Empirical Formulas and Molar Mass:

Part 1-Determination of an Empirical Formula

1

Objectives

• -Explain what empirical formulas are

• -Be able to determine empirical formulas using charges or using experimental data

2

Empirical Formulas

• Empirical formula (“formula unit”) -the lowest whole number ratio of atoms in an ionic compound

• There are two ways to determine the empirical formula for a compound1. Using charges (use your ions and this is

easy) C CRISS O S S

2. Mathematically (yippee!!!)3

Method #1 – Charges

EX: Write the empirical formula for the compound formed by

Na & P c. K & Ne

Sr & Cl d. Cu & Cl

No compoundNa3P

SrCl2 CuCl or CuCl2

4

Method #2 – Mathematically

• Step 1:Use the info given in the problem and DA with the atomic mass of the element (from periodic table- round to 3 SDs) to find moles

• Step 2:Take all the mole values and divide them by the SMALLEST one to figure out a ratio

• Step 3: Use the answers as subscripts in the empirical formula

5

Law of Conservation of Mass

Chemical EquationNa + Cl NaClReactants Products

Mass of Reactants = Mass of Products

Example: Na + Cl NaCl 5.0g + 5.0g = ?

10.0g6

Atomic Mass of an Element = Mole

• Each element has a unique atomic mass and this is a standard for each element

• Atomic mass = mass of 1 mole of an element

• The atomic masses are from the periodic table and we use grams

1 mole O = 6.02 x 1023 atoms O = 15.999 g O1 mole Mg= 6.02 x 1023 atoms Mg= 24.305 g Mg

7

Empirical Formulas 7.06 g of silver combine with an excess of fluorine to produce 8.30 g of a compound

Silver + Fluorine Ag?F?

7.06g 1.24g 8.30g

7.06 g Ag

1.24 g F

X

X

= .0654 moles Ag

= .0653 moles F

= 1.00 or 1

= 1.00 or 1

Found by subtracting!

Ag1F1

ANSWER = AgF 8

Empirical Formulas A compound contains 24.58% K, 35.81%

Mn, and 40.50% O. Find the empirical formula (assume working with 100 grams of the compound and change percentages to

grams)

24.58 g K

35.81 g Mn

40.50 g O

X

X

X

= .629 mole K

= .652 moles Mn

= 2.53 moles O

= 1.00 or 1

= 1.04 or 1

= 4.02 or 4

ANSWER = KMnO49

Uneven Empirical Formulas

• When figuring empirical formulas mathematically, sometimes the resulting numbers don’t come out so clean

• You can’t just assume and round how you choose

10

.05 Rule

• This 0.05 rule allows for experimental error that occurs causing varied number values:– If a value is within .05 of a whole number

(+0.05 or - 0.05), then the value may be rounded

– Ex: 1.96 can be rounded to 2 • 1.07 cannot be rounded to 1• 3.02 could be rounded to 3 • 1.93 cannot be rounded to 2

• If one value is not within .05 of a whole number, all the values must be multiplied by an integer so all values fall within .05 of whole numbers

11

Uneven Empirical Formulas 4.35 g sample of zinc is combined with an

excess of the element phosphorus. 5.72 g of compound are formed. Calculate the empirical

formula.

Zinc + Phosphorous Zn?P?

4.35g 1.37g 5.72gFound by subtracting!

4.35 g Zn

1.37 g P

X

X

= .0665 moles Zn

= .0442 moles P

= 1.50

= 1.00

Not within .05 of a whole number

X 2 = 3.00 or 3

X 2 = 2.00 or 2

ANSWER = Zn3P2

12

Let’s Do It!!!

A compound is found to contain 72.3% Fe and 27.7% O by weight. Calculate the empirical formula.

Assume in 100 g of compound there would be 72.3 g Fe and 27.7 g O

13

72.3g Fe 1 mole FeX —————— 55.8 g Fe

= 1.30 mole Fe

27.7g O 1 mole O X —————— 16.0 g O

= 1.73 mole O

1.30 mole 1.30 mole

1.73 mole 1.30 mole

= 1.00

=1.33

X 3 = 3.00 = 3

X 3 = 3.99 = 4

Fe3O4

14

Objectives

• -Explain what empirical formulas are

• -Be able to determine empirical formulas using charges or using experimental data

15