Self Assembly of a Monolayer Essay Example
Self Assembly of a Monolayer
The purpose of this experiment is to make the condition for molecular monolayer to self-assemble, so we can find Avogadro’s number and to estimate the size of molecule.
Materials and methods * Microliter pipet, oleic acid, distilled water, talcum powder. * Fill container with distilled water * Coat the entire surface of water with talcum powder * Fill microliter pipet with 25µL of oleic acid in alcohol solution * Deliver 25 µL of the sample into the center of dusted water * Take measurements of the monolayer * Find the average diameter * Find area of the monolayer
Results
* 4 measurements of the monolayer 4cm, 3.2cm, 3.4cm, 3cm * Average diameter 3.4cm
To find the average diameter we need to add all and then divide by 4 * Area of the monolayer A=9.1cm² * To find area of the monolayer π(3.4)²4= 9cm² * 1 Liter = 1×10³mL = 1× 106 * 1 mL = 1×10³ * 1µL =1×10-3=1×10-6 a. 1mL of oleic acid was dissolved in 999mL of ethyl alcohol to prepare 1000ml of solution cause the solution is 1:1000 by volume of oleic acid in ethyl alcohol b. 25 µL as .025cm³ ( 25µL1×.001cm³1µL= .025cm³) c. Volume of just oleic acid in sample 2.5×10-5mL (.025mL÷1000=2.5×10-5mL) why divide by 1000 because ration is 1:1000 d. Total mass of oleic acid in sample 2.24×10-5g
(2.5×10-5mL)×.895g/cm³= 2.24×10-5g e. Total moles of oleic acid in the sample 7.9×10-8moles
C18H34O2=282.5g next we going to 2.24×10-5g÷282.5g = 7.9×10-8moles
6)Number of oleic acid molecules in the monolayer 4.3 ×1015cm²
(9.1 cm2 ÷ 21×10-16=4.3×1015cm²)
7) Avogadro’s number 5.49×1022 molecules/mol
(4.3×1015cm²) ÷ (7.9×10-8mol)=5.49×1022 molecules/mol
8) We got close to the value of N (6.02×1023) and our answer was 5.49×1022
Conclusion
We expected for the dusted water to split when dropping the sample oleic sample into it and it did. Also we expected to get the same number as Avogadro’s
The purpose of this experiment is to make the condition for molecular monolayer to self-assemble, so we can find Avogadro’s number and to estimate the size of molecule.
Materials and methods * Microliter pipet, oleic acid, distilled water, talcum powder. * Fill container with distilled water * Coat the entire surface of water with talcum powder * Fill microliter pipet with 25µL of oleic acid in alcohol solution * Deliver 25 µL of the sample into the center of dusted water * Take measurements of the monolayer * Find the average diameter * Find area of the monolayer
Results
* 4 measurements of the monolayer 4cm, 3.2cm, 3.4cm, 3cm * Average diameter 3.4cm
To find the average diameter we need to add all and then divide by 4 * Area of the monolayer A=9.1cm² * To find area of the monolayer π(3.4)²4= 9cm² * 1 Liter = 1×10³mL = 1× 106 * 1 mL = 1×10³ * 1µL =1×10-3=1×10-6 a. 1mL of oleic acid was dissolved in 999mL of ethyl alcohol to prepare 1000ml of solution cause the solution is 1:1000 by volume of oleic acid in ethyl alcohol b. 25 µL as .025cm³ ( 25µL1×.001cm³1µL= .025cm³) c. Volume of just oleic acid in sample 2.5×10-5mL (.025mL÷1000=2.5×10-5mL) why divide by 1000 because ration is 1:1000 d. Total mass of oleic acid in sample 2.24×10-5g
(2.5×10-5mL)×.895g/cm³= 2.24×10-5g e. Total moles of oleic acid in the sample 7.9×10-8moles
C18H34O2=282.5g next we going to 2.24×10-5g÷282.5g = 7.9×10-8moles
6)Number of oleic acid molecules in the monolayer 4.3 ×1015cm²
(9.1 cm2 ÷ 21×10-16=4.3×1015cm²)
7) Avogadro’s number 5.49×1022 molecules/mol
(4.3×1015cm²) ÷ (7.9×10-8mol)=5.49×1022 molecules/mol
8) We got close to the value of N (6.02×1023) and our answer was 5.49×1022
Conclusion
We expected for the dusted water to split when dropping the sample oleic sample into it and it did. Also we expected to get the same number as Avogadro’s