Empirical/Molecular
Formulas
Objective/Warm-Up

SWBAT calculate molar mass of compounds.

What is the molar mass of each of these
elements?
Na
 Cl
C
H

GFM= Gram Formula Mass
Find the gram formula mass of C2H6O
 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

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

Practice problems
Objective/Warm-Up

SWBAT convert using molar mass.

What is the molar mass of each of these
compounds?
NaCl
 CaCl2
 Mg(NO3)2

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?

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?
Calculating Percent
Composition








Example: Na2O
Find the mass of each element:
Na2= 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 %
Practice Problems
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)2
Total Mass  40  (16  1) 2  74 g/mol
32
% of Oxygen  100%  43.2% oxygen
74
Intro to Empirical Formula

http://www.chemcollective.org/stoich/empiri
calformula.php
Empirical Formula
Empirical Formula
A formula that gives the simplest whole-number
ratio of the atoms of each element in a compound.
Molecular Formula
H2O2
Empirical Formula
HO
C6H12O6
CH2O
CH3O
CH3O
C2H4O2
CH2O
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?

Steps to Determine
Empirical Formula
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.
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
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 = CaCl2
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”
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
Mg3N2
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

Wrap-Up
What is the difference between
molecular and empirical formulas?
 Label as molecular or empirical:

C2H4
 Na2O2
 Na2SO4


Explain how to calculate the empirical
formula.
Warm-Up/Objective

SWBAT calculate molecular formulas from
empirical formulas or percent composition.

Label as molecular or empirical:
C2H4
 Na2O2
 Na2SO4


What are the steps to calculate the empirical
formula?
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.
Molecular Formula
The molecular formula gives the actual number of
atoms of each element in a molecular compound.
1.
2.
3.
4.
Steps
Find the empirical formula.
Calculate the Empirical Formula Mass.
Divide the molar mass by the “EFM”.
Multiply empirical formula by factor.
Find the molecular formula for a compound whose molar mass is
~124.06 and empirical formula is CH2O3.
2.
“EFM” = 62.03 g
3.
124.06/62.03 = 2
4.
2(CH2O3) = C2H4O6
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.
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
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.
B.
mole.
Divide each mole by the smallest
N = 0.350 = 1.00 mol N
0.350
O = 0.700 = 2.00 mol O
0.350
Empirical Formula = NO2
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.
Molecular formula
Molar Mass =
92.0 g/mol
Emp. Formula Mass 46.01 g/mol
= 2.00
Molecular Formula = 2 x Emp. Formula = N2O4
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.

Wrap-Up
 Summarize how to find the molecular
formula from the empirical formula.
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?
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
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–3x2=6
O–1x2=2
C3H6O2
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 = C3H6O2
“EFM” = 74.09
Molar mass = 222.24 = ~3
EFM
74.09
3(C3H6O2) = C9H18O6