General

The "dot" between sodium carbonate (Na2CO3) and the water (H2O) does not mean multiply or add. It simply indicates that the sodium carbonate is a hydrated compound. Also, it is unclear whether or not the sodium carbonate is aqueous or solid. Just because it is a hydrate does not necessarily mean it is in an aqueous solution.

Aqueous means it can dissolve in water. It's always nice to state if an element is (Aq), (s), or (L). You will defidently need to write them in when doing net ionic equations and solubility rules. (:

@SmartBasileeImTooLazyToLookBackForName: okay i think I somehow incorporated combustion when you convert to grams, divide by thenmolar mass, than divide by the lowest term for the empirical formula.

[quote=LEGENDairy]@wotsitman: (l) after H2O indicates that H2O is a liquid, (aq) means aqueous, meaning the compound is dissolved in water. For hydrated compounds, the water bit is necessary, dunno about writing in the states.

@novomangle: Uh...well I just remember that you convert the % into grams, then divide those grams by the atomic mass found in the periodic table (also known as molar mass, since its just a single atom of an element) to get the number of moles.[/quote]

oh, this is chem 11 in canada thanks a bunch!

@wotsitman: (l) after H2O indicates that H2O is a liquid, (aq) means aqueous, meaning the compound is dissolved in water. For hydrated compounds, the water bit is necessary, dunno about writing in the states.

@novomangle: Uh...well I just remember that you convert the % into grams, then divide those grams by the atomic mass found in the periodic table (also known as molar mass, since its just a single atom of an element) to get the number of moles. But uh, I'm not sure that you could do it with just the percentage composition of the compound.

[quote=LEGENDairy]@Blazons: Oh yeah, forgot about that. Sorry, my knowledge of mole concept has been a bit dry, doing gas laws right now.

OT: So the answer is actually Na2CO3(aq).10H2O(l)[/quote]
I haven't been in chem in a few years, just curious can you you just use each element/compounds mass when given the % composition? Meh. Ignore my previous post

[quote=LEGENDairy]I've done this before, lemme work it out...

(convert percentage to grams, divide by molar mass (relative atomic mass in periodic table))
Na: 16.1/22.99=0.700mol
C: 4.2/12.01=0.35mol
O:16.8/16.00=1.05mol
H2O:62.9/18.02=3.49mol

Ratio-(divide by smallest no. of moles)
Na: 0.700/0.35=2
C: 0.35/0.35=1
O: 1.05/0.35=3
H2O: 3.49/0.35=9.97=approx 10

Empirical formula: Na2CO3(aq).10H2O(l) (period represents multiply)[/quote]

thanks a lot! i understand how to do this. Also is the aqueous necessary to make this hydrate complete? havent learned to apply aq yet. and what is the (l) after H2O?

@Blazons: Nope, year 11/grade 11 Chemistry.

@Blazons: Oh yeah, forgot about that. Sorry, my knowledge of mole concept has been a bit dry, doing gas laws right now.

OT: So the answer is actually Na2CO3(aq).10H2O(l)

I've done this before, lemme work it out...

(convert percentage to grams, divide by molar mass (relative atomic mass in periodic table))
Na: 16.1/22.99=0.700mol
C: 4.2/12.01=0.35mol
O:16.8/16.00=1.05mol
H2O:62.9/18.02=3.49mol

Ratio-(divide by smallest no. of moles)
Na: 0.700/0.35=2
C: 0.35/0.35=1
O: 1.05/0.35=3
H2O: 3.49/0.35=9.97=approx 10

Empirical formula: Na2CO3(aq).10H2O(l) (period represents that the water molecules are included)

Would it be .629/.042= 15? though it comes out all even number except sodium is 3.8333..... So it's either Na4C041H2O15], water is the hydrate. Or you must multiply the entire compound by 5.....then again it's past 11 o'clock, so I can defidently be off.

Google?

I haven't gotten this far yet.

But ill tell you when we do learn hydrates. If your question hasn't been answered by then.