Research on Density Essay

1684 Words Sep 22nd, 2012 7 Pages
Density: Using Experimental Techniques to Solve an Inquiry based problem

The topic of this experiment is Density. The objective is to find two ways in which the density of a given object can be determined, and to find out which of the two ways is more accurate and hence better to use in such a case. The two methods used in this experiment are finding the dimensions of the object and water displacement. These are two ways of finding the volume of an object, and they were chosen since the density of an object may be found using its mass and its volume. The experiment yielded two different density values, however when error analysis was conducted, the water displacement method was proven to be more accurate.

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Repeat measuring the mass of the object three (3) more times; ensuring that the balance is re-zeroed after each measurement. Tabulate data gathered. Procedure 2: Find the mass of the object given, as done in procedure 1. Ensure that the object is dry when it is being placed on the balance, as this will yield an inaccurate reading. Get a clean 1000milliter beaker. Fill the beaker to approximately half its capacity. Note the exact volume of water placed into the beaker. When reading the volume of the water, ensure that this is done at eye level, and that it is read at the bottom of the meniscus. After noting the initial volume of water, place the object carefully into the water. DO NOT SPLASH. Note the final volume of water in the beaker. Tabulate data gathered.

Table 1: Mass of Given Object for four(4) Separate Trials TRIAL NUMBER 1 2 3 4 MASS(g) 5.32 5.32 5.32 5.32

Table 2: Measurements of Dimensions of the Given Object DIMENSION Length- b Width- w Height- h Table 3: Initial and Final Volumes of Water INTIAL(mL) 500.0 Calculations: Volume1 = area of face of solid x width = (1/2b x h) x w = (1/2 x 10.0cm x5.0cm) x 3.5 = 87.50 cm3 Volume2 = Vf - Vi = 520.0mL- 500.0mL = 20.0mL Density = Mass/Volume D = m/V D1 = m/V1 ; m=5.32g, V1=87.50cm3 FINAL(mL) 520.0 MEASUREMENT(cm) 10.0 3.5 5.0

Therefore D1= 5.32g 87.50cm3 = 0.0608 g/cm3 D2 = m/V2 ; m= 5.32g, V2= 20.00 mL Therefore D2 = 5.32g 20.00mL =0.266g/mL


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