chapter 7 chemical quantities the mole atomic mass and formula mass copyright © 2008 by pearson...
TRANSCRIPT
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Chapter 7 Chemical Quantities
The MoleAtomic Mass and
Formula Mass
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Chapter 7 Slide 2 of 75
The Mole
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Chapter 7 Slide 3 of 75
Collection Terms
A collection term states a specific number of items.
• 1 dozen donuts = 12 donuts
• 1 ream of paper = 500 sheets
• 1 case = 24 cans
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Chapter 7 Slide 4 of 75
A mole (mol) is a collection that contains
• The same number of particles as there are carbon atoms in 12.01 g of carbon.
• 6.022 x 1023 atoms of an element (Avogadro’s number).
1 mol C = 6.022 x 1023 C atoms
1 mol Na = 6.022 x 1023 Na atoms
1 mol Au = 6.022 x 1023 Au atoms
A Mole of Atoms
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Chapter 7 Slide 5 of 75
A mole
• Of a covalent compound has Avogadro’s number of _________________.
1 mol CO2 = 6.022 x 1023 CO2 molecules
1 mol H2O = 6.022 x 1023 H2O molecules
• Of an ionic compound contains Avogadro’s number of ___________ units.
1 mol NaCl = 6.022 x 1023 NaCl formula units
1 mol K2SO4 = 6.022 x 1023 K2SO4 formula units
A Mole of A Compound
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Chapter 7 Slide 6 of 75
Samples of One Mole Quantities
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Chapter 7 Slide 7 of 75
Avogadro’s number 6.022 x 1023 can be written asan equality and two conversion factors.
As an equality:
1 mol particles = 6.022 x 1023 particles
As conversion Factors:
6.022 x 1023 particles and 1 mol particles 1 mol particles 6.022 x 1023 particles
Avogadro’s Number
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Chapter 7 Slide 8 of 75
Using Avogadro’s Number
Avogadro’s number• Converts moles of a substance to
the number of particles.
How many Cu atoms are in 0.50 mol
Cu?
0.50 mol Cu x 6.022 x 1023 Cu atoms
1 mol Cu
= 3.0 x 1023 Cu atoms Copyright © 2008 by Pearson Education, Inc.Publishing as Benjamin Cummings
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Chapter 7 Slide 9 of 75
Using Avogadro’s NumberAvogadro’s number• Converts particles of a substance to
moles.
How many moles of CO2 are
2.50 x 1024 CO2 molecules?
2.50 x 1024 CO2 x 1 mol CO2
6.022 x 1023 CO2
= 4.15 mol CO2
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Chapter 7 Slide 10 of 75
The number of atoms in 2.0 mol Al is
2.0 mol Al x 6.022 x 1023 Al atoms =
1 mol Al
1.2 x 1024 Al atoms
Learning Check
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Chapter 7 Slide 11 of 75
The number of moles of S in 1.8 x 1024 atoms S is
1.8 x 1024 S atoms x 1 mol S =
6.022 x 1023 S atoms
3.0 mol S atoms
Learning Check
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Chapter 7 Slide 12 of 75
Subscripts and Moles
The subscripts in a formula state• The relationship of atoms in the formula.• The moles of each element in 1 mol of compound.
Glucose
C6H12O6
In 1 molecule: 6 atoms C 12 atoms H 6 atoms O
In 1 mol: 6 mol C 12 mol H 6 mol O
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Chapter 7 Slide 13 of 75
Subscripts State Atoms and Moles
1 mol C9H8O4 = 9 mol C 8 mol H 4 mol O
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Chapter 7 Slide 14 of 75
Factors from Subscripts
Subscripts used for conversion factors• Relate moles of each element in 1 mol compound. • For aspirin C9H8O4 can be written as:
9 mol C 8 mol H 4 mol O
1 mol C9H8O4 1 mol C9H8O4 1 mol C9H8O4
and
1 mol C9H8O4 1 mol C9H8O4 1 mol C9H8O4
9 mol C 8 mol H 4 mol O
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Chapter 7 Slide 15 of 75
Learning Check
How many moles O are in 0.150 mol aspirin C9H8O4?
0.150 mol C9H8O4 x 4 mol O = 0.600 mol O
1 mol C9H8O4
subscript factor
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Chapter 7 Slide 16 of 75
Learning Check
How many O atoms are in 0.150 mol aspirin C9H8O4?
