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

##### 1 Answer

#### 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, *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 **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)|)))#