Empirical and Molecular Formulas
Empirical Formula:
Molecular Formula:
-
Note:
CH2O ≠ C2H4O2 ≠ C4H8O4
• Although all of these have the same empirical(lowest common factored)
•
formula they are all ___________________compounds.
• In some cases empirical formula _____the molecular formula.
Simplest Formula Calculations
• Step 1:
Imagine that you have 100 g of the substance.
Thus, % will become mass in grams.
E.g. _________________________________________________________________(Some questions will
give grams right off, instead of %)
•
Step 2:
Calculate the # of moles
Equation:
• Step 3:
Express moles as the simplest ratio by dividing through by the lowest number.
Ex. A= ________mol B= _____________ mol C= _______________ mol
Divide all by ____________ (the lowest number)
A= ______mol/_____=_________ B=________mol/______=______ C=_____mol/_____=______
Therefore, ratio is
:
:
Step 4:
•
Write the simplest formula from mol ratios
Ex. Ratio A:B:C is 1:3:5
Simplest formula calculations
Question:
• A compound is found to contain the following % by mass: 69.58% Ba,
6.090% C, 24.32% O. What is the simplest (i.e. empirical) formula?
Step 1:
Imagine that you have 100 g of the substance.
Thus, % will become mass in grams.
Step 2:
Calculate the # of moles
Moles= mass/molar mass
69.58 g Ba, 6.090 g C, 24.32 g O
Ba:
69.58 g ÷ 137.33 g/mol= ____________mol Ba
C:
6.090 g ÷ 12.01 g/mol= _____________ mol C
O:
24.32 g ÷ 16.00 g/mol= ______________ mol O
- Step 3:
Express moles as the simplest ratio by dividing through by the lowest
number.
Ba
C
O
Mol
Mol
(reduced)
Step 4:
Write the simplest formula from mol ratios
•
Ba:C:O =
:
:
Therefore the simplest formula is
• Mole ratios and simplest formula
Given the following mole ratios for the hypothetical compound AxBy, what would x
and y be if the mol ratio of A and B were:
1234• Question 1
A compound consists of 29.1 % Na, 40.5 % S, and 30.4 % O. Determine the
simplest formula.
Solution:
Step 1: Assume 100 g: 29.1 g Na, 40.5 g S, 30.4 g O
Step 2: Na:
Step 3:
29.1 g ¸ 22.99 g/mol = _____________ mol Na
S:
40.5 g ¸ 32.06 g/mol = _______________ mol S
O:
30.4 g ¸ 16.00 g/mol = _________________ mol O
Na
S
Mol
Mol
(reduced)
Step 4: The simplest formula is ________________________
O
• Question 2
A compound is composed of 7.20 g carbon, 1.20 g hydrogen, and 9.60 g oxygen.
Find the empirical formula for this compound
Solution:
Step 1:
7.20 g C, 1.20 g H, 9.60 g O
Step 2:
C:
7.20 g ¸ 12.01 g/mol = _______________mol C
H:
1.20 g ¸ 1.01 g/mol = _______________ mol H
O:
9.6 g ¸ 16.00 g/mol = ________________ mol O
Step 3:
C
H
O
Mol
Mol
(reduced)
Step 4: The simplest formula is:
Molecular formula calculations
• Now that we have calculated the empirical formulas…
• There is one additional step to solving for a molecular formula.
*****First
E.g. in Q2, the molecular formula can be determined if we know that the
molar mass of the compound is 150 g/mol.
•
First, determine molar mass of the simplest formula.
For CH2O it is 30 g/mol (12+2+16).
•
Divide the molar mass of the compound by this to get a factor:
•
Multiply each subscript in the formula by this factor:
Given:
Subscriptsis the molecular formula.
•
Question 3:
For OF, give the molecular formula if the compound is 70 g/mol.
Solution:
Question 4:
Combustion analysis gives the following:
26.7% C, 2.2% hydrogen, 71.1% oxygen.
If the molecular mass of the compound is 90 g/mol, determine its
molecular formula.
Solution:
Step 1: Assume 100 g total. Thus:
26.7 g C, 2.2 g H, and 71.1 g O
Step 2: C:
26.7 g ¸ 12.01 g/mol = 2.223 mol C
H:
2.2 g ¸ 1.01 g/mol = 2.18 mol H
O:
71.1 g ¸ 16.00 g/mol = 4.444 mol O
Step 3:
C
H
O
Mol
Mol
(reduced)
Step 4: The simplest formula is
Step 5: Factor:
Molecular Formula:
5. What information must be known to determine
a) the empirical formula of a substance?
b) the molecular formula of a substance?
Solution:
For the empirical formula we need to know the moles of each element in the
compound (which can be derived from grams or %).
For the molecular formula we need the above information & the molar mass
of the compound
6. A compound’s empirical formula is CH, and it weighs 104 g/mol. Give the
molecular formula.
•
Molar mass of CH = 13 g/mol
Factor = 104 g/mol ÷ 13 g/mol = _____________
Molecular formula is
Question 7
A substance is decomposed and found to consist of 53.2% C, 11.2% H, and
35.6% O by mass. Calculate the molecular formula of the unknown if its molar mass
is 90 g/mol.
•
Step 1: Assume 100 g total. Thus:
53.2 g C, 11.2 g H, and 35.6 g O
Step 2: C:
53.2 g ¸ 12.01 g/mol = 4.430 mol C
H:
11.2 g ¸ 1.01 g/mol = 11.09 mol H
O:
35.6 g ¸ 16.00 g/mol = 2.225 mol O
Step 3:
C
Mol
H
O
Mol
(reduced)
Step 4: The simplest formula is
Step 5: Factor:
Molecular Formula:
Practice Problems
1. Calculate the percentage composition of each substance:
a) SiH4,
b) FeSO4
2. Calculate the simplest formulas for the compounds whose compositions are
listed:
a) carbon, 15.8%; sulfur, 84.2%
Compounds
Mol
Mol
(reduced)
b) silver,70.1%; nitrogen,9.1%; oxygen,20.8%
Compounds
Mol
Mol
(reduced)
c) K, 26.6%; Cr, 35.4%, O, 38.0%
Compounds
Mol
Mol
(reduced)
3. The simplest formula for glucose is CH2O and its molar mass is 180 g/mol.
What is its molecular formula?
4. Determine the molecular formula for each compound below from the
information listed.
substance
simplest formula
molar mass(g/mol)
a) octane
C4H9
114
b) ethanol
C2H6O
46
c)
naphthalene
C5H4
128
d) melamine
CH2N2
126
a.
b.
c.
d.
a.
b.
c.
5. The percentage composition and approximate molar masses of some
compounds are listed below. Calculate the molecular formula of each:
percentage composition
molar mass(g/mol)
a.
64.9% C, 13.5% H, 21.6% O
74
b.
39.9% C, 6.7% H, 53.4 % O
60
c.
40.3% B, 52.2% N, 7.5% H
80