User:Moira M. Esson/Notebook/CHEM-581/2012/11/09

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2.27x10-6 moles dye x(495.28g dye/1mol dye)=0.001124g dye 2.27x10-6 moles dye x(495.28g dye/1mol dye)=0.001124g dye +
+ '''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 +
+ 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 +
+ '''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. +
+ (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:

Revision as of 15:55, 26 November 2012

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Objectives

1. Prepare solutions to run new PVOH based reactions
2. Perform all calculations necessary for future PVOH reactions
• First PVOH reaction will be the synthesis of polyvinyl acetate
• Second PVOH reaction will be the attachment of 5-(4,6-Dichlorotriazinyl)Aminofluorescein which will be referred to as 5-DTAF

General Formula for reactions

Polyvinyl alcohol + [HAc]→Polyvinyl acetate

General Protocol for Polyvinyl acetate synthesis

1. For a 1:1 ratio of polyvinyl alcohol OH groups to polyvinyl acetate dissolve ~1g PVOH inH2O
2. Add ~1.36g HAc. Add ~2 drops of HCl to reaction mixture to catalyze reaction.
3. Heat the reaction mixture to 40°C.
4. Allow the reaction to stir for 1 hour
5. Add sodium bicarbonate solution to the reaction mixture(titrate the reaction)
6. Rotavap the remaining reaction mixture

Calculations

• Overall, want to determine if have 1g of material how many moles of -OH groups. Once that is determined, will want to determine how many moles of -OH was reacted after the reaction.

Number of moles of OH groups in 1g PVOH: MW of repeating polyvinyl alcohol unit [C2H3OH]: 44.06g/mol

``` 1gPVOH(1mol PVOH unit/44.06g) x (1mol OH/1mol PVOH unit)= 0.022696 mol OH/g
```

Dye attachment calculations:

• Want to react .01% of the OH groups with the dye

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
```

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

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
```

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.

(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: