Transcript Chapter 3
Chem 106, Prof. J.T. Spencer
CHE 106: General Chemistry
1
CHAPTER THREE
Copyright © James T. Spencer 1995 - 1999 All Rights Reserved
Chem 106, Prof. J.T. Spencer
Stoichiometry
2
Chapter Three
Chem 106, Prof. J.T. Spencer
Stoichiometry
3
• Antoine Lavoisier (1734 - 1794) – Law of Conservation of Mass - atoms are neither
created nor destroyed in chemical reactions
– total number of atoms = total number of atoms after reaction
before reaction
– Stoichiometry - quantitative study of chemical
formulas and reactions (Greek; “stoichion”= element, “metron” = measure)
• Chemical Equations - used to describe chemical
reactions in an accurate and convenient fashion 2H 2 + O 2
reactants
2 H 2 O
products
Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
4 To Write and Balance: (Shorthand Communication for a great deal of information) (1) Know Reactants (2) Know ALL Products (3) Balance - Same Number and Kinds of atoms on each side
Chem 106, Prof. J.T. Spencer
Chemical Equations
5
• Chemical Equations – Must have equal numbers of atoms of
each element on each side of the equation = BALANCED EQUATION 2 H 2 + O 2 4 hydrogen 2 oxygen 2 H 2 O 4 hydrogen 2 oxygen N 2 O 5(g) + H 2 nitrogen 6 oxygen 2 hydrogen 2 O 2 HNO 2 nitrogen 6 oxygen 2 hydrogen 3 NOTE The coefficients in front of the formula for a compound refers to the number of molecules (intact) involved while a subscript refers to the ratio of atoms within the molecule Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
6
• Chemical Equations – balancing equations often requires
some trial and error of coefficients PCl 3(l) + 3 H 2 O
(l)
6 hydrogen 3 oxygen 1 phosphorus 3 chloride H 3 PO 3(aq) + 3 HCl 6 hydrogen 3 oxygen 1 phosphorus 3 chlorine C 6 H 12(l) + 9 O 2(g) 6 carbon 18 oxygen 12 hydrogen 6 CO 2(g) + 6 H 2 O
(l)
6 carbon 18 oxygen 12 hydrogen NOTE Never change subscripts in formulas when balancing chemical reactions!
subscripts change compounds; coefficients change amounts Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
7 Sample exercise: Balance the following equations by providing the missing coefficients: C 2 H 4 + O 2 CO 2 + H 2 O Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
8 Sample exercise: Balance the following equations by providing the missing coefficients: C 2 H C 4 + O 2 CO C 2 + H 2 O H O H O Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
9 Sample exercise: Balance the following equations by providing the missing coefficients: C 2 H 4 C 2 + O 2 CO 2 C 1 + H 2 O H 4 O 2 H 2 O 3 Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
10 Sample exercise: Balance the following equations by providing the missing coefficients: C 2 H 4 C 2 + O 2 2 CO 2 + H C (1)2= 2 2 O H 4 O 2 H 2 O 3 5 Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
11 Sample exercise: Balance the following equations by providing the missing coefficients: C 2 H 4 C 2 + O 2 2 CO 2 + 2H C (1)2= 2 2 O H 4 O 2 H (2)2 = 4 O 3 5 6 Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
12 Sample exercise: Balance the following equations by providing the missing coefficients: C 2 H 4 C 2 + 3O 2 2 CO 2 + 2H C (1)2= 2 2 O H 4 O (2)3 = 6 H (2)2 = 4 O 3 5 6 Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
13 Sample exercise: Balance the following equations by providing the missing coefficients: Al + HCl AlCl 3 + H 2 Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
14 Sample exercise: Balance the following equations by providing the missing coefficients: Al + HCl AlCl Al Al 3 + H 2 H Cl H Cl Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
15 Sample exercise: Balance the following equations by providing the missing coefficients: Al + HCl AlCl 3 Al 1 Al 1 + H 2 H 1 Cl 1 H 2 Cl 3 Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
16 Sample exercise: Balance the following equations by providing the missing coefficients: Al + 3HCl AlCl Al 1 Al 1 3 + H 2 H (1)3 = 3 Cl (1)3 = 3 H 2 Cl 3 Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
17 Sample exercise: Balance the following equations by providing the missing coefficients: Al + 6HCl AlCl Al 1 Al 1 3 + 3H 2 H (1)6 = 6 Cl (1)6 = 6 H (2)3 = 6 Cl 3 Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
18 Sample exercise: Balance the following equations by providing the missing coefficients: Al + 6HCl 2AlCl 3 Al 1 + 3H Al (1)2 = 2 2 H (1)6 = 6 Cl (1)6 = 6 H (2)3 = 6 Cl (3)2 = 6 Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Equations
19 Sample exercise: Balance the following equations by providing the missing coefficients: 2Al + 6HCl 2AlCl 3 Al (1)2 = 2 + 3H Al (1)2 = 2 2 H (1)6 = 6 Cl (1)6 = 6 H (2)3 = 6 Cl (3)2 = 6 Chapt. 3.1
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
20
• Chemical Reactions –The course of a chemical reaction can often
be predicted by recognizing general patterns of reactivity through similar reactions previously observed. Elements in same family (column of table) have similar reactions.
–The periodic table is helpful in predicting
products of reactions. Atoms like to assume electron configurations of the Noble Gases.
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
21
• Chemical Reactions
Example, if you know that 2Li + 2H 2 0 2LiOH + H 2 then you should be able to predict the products from the reaction of Na, K and the other members of group 1 (alkali metals) with water. Thus a general reaction would be; 2 M + 2 H 2 O 2 MOH + H 2 Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
22
•Combustion Reactions •Combination Reactions •Decomposition Reactions •Metathesis Reactions
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
23
• Combustion Reactions – Reactions with oxygen (usually from the air) – The complete combustion of hydrocarbons yield
carbon dioxide (CO 2 ) and water (H 2 O) C x H y + (2x+y) / 2 O 2 X CO 2 + Y / 2 H 2 O Generally: Balance Carbon Atoms First Balance Hydrgoens Balance Oxygen Atoms Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
24
• Combustion Reactions – Reactions with oxygen (usually from the air) – The complete combustion of hydrocarbons yield
carbon dioxide (CO 2 ) and water (H 2 O) C x H y + (2x+y) / 2 O 2 X CO 2 + Y / 2 H 2 O Examples octane: C 8 H 18 + 25 / 2 O 2 ethanol: C 2 H 5 OH + 3 O 2 glucose: C 6 H 12 O 6 styrene: C 8 H 8 + 9 O + 10 O 2 2 8 CO 2 2 CO 2 6 CO 2 8 CO 2 + 9 H 2 O + 3 H 2 O + 6 H 2 O + 4 H 2 O Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
25 Sample exercise: Write the balanced equation for the reaction that occurs when ethanol, C 2 H 5 OH(l) is burned in air.
