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EMPIRICAL – simplest whole number ratio of that matches the relative ratio found in a

Empirical Formula from Experimental Data

Determine the Empirical Formula for the compound: C_H_O_

Given the following information:

Chemical Analysis of a compound yields the following results:

60.0 % of the compound’s mass is 13.4 % of the compound’s mass is 26.6 % of the compound’s mass in

STEP 1. Figure out how many grams of each element would be present if you had 100 grams of the substance. (It is the percent written as grams)

If 100 grams of this compound were present there would be:

Carbon 60.0 grams Hydrogen 13.4 grams Oxygen 26.6 grams

STEP 2. Calculate the number of moles of each element. You will need the molar mass for each element.

Atom Mass 1 Moles

Molar _ Mass Carbon 60.0grams 1mole 5.00 moles of C 1 12.00grams Hydrogen 13.4grams 1mole 13.3 moles of H 1 1.01grams Oxygen 26.6grams 1mole 1.66 moles of O 1 16.00grams

STEP 3. Calculate the Subscripts for the Empirical Formula.

5.00 moles of Carbon 13.3 moles of Hydrogen 1.66 moles of Oxygen

These values give us the ratio of the subscripts but are not whole numbers, so we need to do some math to figure out the proper subscripts for our empirical formula.

C5.00H13.3O1.66

These are not all whole numbers as they should be

for a .

DIVIDE ALL NUMBERS BY THE SMALLEST NUMBER

5.00 / 1.66 = 3.01

13.3 / 1.66 = 8.01 THESE VALUES ARE WITHIN OUR ABILITY TO MAKE MEASUREMENTS 1.66 / 1.66 = 1

PLACE THESE VALUES (ROUNDED) WITH THE APPROPRIATE ATOMS

C3H8O1

THE ONE IS NOT USUALLY WRITTEN

C3H8O ANOTHER EXAMPLE: GIVEN THE FOLLOWING INFORMATION, CALCULATE THE EMPIRICAL FORMULA

26.56 % of the compound’s mass is Potassium 35.41 % of the compound’s mass is Chromium 38.03 % of the compound’s mass in Oxygen

STEP 1 – IF 100 grams were present

Potassium 26.56 grams Chromium 35.41 grams Oxygen 38.03 grams

STEP 2. Calculate the number of moles of each element. You will need the molar mass for each element.

1 Mass Moles Molar _ Mass Potassium 26.56grams 1mole .68 moles of K 1 39.10grams Chromium 35.41grams 1mole .68 moles of Cr 1 52.00grams Oxygen 38.03grams 1mole 2.38 moles of O 1 16.00grams

STEP3. DIVIDE ALL NUMBERS BY THE SMALLEST NUMBER (Adjust if necessary)

.68 / .68 = 1 These did not come out to whole numbers. How can we make them whole numbers? .68 / .68 = 1 Multiply them all by 2. (Adjustment) 2.38 / .68 = 3.5

Multiplying all the values by 2 gives 2:2:7 so the formula is

K2Cr2O7