0.150 mol C9H8O4 x 4 mol O x 6.022 x 1023 O atoms
1 mol C9H8O4 1 mol O
subscript Avogadro’s
factor number
= 3.61 x 1023 O atoms
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Chapter 7 Slide 17 of 75
Atomic Mass
Atomic mass is the• Mass of a single atom in
atomic mass units (amu).• Mass of an atom
compared to a 12C atom.• Number below the
symbol of an element. (the average atomic mass)
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Chapter 7 Slide 18 of 75
Periodic Table and Atomic Mass
Ag has atomic mass = 107.9 amu
C has atomic mass = 12.01 amu
S has atomic mass = 32.07 amu
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Chapter 7 Slide 19 of 75
Atomic Mass Factors
The atomic mass • Can be written as an equality.
Example: 1 P atom = 30.97 amu
• Can be written as two ___________ ________.
Example: 1 P atom and 30.97 amu
30.97 amu 1 P atom
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Chapter 7 Slide 20 of 75
Uses of Atomic Mass Factors
The atomic mass factors are used to convert • A specific number of atoms to mass (amu).• An amount in amu to ___________ of atoms.
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Chapter 7 Slide 21 of 75
Using Atomic Mass Factors
1. What is the mass in amu of 75 P atoms?
75 P atoms x 30.97 amu = 2323 amu (2.323 x103 amu)
1 P atom
2. How many Cu atoms have a mass of 4.500 x 105 amu?
4.500 x 105 amu x 1 Cu atom = 7081 Cu atoms
63.55 amu
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Chapter 7 Slide 22 of 75
Learning Check
What is the mass in amu of 75 silver atoms?
75 Ag atoms x 107.9 amu = 8093 amu
1 Ag atom
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Chapter 7 Slide 23 of 75
Learning Check
How many gold atoms have a mass of
1.85 x 105 amu?
1.85 x 105 amu x 1 Au atom = 939 Au atoms
197.0 amu
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Chapter 7 Slide 24 of 75
Formula MassThe formula mass is• The mass in amu of a compound.• The _____ of the atomic masses of the
elements in a formula.
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Chapter 7 Slide 25 of 75
Calculating Formula Mass
To calculate the formula mass of Na2SO4
• Multiply the atomic mass of each element by its subscript, then total the masses of the atoms.
2 Na x 22.99 amu = 45.98 amu 1 Na
1 S x 32.07 amu = 32.07 amu 142.05 amu 1 S
4 O x 16.00 amu = 64.00 amu1 O
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Chapter 7 Slide 26 of 75
Learning Check
Using the periodic table, calculate the formula mass of aluminum sulfide Al2S3.
2 Al x 26.98 amu = 53.96 amu
1 Al
3 S x 32.07 amu = 96.21 amu
1 S 150.17 amu
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Chapter 7 Slide 27 of 75
Molar Mass
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Chapter 7 Slide 28 of 75
Molar Mass
The molar mass • Is the mass of one
mol of an element or compound.
• Is the atomic mass expressed in grams.
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Chapter 7 Slide 29 of 75
Molar Mass from the Periodic Table
Molar mass • Is the atomic
mass expressed in grams.
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Chapter 7 Slide 30 of 75
Give the molar mass for:
A. 1 mol K atoms =
B. 1 mol Sn atoms =
Learning Check
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Chapter 7 Slide 31 of 75
Molar Mass of a Compound
The molar mass of a compound is the sum of the molar masses of the elements in the formula.
Example: Calculate the molar mass of CaCl2.
Element Number of Moles
Atomic Mass Total Mass
Ca 1 40.08 g/mol 40.08 g
Cl 2 35.45 g/mol 70.90 g
CaCl2 110.98 g
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Chapter 7 Slide 32 of 75
Molar Mass of K3PO4
Calculate the molar mass of K3PO4.
