Transcript Calculating Empirical and Molecular Formulas

Calculating Empirical and
Molecular Formulas
What is an Empirical Formula?
• The formula of a compound expressed as the smallest
possible whole-number ratio of subscripts of the
elements in the formula.
What is a Molecular Formula?
It is the formula of a compound in which the subscripts
give the actual number of each element in the formula.
Calculating Empirical Formula
• What if a percent composition is given. How do we find the empirical
formula?
• Example: Chemical analysis of a liquid shows that it is 60.00% C, 13.40%
H, and 26.60% O by mass. Calculate the empirical formula of this
substance.
– Step 1: Convert to grams
• Assume you have a 100.00 g sample, and convert percentages to
grams.
– 60.00% Carbon is the same as 60.00 g of C
– 13.40% of Hydrogen is the same as 13.40 g of H
– 26.60% of Oxygen is the same as 26.60 g of O
Step 2: Convert grams to moles
Step 3: Divide each by the smallest decimal (in moles) to get
whole numbers. * These numbers will be the subscripts.
60.00 g C 1 mole C
12.01 g C
13.40 g H 1 mole H
1.01 g H
26.60 g O 1 mole O
16.00 g O
= 5.00 mole C ÷ 1.66 = 3.01 mol C
= 13.27mole H ÷ 1.66 = 7.99 mol H
Round those
answers to the
nearest whole
number and use
them as the
Subscripts on the
final Empirical
Formula.
= 1.66 mole O ÷ 1.66 = 1.00 mol C
Of the three calculated numbers, this is the
smallest decimal in moles, so you divide all
the other decimals and itself by that number.
THE EMPIRICAL FORMULA
WOULD BE:
C3H8O1  C3H8O
Example 2 for Empirical Formulas
• A compound consists of 72.2% magnesium and 27.8% nitrogen by mass.
What is the empirical formula?
• Step 1: Convert to grams :
72.2 % Mg = 72.2 g Mg
• Step 2: Convert grams to moles
72.2 g Mg 1 mole Mg
27.8 % N = 27.8 g N
= 1.49 mole Mg
= 2.97 mole Mg
24.31 g Mg
27.9 g N
1 mole N
14.01 g N
= 1.99 mole N ÷ 1.99
Step 3: Divide each by the smallest decimal (in moles) to get
whole numbers.
• Still not whole numbers. We have to get rid of the 1.49. If we multiply by 1
we get the same number of course. So, lets multiply by 2 and see what
happens. Remember whatever we do to one side we do it to the other.
1.00 mole N x 2 = 2 mole N
THE EMPIRICAL FORMULA WOULD BE:
Mg3N2
How do we determine the molecular formula from the
empirical formula?
• Example: The empirical formula for a compound is P2O5.
Its experimental molar mass is 284 g/mol. Determine the
molecular formula of the compound.
• Step 1: Find the molar mass of the empirical formula.
• 2 mol P x 30.97 g =
61.94 g P
• 5 mol O x 16.00g = + 80.00 g O
141.94 g/mol
Step 2: Molar mass of a compound divided by the molar mass of
empirical formula.
Experimental molar mass of compound
Molar mass of empirical formula
284 g/mol
= 2.00
141.94 g/ mol
YOUR MOLECULAR FORMULA IS:
P4O10
Assignment
Determine the Empirical Formula for each of the following. Use the
calculated empirical formula to calculate the molecular formula.
1.
A sample compound with a molar mass of 34.00g/mol is found to
consist of 0.44g H and 6.92g O. Calculate both empirical and
molecular formulas.
2.
A compound has a molar mass of 456.18 g/mol and consists of 3.0 g of
Fe and 4.81g of S. Calculate both empirical and molecular formulas
3. A compound consists of 36.48% Na, 25.41% S, and 38.11% O. It
has a molar mass of 252.10 g/mol. Calculate both empirical and
molecular formulas.