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
26 Sample exercise: Write the balanced equation for the reaction that occurs when ethanol, C 2 H 5 OH(l) is burned in air.
C 2 H 5 OH + O 2 CO 2 + H 2 O Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
27 Sample exercise: Write the balanced equation for the reaction that occurs when ethanol, C 2 H 5 OH(l) is burned in air.
C 2 H 5 OH + O 2 CO 2 + H 2 O C C H O H O Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
28 Sample exercise: Write the balanced equation for the reaction that occurs when ethanol, C 2 H 5 OH(l) is burned in air.
C 2 H 5 OH + O 2 CO 2 + H 2 O C 2 C 1 H 6 O 3 H 2 O 3 Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
29 Sample exercise: Write the balanced equation for the reaction that occurs when ethanol, C 2 H 5 OH(l) is burned in air.
C 2 H 5 OH + O 2 2CO 2 + H 2 O C 2 C (1)2 = 2 H 6 O 3 H 2 O 3 5 Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
30 Sample exercise: Write the balanced equation for the reaction that occurs when ethanol, C 2 H 5 OH(l) is burned in air.
C 2 H 5 OH + O 2 2CO 2 + 3H 2 O C 2 C (1)2 = 2 H 6 O 3 H (2)3 = 6 O 3 5 7 Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
31 Sample exercise: Write the balanced equation for the reaction that occurs when ethanol, C 2 H 5 OH(l) is burned in air.
C 2 H 5 OH + 3O 2 2CO 2 + 3H 2 O C 2 C (1)2 = 2 H 6 O 3 7 H (2)3 = 6 O 3 5 7 Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
32
• Combination Reactions – two or more substances react to form a single
product
– especially common in the reactions of pure
elements A + B C Ni(s) + 4 CO(g) BF 3 (g) + NH 3 (g) Ni(CO) 4 (g) BF 3 NH 3 (s) Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
33
• Decomposition Reactions –when one compound reacts to form two
or more products (opposite of combination reactions) C A + B
• (often heat required)
2 NaN 3 (s) B(OH) 3 Air Bag Inflator (J. Chem. Ed. 1990, 67, 61)
(heat) 2 Na(s) + 3 N 2 (g) HBO 2 + H 2 O Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
34
• Metathesis Reactions –when ionic “partners” switch
AB + CD AD + BC
• (often in aqueous solutions)
Ag(NO 3 ) + KCl BaCl 2 + Na 2 SO 4 AgCl(s) + KNO 3 BaSO 4 (s) + 2 NaCl Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
35 Sample Exercise: Write balanced chemical equations for the following reactions: Solid mercury (II) sulfide decomposes into its component elements when heated .
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
36 Sample Exercise: Write balanced chemical equations for the following reactions: Solid mercury (II) sulfide decomposes into its component elements when heated .
Hg +2 S -2 HgS
Hg + S Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
37 Sample Exercise: Write balanced chemical equations for the following reactions: The surface of aluminum metal undergoes a combination reaction with oxygen in air.
Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
38 Sample Exercise: Write balanced chemical equations for the following reactions: The surface of aluminum metal undergoes a combination reaction with oxygen in air.
Al + O 2
Al +3 O -2 Al 2 O 3 Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
39 Sample Exercise: Write balanced chemical equations for the following reactions: The surface of aluminum metal undergoes a combination reaction with oxygen in air.
4Al + 3O 2 Al +3
O -2 2Al 2 O 3 Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
40 Sample Exercise: Write balanced chemical equations for the following reactions: Si 2 H 6 burns when exposed to air . Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
41 Sample Exercise: Write balanced chemical equations for the following reactions: Si 2 H 6 burns when exposed to air . Hint, Si is in the same group as C, and therefore reacts similarly.
Si 2 H 6 + O 2
SiO 2 + H 2 O Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Chemical Reactivity
42 Sample Exercise: Write balanced chemical equations for the following reactions: Si 2 H 6 burns when exposed to air . Hint, Si is in the same group as C, and therefore reacts similarly.
2Si 2 H 6 + 7O 2
4SiO 2 + 6H 2 O Chapt. 3.2
Chem 106, Prof. J.T. Spencer
Atomic and Molecular Weights
43
•Chemical equations indicate exactly the
amounts of two reagents which will react to form an exact amount of products
•Atomic Mass Scale - based upon
standard. 12 C isotope. This isotope is assigned a mass of exactly 12 atomic mass units (amu) and the masses of all other atoms are given relative to this
•Most elements in nature exist as mixtures of
isotopes.
Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic and Molecular Weights
44
• Atomic Mass Scale -
given the following; 100 g of water contains 11.1 g of H and 88.9 g of O and the formula for water is H ( 88.9
/ 11 = 8) 2 O then;
•water has 8 times more O than H by mass •if water has 2 H for 1 O then O atoms must weigh
16 time more than H atoms
•if H is assigned an atomic mass of 1 amu then O
must weigh 16 amu (using the 12 C standard)
•1 amu = 1.66054 x 10
-24 g OR 1 g = 6.02214 x 10 23 amu Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
45
• Direct methods of measuring (separating) mass. • Sample molecules are ionized by e-beam to cations (+1 by
“knocking off” one electron) which are then deflected by magnetic field - for ions of the same charge the angle of deflection in proportional to the ion’s mass vacuum chamber accelerating grid (-) sample beam of pos. ions N Mass Spectrum Hg Int.
ionizing e- beam focusing slits S magnetic field detector 200 mass number (amu) Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
46 Mass Spectrum Cl 35 Int.
Mass Spectrum C 12 Int.
Mass Spectrum P 31 Int.
37 13 mass number (amu) 35 Cl: 75% abundant 37 Cl: 24% abundant mass number (amu) mass number (amu) 12 C: 98.9% abundant 13 C: 1.11% abundant 31 P: 100% abundant
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
47 Unknown white powdery substance injested by unconscious patient. What do you do? Is it Heroin, Cocaine, Caffeine?
Mass Spectrum of Unknown Compound Mass
25 50 75 100 125 150 175 200 225 250 275 300
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
48 MS Library
other peaks at 327 and 369 43 204 215 268 94 146
Mass
25 50 75 100 125 150 175 200 225 250 275 300 194 Caffeine 67 109 42 55 82
MS of Unknown Mass
25 50 75 100 125 150 175 200 225 250 275 300
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
49 MS Library
82 182 303 42 122 150 272
Mass
25 50 75 100 125 150 175 200 225 250 275 300 194 Caffeine 67 109 42 55 82
MS of Unknown Mass
25 50 75 100 125 150 175 200 225 250 275 300
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
50
194
MS Library
67 109 55 82 42
Mass
25 50 75 100 125 150 175 200 225 250 275 300 194 Caffeine 67 109 42 55 82
MS of Unknown Mass
25 50 75 100 125 150 175 200 225 250 275 300
Chem 106, Prof. J.T. Spencer
Mass Spectrometer
51 Unknown white powdery substance injested by unconscious patient. What do you do?
Mass Spectrum O H 3 C N O N CH 3 Caffeine N CH 3 N Mol. Wgt = 194 Mass
25 50 75 100 125 150 175 200 225 250 275 300
Chem 106, Prof. J.T. Spencer
Atomic Weights
52
• Average Atomic Mass (AW)- weighted average (by
% natural abundance) of the isotopes of an element.