Element Number of Moles
Atomic Mass Total Mass in K3PO4
K 3 39.10 g/mol 117.3 g
P 1 30.97 g/mol 30.97 g
O 4 16.00 g/mol 64.00 g
K3PO4 212.3 g
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Chapter 7 Slide 33 of 75
Some One-Mol Quantities
32.07 g 55.85 g 58.44 g 294.20 g 342.30 g
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Chapter 7 Slide 34 of 75
A. K2O 94.20 g/mol
2 mol K (39.10 g/mol) + 1 mol O (16.00 g/mol)
78.20 g + 16.00 g
= 94.20 g
B. Al(OH)3 78.00 g/mol
1 mol Al (26.98 g/mol) + 3 mol O (16.00 g/mol) + 3 mol H (1.008 g/mol) 26.98 g + 48.00 g + 3.024 g = 78.00 g
Learning Check
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Chapter 7 Slide 35 of 75
Calculations Using Molar Mass
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Chapter 7 Slide 36 of 75
Molar mass conversion factors • Are written from molar mass.• Relate grams and moles of an element or
compound.
Example: Write molar mass factors for methane CH4
used in gas cook tops and gas heaters.
Molar mass:
1 mol CH4 = 16.04 g
Conversion factors:
16.04 g CH4 and 1 mol CH4
1 mol CH4 16.04 g CH4
Molar Mass Factors
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Chapter 7 Slide 37 of 75
Acetic acid C2H4O2 gives the sour taste to vinegar.Write two molar mass factors for acetic acid.
Calculate molar mass:
24.02 + 4.032 + 32.00 = 60.05 g/mol
1 mol of acetic acid = 60.05 g acetic acid
Molar mass factors
1 mol acetic acid and 60.05 g acetic acid 60.05 g acetic acid 1 mol acetic acid
Learning Check
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Chapter 7 Slide 38 of 75
Molar mass factors are used to convert between the grams of a substance and the number of moles.
Calculations Using Molar Mass
Grams Molar mass factor Moles
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Chapter 7 Slide 39 of 75
Aluminum is used to build lightweight bicycleframes. How many grams of Al are 3.00 mol Al?
Molar mass equality: 1 mol Al = 26.98 g Al
Setup with molar mass as a factor:
3.00 mol Al x 26.98 g Al = 80.9 g Al1 mol Al
molar mass factor for Al
Moles to Grams
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Chapter 7 Slide 40 of 75
Learning CheckAllyl sulfide C6H10S is a compound that has the odor of garlic. How many moles of C6H10S are in 225 g C6H10S?
Action Plan: Calculate the molar mass, and convert 225 g to moles.
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Grams Molar mass factor Moles
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Chapter 7 Slide 41 of 75
Calculate the molar mass of C6H10S.
(6 x 12.01) + (10 x 1.008) + (1 x 32.07)
= 114.21 g/mol
Set up the calculation using a mole factor.
225 g C6H10S x 1 mol C6H10S
114.21 g C6H10S
molar mass factor(inverted)
= 1.97 mol C6H10S
Solution
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Chapter 7 Slide 42 of 75
Grams, Moles, and Particles
A molar mass factor and Avogadro’s number
convert• Grams to particles molar mass Avogadro’s
number
(g mol particles)• Particles to grams
Avogadro’s molar mass number
(particles mol g)
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Chapter 7 Slide 43 of 75
Learning Check
How many H2O molecules are in 24.0 g H2O?
A) 4.52 x 1023
B) 1.44 x 1025
C) 8.02 x 1023
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Chapter 7 Slide 44 of 75
Learning Check
How many H2O molecules are in 24.0 g H2O?
24.0 g H2O x 1 mol H2O x 6.022 x 1023 H2O molecules
18.02 g H2O 1 mol H2O
= 8.02 x 1023 H2O molecules
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Chapter 7 Slide 45 of 75
Learning CheckIf the odor of C6H10S can be detected from
2 x 10-13 g in one liter of air, how many molecules
of C6H10S are present?
2 x 10-13 g x 1 mol x 6.022 x 1023 molecules
114.21 g 1 mol
= 1 x 109 molecules C6H10S
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Chapter 7 Slide 46 of 75
Percent Composition
and
Empirical Formulas
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Chapter 7 Slide 47 of 75
Percent Composition
Percent composition
• Is the percent by mass of each element in a formula.
Example: Calculate the percent composition of CO2.
CO2 = 1 C(12.01g) + 2 O(16.00 g) = 44.01 g/mol)
12.01 g C x 100 = 27.29 % C 44.01 g CO2
32.00 g O x 100 = 72.71 % O 44.01 g CO2
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Chapter 7 Slide 48 of 75
What is the percent composition of lactic acid, C3H6O3, a compound that appears in the blood after vigorous activity?
