Osmosis in Potatoes Lab. At which concentration of sucrose in water (% mass by sucrose) will the potato core be isotonic, meaning that there is no movement of water in or out of the potato core?

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Table 3: Percent Change in Masses of Potato Cores

Data Processing:

        In order to find the point at which the potato is in an isotonic state, you must go through four different steps of processing the data.

• Step 1: Find the differences in the masses of the potato cell by subtracting the final masses at each time from the initial mass

(Difference= final mass- initial mass)

0% Mass by Sucrose in Solution

• At .5 Hours: 1.619g- 1.506= 0.113g
• At 24 Hours: 1.553g- 1.506= 0.047g
• At 48 Hours: 1.459g- 1.506= -0.047 g

10% Mass by Sucrose in Solution

• At .5 Hours: 1.566g- 1.538= 0.028
• At 24 Hours: 1.664g- 1.538= 0.106g
• At 48 Hours: 1.602g- 1.538= 0.064g

20% Mass by Sucrose in Solution

• At .5 Hours: 1.618g- 1.588= 0.03g
• At 24 Hours: 1.721g- 1.588= 0.133g
• At 48 Hours: 1.654g- 1.588= 0.066g

30% Mass by Sucrose in Solution

• At .5 Hours: 1.439g- 1.578= -0.139g
• At 24 Hours: 0.999g- 1.578= -0.579g
• At 48 Hours: 1.433g- 1.578= -0.145g

40% Mass by Sucrose in Solution

• At .5 Hours: 1.339g- 1.552= -0.213g
• At 24 Hours: 0.807g- 1.552= -0.745g
• At 48 Hours: 1.254g- 1.552= -0.298g

50% Mass by Sucrose in Solution

• At .5 Hours: 1.340g- 1.616= -0.276g
• At 24 Hours: 0.779g- 1.616= -0.837g
• At 48 Hours: 1.082g- 1.616= -0.534g

• Step 2: Find the percent change in mass of each potato core at each time using the formula:

[(final mass- initial mass)/ initial mass] x 100 = % Change in Mass

0% Mass by Sucrose in Solution

• At .5 Hours: [(1.619g- 1.506)/ 1.506] x 100= 7.5%
• At 24 Hours: [(1.553g- 1.506)/ 1.506] x 100= 1.79%
• At 48 Hours: [(1.459g- 1.506)/ 1.506] x 100= -3.12%

10% Mass by Sucrose in Solution

• At .5 Hours: [(1.566g- 1.538)/ 1.538] x 100= 1.82%
• At 24 Hours: [(1.644g- 1.538)/ 1.538] x 100= 6.89%
• At 48 Hours: [(1.602g- 1.538)/ 1.538] x 100= 4.16%

20% Mass by Sucrose in Solution

• At .5 Hours: [(1.618g- 1.588)/ 1.588] x 100= 1.89%
• At 24 Hours: [(1.721g- 1.588)/ 1.588] x 100= 8.38%
• At 48 Hours: [(1.654g- 1.588)/ 1.588] x 100= 4.16%

30% Mass by Sucrose in Solution

• At .5 Hours: [(1.439g- 1.578)/ 1.578] x 100= -8.81%
• At 24 Hours: [(0.999g- 1.578)/ 1.578] x 100= -36.69%
• At 48 Hours: [(1.433g- 1.578)/ 1.578] x 100= -9.19%

40% Mass by Sucrose in Solution

• At .5 Hours: [(1.339g- 1.552)/ 1.552] x 100= -13.72%
• At 24 Hours: [(0.807g- 1.552)/ 1.552] x 100= -48%
• At 48 Hours: [(1.254g- 1.552)/ 1.552] x 100= -19.2%

50% Mass by Sucrose in Solution

• At .5 Hours: [(1.340g- 1.616)/ 1.616] x 100= -17.08%
• At 24 Hours: [(0.779g- 1.616)/ 1.616] x 100= -51.79%
• At 48 Hours: [(1.082g- 1.616)/ 1.616] x 100= -33.04%

• Step 3: Find the average percent change in each potato core using the formula (a= average):  (a1+a2+a3)/ 3= Average Percent Change

0% Mass by Sucrose in Solution

• (7.5%+ 1.79+ -3.12)/ 3= 2.06%

10% Mass by Sucrose in Solution

• (1.82%+ 6.89+ 4.16)/ 3= 4.29%

20% Mass by Sucrose in Solution

• (1.89%+ 8.38+ 4.16)/ 3= 4.81%

30% Mass by Sucrose in Solution

• (-8.81%+ -36.69+ -9.19)/ 3= 18.23%

40% Mass by Sucrose in Solution

• (-13.72%+ -48+ -19.2)/ 3= -26.97%

50% Mass by Sucrose in Solution

• (-17.08%+ -51.79+ -33.04)/ 3= -33.97%

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• Step 4: Put the average changes in mass into a graph, find the line of best fit, and using the equation of the line of best fit, find the point when the solution is isotonic. Then, find the moles of sucrose in the isotonic solution, and find the ratio of moles between the sucrose and water. 