•Example (1);
10 B is 19.78% abundant with a mass of 10.013 amu 11 B is 80.22% abundant with a mass of 11.009 amu therefore the average atomic mass of boron is; (0.1987)(10.013) + (0.8022)(11.009) = 10.82 amu Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic Weights
53
• Average Atomic Mass (AW)- weighted average (by
% natural abundance) of the isotopes of an element.
•Example (2):
194 Pt is 33.90% abundant with a mass of 193.963 amu 195 Pt is 33.80% abundant with a mass of 194.965 amu 196 Pt is 25.30% abundant with a mass of 195.965 amu 198 Pt is 7.210% abundant with a mass of 197.968 amu therefore the average atomic mass of platinum is; (0.3390)(193.963) + (0.3380)(194.965) + (0.2530)(195.965 ) + (0.07210)(197.968)= 195.504 amu Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic and Molecular Weights
54 Sample exercise: Three isotopes of silicon occur in nature: 28 Si (92.21%), which has a mass of 27.97693 amu; 29 Si (4.70%), which has a mass of 28.97659 amu; and the atomic weight of silicon.
30 Si (3.09%), which has a mass of 29.97376 amu. Calculate Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic and Molecular Weights
55 Sample exercise: Three isotopes of silicon occur in nature: 28 Si (92.21%), which has a mass of 27.97693 amu; 29 Si (4.70%), which has a mass of 28.97659 amu; and the atomic weight of silicon.
30 Si (3.09%), which has a mass of 29.97376 amu. Calculate 27.97693(0.9221) + 28.97659(0.0470) + 29.97376(0.0309) = 28.0856 amu Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic and Molecular Weights
56 Sample exercise: Three isotopes of silicon occur in nature: 28 Si (92.21%), which has a mass of 27.97693 amu; 29 Si (4.70%), which has a mass of 28.97659 amu; and the atomic weight of silicon.
30 Si (3.09%), which has a mass of 29.97376 amu. Calculate 27.97693(0.9221) + 28.97659(0.0470) + 29.97376(0.0309) = 28.0856 amu * 3 sig figs 28.1 amu Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Atomic Weights
57
• Sample Problem
[masses for 63 : When a sample of natural copper is vaporized and injected into a mass spectrometer, the results shown are obtained. Use these data to compute the average mass of natural copper. Cu = 62.93 amu and 65 Cu = 64.93 amu] 69.09% 30.91% Given:
•Masses for
63 Cu and 65 C
•Relative abundance of
63 Cu and 65 Cu Find:
•Average Mass of Cu
Mass No. 63 65
Chem 106, Prof. J.T. Spencer
Atomic Weights
58 (.6909 atoms)(62.93 amu) + (.3091 atoms)(64.93amu) = 63.55 amu atom atom.
average mass per atom is; 6355 amu = 63.55 amu/atom 100 atoms 69.09% 30.91% Given:
•Masses for
63 Cu and 65 C
•Relative abundance of
63 Cu and 65 Cu Find:
•Average Mass of Cu
Mass No. 63 65
Chem 106, Prof. J.T. Spencer
Molecular Weights
59
• Formula Weights (FW) - sum of the atomic weights of each
atom in its chemical formula. (note AW is atomic weight)
• formula weight of NaN
3 = 3(AW of N) + 1(AW of Na) 3(14) + 1(23) = 65 amu for sodium azide
• Molecular Weights (MW) - sum of atomic weights of each
atom in its molecular formula
• molecular weight of B
2 H 6 = 2(AW of B) + 6(AW of H) 2(10.8) + 6(1) = 27.6 amu for diborane
• Difference between Molecular and Formula Weights • ionic compounds, with extended arrays, have no well
defined molecules (and no molecular formulas) so we use the formula weights (i.e., NaCl = 58 amu ) Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Molecular Weights
60
• Calculate molecular/formula weights for the following: – P
4 O 10
– BrCl – Ca(NO
3 ) 2 P =31; O = 16; Br = 80; Cl = 35.5; Ca = 40; N = 14
Chem 106, Prof. J.T. Spencer
Molecular Weights
61
• Calculate molecular weights/formula for the following: – P
4 O 10
4(31) + 10(16) = 284 amu
– BrCl
1(80) + 1(35.5) = 115.5 amu
– Ca(NO
3 ) 2
1(40) + 2(14) + 6(16) = 164 amu
P =31; O = 16; Br = 80; Cl = 35.5; Ca = 40; N = 14
Chem 106, Prof. J.T. Spencer
Percentage Composition
62
• Percentage Composition - percentage by mass contributed
by each element in the substance. May be used to verify the purity or identity of a particular compound.
• 100 [
(atoms of an element in formula)(AW) / FW ] = % comp. element Percentage Composition of C 6 H 12 O 6 (FW = 180) % C = 100 (6)(12)/ (180) = 40.0% carbon % O = 100 (6)(16)/ (180) = 53.3% oxygen % H = 100 (12)(1)/ (180) = 6.7% hydrogen 2 example calculations follows Chapt. 3.3
Chem 106, Prof. J.T. Spencer
Percentage Composition
63 Problem : In 1987, the first substance to act as a superconductor at a temperature above that of liquid nitrogen (77 K) was discovered. The approximate formula of the substance is YBa 2 Cu 3 O 7 . Calculate the percent composition by mass of this material.
