Table 1: Density based on dimensional analysis Trial 1 Trial 2 Trial 3 Units
Mass of sample 70.5466 70.5467 70.5465 g
Dimensions of sample Length 4.98 4.98 4.99 cm Width 1.21 1.22 1.21 cm Height 1.21 1.22 1.22 cm
Volum of sample 7.29 7.41 7.37 cm^3
Density of sample (based on dimensional analysis) 9.68 9.52 9.57 g/cm^3 Table 2: Density based on the displacement of water Trial 1 Trial 2 Trial 3 Units
Mass of sample 70.5467 70.5465 70.5466 g
Final volume of the water in the buret (Vf) 29.83 13.11 25.51 ml
Initial volume of the water in the buret (Vo) 14.62 2.61 13.18 ml
Volume of the water dispensed from the buret (Vb=Vf-Vo) 15.21 10.50 12.33 ml
Volume of water level on graduated cylinder: …show more content…
CALCULATIONS
a. Relative Percent Error = IAbsolute ErrorI/True Value X 100%
IAbsolute ErrorI = ITrue Value - Measured ValueI
The Measured Value in my experiment based on table 2 is [(8.50+8.11+9.44)/3]= 8.68 g/cm^3
The True Value is 8.94 g/mL
Absolute Error = 8.94 - 8.68 = 0.26
Relative Percent Error = (0.26/8.94)x100=2.91 Using the correct number of significant figures, gives us the answer: …show more content…
QUESTIONS
"1. The Density formula is D=Mass/Volume. Since we already have the mass = 18.45 g we have to find now the volume of the sphere by finding the difference in volume of the graduated cylinder:
26.8ml-21.7ml=5.10ml
We can now divide the mass by the calculated volume to find the density. 18.45g/5.10ml=3.62g/ml, or 3.62g/cm^3. The Density of the metal is 3.62g/cm^3"
2. The Density formula is D=Mass/Volume. Since we already have the mass = 146 g we now have to find the volume of the solid block using the formula: V=LxWxH. We now plug the values in and we have: V=6.0x3.0x.3.0 V=54cm^3. We can now devide the mass by the volume to find the density: D=146g/54cm^3 D=2.70g/cm^3
3. a. Volume Before NaCl Added: 5.12 mL; Volume After NaCl Added: 7.43 mL
3.b. We determine the volume of NaCl used by finding the difference in volume of the graduated cylinder: 7.43mL-5.12mL=2.31mL (or 2.31cm^3)
3.c. Density=Mass/Volume. The mass given is 5.03g. The volume found is 2.31mL
D=5.03g/2.31mL
D=2.18g/mL
3.d. The Relative Percent Error = IAbsolute ErrorI/True Value X 100%; True Value =