NOTES: 10.3 – Empirical and Molecular Formulas, Lecture notes of Chemistry

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NOTES: 10.3 –

Empirical and Molecular

Formulas

What Could It Be?

Calculating the Empirical

Formula

Example: A compound is found to contain the following… 2.199 g Copper 0.277 g Oxygen Calculate it’s empirical formula.

Calculating the Empirical

Formula

Step 1: Convert the masses to moles.

moles Cu

gCu

moleCu

g Cu 0. 03460

Copper: 2. 199 

Oxygen: moles O g O mole O g O 0. 01731

16. 0 1
17. 277 

Calculating the Empirical

Formula

Step 3: Round off these numbers, they become the subscripts for the elements. Cu 2 O

Empirical Formula Example Problems:

1. A compound is analyzed and found to contain 79.8 g C and 20.2 g H. Determine the empirical formula.
2. What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen?

1. A compound is analyzed and found to contain 79.8 g C and 20.2 g H. Determine the empirical formula. H m H

3. 0
4. 65 mol
5. 2 ol 1.0 C 6.65 mol 6.65 molC  

1. A compound is analyzed and found to contain 79.8 g C and 20.2 g H. Determine the empirical formula. EMPIRICAL FORMULA = CH 3

2. What is the empirical formula of a compound that is 25.9 g nitrogen and 74. g oxygen? O m O

2. 5
3. 85 mol
4. 63 ol 1.0 N 1.85 mol 1.85 mol N  

2. What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen? **EMPIRICAL FORMULA = NO

Is this formula acceptable?**

2. What is the empirical formula of a compound that is 25.9 g nitrogen and 74.1 g oxygen? EMPIRICAL FORMULA: 2 x (NO 2.

) = N

2

O

5

Molecular Formulas

• Indicates the actual formula for a compound; the true formula
• Will be a multiple of the empirical formula
• EXAMPLE : glucose
  • empirical: CH 2

O

• molecular: C 6

H

12

O

6

Molecular Formula Example:

Calculate the molecular formula of the compound whose molar mass is 60.0 g and the empirical formula is CH 4

N.

CH

4 N has a MW of: 12 + 1(4) + 14 = 30.0 g Actual MW is: 60.0 g

Molecular Formula Example:

Calculate the molecular formula of the compound whose molar mass is 60.0 g and the empirical formula is CH 4

N.

So, multiply the empirical form. by 2 2 x (CH 4

N) = C

2

H

8

N

2