M W of YBa 2 Cu 3 O 7 = (88.9) + 2(137.3) + 3(63.6) + 7(16) = 666.0 amu AW: Y = 88.9; Ba =137.3; Cu = 63.6; O = 16.0
Chem 106, Prof. J.T. Spencer
Percentage Composition
64 M W of YBa 2 Cu 3 O 7 = 666.0 amu Y = Ba = Cu O = = 1(88.9) 2(137.3) 3(63.5) 7(16.0) = 100 (88.9) 666.0
= 100 (274.6) 666.0
= 100 (190.5) 666.0
= 100 (112) 666.0
AW: Y = 88.9; Ba =137.3; Cu = 63.6; O = 16.0
= 13.3 % = 41.3 % = 28.6 % = 16.8 %
Chem 106, Prof. J.T. Spencer
Percentage Composition
65 Sample exercise: Calculate the percentage of nitrogen, by mass, in Ca(NO 3 ) 2 .
Chem 106, Prof. J.T. Spencer
Percentage Composition
66 Sample exercise: Calculate the percentage of nitrogen, by mass, in Ca(NO 3 ) 2 .
Formula Mass: 1(40.1) = 40.1
2(14.0) = 28.0
6(16.0) = 96.0
164.1 amu
Chem 106, Prof. J.T. Spencer
Percentage Composition
67 Sample exercise: Calculate the percentage of nitrogen, by mass, in Ca(NO 3 ) 2 .
Formula Mass: part x 100 1(40.1) = 40.1 total 2(14.0) = 28.0
6(16.0) = 96.0
164.1 amu
Chem 106, Prof. J.T. Spencer
Percentage Composition
68 Sample exercise: Calculate the percentage of nitrogen, by mass, in Ca(NO 3 ) 2 .
Formula Mass: part x 100 1(40.1) = 40.1 total 2(14.0) = 28.0
6(16.0) = 96.0 28.0 x 100 = 164.1 amu 164.1
Chem 106, Prof. J.T. Spencer
Percentage Composition
69 Sample exercise: Calculate the percentage of nitrogen, by mass, in Ca(NO 3 ) 2 .
Formula Mass: part x 100 1(40.1) = 40.1 total 2(14.0) = 28.0
6(16.0) = 96.0 28.0 x 100 = 17.1% 164.1 amu 164.1
Chem 106, Prof. J.T. Spencer
Percentage Composition
70
• Example ; determine the elemental
percent composition of strychnine N
•Molecular formula of
strychnine = C 21 H 22 N 2 O 2 O N O
•Molecular weight (MW)
MW = 21(12) + 22(1) + 2(14) + 2(16) = 334 amu
•Percent composition;
% C = 100 (21)(12)/ (334) = 75.4% carbon % N = 100 (2)(14)/ (334) = 6.59% nitrogen % O = 100 (2)(16)/ (334) = 9.59% oxygen % H = 100 (22)(1)/ (334) = 8.38% hydrogen Chapt. 3.3
Mole
Chem 106, Prof. J.T. Spencer 71
• Very small macroscopic samples contain VERY many
atoms, molecules, etc... (e.g. 1 tsp. H 2 O contains 2 x 10 23 mol). [Need convenient counting unit]
• known no. of H atoms in 1 g of H = no. of atoms of O in
16 g of O = no. of C atoms in 12 g of C = etc... (based upon atomic weights)
• Def. = The number of carbon atoms in 12 g of
12 C is called Avogadro’s number Avogadro’s number . One Mole (latin “mole” = a mass) is the amount of material that contains Chapt. 3.4
•Note: a mole refers to a fixed number of any
type of particles!
•Avogadro’s number = 6.023 x 10
23
Chem 106, Prof. J.T. Spencer Avagadro’s Number and the Mole 72
• 1 mole of
12 C atoms = 6.02 x 10 23 atoms
• 1 mole of
11 B atoms = 6.02 x 10 23 atoms
• 1 mol of PCl
3 molecules = 6.02 x 10
• 1 mol of Na+ ions = 6.02 x 10
23 23 molecules Na ions
• 1 mol of toasters = 6.02 x 10
23 toasters
• 1 mole of students = 6.02 x 10
23 students Chapt. 3.4
Chem 106, Prof. J.T. Spencer
Mole
73
•Examples
How many C atoms in 0.5 moles of Carbon?
C atoms = (0.5 moles C)(6.02 x 10 23 mole atoms) = 3.01 x 10 23
How many C atoms are in 0.25 mol of C 6 H 12 O 6 ?
C ato. = (0.25 mol C 6 H 12 O 6 )(6.02 x 10 23 molec) (6 C atoms) mol (1 C 6 H 12 O 6 molec) C atoms = 1.5 x 10 23 atoms Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 74 Sample exercise: How many O atoms are in a) 0.25 moles Ca(NO 3 ) 2 Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 75 Sample exercise: How many O atoms are in a) 0.25 moles Ca(NO 3 ) 2 0.25 mol Ca(NO 3 ) 2 Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 76 Sample exercise: How many O atoms are in a) 0.25 moles Ca(NO 3 ) 2 0.25 mol Ca(NO 3 ) 2 6 mole O 1 mole Ca(NO 3 ) 2 Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 77 Sample exercise: How many O atoms are in a) 0.25 moles Ca(NO 3 ) 2 0.25 mol Ca(NO 3 ) 2 6 mole O 1 mole Ca(NO 3 ) 2 1.5 mol O 6.02 x 10 23 1 mol O atom O Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 78 Sample exercise: How many O atoms are in a) 0.25 moles Ca(NO 3 ) 2 0.25 mol Ca(NO 3 ) 2 6 mole O 1 mole Ca(NO 3 ) 2 1.5 mol O 6.02 x 10 23 1 mol O atom O = 9.03 x 10 23 Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 79 Sample exercise: How many O atoms are in a) 1.50 moles sodium carbonate Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 80 Sample exercise: How many O atoms are in a) 1.50 moles sodium carbonate Na +1 CO 3 -2
Na 2 CO 3 Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 81 Sample exercise: How many O atoms are in a) 1.50 moles sodium carbonate Na +1 CO 3 -2
Na 2 CO 3 1.50 mol Na 2 CO 3 Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 82 Sample exercise: How many O atoms are in a) 1.50 moles sodium carbonate Na +1 CO 3 -2
Na 2 CO 3 1.50 mol Na 2 CO 3 3 mol O 1 mol Na 2 CO 3 Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 83 Sample exercise: How many O atoms are in a) 1.50 moles sodium carbonate Na +1 CO 3 -2
Na 2 CO 3 1.50 mol Na 2 CO 3 3 mol O 1 mol Na 2 CO 3 4.50 mol O 6.02 x 10 23 atom O 1 mol O Chapt. 3.4
Mole
Chem 106, Prof. J.T. Spencer 84 Sample exercise: How many O atoms are in a) 1.50 moles sodium carbonate Na +1 CO 3 -2
Na 2 CO 3 1.50 mol Na 2 CO 3 3 mol O 1 mol Na 2 CO 3 4.50 mol O 6.02 x 10 23 atom O = 2.71 x 10 1 mol O 24 Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Mass 85
• Example - since one
12 C atoms weighs 12 amu and a 24 Mg atom weighs 24 amu (twice as massive) and since a mole always contains the same number of particles, a mole of 24 weigh twice as much as a mole of 12 C.
