Empirical and Molecular Formulas. Empirical vs Molecular Formula The Molecular Formula (MF) gives...
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Empirical and Molecular Formulas
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
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
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.
EF vs. MF
When you have an ionic compound: the EF = MF
For some molecular compounds:the EF = MF, but not always
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
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
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
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
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
Determining the EF
This results in 3 mol C, 8 mol H and 1 mol N
therefore the EF = C3H8N
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
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
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
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.
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
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
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
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
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
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
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
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
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
Timberlake LecturePLUS 25
Why moles?
Why do you need the number of moles of each element in the compound?
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
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
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.
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
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
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?_____
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?_____
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
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
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?
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