CHM 1112 (General Chemistry Lab I)

Introduction

A spectrophotometer measures the amount of light absorbed by a solution at different wavelengths of light emitted. Beer’s Law says that absorbance is equal to molar absorptivity times the thickness of the sample times the concentration of the sample. Beer’s law also states that conformity of a solution is able to be determined by plotting its absorbances versus its concentrations, and if a straight line results crossing through the origin, the solution has conformity. Using this information, it is possible to determine an unknown concentration of a solution by finding its absorbance, or if given its concentration, its absorbance can be found without the use of a spectrophotometer.

Experimental

First, a spectrophotometer was turned on, allowed to warm up for about 15 minutes, and was set at a wavelength 400 nm. A cuvette filled with deionized water was used for blanking the spectrophotometer. A second cuvette was filled with a solution of potassium permanganate which was provided. Each cuvette was wiped with a Kimwipe before being placed in the spectrophotometer in order to eliminate smudges which could affect the light passing through. The spectrophotometer was blanked at 400 nm and the cuvette with the potassium permanganate solution was placed in, and its absorbance was read and recorded. It was taken out, and the spectrophotometer was then blanked at 410 nm. The cuvette with the potassium permanganate solution was once against placed in the spectrophotometer. Its absorbance was read and recorded again. This process was repeated, increasing the wavelength of the spectrophotometer by 10 nm until it reached 640 nm when recording ceased. The wavelength with the highest absorbance was used for the rest of the experiment.

Four volumetric flasks were then used to make solutions of KMnO₄. Flask 1 was a 100 mL volumetric flask that contained 10 mL of 3.170 x 10^-4 M KMnO₄, which was dispensed into the flask using a buret. Flasks 2 through 4 were all 50 mL volumetric flasks that contained 20 mL, 30 mL, and 40 mL respectively of 3.170 x 10^-4 M KMnO₄. All four volumetric flasks were filled to the line on the neck with deionized water. All the flasks were agitated, and cuvettes were filled with each sample. Each cuvette was placed in the spectrophotometer and their absorbances were all recorded.

Next the unknown was placed into a 250 mL beaker and 10 mL of concentrated nitric acid was added to it. Then 0.5 g of potassium periodate was dissolved in 40 mL of deionized water. This solution was heated with a hot plate in order to aid the dissolving process. The contents of the 250 mL beaker were emptied into this solution and were heated for about 10 minutes, but the solution was never brought to a boil. After heating, the solution was put on ice and brought back to room temperature. A cuvette was then filled with this solution and its absorbance was determined and recorded using the spectrophotometer.

Results

Absorption vs. Wavelength for Maximum Absorbance Determination:

Standard Solutions:

Unknown Number: 14

Wavelength: 530

Calculations

To find the concentration of the standards, I figured out how much the KMnO_4­ was diluted in each volumetric flask. I did this by taking the amount of KMnO_4­ added, then divided by the total volume on the volumetric flask. I then multiplied this percentage by the original concentration of KMnO₄, which was 3.170 x 10^-4. To find the concentration of the unknown solution, I first got the equation of the standard curve line, which was y = 2701.2x – 0.048. I then substituted the absorbance I found for the unknown, which was 0.322, for y. I could then find the value of x, which was the concentration.

Discussion/Conclusions

Potassium permanganate does indeed seem to follow Beer’s Law. When I plotted the absorbances found against the concentrations, I was left with nearly a straight line that goes almost directly through the origin. It is only 0.048 absorbances away from going through the origin, and the best fit line is very close to hitting every point plotted. This is one way to prove conformity and Beer’s Law.

Sources of error in this experiment could occur many different ways. If the cuvettes are not wiped off before being placed in the spectrophotometer, there could be smudges or fingerprints that would cause error. The wavelength on the spectrophotometer had to be set by eye, so there is some room for error there, too. If the dilutions are made inaccurately, that would also cause error in absorption readings. Overall, if anything measured in this experiment was measured inaccurately, that would cause error. Also, if the solution with the unknown in it was boiled, that may cause it to form something different than we wanted to measure and that would cause error, too.

