# 3.10 Determining a Chemical Formula from Experimental Data

```3.10 Determining a Chemical
Formula from Experimental Data
Analysis of a Compounds Constituent
Elements
• The mass of elements that make up a compound can be
used to determine the compounds chemical formula.
• When determining a compounds empirical formula
from mass data:
– 1st convert mass to moles. (If % comp. is given, convert to
mass.)
– 2nd divide all subscripts (number of moles) by the smallest
one to determine the number of atoms of each element.
– 3rd we must make the number of atoms of each element a
simple whole number.
Let’s Try a Sample Problem
A sample of a compound is decomposed in a laboratory and produces 165 g carbon,
27.8 g hydrogen, and 220.2 g oxygen. Calculate the empirical formula of the
compound.
Step 1:
165 g C
12.01 g C
------------ = --------------X
1.00 mol
X = 13.7 mol C
27.8 g H
1.01 g H
------------ = --------------X
1.00 mol
X = 27.5 mol H
220.2 g O
16.00 g O
------------- = ----------------- X = 13.7 mol O
X
1.00 mol O
Sample Problem 1 (Continued)
A sample of a compound is decomposed in a laboratory and produces 165 g carbon,
27.8 g hydrogen, and 220.2 g oxygen. Calculate the empirical formula of the
compound.
13.7 mol C
--------------- = 1 mol C
13.7
27.5 mol H
--------------- = 2 mol H
13.7
13.7 mol O
--------------- = 1 mol O
13.7
This compounds empirical formula is: CH2O
Calculating Molecular Formulas for
Compounds
• If we know the empirical formula of a compound and its molar mass, we
can calculate the molecular formula.
• Since a molecular formula is a whole number multiple of its empirical
formula, we can calculate the simple whole number multiple (n) by:
molar mass
n = -----------------------------------------empirical formula molar mass
• Next, all we have to do is multiple the empirical formula by (n) to determine
the molecular formula of the compound.
Molecular formula = empirical formula X n
Let’s Try a Sample Problem
A compound with the percent composition shown next has a molar mass of 60.10 g/mol. Determine its molecular
formula.
C = 39.97% , H= 13.41%, and N= 46.62%
Step 1: Convert % to mass
39.97% C = 39.97 g
13.41% H = 13.41 g
46.62% N = 46.62 g
Step 2: Convert mass to moles (I will use dimensional analysis this time)
1.00 mol C
39.97 g C X --------------- = 3.33 mol C
12.01 g C
1.00 mol H
13.41 g H X --------------- = 13.3 mol H
1.01 g H
1.00 mol N
46.62 g N X --------------- = 3.33 mol N
14.01 g N
Sample Problem 2 (Continued)
A compound with the percent composition shown next has a molar mass of 60.10 g/mol.
Determine its molecular formula.
Step 3: Convert mol to atoms
3.33 mol C
-------------- = 1
3.33 mol
13.3 mol H
-------------- = 4
3.33 mol
3.33 mol N
-------------- = 1
3.33 mol
The empirical formula of this compound is: CH4N
(Now we must determine its molecular formula)
Sample Problem 2 (Continued)
A compound with the percent composition shown next has a
molar mass of 60.10 g/mol. Determine its molecular formula.
Step 4: Determine molecular formula
molar mass
60.10 g/mol
n = ----------------------------------------- = ----------------- = 2
empirical formula molar mass
30.06 g/mol
Molecular formula = empirical formula X n = (CH4N)2 = C2H8N2
Chapter 3 pg. 133 #’s 86, 87, 91 and 94 (only do the a’s)
Read 3.11 and 4.2 pgs. 119-122 and pgs. 140-145
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