Graph 1: Average Change in Mass of Potato Cores

At what percent mass by sucrose is the solution isotonic?

• y= -0.8485x+9.8771

• Plug “0” in for “y”: 0= -0.8485x+9.8771

• Work out the equation, and find the value for “x”:

0= -0.8485x+9.8771

-9.8771= -0.8485x

x= 11.64

When the solution has a 11.64 percent mass by sucrose, then it will be isotonic because then the average percent change would be 0%, meaning that no water is moving in or out of the potato.

Finding the number of moles:

        In order to find the number of moles of sucrose in the isotonic solution, we have to find the molar mass, and mass of the solution.

11.64% sucrose is the same thing as saying that there are 11.64 grams of sucrose for every 100 grams of water in the solution, so the mass would be 11.64 grams.

 #of moles= m (mass)/M (Molar Mass)

Sucrose:

# of moles= 11.64g / 342.34g/mol

#of moles = .034

Water:

#of moles= 100g / 18.02g/mol

# of moles= 5.549

Ratio:

.034/ 5.549= .0061

So, for every .0061 moles of sucrose in the solution, there will be 1 mole of water.

Conclusion:

        For this lab, I would say that my hypothesis is accepted. It stated that the solution would be isotonic when its concentration of sucrose is somewhere between 10 and 30 percent, and the concentration when the solution was isotonic was 11.64%, very clearly in between 10 and 30.  In order to determine whether or not my hypothesis was correct, I had to go through a series of steps. The very first thing I had to do was take five cores of equal size from the same potato, record their initial masses, and put them each in a separate test tube containing water with different concentrations of sucrose (0, 10, 20, 30, 40, and 50% by mass). After letting the cores soak in each solution for 30 minutes, we took them out with the tweezers, and recording their masses again. We repeated these steps after 24 hours and then again after 48 hours. By recording these masses, we could then find the change between the initial mass, and the final mass at each time. With this information, we could determine when diffusion was occurring in the potato, and when osmosis was occurring. Diffusion, a net transport of molecules from a region of higher concentration to one of lower concentration by random molecular motion, occurred when the masses of the potato core decreased, meaning that water had traveled out of them (30-50% mass by sucrose) thus making the solution hypertonic. Osmosis, the movement of water through a semipermiable membrane from a place of higher concentration to a place of lower concentration, occurred when the change in masses was positive (first part of 0 to 20% mass by sucrose), meaning that the mass of the potato increased thus making the solution hypotonic. Then, we found the percent changes of the masses of each solution at each time, and averaged the changes so there would be one percent change for each solution. Then, we put these percent changes on a graph, found the line of best fit and equation of the graph, and then substituted “0” as the y value in order to find the concentration of the solution when there was no fluctuation of the mass (11.64% mass by sucrose), meaning that no water moved in or out of potato (isotonic stage). Once the concentration of the isotonic stage was found, then we are able to find the ratio of moles of sucrose to moles of water that is in the solution. In order to this, we found the molar masses of each (sucrose, water) and plugged everything into the equation #of moles= m/M. The ratio of sucrose to water moles was .0061 moles of sucrose for every mole of water. In conclusion the hypothesis was accepted because it stated that the solution would be isotonic when it has a sucrose concentration of 10-30%. Proof of this is that the solution had 0 percent change in mass when the sucrose concentration was 11.64% (a number in between 10 and 30) or when there were .0061 moles of sucrose for every 1 mole of water in the solution.

Evaluation:

        If there is one thing I have learned, it is that there is no such thing as a perfect answer. This may well be the case with the answer of our lab. Even though I feel confident, and sure with the answers I received, there is always a factor of human error and uncontrollable events that could have prevented us from getting accurate results. Because of my inexperience with a cork borer, I may not have cut the potatoes to exactly the same size, thus affecting the size, water capacity, and the fluctuation of mass of the potato. A simple thing like how well you handle a cork borer could throw off your entire data collection for a certain core, and in the end alter the line of best fit and the answer to the question. Another problem we encountered was the number of dead ants in our solution. It could have been that the ants ate part of the potato, or they absorbed some of the sucrose, thus making the labeled percent of sucrose inaccurate. Once again, we head back to the possibility of imperfections in things such as measurements, and how there is a very slim chance that each test tube contained an EXACT amount of sucrose by mass, meaning that it is highly unlikely that the test tube labeled with 10% concentration of sucrose had exactly 10% sucrose by mass. These inaccurate measurements could have affected what the equation for our line of best fit was, and thus, again, affecting the outcome of our final answer. Another problem encountered was how the potato cores in solutions with less concentrations of sucrose would float to the top, sometimes with parts of the core above the surface of the water. This could affect our recordings of mass because some of the water that was stored in the top of our potato never moved out because it was never completely submerged, or vice versa. One last thing I noticed is that when we took the potato cores out to measure the masses, we drained out some of the water, and didn’t place it back into the test tube. This could maybe affect the amount of water that is left to move into the potato, or maybe even effect the final measuring of mass. The point is that there are many little factors, events, and disturbances that may not seem like much, but when they are all added up, they can change, sometimes drastically, your final answer.