Introduction

Enthalpy of hydration is the energy change for converting 1 mol of an anhydrous substance to 1 mol of the hydrated substance. In order to find this number, it is necessary to first calculate the enthalpy of dissolution for each substance separately, and then find the different between the two. The enthalpy of dissolution is the energy change of dissolving 1 mol of a substance in water. It is calculated using temperature changes in the water, heat capacity of the substance, and the weight of the mixture. For this experiment, MgSO₄ and MgSO₄ ∙ 7 H₂O were used and the enthalpy of hydration between the two was calculated.

Experimental

A Styrofoam cup and stirring bar were first obtained and weighed together. This mass was recorded. 100.0 mL of deionized water was measured with a graduated cylinder and then put into the cup with the stirring bar. The cup was again weighed and this new mass was recorded. The cup was then placed on a mixing plate set on medium to high and its temperature was recorded every 30 second for 4.5 minutes. An unknown amount of MgSO₄ salt was added to the cup. The cup kept on the mixing plate set on medium to high and its temperature was recorded every minute for 15 minutes. Finally, the cup was weighed and its final mass was recorded. This process was repeated placing the MgSO₄ with MgSO₄ ∙ 7 H₂O.

Results

Enthalpy of Hydration: -106.6 kJ

Calculations

To find the mass of water used, I subtracted the weight of the cup with just the stirring rod from the weight of the cup with the stirring rod and water. To find the weight of the salt used, I subtracted the weight of the cup, stirring rod, and water from the final weight of the cup. In order to find the moles of solute used, I divided the mass of the salt by its molar mass. To find the change in temperature, I subtracted the initial temperature from the final temperature. In order to find Q, the heat capacity of the reaction mixture, I used the equation Q = – (mass of mixture) * (heat capacity of mixture) * (ΔT). To find the ΔH_dissolution, I used the equation ΔH = Q / (number of moles of solute). Lastly, to calculate the enthalpy of hydration, I subtracted the ΔH_dissolution of the MgSO₄ ∙ 7 H₂O from the ΔH_dissolution of the MgSO₄.

Discussion/Conclusions

I was surprised that while the MgSO₄ salt heated the water, the MgSO₄ ∙ 7 H₂O salt cooled the water down. It was interesting that two substances very close in chemical makeup could have such different reactions in water. My graph for the temperature change of water with MgSO₄ seems to only gradually jump in temperature after adding the salt. I believe this is because my lab partner forgot to turn the mixer on, so the salt was not completely mixing at first. Other than that, the procedure went well. The enthalpy of hydration of -106.6 kJ seems fairly high. Water takes 4.184 kJ to be raised only 1 ºC, so 106.6 kJ seems like a lot of energy.

Introduction

One gram of water takes 4.184 joules of energy to increase its temperature 1 ºC. This is the most energy any substance takes to raise its temperature 1 ºC. In contrast to taking the most energy to raise its temperature 1 ºC, this means that it also takes the longest to cool down. This means that water has the highest heat capacity. It must release 4.184 joules of energy in order to decrease its temperature by just 1 ºC. The heat water releases is absorbed by its environment. Knowing the heat capacity of water, it is possible to find how well its environment insulates it. Also using the heat capacity of water, one can figure out the heat capacity of an unknown substance by putting it in water and measure the temperature change of the water and the unknown substance. In this experiment, this is exactly what was performed.

Experimental

First, an empty Styrofoam cup and lid were weighed and its mass was recorded. 70 mL of room temperature water was then added to the Styrofoam cup and it was reweighed and recorded. The temperature of the water was also recorded. Next, 30 mL of water was heated until boiling and this temperature was also recorded. The boiling water was poured into the Styrofoam cup and the final temperature of the combined water was measured and recorded. The final mass of the cup was also recorded.

