Question

Density measurements were conducted on a 22.5oC sample of water which had a theoretical density of 0.997655 g/ml. A volume of 10.00 ml of the water had a mass of 9.98 g. A volume of 15.00 ml of the water had a mass of 15.61 g. A volume of 25.00 ml of the water had a mass of 25.65 g. (1) Show the calculation of the density of each volume. (2) Show the calculation of the average density. (3) Show the calculation of the percent error based on the theoretical density of 0.997655 g/ml.

Answer 1

Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).

• (1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:

$p_{A} =9.98g/10.00mL=0.998g/mL\\p_{B} =15.61g/15.00mL=1.041g/mL\\p_{C} =25.65g/25.00mL=1.026g/mL$

• (2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):

$p_{average}=\frac{p_{1}+p_{2}+p_{3}}{3}   \\p_{average}=\frac{(0.998+1.041+1.026)g/mL}{3}\\p_{average}=1.022g/mL$

• (3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:

%error=$\frac{|p_{average}-p_{theoretical}|}{p_{theoretical}} *100$

%error=$\frac{|1.022g/mL-0.997655g/mL|}{0.997655g/mL}*100$

%error=2.44 %