DATA AND OBSERVATIONS:
Code of unknown liquid: C
Appearance of unknown liquid:
Green, translucent liquid.
Mass of aluminum cylinder:
54.494 g
Initial water level in graduate (for aluminum):
44.1 mL
Final water level in graduate (for aluminum):
64.9 mL
Mass of five rubber stoppers:
35.531 g
Initial water level in graduate (for stoppers):
49.2 mL
Final water level in graduate (for stoppers):
77.9 mL
Volume of unknown liquid dispensed:
9.22 mL
Mass of empty 100 milliliter beaker:
57.858 g
Mass of beaker and unknown liquid:
67.231 g
CALCULATIONS:
Volume of the aluminum cylinder:
64.9 mL - 44.1 mL = 20.8 mL
Volume of the five rubber stoppers:
77.9 mL - 49.2 mL = 28.7 mL
Mass of unknown liquid:
67.231 g - 57.858 g = 9.373 g
Density of the aluminum cylinder:
D = 54.494 g = 2.62 g/mL
20.8 mL
Density of the rubber stoppers:
D = 35.531 g = 1.24 g/mL
28.7 mL
Density of unknown liquid:
D = 9.373 g = 1.02 g/mL
9.22 mL
Percent error for density of aluminum:
(2.62 g/mL - 2.70 g/mL) x 100 = -2.96%
2.70 g/mL
Percent error for density of rubber:
(1.24 g/mL - 1.26 g/mL) x 100 = -1.59%
1.26 g/mL
Percent error for density of unknown liquid:
(1.02 g/mL - 1.000 g/mL) x 100 = 2.00%
1.000 g/mL
QUESTIONS:
1. Measurement A is more precise because it has divisions on the graduated cylinder of
0.5 mL rather than divisions of 1 mL like the larger graduated cylinder. Since the volume
of the object is only 10 mL it will not exceed the 25 mL graduated cylinder’s capacity
and thus give you a more precise measurement.
2. If only one rubber stopper was used in the measurement of volume you would not
achieve enough significant figures to give you a reasonably precise answer. Using more
stoppers increases the amount of significant figures we can have due to the lack of
precision with a 100 mL graduated cylinder.
3. Volume of object:
Density of object:
51.7 mL - 42.1 mL = 9.60 mL
56.238 g = 5.86 g/mL
9.60 mL
5. 1.25 L 1000 mL = 1250 mL 1.59 g = 1990 g or 1.99 kg
1L
1 mL