For the next part of the experiment, an empty Styrofoam cup and lid were again weighed and its mass was recorded. 100 mL of room temperature water was added to the cup and it was reweighed and recorded. The temperature of the water was also recorded. Next, an unknown metal was heated to about 100 ºC and then poured into the Styrofoam cup. The final temperature of the water was measured and recorded, as was the final mass of the cup.

Results

Identification of Metal: 12

Determination of Calorimeter Constant, B:

Determination of the Heat Capacity of a Metal:

Calculations

For the determination of the calorimeter constant, to find the mass of cool water in the cup, I simply subtracted the mass of the empty cup from the mass of the cup with 70 mL of water. To find the mass of hot water added, I subtracted the mass the cup with 70 mL of water from the mass of the cup with the 70 mL of cool water and 30 mL of hot water. To find the temperature changes, I found the difference in temperatures between the final and the initial readings. I then converted those temperatures to Kelvin from Celsius by adding 273. To find the calorimeter constant, I used the equation B = -C_P(m_CW ΔT_CW + m_HW ΔT_HW) / ΔT_CW.

For the determination of the heat capacity of a metal, I performed the same operation as I did for the determination of the calorimeter constant, only replacing the mass of hot water with the mass of the hot metal. The equation for the heat capacity of the metal was also different. It was C_{P, M} = – ΔT_CW (B + m_CW C_P) / (m_HM ΔT_HM). Lastly, to find the molar mass of the metal, I then divided 25 J/mole K by the heat capacity of the metal, which is the Law of Dulong and Petit.

Discussion/Conclusions

My final results do not seem very accurate. The calorimeter constants seem fairly close, but that difference affected the result of the heat capacity of my metal greatly. If I had gotten the same calorimeter constant for both trials, then the heat capacity of the metals would have came out nearly equal. As a result of the heat capacity of the metals being different, their molar masses were also thrown off. In conclusion, I must have made a mistake in a reading while finding the calorimeter constant.

Something that surprised me was how low the heat capacity of the metal was. Normally I think of metals as being very hot and staying hot, but this experiment proved how metals actually cool very quickly. People usually make this generalization, but they are wrong. I think it is because metals heat up more quickly, so they typically think they stay hot. In actuality, hot water is more dangerous than hot metal.

Introduction

Acids have a pH between 0 and 7, while bases have a pH between 7 and 14. A solution with a pH of 7 is said to be neutral; it is neither an acid nor a base. When given an acidic solution, it is possible add basic solution in order to neutralize it. To tell if the solution has been neutralized, an indicator such as phenolphthalein is used. The indicator will make the solution change color when it has become basic. In this experiment, a 10% solution of unknown molarity HCl was titrated with 1.00 * 10^-1 M NaOH in order to neutralize the HCl. The molarity of the HCl was then able to be calculated knowing the molarity of NaOH, the volume of NaOH used, and the volume of HCl.

Experimental

First, 25.00 ml of unknown sample of acid was delivered to a 250 mL volumetric flask using a volumetric pipette. The volumetric flask was then filled with de-ionized water to the mark in order to complete the 10% solution. A buret was then filled with 1.00 * 10^-1 M NaOH solution and its starting point was recorded. Next, 25.00 mL of the acid solution was delivered to 300 mL Erlenmeyer flask and 3 drops of phenolphthalein were added. The Erlenmeyer flask was put under the buret and the NaOH solution was dispensed into the Erlenmeye flask until the indicator turned the solution pink for around 15 seconds. The final recording on the buret was recorded, and the process was performed 3 times in order to reduce error.

Results

Molarity of NaOH solution: 1.00 * 10^-1 M

Identification of acid: I

Mean Molarity of Undiluted acid solution, M: 1.01 M

Calculations

In order to find the volume of titrant used in mL, I simply subtracted the initial buret reading from the final buret reading. To convert that volume in mL to L, I divided by 1000, as there are 1000 mL in 1 L. To find the molarity of diluted acid solution, I used the equation M₁V₁ = M₂V₂. M₁ equals the molarity of the NaOH solution used (1.00 * 10^-1 M), V₁ equals the volume of titrant used in L, V₂ equals the volume of acid solution used (0.02500 L), and M₂ is the molarity of the acid solution, which was solved for. To find the molarity of the undiluted acid solution, I knew that a 10% solution was used, so I multiplied the molarity of the diluted solution by 10 to get the molarity of the undiluted solution. Lastly, to find the mean molarity of undiluted solution, I added the molarity of undiluted acid solution from the 3 trials and then divided by 3.

