Density Lab Report Example | Topics and Well Written Essays - 1000 words

Density - Lab Report Example By means of displacement, the difference in the initial and final volumes of liquid used would be the volume displaced through the solid object plunged into the water, and, therefore, it serves as the volume of the material itself. At this stage, given the mass, it may be hypothesized that the mass of pennies divided by the volume displaced or occupied which causes rise in the water level within the graduated cylinder yields to the penny’s density. Methods and Materials Type of Metal Used: Early-1985 penny with an accepted density value of 7.18 g/ml. Initially, a 50-ml graduated cylinder was filled with 20.00 ml of water, and it was carefully recorded. The initial mass of cylinder and water (combined) was acquired using a top-loading balance and then recorded as initial reading for the cylinder-balance setup. Pennies were dropped by increments of two where the new volume was read and the new mass was measured using the same balancing equipment each time. This step was carried out repetitively to make a total of five data sets, which includes noting of final volumes along and masses (7th and 8th columns) through addition of previous differences with actual volumes and masses, correspondingly. There were ten pennies dropped all in all, and in order to obtain the experimental value of density for each set of pennies, the following equation was applied: Density, ? = [ Mass(2) - Mass(1) ] / [ Volume(2) - Volume(1) ] (in g/ml) Results Initial Volume of Water (ml): 20.0 ml_ Initial Mass of Cylinder + Water: 105.06 g_ density, g/ml % difference Rep A 5.03 29.94 Rep B 9.82 36.77 Rep C 10.06 40.11 Rep D 5.00 30.36 Rep E 9.98 39.00 Sample Calculations (using Reps A & B of the table): Actual Volume = 21.5 ml - 21.0 ml = 0.5 ml Actual Mass = 115.0 g - 110.09 g = 4.91 g Density = Actual Mass / Actual Volume = 4.91 g / 0.5 ml = 9.82 g/ml Final Volume = Actual V1 + Actual V2 = 1.0 ml + 0.5 ml = 1.5 ml Final Mass = Actual M1 + Actual M2 = 5.03 g + 4.91 g = 9.94 g Then us ing the given theoretical value = 7.18 g/ml and the formula % difference = | 9.82 - 7.18 | / 7.18 x 100% = 36.77% volume, ml mass, g 20 105.06 21 110.09 21.5 115 22 120.03 23 125.03 23.5 130.02 Based on the 3rd and 4th columns of the first table, beginning with a volume (water) of 20.0 ml and a mass (cylinder + water) of 105.06 g, the amounts (volume and mass) of each succeeding row are subtracted from the corresponding amounts of the preceding row to generate the 5th and 6th column outputs showing actual entries specific for every two-penny increment. Since these densities appear to be significantly different as compared to the literature value of 7.18 g/ml, the average density was estimated from the graph of mass vs. volume of pennies. Considering the best-fit line drawn (via MS Excel program) fairly between the plotted coordinates, the slope would be 7.186 g/ml according to the resulting equation m = 7.186v – 39.36, so that percent difference equals (7.186 - 7.18) / 7.18 x 100% or 0.0836%, which is appreciably lower than the % difference solved individually, as shown prior. Discussion / Conclusion Though the outcomes reflect inconsistent values of density on the basis of the actual volumes and actual masses which had been arrived at through the displacement method,