User:Moira M. Esson/Notebook/CHEM-581/2012/11/09: Difference between revisions
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2.27x10<sup>-6</sup> moles dye x(495.28g dye/1mol dye)=0.001124g dye | 2.27x10<sup>-6</sup> moles dye x(495.28g dye/1mol dye)=0.001124g dye | ||
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'''Polyvinyl acetate synthesis calculations''': | |||
*For all reaction mixtures, the total volume of the reaction mixtures will be 7mL | |||
* HAc will be the abbreviation for acetic acid | |||
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Density of HAc= 1.05g/mL | |||
(A.) 1:1 ratio | |||
1g PVOH x (1mol PVOH/44g PVOH) x (1mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 1.36g HAc | |||
1.36g HAc x (1mL/1.05g HAc)= 1.295mL HAc | |||
(B.) 20% ratio | |||
1gPVOH x (1mol PVOH/44g PVOH) x (0.2mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 0.273g HAc | |||
0.273g HAc x (1mL/1.05g HAc)=0.26mL HAc | |||
(C.) 40% ratio | |||
1gPVOH x (1mol PVOH/44g PVOH) x (0.4mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 0.545g HAc | |||
0.5195g HAc x (1mL/1.05g)= 0.5195mL HAc | |||
(D.) 60% ratio | |||
1gPVOH x (1mol PVOH/44g PVOH) x (0.6mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 0.818g HAc | |||
0.818g HAc x (1mL HAc/ 1.05g)= 0.779mL HAc | |||
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'''Titration calcualtions''': | |||
(A.) Sodium bicarbonate solution preparation | |||
0.2L x (1mol/1L) x (84.01g/1mol)= 16.802g sodium bicarbonate. | |||
*Note: The actual amount of sodium bicarbonate used was 16.8046g. Also, it is not necessary to prepare this solution in a volumetric flask. Will standardize the solution to know the actual concentration. | |||
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(B.)Potassium hydrogen phthalate(PHP) needed | |||
MW PHP= 204.24g/mol | |||
1M PHP= x/0.1L | |||
x=0.1mol x 204.24g/mol | |||
x= 20.424g PHP needed | |||
*Note: | |||
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Revision as of 13:55, 26 November 2012
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Objectives
General Formula for reactionsPolyvinyl alcohol + [HAc]→Polyvinyl acetate
General Protocol for Polyvinyl acetate synthesis
Calculations
Number of moles of OH groups in 1g PVOH: MW of repeating polyvinyl alcohol unit [C2H3OH]: 44.06g/mol Polyvinyl alcohol repeating unit picture 1gPVOH(1mol PVOH unit/44.06g) x (1mol OH/1mol PVOH unit)= 0.022696 mol OH/g Dye attachment calculations:
MW 5-DTAF dye: 495.28g/mol
(0.0227mol OH/g) x (.0001)= 2.27x10-6 moles dye
2.27x10-6 moles dye x(495.28g dye/1mol dye)=0.001124g dye
1g PVOH x (1mol PVOH/44g PVOH) x (1mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 1.36g HAc 1.36g HAc x (1mL/1.05g HAc)= 1.295mL HAc (B.) 20% ratio 1gPVOH x (1mol PVOH/44g PVOH) x (0.2mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 0.273g HAc 0.273g HAc x (1mL/1.05g HAc)=0.26mL HAc (C.) 40% ratio 1gPVOH x (1mol PVOH/44g PVOH) x (0.4mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 0.545g HAc 0.5195g HAc x (1mL/1.05g)= 0.5195mL HAc (D.) 60% ratio 1gPVOH x (1mol PVOH/44g PVOH) x (0.6mol HAc/1mol PVOH) x (60g HAc/1mol HAc)= 0.818g HAc 0.818g HAc x (1mL HAc/ 1.05g)= 0.779mL HAc
0.2L x (1mol/1L) x (84.01g/1mol)= 16.802g sodium bicarbonate.
1M PHP= x/0.1L x=0.1mol x 204.24g/mol x= 20.424g PHP needed
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