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Empirical and Molecular Formulas
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Empirical vs Molecular Formula
• The Molecular Formula (MF) gives the actual number of each type of atom present.
• The Empirical Formula (EF) gives the lowest whole-number ratio of the atoms present.
• Example: C2H6 and C3H9
– they have the same EF CH3 yet have very different MF
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Timberlake LecturePLUS 3
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 acetyleneCH C6H6 benzeneCO2 CO2 carbon dioxideCH2O C5H10O5 ribose
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Timberlake LecturePLUS 4
• An empirical formula represents the simplest whole number ratio of the atoms in a compound.
• The molecular formula is the true or actual ratio of the atoms in a compound.
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EF vs. MF
When you have an ionic compound: the EF = MF
For some molecular compounds:the EF = MF, but not always
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Determining the EF
Remember: The EF is the lowest whole-number ratio of the moles of each atom present.
Example: CH4 has 1 mol C atoms
4 mol H atoms
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Determining the EF
If a compound consists of:62.1% C 13.8% H 24.1% N
The percentages are based on MASS not MOLES
We can compare moles, not masses
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Determining the EF
In order to determine the ratio of C:H:N, we need to know the mole ratio
Step 1: Convert % of each into gramsMake it easy on yourself, assume a sample size of 100.00g
62.1g C 13.8g H 24.1g N
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Determining the EFNow that you know how many grams of each atom
you have:Step 2: Convert grams to moles using the molar
mass of each62.1g C
13.8g H
24.1 g N
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Determining the EF
Now that you have the mol ratio, you need to make them Whole-Numbers.
Step 3: Divide each mol by the smallest mol value from step #2
5.17 mol C13.7 mol H1.72 mol N
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Determining the EF
This results in 3 mol C, 8 mol H and 1 mol N
therefore the EF = C3H8N
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Timberlake LecturePLUS 12
Learning Check EF-1
A. What is the empirical formula for C4H8?
1) C2H4 2) CH2 3) CHB. 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
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Timberlake LecturePLUS 13
Solution EF-1
A. What is the empirical formula for C4H8?
2) CH2
B. What is the empirical formula for C8H14?
1) C4H7
C. What is a molecular formula for CH2O?
1) CH2O 2) C2H4O2 3) C3H6O3
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Timberlake LecturePLUS 14
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) SN2) SN4 3) S4N4
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Timberlake LecturePLUS 15
Solution 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
If the actual formula has 4 atoms of N, and S is related 1:1, then there must also be 4 atoms of S.
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Timberlake LecturePLUS 16
Empirical and Molecular Formulas
molar mass = a whole number = nsimplest 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
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EF to MFTo go from the EF to the MF, you need two additional
pieces of information:1 – calculate the mass from your EF2 – You must be given the mass of the MF
(X) EFmass = MFmass
X = MFmass / EFmass
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EF to MF
Example: You found the EF to be HO, and the MFmass = 34.02g
1. Calculate the EFmass = 17.01g
2. Calculate X = 34.02g / 17.01g3. X = 24. EF = HO MF = H2O2
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Timberlake LecturePLUS 19
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
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Timberlake LecturePLUS 20
Solution EF-3
A compound has a formula mass of 176.0 and an empirical formula of C3H4O3. What is the molecular formula?2) C6H8O6
C3H4O3 = 88.0 g/EF
176.0 g = 2.00 88.0
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Timberlake LecturePLUS 21
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
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Timberlake LecturePLUS 22
Solution EF-4
If there are 192.0 g of O in the molecular formula, what is the true formula if the EF is C7H6O4?
3) C21H18O12
192 g O = 3 x O4 or 3 x C7H6O4
64.0 g O in EF
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Timberlake LecturePLUS 23
Finding the Molecular Formula
A compound is Cl 71.65%, C 24.27%, and H 4.07%. What are the empirical and molecular formulas? The molar mass is known to be 99.0 g/mol.
1. State mass percents as grams in a 100.00-g sample of the compound.
Cl 71.65 g C 24.27 g H 4.07 g
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Timberlake LecturePLUS 24
2. Calculate the number of moles of each element. 71.65 g Cl x 1 mol Cl = 2.02 mol Cl
35.5 g Cl
24.27 g C x 1 mol C = 2.02 mol C 12.0 g C
4.07 g H x 1 mol H = 4.04 mol H 1.01 g H
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Timberlake LecturePLUS 25
Why moles?
Why do you need the number of moles of each element in the compound?
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Timberlake LecturePLUS 26
3. Find the smallest whole number ratio by dividing each mole value by the smallest mole values:Cl: 2.02 = 1 Cl
2.02
C: 2.02 = 1 C 2.02
H: 4.04 = 2 H 2.024. Write the simplest or empirical formula
CH2Cl
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Timberlake LecturePLUS 27
5. EM (empirical mass)= 1(C) + 2(H) + 1(Cl) = 49.5
6. n = molar mass/empirical mass
Molar mass = 99.0 g/mol = n = 2 E M 49.5 g/EM
7.Molecular formula(CH2Cl)2 = C2H4Cl2
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Timberlake LecturePLUS 28
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.
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Timberlake LecturePLUS 29
Solution EF-5
60.0 g C x ___________= ______ mol C
4.5 g H x ___________ = _______mol H
35.5 g O x ___________ = _______mol O
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Timberlake LecturePLUS 30
Solution EF-5
60.0 g C x 1 mol C = 5.00 mol C 12.0 g C
4.5 g H x 1 mol H = 4.5 mol H 1.01 g H
35.5 g O x 1mol O = 2.22 mol O 16.0 g O
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Timberlake LecturePLUS 31
Divide by the smallest # of moles.5.00 mol C = ________________
______ mol O
4.5 mol H = ______________________ mol O
2.22 mol O = ______________________ mol OAre are the results whole numbers?_____
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Timberlake LecturePLUS 32
Divide by the smallest # of moles.5.00 mol C = ___2.25__
2.22 mol O
4.5 mol H = ___2.00__2.22 mol O
2.22 mol O = ___1.00__2.22 mol OAre are the results whole numbers?_____
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Timberlake LecturePLUS 33
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) 0.5 x 2 = 1(1/3) 0.333 x 3 = 1 (1/4) 0.25 x 4 = 1(3/4) 0.75 x 4 = 3
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Timberlake LecturePLUS 34
Multiply everything x 4
C: 2.25 mol C x 4 = 9 mol CH: 2.0 mol H x 4 = 8 mol HO: 1.00 mol O x 4 = 4 mol O Use the whole numbers of mols as the subscripts in the simplest formula
C9H8O4
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Timberlake LecturePLUS 35
Learning Check EF-6
A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g/mol, what is the molecular formula?
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Timberlake LecturePLUS 36
Solution EF 6
0.853 mol S /0.853 = 1 S0.857 mol N /0.853 = 1 N1.71 mol Cl /0.853 = 2 ClEmpirical formula = SNCl2 = 117.1 g/EF
Mol. Mass/ Empirical mass 351/117.1 = 3
Molecular formula = S3N3Cl6