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

1 Answer
Aug 6, 2016

Answer:

#"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)|)))#