Results and discussion:
The purpose of the experiment was to determine the densities of the unknown liquid and solid as precisely and accurately as possible in order to identify them. It was carried out first on distilled water in order to rule out systematic errors. A 10 mL beaker was placed on a top pan balance and “tared”. The beaker was then removed and 10 mL of distilled water was pipetted into it before it was placed on the balance again. Thus, the mass of the water was obtained. Five experimental trials were performed. The data and calculated results were tabulated (Table 1). Table 1: Mass and Volume of Water and Density Calculations
Trial #Mass of water (g)Volume of water (mL)Density of water (g/ mL ) 19.9510.000.995
The densities were calculated using the relationship
Density (g/ mL) = (Mass (g))/(Volume (mL))
An example of such a calculation using the trial #1 data:
Density (g/ mL) = (9.95 (g))/( 10.00 (mL)) = 0.995 g/mL
The average of these five determined density values,
(0.995 g/( mL) + 0.998 g/mL + 0.995 g/mL + 0.995 g/mL + 0.993 g/mL)/5 = 0.995 g/mL was considered to be the density of the distilled water.
Given the relatively small average deviation of the calculated density values, (|0.995 g/( mL)-0.995 g/( mL)| +|0.998 g/mL- 0.995 g/( mL)|+|0.995 g/mL-0.995 g/( mL)| +|0.995 g/mL- |0.995 g/( mL)| +|0.993 g/mL- 0.995 g/( mL) |)/5 = 0.001 g/mL
the results were determined to contain an adequate degree of precision. Given the relatively small percentage error of the calculated density, Error = 1 g/mL – 0.995 g/mL = 0.005 g/mL
Percentage error = (0.005 g/mL)/( 1 g/mL) * 100% = 0.5 %
the results were determined to contain an adequate degree of accuracy.
Next, the experiment was carried out on Liquid 3. A 10 mL beaker was placed on a top pan balance and “tared”....