User:Moira M. Esson/Notebook/CHEM-581/2012/11/09: Difference between revisions
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(C.)Reaction mixture preparations | (C.)Reaction mixture preparations | ||
#~1g of PVOH was measured and placed in a 10mL beaker | # ~1g of PVOH was measured and placed in a 10mL beaker | ||
# | # The necessary amount of deionized H<sub>2</sub>O was added to the beaker(7mL - the amount of HAc that would be added) and a small stir bar was placed in the beaker | ||
# The beaker was parafilmed and set in a drawer. | |||
# The following process was followed for each of the 4 polyvinyl acetate reactions | |||
Actual amount of PVOH measured: | |||
{| {{table}} | |||
| align="center" style="background:#f0f0f0;"|'''Ratio of polyvinyl acteate and polyvinyl alcohol''' | |||
| align="center" style="background:#f0f0f0;"|'''Amount of PVOH(MW 22K) used (g)''' | |||
|- | |||
| 1 to 1 reaction||1.0062 | |||
|- | |||
| 20%||1.0043 | |||
|- | |||
| 40%||1.0052 | |||
|- | |||
| 60%||1.0027 | |||
|} | |||
Revision as of 14:43, 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 Solution Preparations
Actual amount of PVOH measured:
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