Mg must 1 12 C atom weighs 12 amu; 1 mol 12 C weigh 12 g 1 24 Mg atom weighs 24 amu; 1 mol 24 Mg weigh 24 g 1 238 U atom weighs 238 amu; 1 mol 238 U weighs 238 g Molar Mass - (in grams) of any substance is always numerically equal to its formula weight (in amu). Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 86 Name Formula Formula Mass of 1 mol Number and kind weight of form units of particles in 1 mol atomic nitrogen N 14.0
14.0
6.02 X 10 23 N atoms molec. nitrogen N scandium ScCl 3 151.5
151.5
6.02 X 10 23 chloride 6.02 X 10 23 ScCl 3 Sc 3+ units ions 3(6.02 X 10 23 ) Cl- ions glucose C 6 H 2 12 O 6 28.0
180.0
28.0
180.0
Chapt. 3.4
6.02 X 10 23 N 2 2(6.02 X 10 23 molec.
) N atoms 6.02 X 10 23 gluc. molec.
6(6.02 X 10 23 ) C atoms 12(6.02 X 10 23 ) H atoms
Chem 106, Prof. J.T. Spencer Molar Relationships 87 (1) How many moles of phosphorus trichloride, PCl
3 , are in
50 g of the substance? (MW = 137.4 amu) Moles of PCl 3 = 1mol PCl 3 (50 g PCl 3 ) = 0.36 moles PCl 3 137.4 g (2) How many molecules of PCl molecules of PCl 3
3 are in 50 g?
= (0.36 moles )(6.023 x 10 23 = 2.2 x 10 23 PCl 3 molecules molecules) 1 mole (3) How many grams of PCl
3 are in 0.75 moles?
x grams = (0.75 mole)(137.4 g PCl 3 ) = 103 g of PCl 3 1 mole PCl 3 Chapt. 3.4
Grams Chem 106, Prof. J.T. Spencer Molar Relationships 88 use molar mass use molar mass Moles use Avagadro’s number use Avagadro’s number Items (molecules, atoms, etc...) Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 89 Sample exercise: How many moles of NaHCO 3 are present in 508 g of this substance?
Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 90 Sample exercise: How many moles of NaHCO 3 are present in 508 g of this substance?
508 g Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 91 Sample exercise: How many moles of NaHCO 3 are present in 508 g of this substance?
508 g 1 mol 84 g Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 92 Sample exercise: How many moles of NaHCO 3 are present in 508 g of this substance?
508 g 1 mol = 6.05 g NaHCO 3 84 g Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 93 Sample exercise: What is the mass, in grams, of a) 6.33 mol NaHCO 3 Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 94 Sample exercise: What is the mass, in grams, of a) 6.33 mol NaHCO 3 6.33 mol NaHCO 3 84 g NaHCO 3 1 mol NaHCO 3 = 532 g NaHCO 3 Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 95 Sample exercise: What is the mass, in grams, of b) 3.0 x 10 -5 mol sulfuric acid Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 96 Sample exercise: What is the mass, in grams, of b) 3.0 x 10 -5 mol sulfuric acid 3.0 x 10 -5 mol H 2 SO 4 98 g H 2 SO 4 1 mol H 2 SO 4 = 2.9 x 10 -3 g Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 97 Sample exercise: How many nitric acid molecules are in 4.20 g of HNO 3 ?
Chapt. 3.4
Chem 106, Prof. J.T. Spencer Molar Relationships 98 Sample exercise: How many nitric acid molecules are in 4.20 g of HNO 3 ?
4.20 g HNO 3 6.02 x 10 23 molec HNO 3 63 g HSO 3 = 4.01 x 10 22 molec Chapt. 3.4
Chem 106, Prof. J.T. Spencer Empirical Formulas 99
• Empirical Formula - Relative number of each element in a
compound.
• Using moles and percent weight (elemental analysis by
chemical means), we can calculate an empirical formula
• Steps; » assume a 100 g (convenient since working with %
because the elements % can be thought of as g)
» calculate moles of element present in 100g sample » find ratios of moles (approx) to lead to integral
formula subscripts.
Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas 100 Mass Percent of Elements assume 100 g sample Grams of each Element Use atomic weights Empirical Formula calculate mole ratio Moles of each Elements Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 101 Determine the empirical formula for a compound which contains 87.5 % N and 12.5% H by mass.
% g in 100g moles ratio 87.5 % N = 87.5 g N 1 mole = 6.25 moles N = 1 14 g 12.5% H = 12.5 g H 1 mole H = 12.5 moles H = 2 1 g Empirical Formula = NH 2 Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 102 Sample exercise: A 5.325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?
Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 103 Sample exercise: A 5.325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?
3.758 g x 100 = 70.58% C -> 70.58 g C 1 mol C = 5.88 mol C 5.325 g 12 g C Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 104 Sample exercise: A 5.325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?
0.316 g x 100 = 5.93% H -> 5.93 g H 1 mol H = 5.93 mol H 5.325 g 1 g H Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 105 Sample exercise: A 5.325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?
1.251 g x 100 = 23.49% O -> 23.49 g O 1 mol O = 1.47 mol O 5.325 g 16 g O Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 106 Sample exercise: A 5.325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?
5.88 mol C ; 5.93 mol H ; 1.47 mol O Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 107 Sample exercise: A 5.325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?
5.88 mol C ; 5.93 mol H ; 1.47 mol O 1.47 mol 1.47 mol 1.47 mol 4 4 1 Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 108 Sample exercise: A 5.325 g sample of methyl benzoate, a compound used in the manufacture of perfumes, is found to contain 3.758 g of carbon, 0.316 g of hydrogen, and 1.251 g of oxygen. What is the empirical formula of this substance?
5.88 mol C ; 5.93 mol H ; 1.47 mol O C 4 H 4 O 1.47 mol 1.47 mol 1.47 mol 4 4 1 Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 109 A white unknown substance (mass spec. problem) found on an unconscious patient is suspected by a forensic chemist of being either cocaine or caffeine. Combustion of a 50.86 mg sample yielded 150.0 mg of CO of cocaine is C 17 H 21 NO 4 2 and 46.05 mg of water. Further analysis showed the compound contained 9.39% N by mass. The formula . Can the substance be cocaine?
Unknown % C, H, and O.
Percent Composition Known 50.86 mg of cmpd gave 150.0 mg of CO 2 46.05 mg of H 2 O.
compound contains 9.39% N.
formula of Cocaine is C 17 H 21 NO 4 .
and Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 110
•Combustion Reactions –Reactions with oxygen (usually from the air) –The complete combustion of hydrocarbons yield carbon
dioxide (CO 2 ) and water (H 2 O) C x H y + (2x+y) / 2 O 2 X CO 2 + Y / 2 H 2 O Known 50.86 mg of cmpd gave 150.0 mg of CO 2 46.05 mg of H 2 O.
compound contains 9.39% N.
formula of Cocaine is C 17 H 21 NO 4 .
and Chapt. 3.5
Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 111 Compute % C and % H (from combustion).