Discussion/Conclusions

The final result of the molarity of the undiluted acid came very close to a whole number, which should mean my results are valid. It is also a very plausible number for the molarity of a solution. The volume of titrant used in my last two trials is nearly exact, so I must have performed them very well. The first trial is probably slightly off, as it was my first time doing a titration. On my first trial I added drops of NaOH too quickly towards the end. It is necessary to add the final drops and half drops very carefully towards the end, as it is a fine line between neutralizing the solution and making it basic. Overall, it was a very tedious experiment that relied on precision in order to achieve viable results.

Introduction

When two aqueous solutions are mixed together, they often react chemically and form products. The chemical reaction may be visible as a change in color from the reactants to the products, the release of gas in the product, or the formation of a precipitate. A precipitate forms when one of the resulting products is aqueous and the other product is insoluble. The insolubility means that the product will be a solid, which usually settles to the bottom of the resulting aqueous solution.

It is possible to determine how much precipitate will form before actually combining the reactants. Using stoichiometry, one can figure out how much precipitate theoretically will be produced. In the lab there are outside factors, which can affect how much precipitate is actually formed. Only under ideal conditions would the actual amount of precipitate formed match the theoretically amount of precipitate formed.

In this lab, strontium iodate monohydrate was synthesized using the following equation: Sr(NO₃)₂ + 2KIO₃ —-> Sr(IO₃)₂ + 2KNO₃. With the knowledge of the starting amount of Sr(NO₃)₂ and KIO₃ used, it was possible to figure out how much strontium iodate monohydrate should be theoretically produced. Then after finding out how much strontium iodate monohydrate actually was produced, the percent yield was found.

Experimental

First, approximately 40.00 ml of 5.00 x 10^-2 M Sr(NO₃)₂ and 50.00 mL of 1.00 x 10^-1 M KIO₃ were put into separate graduated cylinders. They were then combined in a beaker sitting in ice. This was to prevent the strontium iodate monohydrate from becoming soluble in water, as its solubility in water goes up with its temperature. The mixture was then stirred for about 10 minutes, so that all the precipitate could form. The precipitate and supernate were then poured into a vacuum filter, and the beaker was rinsed with ice cold distilled water to get all the precipitate out. The filter paper used in the vacuum filter was first weighed before filtration, then once the filtered precipitate was dry, the filter paper and precipitate was weighed. This whole procedure was performed twice.

Results

Volume of Sr(NO₃)₂ solution used: 39.99 mL (1^st run), 40.25 ml (2^nd run)

Molarity of Sr(NO₃)₂ solution used: 5.00 x 10^-2 M

Volume of KIO₃ solution used: 49.98 mL (1^st run), 49.50 mL (2^nd run)

Molarity of KIO₃ solution used: 1.00 x 10^-1 M

Mass of product, watch glass, and filter paper: 32.04 g (1^st run), 27.32 g (2^nd run)

Mass of watch glass and filter paper: 31.33 g (1^st run), 26.47 g (2^nd run)

Mass of product: 0.71 g (1^st run), 0.85 g (2^nd run)

Mass of Sr(NO₃)₂: 0.68 g (1^st run), 0.82 g (2^nd run)

Number of moles of Sr(NO₃)₂ used: 0.00200 moles (1^st run), 0.00201 moles (2^nd run)

Number of moles of KIO₃ used: 0.00500 moles (1^st run), 0.00495 moles (2^nd run)

Limiting reagent: Sr(NO₃)₂

Theoretical yield of product, moles: 0.00200 moles (1^st run), 0.00201 moles (2^nd run)

