Calculate the empirical formula for the compound found to contain 2.82g/Na, 4.35g/Cl and 7.83g/O. My biggest question is, when you get to the step where you divide all three subscripts, by the smallest number, how do I get rid of the decimal?

1 Answer

Answer:

You round to the nearest integer.

Explanation:

You know that your compound contains 2.82 g of sodium, 4.35 g of chlorine, and 7.83 g of oxygen.

What you do next is use each element's molar mass to try and find how many moles of each you get in that sample of the compound.

In this case, you have

#"For Na: " (2.82color(red)(cancel(color(black)("g"))))/(23.0color(red)(cancel(color(black)("g")))/"mol") = "0.1226 moles Na"#

#"For Cl: "(4.35color(red)(cancel(color(black)("g"))))/(35.45color(red)(cancel(color(black)("g")))/"mol") = "0.1227 moles Cl"#

#"For O: "(7.83color(red)(cancel(color(black)("g"))))/(16.0color(red)(cancel(color(black)("g")))/"mol") = "0.4894 moles O"#

Now, the reason you divide these numbers by the smallest one is so that you ahve an idea of the mole ratios that exist between the elements that make up the compound.

In this case, you have

#"For Na: " (0.1226color(red)(cancel(color(black)("moles"))))/(0.1226color(red)(cancel(color(black)("moles")))) = 1#

#"For Cl: " (0.1227color(red)(cancel(color(black)("moles"))))/(0.1226color(red)(cancel(color(black)("moles")))) =1.001#

#"For O: " (0.4894color(red)(cancel(color(black)("moles"))))/(0.1226color(red)(cancel(color(black)("moles")))) = 3.992#

So, what do you do with the numbers that hav a decimal? You round them to the nearest integer.

For #1.001#, the nearest integers is #1#, so you have

#"For Cl: " 1.001 ~~ 1#

For #3.992#, the nearest integer is #4#, so you have

#"For O: " 3.992 ~~ 4#

This means that the empirical formula, which tells you what the smallest integer ratio of atoms you get in a compound, will be

#"Na"_1"Cl"_1"O"_4 " "#, or #" " "NaClO"""_4#