C: H: N: O: 0.150 g CO 2 = 0.00341 mol CO 2 = 0.00341 mol C = 40.9 mg C 0.04605 g H 2 O = 0.00256 mol H 2 O = 0.00512 mol H = 5.12 mg H (0.0939)(50.86) = 4.77 mg N = 0.000341 mol N (50.86 g sample)-(40.9 mg C + 5.12 mg H + 4.77 mg N) = = 0.08 mg O = 0.000006 mol O Known 50.86 mg of cmpd gave 150.0 mg of CO 2 46.05 mg of H 2 O.
compound contains 9.39% N.
formula of Cocaine is C 17 H 21 NO 4 .
and
C: H: N: O: Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 112 mg in sample 40.9 mg C 5.12 mg H 4.77 mg N 0.08 mg O % in sample calc’n 100 (40.9mg/50.86 mg) 100 (5.11mg/50.86mg) 100 (4.77mg/50.86 mg) 100 (0.08mg/50.86mg) % sample % cocaine 80.5 % C 67.3% C 10.1% H 9.4% N 0.02% O 6.9% H 4.6% N 21.2% O Known 50.86 mg of cmpd gave 150.0 mg of CO 2 46.05 mg of H 2 O.
compound contains 9.39% N.
formula of Cocaine is C 17 H 21 NO 4 .
and
C: H: N: O: Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 113 g in cocaine % in sample calc’n (303 grams) 17(12) = 204 g C 100 (204/303) 21(1) g H 100 (21/303) 1(14) g N 4(16) g O 100 (14/303) 100 (64/303) % sample % cocaine 80.5 % C 67.3% C 10.1% H 9.4% N 0.02% O 6.9% H 4.6% N 21.2% O Known: formula of Cocaine is C 17 H 21 NO 4 .
MW = 17(12) + 21(1) + 1(14) + 4(16) = 303 amu
C: H: N: O: Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 114 mg in sample 40.9 mg C 5.12 mg H 4.77 mg N 0.08 mg O % in sample calc’n 100 (40.9mg/50.86 mg) 100 (5.11mg/50.86mg) 100 (4.77mg/50.86 mg) 100 (0.08mg/50.86mg) % sample % cocaine 80.5 % C 67.3% C 10.1% H 9.4% N 0.02% O 6.9% H 4.6% N 21.2% O
C: H: N: O: Chem 106, Prof. J.T. Spencer Empirical Formulas, Examples 115 mg in sample 40.9 mg C 5.12 mg H 4.77 mg N 0.08 mg O % in sample calc’n 100 (40.9mg/50.86 mg) 100 (5.11mg/50.86mg) 100 (4.77mg/50.86 mg) 100 (0.08mg/50.86mg) % sample % cocaine 80.5 % C 67.3% C 10.1% H 9.4% N 0.02% O 6.9% H 4.6% N 21.2% O
Compound is Not Cocaine from analysis
Chem 106, Prof. J.T. Spencer
Combustion Reactions
116
Reaction of Hydrogen with Oxygen [COMBUSTION] (note precautions)
» »
2 H 2 (g) + O 2 (g) 2 H 2 O(g)
H = -232 kJ/mol H 2 O Ignition temperature = 580 ° - 590° C
»
Explosive [“when stuff gets really big really
fas
t” Beakman’ World]
»
The rapid release of energy [-232 kJ/mol H 2 O] into the surrounding air causes the air to very quickly expand. the explosion from pure H 2 sound quiter because the air expansion is slower.
[Video No. 20-21; 4:42 +1:29 m]
[Video No. 22; 2:50 m] Chem 106, Prof. J.T. Spencer
Combustion Reactions
117
Combustion of Alcohol (ethanol): C 2 H 5 OH(g) + 3 O 2 (g) 2CO 2 (g) + 3 H 2 O(g)
H =1366.2 kJ mol -1
Tesla coil produces a high voltage electric spark. The spark is required to initiate this reaction.
Conversion of chemical energy (PE stored in bonds) to mechanical energy.
Questions for After Demonstration Are other types of energy are produced besides mechanical energy?
Why can the reaction not be repeated without flushing the bottle with air first?
O 2 flow Chem 106, Prof. J.T. Spencer Combustion Analysis 118
• Empirical Formula from reaction with oxygen • Organic Compounds - C to CO
2 and H to H 2 O
• Use CO
2 and H and H in original sample furnace 2 O to determine the amount of C H 2 O absorbant (Mg(ClO 4 ) 2 ) sample contaminant catalyst (CuO); oxidizes traces of CO and C to CO 2 CO 2 absorbant (NaOH) Chapt. 3.5
Chem 106, Prof. J.T. Spencer Combustion Analysis 119 The combustion of 5.00g of an organic compound containing C, H, and O yields 9.57 g CO 2 and 5.87 g H 2 O. What is the empirical formula?
Chapt. 3.5
Chem 106, Prof. J.T. Spencer Combustion Analysis 120 The combustion of 5.00g of an organic compound containing C, H, and O yields 9.57 g CO 2 and 5.87 g H 2 O. What is the empirical formula?
9.57 g CO 2 = 0.217 mol CO 2 = 0.217 mol C = 2.60 g C 5.87 g H 2 O = 0.326 mol H 2 O = 0.652 mol H = 0.652 g H whatever is left over must be the amt. of O originally present; 5.00 g - (2.60 g C + 0.652 g H) = 1.75 g of O = 0.109 mol O thus; 0.217 mol C 0.109 mol O 0.652 mol H divide each by 0.109
1.99
1.00 thus C 2 H 6 O 5.98
(ethanol)
Chapt. 3.5
Chem 106, Prof. J.T. Spencer Combustion Analysis 121 The combustion of 0.596 g of a compound containing only B and H yields 1.17 g H 2 O and all the boron is recovered as B 2 O 3 . What is the empirical formula?
Chapt. 3.5
Chem 106, Prof. J.T. Spencer Combustion Analysis 122 The combustion of 0.596 g of a compound containing only B and H yields 1.17 g H 2 O and all the boron is recovered as B 2 O 3 . What is the empirical formula?
(1) Chem Equation; B x H y + O 2 H 2 O + B 2 O 3 Chapt. 3.5
Chem 106, Prof. J.T. Spencer Combustion Analysis 123 The combustion of 0.596 g of a compound containing only B and H yields 1.17 g H 2 O and all the boron is recovered as B 2 O 3 . What is the empirical formula?