Theoretical yield of product, g: 0.875 g (1^st run), 0.880 g (2^nd run)

Percent yield: 78.% (1^st run), 93.% (2^nd run)

Mean percent yield: 86.%

Calculations

In order to find the mass of the precipitate, took the mass of the product, watch glass, and filter paper minus the mass of the watch glass and filter paper. To find the moles of the reactants used, I used the equation Molarity = Moles/Liters and rearranged it to the equation Moles = Molarity x Liters. Before subbing the volume in, I had to convert mL to L by dividing by 1000. To find the limiting reagent, I first had to look at the balanced equation of the chemical reaction, which was Sr(NO₃)₂ + 2KIO₃ —-> Sr(IO₃)₂ + 2KNO₃. Since I knew for every 2 moles of KIO₃ used, 1 mole of Sr(NO₃)₂ was used, I could substitute the actual number of moles of each that were used in the experiment to find the limiting reagent. If KIO₃ was the limiting reagent, then 0.00250 moles of Sr(NO₃)₂ would be needed for the 0.00500 moles of KIO₃, but there were only 0.00200 moles of Sr(NO₃)₂ used, so that made it the limiting reagent.

In order to find the percent yield, I first took the number of moles of Sr(NO₃)₂ used because from the balanced equation, I knew that there would be an equal number of moles of Sr(IO₃)₂ produced. I then found the molar mass of Sr(IO₃)₂, which is 437.43 g, then multiplied by 0.00200 moles, which is the number of moles of Sr(IO₃)₂ theoretically produced, to find the number of grams of Sr(IO₃)₂, theoretically produced (0.875 g). I then took mass of the precipitate strontium iodate monohydrate produced (0.71 g) and needed to find the mass of Sr(IO₃)₂ produced. So I found the percent of Sr(IO₃)₂ that makes up strontium iodate monohydrate by taking the molar mass of Sr(IO₃)₂ (437.43 g) and divided by the molar mass of strontium iodate monohydrate (455.44 g) to get 96.046%. I then multiplied the mass of the product by this to find the mass of Sr(IO₃)₂ actually formed (0.71 g x 0.96046 g= 0.68 g). I then took 0.68 g and divided by the mass of Sr(IO₃)₂ theoretically produced (0.875 g) and multiplied by 100 to find get the percent yield of 78.%. I repeated this for the values found in the second run. For the mean percent yield, I took the percent yields found for each run, added, then up, and divided by 2.

Discussion/Conclusions

My results were fairly close to what they should have been. A mean percent yield of 86.% is probably good considering all the factors that can cause the percent yield to be less than 100%. The Sr(IO₃)₂ could get too warm and wash away in the water, the precipitate could not completely form when mixing the reactants, and some precipitate could become stuck in the beaker and not wash out. Those are factors that are not all easily controlled, so overall my percent yield of 86.% seems plausible when looking at those factors that could affect the results.

If I were to repeat this experiment, I would probably take more time in letting the reactants mix and form the precipitate. That way I could be sure almost all the precipitate actually formed. I would also keep the wash bottle in colder conditions to make sure none of the precipitate washed away.

Introduction

For this lab, seven small experiments were performed. First, solid Mg was combined with a HCl solution. The reactants started to bubble and give off a colorless gas was given all while the Mg was dissolving. Next a Pb(NO₃)₂ solution was mixed with a KI solution. These two clear, colorless liquids produced a transparent yellow liquid and yellow sediment at the bottom of the test tube. After that, CuSO₄ ∙ H₂O, a fine blue crystal solid, was heated to leave behind a white powdery solid. Next a HCl solution and NaOH solution with phenolphthalein were combined. When the HCl solution was added to the NaOH, its color changed from purple to clear. Following that experiment, Cu was heated in the presence of O₂. The shiny, copper colored metal turned dark gray and lost its luster. After that, a CuSO₄ solution was combined with solid Fe to produce rust. Lastly, a FeCl₃ solution (yellow in color) was mixed with a NaOH solution (colorless in color) to yield a cloudy orange liquid.

