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To find percent composition, you must work with the atomic masses. Look on a periodic chart to find the atomic masses of F and Na. You will find that Na has an Atomic Mass of 22.990 grams per mole of Na and that F has an atomic mass of 18.998 grams per mole of F.
So, add these both together, and that will tell you how much mass NaF will be. NaF will have a mass of 41.988 grams per mole of NaF.
Now to take the percent composition of F in NaF, simply divide the mass of F by the mass of NaF. That will give you [18.998/41.988]=.452 which will be about 45%.
So your answer is that F has a 45% composition of NaF.
PS: Since Na and F only had a stoichiometry of 1 (meaning there is 1 Na and 1 F in the molecule), no multiplication of the atomic masses was necessary when calculating the total mass of the molecule.
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Sodium sulfate has Na2SO4 as its chemical formula. This is the anhydrous form (without water), and it is also seen written as Na2SO4 + 10 H2O, which is its decahydrate. Wikipedia has additional information on this idustrial chemical, and a link is provided below so you can surf on over.
Why is an aqueous solution of NaF basic?
I will dissolve in water to form Na+ and F- NaF <-> Na+ + F- ( NaF is not particularly soluble at room temperature so it might need some heating. After this step the F- anion will react with an H2O molecule to create an HF molecule. F- + H2O <-> FH + OH- Note that F- like other halogens is not likely to attack an hydrogen molecule, however what make F- unique is that the bond it creates is so strong that once it is made, it is very difficult to break. As a result the FH form will exist in solution and accumulate up to a certain amount(the bond can still be broken however, so it is not a strong base), hence why F- is a weak base unlike other halogen ions which are neutral.
