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

Aug 6, 2016

${\text{Na"_2"S"_2"O}}_{3}$

#### Explanation:

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

Grab one and look for the molar mass of sodium, $\text{Na}$, sulfur, $\text{S}$, and oxygen, $\text{O}$. 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

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\text{Na"_2"S"_2"O}}_{3}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$