An insecticide has the following weight percents C=55.6%, H=4.38%, Cl=30.8% and O=9.26%. WHat is the empirical

Question

formula of the insecticide?

Answer 1

if you had a 100g sample, there would be:
55.6g C, 4.38g H, 30.8 g Cl, and 9.26g O

55.6 / 12.01 = 4.629 mol C
4.38 / 1.01 = 4.337 mol H
30.8 / 36.45 = .869 mol Cl
9.26 / 16 = .579 mol O
now divide by the smallest number

4.629/.579 = 8.00 C
4.337/.579 = 7.50 H
.869/.579 = 1.50 Cl
.579/.579 = 1.00 O
multiply everything by two to get all whole numbers

8 x 2 = 16 C
7.5 x 2 = 15 H
1.5 x 2 = 3 Cl
1 x 2 = 2 O

the empirical formula is thus C16H15Cl3O2

Answer 2

Start by assumng that yu have a 100 gram sample. SO you have 55.6 grams of carbon, 4.38 grams of hydrogen, etc.

Now divide each quantity by the respective molecular mass of each element. Carbon is 12, hydrogen is 1, etc.

This now gives you the molar ratio. Select the smallest of the numbers and divide all by the smallest. Thgis wil give you whole number ratios and you can easily write the empirical formula.

Answer 3

C16H15Cl3O2