Post on 27-Dec-2015
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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 acetylene
CH C6H6 benzene
CO2 CO2 carbon dioxide
CH2O C5H10O5 ribose
• An empirical formula represents the simplest whole number ratio of the atoms in a compound.
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• 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
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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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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
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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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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) C9H12O911
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) C9H12O912
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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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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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.
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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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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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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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Homework
• Worksheet C: #1-7