(1) Chem Equation; B x H y + O 2 (2) 1.17 g H 2 O = 1.17g = 0.065 mol H 2 O H 2 O + B 2 O 3 18 g/mol (3) g H = (0.065 mol H 2 O) (1 g H 2 O) (2 H mol H) = 0.130 mol H 1 mol H 1 mol H 2 O 0.130 mol H = 0.130 g H Chapt. 3.5
Chem 106, Prof. J.T. Spencer Combustion Analysis 124 The combustion of 0.596 g of a compound containing only B and H yields 1.17 g H 2 O and all the boron is recovered as B 2 O 3 . What is the empirical formula?
(1) Chem Equation; B x H y + O 2 (2) 1.17 g H 2 O = 1.17g = 0.065 mol H 2 O H 2 O + B 2 O 3 18 g/mol (3) g H = (0.065 mol H 2 O) (1 g H 2 O) (2 H mol H) = 0.130 mol H 1 mol H 1 mol H 2 O 0.130 mol H = 0.130 g H (4) (0.596 g tot)-(0.130 g H) = 0.466 g B; 0.466 g B = 0.043 mol B 10.8 g/ mol (5) B = 0.0431 mol = 1.00
H = 0.130 mol = 3.01
0.0431
BH 3 0.0431
Chapt. 3.5
Chem 106, Prof. J.T. Spencer Chemical Problem Solving 125
• Read and UNDERSTAND the problem - determine what
is being given and what is required.
• Identify the Unknown and Given data. • Set up the problem - determine what kinds of
information bear upon the problem, what solution pathways may be available, what chemical principles should give guidance, etc...
• Solve the problem - Use the data given and the
appropriate relationships or equations to work throught the problem.
• Check your work - not just the mathematical functions
but ask if the answer makes sense and provides what is being asked for! (sig. figs) Chapt. 3.5
Chem 106, Prof. J.T. Spencer Equations and the Mole 126
• Coefficients in a balanced chemical equation refer to
both the relative number of molecules involved in a reaction and the relative number of moles.
Chapt. 3.6
• Stoichiometric equivalence - from coefficients in a
chemical equation; B 2 H 6 + 3 O 2 3 H 2 O + B 2 O 3
• 1 mol B
2 H 6 equiv. to 3 moles O 2 equiv. to 3 mol H 2 0, ...
• Used to calculate quantities involved in a reaction
grams of compound A grams of compound B use molar mass of A moles of compound A use coeff of A and B from balanced eqn.
use molar mass of B moles of compound B
Chem 106, Prof. J.T. Spencer Mole Calculations 127
• Given the reaction for the formation of B
of LiH?
2 H 6 (diborane), how many grams of diborane can be prepared from 3.0 g 6 LiH + 8BF 3 [B 2 H 6 6 LiBF 4 + B 2 H MW = 27.6 and LiH MW = 7.9] 6 3.0 g LiH (1 mol LiH) = 0.38 mol LiH 7.9 g LiH 0.38 mol LiH (1 mol B 2 H 6 6 mol LiH ) 27.6 g B 2 H 6 1 mol B 2 H 6 = 1.7 g B 2 H 6 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 128
• Given the reaction for the formation of B
2 H 6 how many grams of BF 3 LiH ?
(diborane), are required to react with 3.0 g of 6 LiH + 8BF 3 [B 2 H 6 MW = 27.6, BF 3 6 LiBF 4 + B 2 H 6 = 67.8 and LiH MW = 7.9] 3.0 g LiH (1 mol LiH) (8 mol BF 3 ) (67.8 g BF 3 ) = 34 g BF 3 7.9 g LiH 6 mol LiH 1 mol BF 3 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 129 Sample exercise: A common laboratory method for preparing small amounts of O 2 involves the decomposition of KClO 3 : 2KClO 3
2KCl + 3O 2 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 130 Sample exercise: A common laboratory method for preparing small amounts of O 2 involves the decomposition of KClO 3 : 2KClO 3 4.50 g KClO
How many grams of oxygen is produced from 4.50 g KClO 3 2KCl + 3O 3 ?
2 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 131 Sample exercise: A common laboratory method for preparing small amounts of O 2 involves the decomposition of KClO 3 : 2KClO 3 4.50 g KClO
How many grams of oxygen is produced from 4.50 g KClO 3 2KCl + 3O 3 ?
1 mol KClO 3 122.6 g KClO 3 2 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 132 Sample exercise: A common laboratory method for preparing small amounts of O 2 involves the decomposition of KClO 3 : 2KClO 3 4.50 g KClO
How many grams of oxygen is produced from 4.50 g KClO 3 2KCl + 3O 3 ?
1 mol KClO 3 2 3 mol O 2 122.6 g KClO 3 2 mol KClO 3 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 133 Sample exercise: A common laboratory method for preparing small amounts of O 2 involves the decomposition of KClO 3 : 2KClO 3 4.50 g KClO
How many grams of oxygen is produced from 4.50 g KClO 3 2KCl + 3O 3 ?
1 mol KClO 3 2 3 mol O 2 32 g O 2 122.6 g KClO 3 2 mol KClO 3 1 mol O 2 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 134 Sample exercise: A common laboratory method for preparing small amounts of O 2 involves the decomposition of KClO 3 : 2KClO 3 4.50 g KClO
How many grams of oxygen is produced from 4.50 g KClO 3 2KCl + 3O 3 ?
1 mol KClO 3 2 3 mol O 2 32 g O 2 122.6 g KClO 3 2 mol KClO 3 1 mol O 2 = 1.76 g O 2 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 135 Sample exercise: Propane, C What mass of O 2 3 H 8 , is a common fuel used for cooking and home heating. is consumed in the combustion of 1.00 g of propane?
Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 136 Sample exercise: Propane, C What mass of O 2 3 H 8 , is a common fuel used for cooking and home heating. is consumed in the combustion of 1.00 g of propane?
C 3 H 8 + O 2
CO 2 + H 2 O Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 137 Sample exercise: Propane, C What mass of O 2 3 H 8 , is a common fuel used for cooking and home heating. is consumed in the combustion of 1.00 g of propane?
C 3 H 8 + 5O 2
3CO 2 + 4H 2 O Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 138 Sample exercise: Propane, C What mass of O 2 3 H 8 , is a common fuel used for cooking and home heating. is consumed in the combustion of 1.00 g of propane?
C 3 H 8 + 5O 2
3CO 2 + 4H 2 O 1.00 g C 3 H 8 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 139 Sample exercise: Propane, C What mass of O 2 3 H 8 , is a common fuel used for cooking and home heating. is consumed in the combustion of 1.00 g of propane?
C 3 H 8 + 5O 2
3CO 2 + 4H 2 O 1.00 g C 3 H 8 1 mol C 3 H 44 g C 3 H 8 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 140 Sample exercise: Propane, C What mass of O 2 3 H 8 , is a common fuel used for cooking and home heating. is consumed in the combustion of 1.00 g of propane?
