Empirical-Molecular formula(Empirical-Molecular formula).ppt

Empirical-Molecular formula

1. Empirical/Molecular
Formulas

2. Objective/Warm-Up
SWBAT calculate molar mass of compounds.
What is the molar mass of each of these
elements?
Na
Cl
C
H

3. GFM= Gram Formula Mass
Find the gram formula mass of C
2H
6O
C = 12.01 x 2 = 24.02
H = 1.01 x 6 = 6.06
O = 16.00 x 1 = 16.01
Total = 46.09 g/mol
Keep 2 decimal places.
Unit is g/mol

4. Molar Mass
Example: Ca(OH)
2
Ca = 40.01 g/mol x 1 = 40.08 g/mol
O = 16.00 g/mol x 2 = 32.00 g/mol
H = 1.01 g/mol x 2 = 2.02 g/mol
Total = 74.10 g/mol

5. Practice problems

6. Objective/Warm-Up
SWBAT convert using molar mass.
What is the molar mass of each of these
compounds?
NaCl
CaCl
2
Mg(NO
3)
2

8. Objective/Warm-Up
SWBAT calculate percent composition
by mass.
SWBAT distinguish between empirical
and molecular formulas.
How do you find the percent of
something? For example, how would
you find the percent of girls or boys in
this class?

9. SWBAT calculate percent composition
by mass.
How do you find the percent of
something? For example, how would
you find the percent of girls or boys in
this class?

11. Calculating Percent
Composition
Example: Na
2O
Find the mass of each element:
Na
2= 22.99 x 2 = 45.98 g/mol
O = 16.00 g/mol
Take the part divided by the whole:
% Na = (45.98 g/mol) / (61.98 g/mol) = 74.2 %
% O = (16.00 g/mol) / (61.98 g/mol) = 25.8 %
The total should add up to 100 %

12. Practice Problems

13. Objectives/Warm-Up
SWBAT distinguish between empirical and
molecular formulas.
SWBAT determine the empirical formula of a
compound from percent composition.
Find the percent of oxygen in Ca(OH)
2oxygen %2.43%100
74
32
Oxygen of %
g/mol 74 2 1) (16 40 Mass Total



14. Intro to Empirical Formula
http://www.chemcollective.org/stoich/empiri
calformula.php

15. CH
2OC
2H
4O
2
CH
3OCH
3O
Empirical Formula
Empirical Formula
A formula that gives the simplest whole-number
ratioof the atoms of each element in a compound.
Molecular Formula Empirical Formula
H
2O
2 HO
C
6H
12O
6 CH
2O

16. Wrap-Up
Give three new examples of a
molecular formula and give the
corresponding empirical formula.
Why is it important to know the
difference between molecular and
empirical formulas?

17. Determine the empirical formula for a compound
containing 2.128 g Cl and 1.203 g Ca.
Steps
1.Find mole amounts.
2.Divide each mole by the smallest mole.
Steps to Determine
Empirical Formula

18. Determine the empirical formula for a
compound containing 2.128 g Cl and
1.203 g Ca.
1.Find mole amounts.
2.128 g Cl x 1 mol Cl = 0.0600 mol Cl
35.45 g Cl
1.203 g Ca x 1 mol Ca= 0.0300 mol Ca
40.08 g Ca

19. Determine the empirical formula for a
compound containing 2.128 g Cl and
1.203 g Ca.
2.Divide each mole by the
smallest mole.
Cl = 0.0600 mol Cl= 2.00 mol Cl
0.0300
Ca = 0.0300 mol Ca= 1.00 mol Ca
0.0300
Ratio –1 Ca: 2 Cl
Empirical Formula = CaCl
2

20. A compound weighing 298.12 g consists
of 72.2% magnesium and 27.8% nitrogen
by mass. What is the empirical formula?
Hint
“Percent to mass
Mass to mole
Divide by small
Multiply ‘til whole”

21. A compound weighing 298.12 g consists of 72.2%
magnesium and 27.8% nitrogen by mass. What is
the empirical formula?
Percent to mass: Mg –(72.2%/100)*298.12 g = 215.24 g
N –(27.8%/100)*298.12 g = 82.88 g
Mass to mole: Mg –215.24 g * ( 1 mole) = 8.86 mole
24.3 g
N –82.88 g * (1 mole ) = 5.92 mole
14.01 g
Divide by small: Mg -8.86 mole/5.92 mole = 1.50
N -5.92 mole/5.92 mole = 1.00
Multiply ‘til whole:Mg –1.50 x 2 = 3.00
N –1.00 x 2 = 2.00
Mg
3N
2

22. Practice
If the problem does not give you how
many grams, assume 100 grams of
the sample.
http://www.chemcollective.org/stoich/e
f_analysis.php

23. Wrap-Up
What is the difference between
molecular and empirical formulas?
Label as molecular or empirical:
C
2H
4
Na
2O
2
Na
2SO
4
Explain how to calculate the empirical
formula.

