What's The Mass Of Al2(CO3)3.9H2O Needed To Contain 8.55*10 Raised To 25 Atoms Of Carbon?

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Naja Ellis Profile
Naja Ellis answered
The way that I would tackle this one would first find the Molecular weight of Al2(CO3)3.9H2O:
In this molecule we have:
2 Al atoms:  2 Al atoms = 2 x 26.93 gm. = 53.86
3 C atoms: 3 Carbon atoms = 3 x 12.01 = 36.03
18 O atoms: 18 Oxygen atoms = 18 x 16.00 = 288.00
18 H atoms: 18 hydrogen atoms = 18 x 1.008 = 18.144
Total molecular Weight is: 53.86 + 36.03 + 288.00 + 18.14 =
396.03grams Al2(CO)3.9H20/mole

Now we can determine the mass of the Al2(CO3)3.9H2O needed to contain 8.55 x 10+25 Carbon C atoms?

We need 1/3 the number of Al2(CO3)3.9H2O molecules to supply 8.55 x 10+25 C carbon molecules

2.85 x 10+25  molecules of Al2(CO3)3 x 1mole molecules/6.022 x 10 26 molecules
= 0.047 moles of Al2(CO3)3.9H2O x 396.03 gm/mole of Al2(CO3)3.9H2O = 18.61 grams of Al2(CO3)3.9H2O needed to produce 8.55 x 10+25 carbon atoms.

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