empirical formula and molecular formula

12
EMPIRICAL FORMULA AND MOLECULAR FORMULA

Upload: todd

Post on 13-Jan-2016

136 views

Category:

Documents


4 download

DESCRIPTION

Empirical Formula and Molecular Formula. For Ionic cmpds we can predict formulas based on common ion charges Sodium + Oxygen together Na + O 2-  Na 2 O For molecular cmpds, charges are not involved, and this makes it harder to predict. If you know the % comp it can be done!!. - PowerPoint PPT Presentation

TRANSCRIPT

Page 1: Empirical Formula and Molecular Formula

EMPIRICAL FORMULA AND MOLECULAR FORMULA

Page 2: Empirical Formula and Molecular Formula

For Ionic cmpds we can predict formulas based on common ion charges

Sodium + Oxygen togetherNa+ O2- Na2O

For molecular cmpds, charges are not involved, and this makes it harder to predict. If you know the % comp it can be done!!

Page 3: Empirical Formula and Molecular Formula

EMPIRICAL FORMULA

Smallest whole number ratio of elements in a cmpd. (not necessarily the real formula!!)

Page 4: Empirical Formula and Molecular Formula

Real Formula E.F. H2O H2O

C6H12O6 CH2O Can be same!

Page 5: Empirical Formula and Molecular Formula

STEPS

1) Change % to grams2) Change grams to mole3) Divide all by smallest number of moles4) Use answers from step 3 as subscripts

Page 6: Empirical Formula and Molecular Formula

WHAT IS THE EMPIRICAL FORMULA FOR A COMPOUND THAT IS 79.9% C AND 20.1% H

1) 79.9% 79.9g+20.1% +20.1g100% 100.0g2) 3) 79.9g x 1 mol = 6.66mol C 6.66 = 1

12gC 20.1 g x 1 mol = 20.1 mol H 6.66= 3

1gH

4) CH3

Page 7: Empirical Formula and Molecular Formula

Find the empirical formula for a cmpd that is 54.5gC, 13.6gH, 31.8gN

Page 8: Empirical Formula and Molecular Formula

Find the empirical formula for a cmpd that is 54.5gC, 13.6gH, 31.8gN

54.5gC x 1molC = 4.54molC 12g

13.6gH x 1 mol H = 13.6molH 1g

31.8gN x 1mol N = 2.27mol N 14g

2.27 =2

2.27 =6

2.27 =1

C2H6N

Page 9: Empirical Formula and Molecular Formula

MOLECULAR FORMULA

This formula has the actual number of atoms in the cmpd. (the EF could be the same or have less atoms)

Page 10: Empirical Formula and Molecular Formula

STEPS

1) Find molar mass of the E. F. 2) Divide

M.F.’s molar massE. F. molar mass

3) If greater than 1, multiply all the subscripts in the E.F. by your answer in # 2

Given in problem

Page 11: Empirical Formula and Molecular Formula

EX.

What is the M.F. if: NO2 = E.F

Molar mass is 92g/mol

Page 12: Empirical Formula and Molecular Formula

M.F.’S MOLAR MASSE. F. MOLAR MASS

92 (MF)(14+2(16)= 46 (EF)

=2

NO2 becomes N2O4