Analysing results

The investigation cannot be based on changes in mass from just one potato cylinder.

In scientific tests, we must ensure a number of measurements are made to allow for uncertainty and error in data and anomalous results.

Percentage changes in mass must be calculated for each cylinder. A mean value for the change in mass of potato cylinders at each concentration - measured in moles of sucrose per dm3 of solution, or mol dm−3 – should be calculated.

A graph is plotted of change in mass, in per cent, against concentration of sucrose.

Where potato cylinders have gained in mass, the change will be positive.

Where potato cylinders have decreased in mass, the change will be negative.

Concentration of sucrose (mol dm−3)Average change in mass (%)
A graph showing the change of mass and the concentration of sucrose.

Where the plotted line crosses the horizontal axis at 0 per cent change in mass, the sucrose concentration is equal to the concentration of dissolved substances in the potato cells.

This can be identified on the graph as the point which shows no change in mass, and therefore represents no net movement of water by osmosis.


What is the concentration of solutes in the cells of the potato in this investigation?

0.27 mol dm−3.

This value is where the line crosses the x-axis.


The concentration of dissolved solutes in the cells of different potatoes will vary slightly from potato to potato. If we have a set of data for a range in concentrations, we can look at the range, and the mean - but these do not tell us whether data are evenly spread or whether they are clustered together within a certain range.

Scientists use percentiles to divide a set of data into 100, and look to see where the data lie within these divisions.

The median is the point in a set of data where 50 per cent of the data fall above this value, and 50 per cent below it. This is the 50th percentile.

The 75th percentile is where 75 percent of the data fall below this value.

There are several methods of finding a percentile. The simplest is the nearest rank method. As with finding the median of a set of data, begin by putting the data into order.

For a range of values for the concentration of potato cell sap:


Arranged in order:

To find, for example the 50th percentile, first find the rank:

\[\text{ordered rank} = \frac{\text{required percentile}}{100} \times \text{number of entries in data set}\]

\[\text{ordered rank} = \frac{50}{100} \times 32 = 16\]

So the 50th percentile will be the 16th number in the ordered data set, starting from the left. The 50th percentile is 0.27.

In instances where the ordered rank is not a whole number, you should round the number up.


Find the 90th percentile for the same set of data.

0.32 mol dm−3.

\[\text{ordered rank} = \frac{\text{percentile}}{100} \times \text{number in data set}\]

\[\text{ordered rank} = \frac{90}{100} \times 32 = 29\]

This method will only give percentiles as numbers that exist in the data set.

In other methods, percentiles can be interpolated for values that don't exist in the data set.

The percentile rank will tell us the position of a value within a range.

To find the percentile rank of a value, \(x\):

\[\text{percentile rank} = \frac{(B + 0.5E)}{n} \times 100\]


  • \(B\) = number of values below \(x\)
  • \(E\) = number of values equal to \(x\)
  • \(n\) = number of values

So, for the student's value of 0.27 mol dm−3:

\(\frac{(13 + 2)}{32} \times 100 = \frac{15}{32} \times 100\) = 47th percentile


For another student's value of 0.29 mol dm−3, find the percentile rank.

75th percentile.

\(\frac{(B + 0.5E)}{n} \times 100 = \frac{(21 + 3)}{32} \times 100 = \frac{24}{32} \times 100\) = 75th percentile