Introduction
When given an unknown substance, there are only a few ways to determine what it is. One way is to measure its density at a given temperature. Any pure substance has a specific density at a specific temperature. Density is defined as being equal to an object’s mass divided by its volume. The task for our lab was to determine the density of water and compare our recordings to the actual density of water listed in our lab packet. Then we were to find the density of an unknown liquid and find out what the substance was by matching its density with densities of substances listed in our packet.
Experimental
In order to find the density of water, one must know its mass and volume first. To find the water’s mass, we first weighed an empty Erlenmyer flask and rubber stopper. The rubber stopper was needed to insure no water would evaporate from the flask. This mass was recorded. The Erlenmyer flask was then filled with about 30 mL of deionized water dispensed from a buret. The flask and rubber stopper were reweighed, and the difference between the intial and final masses was the mass of the water. In order to find the volume of the water, I took note of the starting point of the water in the buret, then took note of the ending point of the water after about 30 mL were into the flask. That difference was the volume of water emptied into the flask. This process was performed three times in order to eliminate any error. Then the whole process was performed again unknown substance “Q” three times.
Results
Data for determination of the density of water:
| Trial | Mass of empty flask and stopper | Mass of full flask and stopper | Mass of water | Initial volume | Final volume | Volume of water | Density |
| 1 | 46.9959 g | 76.9994 g | 30.0035 g | 1.325 mL | 31.75 mL | 30.425 mL | 0.9861 g/mL |
| 2 | 46.9728 g | 77.2444 g | 30.2716 g | 2 mL | 32.4 mL | 30.4 mL | 0.9958 g/mL |
| 3 | 47.2037 g | 77.5281 g | 30.3244 g | 0.45 mL | 31.875 mL | 31.425 mL | 0.9650 g/mL |
Temperature of water: 19.5 °C
Average density of water: 0.9823 g/mL
Precision: 31.3544 ppt
Actual density of water (at 19.5 °C): 0.99834 g/mL
Error: 0.01604 g/mL
Data for determination of the density of an unknown liquid:
| Trial | Mass of empty flask and stopper | Mass of full flask and stopper | Mass of unknown | Initial volume | Final volume | Volume of unknown | Density |
| 1 | 47.0284 g | 70.7437 g | 23.7153 g | 0.6 mL | 30.975 mL | 30.375 mL | 0.7808 g/mL |
| 2 | 47.0235 g | 70.6998 g | 23.6763 g | 0.225 mL | 30.5 mL | 30.275 mL | 0.7820 g/mL |
| 3 | 47.0176 g | 70.6753 g | 23.6577 g | 0.4 mL | 30.65 mL | 30.25 mL | 0.7821 g/mL |
Temperature of unknown: 19.5 °C
Average density of unknown: 0.7816 g/mL
Precision: 1.6632 ppt
Unknown code: Q
Name of unknown: 2-propanol
Actual density of 2-propanol (at 20 °C): 0.786 g/mL
Error: 0.0044 g/mL
Calculations
In order to find the mass of the water and unknown, a simple subtraction problem was used. I simply subtracted the mass of the empty flask and rubber stopper from the mass of the full flask and rubber stopper. A sample equation would be 76.9994 g – 46.9959 g = 30.0035 g. This same method was used to find the volume of water in the flask. I subtracted the intial amount of liquid in the buret from the final amount of liquid in the buret. This difference was how much liquid was dispensed. An example would be 31.75 mL – 1.325 mL = 30.425 mL.
In order to find the density, I simply divided the mass found by the volume found. For example, 30.0035 g divided by 30.425 mL equals 0.9861 g/mL. To find the average density, I added the three densities I found, then divided that total by three to find the average. The equation for the water was (0.9861 g/mL + 0.9958 g/mL + 0.9650 g/mL) / 3 = 0.9823 g/mL.
Precision was found by taking the absolute value of the highest density minus the lowest density, dividing that difference by the average density, and then multiplying that answer by 1000. For example, the precision for the water was found by this equation: |(0.9958 g/mL – 0.9650 g/mL)| / 0.9823 g/mL x 1000. This gave me an answer of 31.3544 ppt (parts per thousand).
Finally to find the error, I found the absoulte value of my measured density minus the actual density. With my data, my equation for the water was |0.9823 g/mL – 0.99834 g/mL| = 0.01604 g/mL.
Discussion/Conclusions
My results for water turned out fairly well. My accuracy was very high, but my precision was not quite as good. A precision of 31.3544 ppt is a lot higher than 4 ppt, which is what is typically required to make sure my measurements were precise. However, my accuracy turned out to be very high, as my error was very low. It seems that I was lucky to have gotten such good accuracy with bad precision.
My data from the unknown substance turned out incredibly well. My precision was 1.6632 ppt, which is well under 4 ppt which is typically required. I assume 0 ppt would be perfect precision, so 1.6632 ppt is very good. My accuracy was also high, as my error was only 0.0044 g/mL. I don’t think I could have done the procedure much better than that.
Some of my error can be accounted for by a leak in my buret. A few drops of the liquid inside seemed to drip out from right above the bottom where it was supposed to come out. Also, I may have gotten finger prints on the flask, which would have added a slight bit of extra weight that could throw my calculations off. Lastly, the density given for my unknown was listed at 20 °C, but I measured the room temperature to be 19.5 °C. Therefore, I should be even closer to the actual density. For the water, I took the average of the density of water at 19 °C and 20 °C to find its density at 19.5 °C, but that may be incorrect if the density follows a curve. In the end, my data was fairly accurate with the actual data, so the experment was a success. My results showed that density is really equal to a substance’s mass divided by its volume.
