A mass-change dataset records the mass of tissue before and after it sits in solutions of different concentrations. To interpret it, calculate the percentage change in mass for each piece, then link the sign and size of the change to the direction of water movement.
This lesson puts predicting water movement in an osmosis model to work on numbers. It is part of movement across membranes.
How do you turn a table into a conclusion?
Follow the same routine every time:
- Calculate the change: final mass minus initial mass.
- Convert to percentage change: change ÷ initial mass × 100.
- Read the sign: positive means water was gained, negative means water was lost.
- Compare with concentration: as the solution gets more concentrated, expect the percentage change to decrease.
- Link to the model: explain the direction of water movement using water potential.
Why percentage? If two pieces start at different masses, a raw gain is not a fair comparison. Piece A starts at 2.0 g and gains 0.30 g, which is 0.30 ÷ 2.0 × 100 = 15%.
Piece B starts at 3.0 g and gains 0.36 g, which is 0.36 ÷ 3.0 × 100 = 12%. The raw gain is bigger for B, but A gained proportionally more.
Worked example
The data are invented for practice. Potato cylinders were weighed, left in sucrose solutions for one hour, blotted dry and weighed again.
| Sucrose (mol/dm³) | Initial mass (g) | Final mass (g) | Change (g) | Change (%) |
|---|---|---|---|---|
| 0.0 | 2.50 | 2.85 | +0.35 | +14.0 |
| 0.2 | 2.40 | 2.58 | +0.18 | +7.5 |
| 0.4 | 2.50 | 2.45 | −0.05 | −2.0 |
| 0.6 | 2.60 | 2.34 | −0.26 | −10.0 |
| 0.8 | 2.50 | 2.15 | −0.35 | −14.0 |
Step 1, check the calculations. At 0.2: change is 2.58 − 2.40 = +0.18 g, and 0.18 ÷ 2.40 × 100 = 7.5%. At 0.6: change is 2.34 − 2.60 = −0.26 g, and −0.26 ÷ 2.60 × 100 = −10.0%.
Step 2, describe the trend. As the sucrose concentration increases, the percentage change in mass decreases from +14.0% to −14.0%.
Step 3, explain. In pure water (0.0), the solution outside has a higher water potential than the cell contents, so water enters by osmosis and the mass rises. In 0.8 mol/dm³ sucrose the outside has the lower water potential, so water leaves and the mass falls.
Step 4, estimate where there is no change. The line crosses zero between 0.2 (+7.5%) and 0.4 (−2.0%). The total fall across that step is 9.5 percentage points. The zero point is 7.5 ÷ 9.5 = 0.79 of the way along, so 0.2 + 0.79 × 0.2 = 0.358, which is about 0.36 mol/dm³.
Step 5, interpret the estimate. At about 0.36 mol/dm³, the solution has about the same water potential as the potato cells, so there is no net movement of water. This is an estimate from a straight line between two points.
The mistake to watch for
One common error is to compare raw changes when the starting masses differ.
Mistaken answer: Cylinder B gained 0.36 g and cylinder A gained only 0.30 g, so B sat in the more dilute solution.
The starting masses were 3.0 g and 2.0 g, so the raw gains are not a fair comparison. A gained 15% and B gained 12%.
The correction is to use percentage change, and to keep the claim modest: A gained proportionally more than B. A second common slip is to drop the negative sign, which turns a loss into a gain.
Check yourself
Work these out first, then compare.
1. A cylinder has an initial mass of 3.20 g and a final mass of 3.68 g. Calculate the percentage change in mass.
Show answer
Change = 3.68 − 3.20 = 0.48 g. Percentage change = 0.48 ÷ 3.20 × 100 = +15%.
2. A cylinder has an initial mass of 2.80 g and a final mass of 2.52 g. Calculate the percentage change and say what it shows.
Show answer
Change = 2.52 − 2.80 = −0.28 g. Percentage change = −0.28 ÷ 2.80 × 100 = −10%. The cylinder lost water by osmosis, so the solution was more concentrated than the cell contents.
3. Two points on a graph are +6% at 0.2 mol/dm³ and −4% at 0.4 mol/dm³. Estimate the concentration with no change in mass.
Show answer
The total difference is 10 percentage points. Zero is 6 ÷ 10 = 0.6 of the way along, so 0.2 + 0.6 × 0.2 = 0.32 mol/dm³ (an estimate).
Where this leads next
Next, look at the limits of what such data can tell you in separating a cell response from a whole-organism claim. Then use the movement across membranes practice set and log any slips in the mistake log and retest queue.
Students often know the formula but lose marks when the table changes shape. A teacher in online one-to-one Biology tuition can give you new tables and watch how you approach them.