# What is empirical formula for Chloroform (89.1% Chlorine, 0.84% Hydrogen, 10.06% Carbon)?

Dec 5, 2015

${\text{CHCl}}_{3}$

#### Explanation:

Your strategy when dealing with mass percentages is to pick a sample of your compound and determine how many moles of each element it contains.

To make the calculations easier, you can pick a $\text{100.0-g}$ sample of chloroform and use its percent concentration by mass to find that it contains

• $\text{89.1 g } \to$ chlorine
• $\text{0.84 g } \to$ hydrogen
• $\text{10.06 g } \to$ carbon

Next, use the molar masses of the three elements to find how many moles of each you have in this sample

$\text{For Cl: " 89.1 color(red)(cancel(color(black)("g"))) * "1 mole Cl"/(35.453 color(red)(cancel(color(black)("g")))) = "2.513 moles C}$

$\text{For H: " 0.84 color(red)(cancel(color(black)("g"))) * "1 mole H"/(1.00794color(red)(cancel(color(black)("g")))) = "0.8334 moles H}$

$\text{For C: " 10.06 color(red)(cancel(color(black)("g"))) * "1 mole C"/(12.011 color(red)(cancel(color(black)("g")))) = "0.8376 moles C}$

To find the mole ratios that exist between the elements in the compound, divided these values by the smallest one

"For Cl: " (2.513 color(red)(cancel(color(black)("moles"))))/(0.8334 color(red)(cancel(color(black)("moles")))) = 3.0154 ~~ 3

"For H: " (0.8334 color(red)(cancel(color(black)("moles"))))/(0.8334color(red)(cancel(color(black)("moles")))) = 1

"For C: " (0.8376 color(red)(cancel(color(black)("moles"))))/(0.8334color(red)(cancel(color(black)("moles")))) = 1.005 ~~ 1

The compound's empirical formula, which tells you what the smallest whole number ratio that exists between the elements that make up a compound is, will thus be

"C"_1"H"_1"Cl"_3 implies color(green)("CHCl"_3)