→
Learning Check
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Chapter 7 Slide 49 of 75
STEP 1
3C(12.01) + 6H(1.008) + 3O(16.00) = 90.08 g/mol
36.03 g C + 6.048 g H + 48.00 g O = 90.08 g/mol
STEP 2
%C = 36.03 g C x 100 = 40.00% C 90.08 g cpd
%H = 6.048 g H x 100 = 6.714% H 90.08 g cpd
%O = 48.00 g O x 100 = 53.29% O 90.08 g cpd
Solution
C3H6O3
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Chapter 7 Slide 50 of 75
Learning Check
The chemical isoamyl acetate C7H14O2 contributes to the odor of pears. What is the percent carbon in isoamyl acetate?
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Chapter 7 Slide 51 of 75
Molar mass C7H14O2 = 7C(12.01) + 14H(1.008)
+ 2O(16.00) = 130.18 g/mol
Total C = 7C(12.01) = g
% C = total g C x 100 total g cpd
% C = 84.07 g C x 100 = 64.58 % C 130.18 g cpd
Solution
C7H14O2
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Chapter 7 Slide 52 of 75
The empirical formula
• Is the simplest whole number ratio of the atoms.
• Is calculated by dividing the subscripts in the actual (molecular) formula by a whole number to give the lowest ratio.
C5H10O5 5 = C1H2O1 = CH2Oactual (molecular) empirical formula
formula
Empirical Formulas
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Chapter 7 Slide 53 of 75
Some Molecular and Empirical Formulas
• The molecular formula is the same or a multiple of the empirical.
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Chapter 7 Slide 54 of 75
1. What is the empirical formula for _________?
A) C2H4 B) CH2 C) CH
2. What is the empirical formula for _________?
A) C4H7 B) C6H12 C) C8H14
3. Which is a possible molecular formula for ______?
A) C4H4O4 B) C2H4O4 C) C3H6O3
Learning Check P-1
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Chapter 7 Slide 55 of 75
A compound contains 7.31 g Ni and 20.0 g Br. Calculate its empirical (simplest) formula.
Lets ask, what does the empirical formula tell us?
The ratio of atoms of each element,
ratio of moles of each element.
Action Plan: Convert mass (g) to moles.
Learning Check
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Chapter 7 Slide 56 of 75
Convert 7.31 g Ni and 20.0 g Br to moles.
7.31 g Ni x 1 mol Ni = 0.125 mol Ni
58.69 g Ni
20.0 g Br x 1 mol Br = 0.250 mol Br
79.90 g Br
Divide by smallest:
0.125 mol Ni = 1 Ni 0.250 mol Br = 2 Br
0.125 0.125
Write ratio as subscripts: NiBr2
Solution
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Chapter 7 Slide 57 of 75
Converting Decimals to Whole Numbers
When the number of moles for an element is adecimal, all the moles are multiplied by a small
integer to obtain whole number.
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Chapter 7 Slide 58 of 75
Aspirin is 60.0% C, 4.5 % H and 35.5 % O. Calculate its empirical (simplest) formula.
We have percentages, not grams.
However, the percentages are really an expression of grams of an element in 100. (exact) grams of compound.
Learning Check
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Chapter 7 Slide 59 of 75
STEP 1. Calculate the moles of each element in 100. g of the compound.
100. g aspirin contains 60.0% C or 60.0 g C, 4.5% H or 4.5 g H, and 35.5% O or 35.5 g O.
60.0 g C x 1 mol C = 5.00 mol C 12.01 g C
4.5 g H x 1 mol H = 4.5 mol H 1.008 g H
35.5 g O x 1mol O = 2.22 mol O
16.00 g O
Solution
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Chapter 7 Slide 60 of 75
Solution (continued)
STEP 2. Divide by the smallest number of mol.
5.00 mol C = 2.25 mol C (decimal)
2.22 4.5 mol H = 2.0 mol H2.222.22 mol O = 1.00 mole O2.22
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Chapter 7 Slide 61 of 75
Solution (continued)
3. Use the lowest whole number ratio as subscripts
When the moles are not whole numbers, multiply by a factor to give whole numbers, in this case 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
Using these whole numbers as subscripts the simplest formula is
C9H8O4
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Chapter 7 Slide 62 of 75
Molecular Formulas
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Chapter 7 Slide 63 of 75
A molecular formula• Is a multiple (or equal) of its empirical formula.
