Question

What is the empirical formula for the compound that has 46 grams of sodium, 64 grams of sulfur, and 48 grams of oxygen?

Answer 1

`"Na"_2"S"_2"O"_3`

Explanation:

Your tool of choice here will be the Periodic Table of Elements.

Grab one and look for the molar mass of sodium, `"Na"`, sulfur, `"S"`, and oxygen, `"O"`. Notice that you have

`M_("M Na") ~~ "23 g mol"^(-1)`

`M_("M S") ~~ "32 g mol"^(-1)`

`M_("M O") ~~ "16 g mol"^(-1)`

This tells you that the sample given to you contains

`46 color(red)(cancel(color(black)("g"))) * "1 mole Na"/(23color(red)(cancel(color(black)("g")))) = "2 moles Na"`

`64 color(red)(cancel(color(black)("g"))) * "1 mole S"/(32color(red)(cancel(color(black)("g")))) = "2 moles S"`

`48 color(red)(cancel(color(black)("g"))) * "1 mole O"/(16color(red)(cancel(color(black)("g")))) = "3 moles O"`

Now, in order to write the empirical formula of the compound, you must find the smallest whole number ratio that exists between its constituent elements.

Since the `2:2:3` ratio is already in its smallest whole number form, you can say that the empirical formula of the compound is

`color(green)(|bar(ul(color(white)(a/a)color(black)("Na"_2"S"_2"O"_3)color(white)(a/a)|)))`