mass multiplied by the Atomic Mass Number, [b]u[/b] = 1.66E-27 (kg)
So, you can work out the mass of aluminium in the sample just by multiplying the number of Al atoms by [b]u[/b] then multiplying by the Atomic Mass of Al.

The number of Oxygens is 1.5x the number of aluminiums. Work that out, then multiply by [b]u[/b] then by the atomic mass of Oxygen.

Add together the two masses and you have the mass of the sample [i]in Kilograms[/i]

Divide by 1,000,000 to convert this into milligrams.

I hope that all makes sense
(There are quite a few ways of going about this question - you could use moles if you wanted to. I Just took a slightly more physics based approach :?)

Thank-you very much.

answer

hi.
you have to find the mole of the compound(Al2O3) for this you should divide by 4,27*10E25/6,02*10E23 =68.87
68.87 mnas this compound contains 68,87 moles Al
1 mole Al2O3 contains 2 mole Al
? 68.87 mole Al
-------------------------------------------------
?=34,43 mole Al2O3

1 mole Al2O3 contains (27*2 + 16*3)102 g
34,43 mole Al2O3 contains ?
-------------------------------------------------
?= 3512,41 g Al2O3

in order to change miligram *10E3
the answer is 3512,41*10E3 =0,351241*10E7

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