• Has a molar mass that is the empirical formula mass multiplied by a whole number.
molar mass = a whole number empirical mass
• Is obtained by multiplying the empirical formula by a whole number.
Relating Molecular and Empirical Formulas
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Chapter 7 Slide 64 of 75
Diagram of Molecular and Empirical Formulas
A small integer links• A molecular formula and its empirical
formula.• A molar mass and its empirical formula
mass.
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Chapter 7 Slide 65 of 75
Determine the molecular formula of compound thathas a molar mass of 78.11 g and an empiricalformula of CH.
STEP 1. Empirical formula mass of CH = 13.02 g
STEP 2. Divide the molar mass by the empirical mass.
78.11 g = 5.999 ~ 6 13.02 g
STEP 3. Multiply each subscript in C1H1 by 6.
molecular formula = C1x 6 H1 x 6 = C6H6
Finding the Molecular Formula
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Chapter 7 Slide 66 of 75
Some Compounds with Empirical Formula CH2O
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formaldehyde
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Chapter 7 Slide 67 of 75
A compound has a molar mass of 176.1g and an empirical formula of C3H4O3. What is the molecular formula?
What do we need to do?
From the empirical formula, determine the
empirical mass.
Then determine the whole number integer.
Learning Check
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Chapter 7 Slide 68 of 75
A compound has a formula mass of 176.1 and an empirical formula of C3H4O3. What is the molecular formula?
C3H4O3 = 88.06 g/EF
176.1 g (molar mass) = 2.00 88.06 g (empirical mass)
Molecular formula = 2 x empirical formulaC3 x 2H4 x 2O3 x 2 = C6H8O6
Solution
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Chapter 7 Slide 69 of 75
A compound contains C 24.27%, H 4.07%, and Cl 71.65%. The molar mass is about 99 g. What are the empirical and molecular formulas?
STEP 1. Calculate the empirical formula.
Write the mass percents as the grams in a 100.00-g sample of the compound.
C 24.27 g H 4.07 g Cl 71.65 g
Molecular Formula
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Chapter 7 Slide 70 of 75
Finding the Molecular Formula (Continued)Calculate the number of moles of each element.
24.27 g C x 1 mol C = 2.021 mol C 12.01 g C
4.07 g H x 1 mol H = 4.04 mol H 1.008 g H
71.65 g Cl x 1 mol Cl = 2.021 mol Cl 35.45 g Cl
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Chapter 7 Slide 71 of 75
Finding the Molecular Formula (Continued)Divide by the smallest number of moles:
2.021 mol C = 1 mol C
2.021
4.04 mol H = 2 mol H
2.021
2.02 mol Cl = 1 mol Cl
2.021
Empirical formula = C1H2Cl1 = CH2Cl
Calculate empirical mass (EM) CH2Cl = 49.48 g
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Chapter 7 Slide 72 of 75
Finding the Molecular Formula (Continued)
STEP 2. Divide molar mass by empirical mass.Molar mass = 99 g = 2
Empirical mass 49.48 g
STEP 3. Multiply the empirical formula by the small integer to determine the molecular formula.
2 x (CH2Cl)
C1 x2 H2 x 2 Cl1x 2 = C2H4Cl2
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Chapter 7 Slide 73 of 75
A compound is 27.4% S, 12.0% N and 60.6 % Cl. If the compound has a molar mass of 351 g, what is the molecular formula?
Learning Check
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Chapter 7 Slide 74 of 75
In 100. g, there are 27.4 g S, 12.0 g N, and 60.6 g Cl.
27.4 g S x 1 mol S = 0.854 mol S 32.07 g S
12.0 g N x 1 mol N = 0.857 mol N 14.01 g N
60.6 g Cl x 1mol Cl = 1.71 mol Cl 35.45 g Cl
Solution
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Chapter 7 Slide 75 of 75
Divide by the smallest number of moles
0.854 mol S /0.854 = 1.00 mol S
0.857 mol N/0.854 = 1.00 mol N
1.71 mol Cl/0.854 = 2.00 mol Cl
empirical formula = SNCl2 = 116.98 g
Molar Mass/ Empirical mass
351 g = 3
116.98 g molecular formula = (SNCl2)3 = S3N3Cl6
Solution (continued)