*Please write the identity of the base you will use in the table. 2-bromo-2-methylbutane HON 1-But KotBu 1-Prop Base: Molar mass (g/mol) 151.04 g/mol 54.106 74.12112.21 60.1 g/mol Density or concentration 1.18 g/mL 2.04 0.81 0.902 0.803 Mass or volume used (g or mL) 1.3mL 12.5 12.5 12.5 12.5 mL Moles 0.00976 0.5 0.137 0.10 0.167 Equivalents** 1.0 1:50 14 1: **Equivalents: The reagent present in the smallest molar amount is set to 1.0 equivalent (eq). For each other reagent, #eq equals the ratio of its #moles divided by the #moles of the minimum reagent. The ratio of equivalents is equal to the molar ratio. 2. Present a calculation of the number of moles of BOTH of the reactants: 2-bromo-2-methylbutane] [KOH DEM/V - 204ginal= m/ 12.5 mx Dm=25.5g n= 25.5%/56.106gimo n=0.45 Density= m/v 1.18 glmx = m / 1.3 m m= 1.534 g les (n)= SS 1-Butanol Dam/v 0.81 gimm n= 10.1250 n = 1.5348/(51.04 g/mo armassn= 0.00976 KOLBU D=m/v 0.902 glmm/12.5 myl m11.2759 12.5 74.12 g/mol =P m = 10.125 g Pin = 0.137 mol 1-Propanol D=m/v = 0.803 g/m2 = m/12.5mx n = 11.2759/112.21 g/mol +10.10 mol/ m= 10.03159 n= 10.0375 \n = 0.167mol 60.18 Imo
*Please write the identity of the base you will use in the table. 2-bromo-2-methylbutane HON 1-But KotBu 1-Prop Base: Molar mass (g/mol) 151.04 g/mol 54.106 74.12112.21 60.1 g/mol Density or concentration 1.18 g/mL 2.04 0.81 0.902 0.803 Mass or volume used (g or mL) 1.3mL 12.5 12.5 12.5 12.5 mL Moles 0.00976 0.5 0.137 0.10 0.167 Equivalents** 1.0 1:50 14 1: **Equivalents: The reagent present in the smallest molar amount is set to 1.0 equivalent (eq). For each other reagent, #eq equals the ratio of its #moles divided by the #moles of the minimum reagent. The ratio of equivalents is equal to the molar ratio. 2. Present a calculation of the number of moles of BOTH of the reactants: 2-bromo-2-methylbutane] [KOH DEM/V - 204ginal= m/ 12.5 mx Dm=25.5g n= 25.5%/56.106gimo n=0.45 Density= m/v 1.18 glmx = m / 1.3 m m= 1.534 g les (n)= SS 1-Butanol Dam/v 0.81 gimm n= 10.1250 n = 1.5348/(51.04 g/mo armassn= 0.00976 KOLBU D=m/v 0.902 glmm/12.5 myl m11.2759 12.5 74.12 g/mol =P m = 10.125 g Pin = 0.137 mol 1-Propanol D=m/v = 0.803 g/m2 = m/12.5mx n = 11.2759/112.21 g/mol +10.10 mol/ m= 10.03159 n= 10.0375 \n = 0.167mol 60.18 Imo
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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sample weight:
KOH: 1g
KOtBu 1g
1-propanol: 1g
1-butanol 0.742g
calculate the thoeretical and percent yields of methylbutenes
![*Please write the identity of the base you will use in the table.
2-bromo-2-methylbutane
HON
1-But
KotBu
1-Prop
Base:
Molar mass (g/mol)
151.04 g/mol
54.106
74.12112.21 60.1
g/mol
Density or concentration
1.18 g/mL
2.04
0.81 0.902 0.803
Mass or volume used
(g or mL)
1.3mL
12.5
12.5 12.5 12.5
mL
Moles
0.00976
0.5 0.137 0.10 0.167
Equivalents**
1.0
1:50 14 1:
**Equivalents: The reagent present in the smallest molar amount is set to 1.0 equivalent (eq). For each
other reagent, #eq equals the ratio of its #moles divided by the #moles of the minimum reagent. The
ratio of equivalents is equal to the molar ratio.
2. Present a calculation of the number of moles of BOTH of the reactants:
2-bromo-2-methylbutane] [KOH DEM/V - 204ginal= m/ 12.5 mx Dm=25.5g
n= 25.5%/56.106gimo n=0.45
Density= m/v
1.18 glmx = m / 1.3 m
m= 1.534 g
les (n)=
SS
1-Butanol Dam/v
0.81 gimm
n= 10.1250
n = 1.5348/(51.04 g/mo
armassn= 0.00976
KOLBU D=m/v
0.902 glmm/12.5 myl
m11.2759
12.5
74.12 g/mol
=P m = 10.125 g
Pin = 0.137 mol
1-Propanol D=m/v = 0.803 g/m2 = m/12.5mx
n = 11.2759/112.21 g/mol +10.10 mol/
m= 10.03159
n= 10.0375 \n = 0.167mol
60.18 Imo](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F696c7732-f5ff-44db-9d58-26f18bb6317e%2F1e6719e2-bd72-4eb4-bcb6-bc443ab46102%2F5ff644_processed.png&w=3840&q=75)
Transcribed Image Text:*Please write the identity of the base you will use in the table.
2-bromo-2-methylbutane
HON
1-But
KotBu
1-Prop
Base:
Molar mass (g/mol)
151.04 g/mol
54.106
74.12112.21 60.1
g/mol
Density or concentration
1.18 g/mL
2.04
0.81 0.902 0.803
Mass or volume used
(g or mL)
1.3mL
12.5
12.5 12.5 12.5
mL
Moles
0.00976
0.5 0.137 0.10 0.167
Equivalents**
1.0
1:50 14 1:
**Equivalents: The reagent present in the smallest molar amount is set to 1.0 equivalent (eq). For each
other reagent, #eq equals the ratio of its #moles divided by the #moles of the minimum reagent. The
ratio of equivalents is equal to the molar ratio.
2. Present a calculation of the number of moles of BOTH of the reactants:
2-bromo-2-methylbutane] [KOH DEM/V - 204ginal= m/ 12.5 mx Dm=25.5g
n= 25.5%/56.106gimo n=0.45
Density= m/v
1.18 glmx = m / 1.3 m
m= 1.534 g
les (n)=
SS
1-Butanol Dam/v
0.81 gimm
n= 10.1250
n = 1.5348/(51.04 g/mo
armassn= 0.00976
KOLBU D=m/v
0.902 glmm/12.5 myl
m11.2759
12.5
74.12 g/mol
=P m = 10.125 g
Pin = 0.137 mol
1-Propanol D=m/v = 0.803 g/m2 = m/12.5mx
n = 11.2759/112.21 g/mol +10.10 mol/
m= 10.03159
n= 10.0375 \n = 0.167mol
60.18 Imo
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