C 3 H 8 + 5O 2
3CO 2 + 4H 2 O 1.00 g C 3 H 8 1 mol C 3 H 44 g C 3 H 8 5 mol O 2 1 mol C 3 H 8 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 141 Sample exercise: Propane, C What mass of O 2 3 H 8 , is a common fuel used for cooking and home heating. is consumed in the combustion of 1.00 g of propane?
C 3 H 8 + 5O 2
3CO 2 + 4H 2 O 1.00 g C 3 H 8 1 mol C 3 H 44 g C 3 H 8 5 mol O 2 1 mol C 3 H 8 32 g O 2 1 mol O 2 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Mole Calculations 142 Sample exercise: Propane, C What mass of O 2 3 H 8 , is a common fuel used for cooking and home heating. is consumed in the combustion of 1.00 g of propane?
C 3 H 8 + 5O 2
3CO 2 + 4H 2 O 1.00 g C 3 H 8 1 mol C 3 H 44 g C 3 H 8 5 mol O 2 1 mol C 3 H 8 32 g O 2 1 mol O 2 = 3.64 g O 2 Chapt. 3.6
Chem 106, Prof. J.T. Spencer Limiting Reagent 143
• Sometimes after one reagent is completely
consumed in the reaction some of another reagent is left over. The reagent which is completely consumed limits the extent of the reaction = LIMITING REAGENT.
Limiting Reagent Chapt. 3.7
+
Chem 106, Prof. J.T. Spencer Limiting Reagent Calculations 144
• Given the reaction for the formation of B
5.0 g of LiH and 5.0 g of BF 3 2 H 6 (diborane), if were reacted how much of which reagent would be left over?
6 LiH + 8BF 3 [B 2 H 6 MW = 27.6. BF 3 6 LiBF 4 + B 2 H 6 = 67.8 and LiH MW = 7.9] Know : Quantities (g and moles) of starting materials Molar ratios between all the starting materials and products.
Find : Which reagent is completely consumed ( limiting reagent ) and which is left over Chapt. 3.6
Chem 106, Prof. J.T. Spencer Limiting Reagent Calculations 145
• Given the reaction for the formation of B
5.0 g of LiH and 5.0 g of BF 3 2 H 6 (diborane), if were reacted how much of which reagent would be left over?
6 LiH + 8BF 3 [B 2 H 6 MW = 27.6. BF 3 6 LiBF 4 + B 2 H 6 = 67.8 and LiH MW = 7.9] 5.0 g LiH (1 mol LiH) = 0.63 mol AND 5.0 g BF 3 (1 mol BF 3 ) = 0.074 mol 7.9 g LiH 67.8 g BF 3 If all the LiH were consumed, then 0.84 mol BF 3 would be required [(0.63 mol LiH)(8 mol BF 3 )] = 0.84 mol BF 3 6 mol LiH) Chapt. 3.6
Chem 106, Prof. J.T. Spencer Limiting Reagent Calculations 146
• Given the reaction for the formation of B
5.0 g of LiH and 5.0 g of BF 3 2 H 6 (diborane), if were reacted how much of which reagent would be left over?
6 LiH + 8BF 3 [B 2 H 6 MW = 27.6. BF 3 6 LiBF 4 + B 2 H 6 = 67.8 and LiH MW = 7.9] 5.0 g LiH (1 mol LiH) = 0.63 mol AND 5.0 g BF 3 (1 mol BF 3 ) = 0.074 mol 7.9 g LiH 67.8 g BF 3 If all the LiH were consumed, then 0.84 mol BF 3 would be required Since only 0.074 mol of BF 3 is available, BF 3 is the limiting reagent (all consumed).
0.074 mol BF 3 (6 mol LiH) = 0.056 mol LiH consumed 8 mol BF 3 therefore remaining LiH = (0.63 mol - 0.056 mol)(7.9 g/mol) = 4.53 g Chapt. 3.6
Chem 106, Prof. J.T. Spencer Limiting Reagent Problems 147
• Equal weights (5.00 g) of Zn(s) and I
2 (s) are mixed together to form ZnI 2 . How much ZnI 2 much of each reactant remains at the end of the reaction and which is the limiting reagent?
is formed? How Zn (AW = 65.4 amu) I 2 (MW = 253.8 amu) Zn(s) + I 2 (s) ZnI 2 (s) Chapt. 3.7
Chem 106, Prof. J.T. Spencer Limiting Reagent Problems 148
• Equal weights (5.00 g) of Zn(s) and I
2 (s) are mixed together to form ZnI 2 . How much ZnI 2 much of each reactant remains at the end of the reaction and which is the limiting reagent?
is formed? How Zn (AW = 65.4 amu) I 2 (MW = 253.8 amu) Zn = 5.0 g Zn (1 mol Zn) = 0.076 mol Zn 65.4 g Zn I 2 = 5.0 g I 2 I 2 (1 mol I 2 Zn(s) + I 2 (s) ) = 0.020 mol I 253.8 g I 2 is the limiting reagent.
2 ZnI 2 (s) Zn remaining = (0.076 Zn - 0.020 mol Zn) (64.5 g Zn) = 3.66 g Zn 1 mol Zn Chapt. 3.7
Chem 106, Prof. J.T. Spencer Theoretical Yields 149
• Theoretical Yield - quantity of product
calculated to form when all the limiting reagent is consumed (calculated from molar ratios).
• Actual Yield - the amount of product
experimentally obtained from a reaction
• Percent Yield - describes relationship between
theoretical and actual yields; percent yield = actual yield (100) theoretical yield Chapt. 3.6
Chem 106, Prof. J.T. Spencer Percent Yields 150
• Given the reaction of 2.05 g of hydrogen sulfide with 1.84 g
of sodium hydroxide, calculate how the theoretical yield of Na 2 S. What is the percent yield if the amt. of Na 2 S obtained was 3.65 g. [H 2 S (MW = 34.1); Na 2 S (MW = 78.1)] H 2 S(g) + 2 NaOH(aq) Na 2 S(aq) + 2 H 2 O (2.05 g H 2 S)(1 mol H 2 S)(1 mol Na 2 S)(78.1 g Na 2 S) = 4.70 g Na 2 S 34.1 g H 2 S 1 mol H 2 S 1 mol Na 2 S theoretical yield % yield = 3.65 g (actual yield) (100) = 77.7 % yield 4.70 g (theoretical yield) Chapt. 3.6
Chem 106, Prof. J.T. Spencer
End of Chapter 3
151 Major Topics (not exhaustive list): (1) Chemical Equations (2) Periodic Table and Reaction Types (3) Atomic and Molecular Weights (formula weights, % compositions, etc...) (4) Molar Concepts (5) Empirical Formulas (6) Info from Balanced Eqns.
(7) Limiting Reagents (8) Percent Yields