# How many sodium atoms in 5.31 grams of sodium carbonate? Na2CO3

This just brings me back to my AP Chemistry days, which is a good and bad thing (good because I loved the class, bad because I got a 4 on the test).

So first you need to find the molar weight of the Na2CO3. To do this, look at the periodic table. Na's weight is 22.990 g/mol, C's weight is 12.0111 g/mol, and O's weight is 15.999 g/mol. Therefore, the total weight of one mole of sodium carbonate is:

22.990 g/mol * 2 + 12.011 g/mol + 15.999 g/mol * 3 = 105.988 g/mol

Now you need to find how many moles of sodium carbonate are in 5.31 grams. How you do this is simply dividing 5.31 by 105.988 so you end up with moles as your unit.

(5.31 g) / (105.988 g/mol) = 0.0501 mols

The 2 subscript of Na tells you that for every one mole of sodium carbonate, there are two moles of sodium. Therefore, there are 2* 0.0501 mols = 0.100 mols Na.

To convert from moles to atoms, multiply the amount of moles of sodium to Avogadro's number, which is about 6.02*10^23 mol^-1. The result will be 6.03*10^22 atoms.

It's been a while since I've done this so you should double check. I hope this is right. If not, then I apologize.

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