24. Warm-Up/Objective
SWBAT calculate molecular formulas from
empirical formulas or percent composition.
Label as molecular or empirical:
C
2H
4
Na
2O
2
Na
2SO
4
What are the steps to calculate the empirical
formula?

25. You are a Forensic Scientist
 The victim in the
following case is a 35-year old
white male named Tony
DeMoy.Initial investigators
say they found several signs
around the death site that
suggest foul play.Four
possible causes of his untimely
death have been suggested by
his wife who has been ruled out
as a suspect because of a
proven alibi.Your task is to
identify who and what killed
Tony DeMoy.

26. Molecular Formula
The molecular formula gives the actualnumber of
atoms of each element in a molecular compound.
Steps
1. Find the empirical formula.
2. Calculate the Empirical Formula Mass.
3. Divide the molar mass by the “EFM”.
4. Multiply empirical formula by factor.
Find the molecular formula for a compound whose molar mass is
~124.06 and empirical formula is CH
2O
3.
2. “EFM” = 62.03 g
3. 124.06/62.03 = 2
4. 2(CH
2O
3) = C
2H
4O
6

27. Find the molecular formula for a compound that
contains 4.90 g N and 11.2 g O. The molar mass
of the compound is 92.0 g/mol.
Steps
1.Find the empirical formula.
2.Calculate the Empirical Formula Mass.
3.Divide the molar mass by the “EFM”.
4.Multiply empirical formula by factor.

28. Find the molecular formula for a compound
that contains 4.90 g N and 11.2 g O. The
molar mass of the compound is 92.0 g/mol.
Empirical formula.
A.Find mole amounts.
4.90 g N x 1 mol N= 0.350 mol N
14.01 g N
11.2 g O x 1 mol O= 0.700 mol O
16.00 g O

29. B.Divide each mole by the smallest
mole.
N = 0.350= 1.00 mol N
0.350
O = 0.700= 2.00 mol O
0.350
Empirical Formula = NO
2
Empirical Formula Mass = 46.01 g/mol
Find the molecular formula for a compound
that contains 4.90 g N and 11.2 g O. The
molar mass of the compound is 92.0 g/mol.

30. Molecular formula
Molar Mass= 92.0 g/mol= 2.00
Emp. Formula Mass 46.01 g/mol
Molecular Formula = 2 x Emp. Formula = N
2O
4
Find the molecular formula for a compound
that contains 4.90 g N and 11.2 g O. The
molar mass of the compound is 92.0 g/mol.

31. Solving the Crime
With a partner, analyze each piece of
evidence in the lab area.
There are 4 suspected compounds.
Find the molecular formula of each
compound, then see the teacher for
possible identity of those compounds.

32. Wrap-Up
Summarize how to find the molecular
formula from the empirical formula.

33. A 528.39 g compound containing only carbon,
hydrogen, and oxygen is found to be 48.38%
carbon and 8.12% hydrogen by mass. The
molar mass of this compound is known to be
~222.25 g/mol. What is its molecular formula?

34. A 528.39 g compound containing only carbon,
hydrogen, and oxygen is found to be 48.38%
carbon and 8.12% hydrogen by mass. The molar
mass of this compound is known to be ~222.25
g/mol. What is its molecular formula?
g C–(48.38/100)*528.39 g = 255.64 g
g H–(8.12/100)*528.39 g = 42.91 g
g O–(43.5/100)*528.39 g = 229.85 g
mole C-255.64 g * (1 mole ) = 21.29 mol
12.01 g
mole H–42.91 g * (1 mole ) = 42.49 mol
1.01 g
mole O–229.85 g * (1 mole ) = 14.37 mol
16.00 g

35. A 528.39 g compound containing only carbon,
hydrogen, and oxygen is found to be 48.38%
carbon and 8.12% hydrogen by mass. The molar
mass of this compound is known to be ~222.25
g/mol. What is its molecular formula?
From last slide: 21.29 mol C, 42.49 mol H, 14.27 mol O
C –21.29/14.27 = 1.49
H –42.49/14.27 = 2.98 (esentially 3)
O –14.27/14.27 = 1.00
C –1.49 x 2 = 3
H –3 x 2 = 6
O –1 x 2 = 2
C
3H
6O
2

36. A 528.39 g compound containing only carbon,
hydrogen, and oxygen is found to be 48.38%
carbon and 8.12% hydrogen by mass. The molar
mass of this compound is known to be ~222.25
g/mol. What is its molecular formula?
From last slide: Empirical formula = C
3H
6O
2
“EFM” = 74.09
Molar mass =222.24 = ~3
EFM 74.09
3(C
3H
6O
2) = C
9H
18O
6