Transcript Empirical Formula

Empirical Formula
From percentage to formula
Types of Formulas
The formulas for compounds can be expressed as an
empirical formula and as a molecular(true) formula.
Empirical
Molecular (true) Name
CH
C2H2
CH
C6H6
CO2
CO2
CH2O
C5H10O5
acetylene
benzene
carbon dioxide
ribose
• An empirical formula represents the simplest
whole number ratio of the atoms in a
compound.
3
• It is not just the ratio of atoms, it is also the
ratio of moles of atoms
• In 1 mole of CO2 there is 1 mole of carbon and
2 moles of oxygen
• In one molecule of CO2 there is 1 atom of C
and 2 atoms of O
• The molecular formula is the true or actual
ratio of the atoms in a compound.
Learning Check EF-1
A. What is the empirical formula for C4H8?
1) C2H4
2) CH2
3) CH
B. What is the empirical formula for C8H14?
1) C4H7
2) C6H12
3) C8H14
C. What is a molecular formula for CH2O?
1) CH2O
2) C2H4O2
3) C3H6O3
6
Learning Check EF-1
A. What is the empirical formula for C4H8?
1) C2H4
2) CH2
3) CH
B. What is the empirical formula for C8H14?
1) C4H7
2) C6H12
3) C8H14
C. What is a molecular formula for CH2O?
1) CH2O
2) C2H4O2
3) C3H6O3
7
Learning Check EF-2
If the molecular formula has 4 atoms of N,
what is the molecular formula if SN is the
empirical formula? Explain.
1) SN
2) SN4
3) S4N4
8
Learning Check EF-2
If the molecular formula has 4 atoms of N,
what is the molecular formula if SN is the
empirical formula? Explain.
3) S4N4
9
Empirical and Molecular Formulas
molar mass
= a whole number = n
simplest mass
n = 1 molar mass = empirical mass
molecular formula = empirical formula
n = 2 molar mass = 2 x empirical mass
molecular formula =
2 x empirical formula
molecular formula = or > empirical formula
10
Learning Check EF-3
A compound has a formula mass of 176.0 and an
empirical formula of C3H4O3. What is the
molecular
formula?
1) C3H4O3
2) C6H8O6
3) C9H12O9
11
Learning Check EF-3
A compound has a formula mass of 176.0 and an
empirical formula of C3H4O3. What is the
molecular
formula?
1) C3H4O3
2) C6H8O6
3) C9H12O9
12
Learning Check EF-4
If there are 192.0 g of O in the molecular
formula, what is the true formula if the EF is
C7H6O4?
1) C7H6O4
2) C14H12O8
3) C21H18O12
13
Learning Check EF-4
If there are 192.0 g of O in the molecular
formula, what is the true formula if the EF is
C7H6O4?
1) C7H6O4
2) C14H12O8
3) C21H18O12
14
Calculating Empirical Formula
•
•
•
•
•
•
Pretend that you have a 100 gram sample of the
compound.
That is, change the % to grams.
Convert the grams to mols for each element.
Write the number of mols as a subscript in a
chemical formula.
Divide each number by the least number.
Multiply the result to get rid of any fractions.
Example
• Calculate the empirical formula of a compound
composed of 38.67 % C, 16.22 % H, and 45.11
%N.
• Assume 100 g so
• 38.67 g C x 1mol C = 3.220 mole C
12.01 gC
• 16.22 g H x 1mol H = 16.09 mole H
1.01 gH
• 45.11 g N x 1mol N = 3.219 mole N
14.01 gN
• If we divide all of these by the smallest
• one it will give us the subscripts for the
empirical formula
• 3.220 mol C = 1
3.219 mol N
• 16.09 mol H = 5
3.219 mol N
• 3.219 mole N = 1
3.219 mol N
• Empirical formula: CH5N
Learning Check EF-5
Aspirin is 60.0% C, 4.5 % H and 35.5 O.
Calculate its simplest formula. In 100 g of
aspirin, there are 60.0 g C, 4.5 g H, and
35.5 g O.
18
Solution EF-5
60.0 g C x ___________= ______ mol C
4.5 g H x ___________ = _______mol H
35.5 g O x ___________ = _______mol O
19
Divide by the smallest # of moles.
5.00 mol C
=
________________
______ mol O
4.5 mol H
=
______ mol O
________________
2.22 mol O = ________________
______ mol O
Are are the results whole numbers?_____
20
Finding Subscripts
A fraction between 0.1 and 0.9 must not be
rounded. Multiply all results by an integer to
give whole numbers for subscripts.
(1/2)
(1/3)
(1/4)
(3/4)
(1/5)
0.5
x2
0.333
x3
0.25
x4
0.75 x 4 =
0.20
x5 =
=
=
=
3
5
1
1
1
21
Homework
• Worksheet C: #1-7