What is the empirical formula of a substance with 1.587 g arsenic and 3.755 g of chlorine?

1 Answer
Jul 30, 2016

Answer:

#"AsCl"_5#

Explanation:

In order to find a compound's empirical formula, you must find the smallest whole number ratio that exists between its constituent elements.

In your case, the compound is said to contain arsenic, #"As"#, and chlorine, #"Cl"#. The problem provides you with grams of arsenic and grams of chlorine, so your first goal will be to convert these to moles.

To do that, use the molar masses of the two elements

#M_("M As") = "74.922 g mol"^(-1)#

#M_("M Cl") = "35.453 g mol"^(-1)#

Your sample will thus contain

#"For As: " 1.587 color(red)(cancel(color(black)("g"))) * "1 mole As"/(74.922 color(red)(cancel(color(black)("g")))) = "0.02118 moles As"#

#"For Cl: " 3.755 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453color(red)(cancel(color(black)("g")))) = "0.1059 moles Cl"#

To find the mole ratio that exists between the two elements, divide both values by the smallest one

#"For As: " (0.02118 color(red)(cancel(color(black)("moles"))))/(0.02118color(red)(cancel(color(black)("moles")))) = 1#

#"For Cl: " (0.1059color(red)(cancel(color(black)("moles"))))/(0.02118color(red)(cancel(color(black)("moles")))) = 5#

Since a #1:5# already represents the smallest whole number ratio that can exist between arsenic and chlorine in the compound, you can say that the empirical formula will be

#"As"_1"Cl"_5 implies color(green)(|bar(ul(color(white)(a/a)color(black)("AsCl"_5)color(white)(a/a)|)))#