Introduction

Salts are substances that have the property of readily allowing water molecules to attach themselves to its crystalline structure. The water molecules that combine with the salt can result from exposure to humid conditions. When a substance becomes saturated with water, it is said to be hydrated. A salt that becomes hydrated is called an ionic hydrate. The forces holding the water molecules and crystal lattice together can either be hydrogen bonds, coordinate covalent bonds, or ion-dipole forces. Through heat, these bonds can be separated and the original salt can be recovered. In this experiment, an unknown ionic hydrate was heated in order break any bonds between water molecules and the salt’s crystal lattice to recover the original salt. Then when given the molar mass of the unknown salt, the ratio of H₂O to salt in the original hydrate was able to be figured out.

Experimental

An empty crucible was first heated with a Bunsen burner to get rid of any moisture that would add weight to the crucible. It was then placed in a desiccator to cool down, while not absorbing any moisture from the air. Once cool, the empty crucible was then weighed and the mass was recorded. Next 1 g of unknown ionic hydrate was placed into the crucible and this new mass was recorded. The crucible with ionic hydrate inside was then heated with its lid partially on for 20 minutes. The heat started off low and was steadily increased throughout the 20 minutes. The crucible and ionic hydrate were then allowed to cool in the desiccator. Once cool, they were weighed and this mass was recorded. The crucible and ionic hydrate were heated once again to remove any excess moisture left from the hydrate. The hydrate and crucible were cooled and weighed as before, and if its new mass was less than 0.003 g different than the previous weigh, it was considered to be completely free of water. The remaining product was the anhydrous salt.

Results

Identification code of unknown: B

Mass of empty crucible: 8.766 g

Mass of hydrate sample: 1.057 g

Mass of crucible plus sample before heating: 9.823 g

Mass of crucible and sample…

• After first heating and cooling: 9.674 g
• After second heating and cooling: 9.674 g

Mass of water lost: 0.149 g

Mass percent of water in sample: 14.1 %

Molar mass of anhydrous salt: 110. g/mol

Mass of anhydrous salt: 0.908 g

Number of moles of anhydrous salt remaining in the crucible: 0.00825 mol

Number of moles of water lost: 0.00827 mol

Average number of moles of water per mole of hydrate: 1 mol

Formula of hydrate: X ∙ H₂O

Calculations

In order to find the mass of the crucible and sample together, I simply added the mass of the empty crucible and mass of the hydrate alone. To find the mass of water lost, I subtracted the mass of the crucible and sample after the second heating from the mass of the crucible and sample before heating. In order to find the mass percent of water in the sample, I divided the mass of water lost (0.149 g) by the original mass of the hydrate sample (1.057 g) and then multiplied that answer by 100 to find the percent. To find the mass of anhydrous salt, I subtracted the mass of water lost from the mass of the hydrate sample. To find the number of moles of anhydrous salt remaining in the crucible, I took the mass of anhydrous salt (0.908 g) and divided that by its molar mass (110. g/mol). I performed that same operation to find the number of moles of water lost using a molar mass of 18.01528 g/mol for H₂O. Lastly, to find the number of moles of water per mole of hydrate, I divided the number of moles of anhydrous salt remaining by itself and by the number of moles of water lost. This came out to a 1:1.002 ratio, which is very close to 1:1.

Discussion/Conclusions

My final equation for my hydrate came out to almost exactly a whole number ratio. This could mean I performed the experiment very well, or that I simply got lucky and my numbers ended up coming out even. There is no way for me to check this because I don’t know the formula of the anhydrous salt, which would tell me what ionic hydrate it usually forms. If I knew the formula of the anhydrous salt, then I would be able to check my work and make sure my results came out correctly.

Overall, I think the experiment went smoothly. I was able to heat the crucible without turning it red, which could have caused some of the hydrate to burn away. I only had to reheat the crucible once before I got a constant mass. I did not have any trouble with that step, as I know some people did. Everything went as planned and I am fairly confident my formula is correct.

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:

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:

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.