Empirical Formula
description
Transcript of Empirical 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 acetylene
CH C6H6 benzene
CO2 CO2 carbon
dioxide
CH2O C5H10O5 ribose
• 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
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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
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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
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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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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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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
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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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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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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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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
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
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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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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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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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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?_____
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Divide by the smallest # of moles.
5.00 mol C = ___2.25_=2 ¼ = 9/4
2.22 mol O
4.5 mol H = ___2.00__
2.22 mol O
2.22 mol O = ___1.00__
2.22 mol O
Are the results whole numbers? no
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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
(1/5) 0.20 x 5 = 5
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Multiply everything x 4 to clear the denominator
C: 9/4 mol C x 4 = 9 mol C
H: 2.0 mol H x 4 = 8 mol H
O: 1.00 mol O x 4 = 4 mol O
Use the whole numbers of mols as the subscripts in the simplest formula
C9H8O4