CHEM 111 Molar mass of a volatile liquid PDF

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Destiny Cambero CHEM 111 – Farnum MTWR 9:40 AM 08/15/18

Experiment: 12 Molar Mass of a Volatile Liquid Conclusion: In conclusion, my unknown sample was #6. The average molar mass of my sample was 78.24g/mol, my standard deviation of my molar mass was 15.33, and my relative standard deviation was 19.59%. Due to both my standard deviation and my relative standard deviation being high, I was neither precise not accurate when performing this experiment.

Experiment: Molar Mass of a Volatile Liquid – Abstract The purpose of this was to perform an experiment that allowed us to apply and measure the physical properties of pressure, volume, and temperature, for our unknown volatile liquid substance. Volatile liquids boil at low temperatures and as a result usually have a low molar mass. By Utilizing these properties allowed us to determine the molar mass of our unknown liquid, which is the number of grams per mole of an element or compound. To do this we continuously weighed, boiled, and reweighed our substance in a fixed volume at a measured temperature and pressure. To calculate this, we will first use the ideal gas law and rearrange the equation to solve for the amount of moles. To determine the number of grams we simply subtracted the mass of our liquid sample and its container by the mass of the container empty dry container. The molar mass is simply our acquired mass in grams divided by the number of moles. The major results in this experiment were the average molar mass, which turned out to be 78.24g/mol. As for my standard deviation, I got a result of 15.33 and a relative standard deviation of 19.59%. Although I had believed that the experiment went well, based on my results my experiment lacked both accuracy and precision. There are many possible factors that could have contributed to this, most likely due to the fluctuation during my mass readings.

Experiment: Molar Mass of a Volatile Liquid – Introduction The goal of this experiment was to determine the molar mass of a volatile liquid using the physical properties of pressure, volume, mass, and temperature. When identifying a compound chemists use physical properties, these can consist of things such as melting point, color, density and other factors. In this experiment, we will be identifying the physical property of molar mass, this is the mass in grams per one mole of an element or compound. Different analytical methods can be used to identify the molar mass, these methods vary based on the compounds properties. Since we are determining the mass of a volatile liquid we will use the Dumas method and the ideal gas law. The Dumas method, was named after a French chemist named Jean-Baptiste Dumas, this method consisted of filling a glass bulb with the volatile liquid and vaporizing the sample by heating it while immersed in a hot water bath. This allows us to determine these physical properties in vapor phase. Volatile liquids have low boiling points due to their relatively low molar masses, because of this, heating our unknown substance to its boiling point is an effective way of identifying our unknown molar mass. The Dumas method utilizes the ideal gas law to determine the number of moles in a substance which leads to the determination of an unknown molar mass. The ideal gas law is an equation that is used to determine the state of a hypothetical gas usually at STP conditions, which stands for standard temperature and pressure. This law contains all the simple gas laws within it, Boyle’s, Charles’s, and Avogadro’s Law. Boyles Law assumes a constant temperature and amount of a gas, where as pressure increases, volume decreases. Charles law assumes a constant pressure and amount of gas, where as temperature increases volume increases. Lastly, Avogadro’s Law assumes a constant temperature and pressure and is independent of the normal nature of a gas. This law can be used to determine the volume and change in the gas amount if these conditions are kept constant. The ideal gas law is represented by the equation pV=nRT, where p stands for pressure, V for volume, n for the number of moles, T for temperature, and R

is the ideal gas constant 0.08206 L•atm/moles• Kelvin. This equation can be rearranged to determine missing properties. These equations can be organized together as, M= mRT/PV, where M represents the molecular mass of the unknown gas. To better understand the concept of the ideal gas law we should understand the fundamentals of the kinetic molecular theory of gases, which is a model used to describe how gases behave due to their microscopic and macroscopic properties. There are five basic assumptions of gases made from this model, such as gases are made up of particles that have no defined volume but a defined mass, they undergo no intermolecular forces, are in constant random motion, have elastic collisions, and the kinetic energy is the same for all gases at a given temperature. This model implies the ideal gas law, which we can derive from the equation P= Ftotal/A. Where P stands for the pressure on the wall of the container which is equal to the total force of particle collisions divided by the total area. Equations: pV=Nrt P= Ftotal/A M= mRT/PV n 2 ]×[V−nb]=nRT [P+a( ) V

The neutralization of an acid base reaction is defined as a reaction that involves an acid and a base reacting to form a water and a salt with a combination of H+ and OH- ions to produce water. For example the neutralization of a strong acid and a strong base naturalizes to equal 7, which is a neutral pH. It is determined by density and specific heat, and they are exothermic reactions. The net ionic equation that expresses this is H+(aq) + OH-(aq)= H2O, based on this equation our reactions with our additional HCl and HNO3 we should hypothesize that our ∆H of the reaction will be similar. We were required to enthalpy of neutralization values for both HCl and HNO3 with NaOH, based on the following website, USA: 1-585-535-1023UK: 44-208-133-5697AUS: 61-280-07-5697. (n.d.). Retrieved from http://www.chemistry-assignment.com/standardenthalpy-of-neutralization, I found that the ∆Hreact. of HCl + NaOH and ∆Hreact. of HCl + HNO3 were both -57.1. By finding this I was able to calculate my percent error for my reactions. This also allows us to determine ∆Hreact/moles of H2O. Methods In this experiment, we prepared an unknown dry metal to observe a physical reaction by obtaining its mass and boiled it in a tube that was placed in a beaker of water to heat it up while maintaining the dryness of a sample. We then prepared the water in the calorimeter by obtaining two Styrofoam coffee cups to prepare the apparatus followed by a Styrofoam lid, stirrer, and a 110° thermometer, and then collected the mass of the calorimeter. The reason we use Styrofoam is because it is a good insulator and it insulates the inside from the outside. If we had used a different material or possibly a single cup instead of two, then there is a higher chance that there will not be enough insulation and some of the heat will be lost. So, for this experiment we are to assume that our coffee cup calorimeter is a “perfect” insulator. We then added 20ml of DI water and measured the mass of the calorimeter with the additional water. We only add 20ml of water because water has a high heat capacity and if we have too high of a volume of water then it will take a significantly larger amount of heat to view a visible temperature difference. Once the

metal has boiled for 10minutes and reached equilibrium we then take the temperature of the boiling water in the beaker and the temperature in the calorimeter. It’s important to heat for 10minutes because we want it to reach equilibrium, that means that we heat it that period of time because we want to make sure that the metal is heated to its full potential. We need to make sure that the water is heated and that then the metal is heated. If it was heated less than that then there is a possibility the metal isn’t sufficiently heated. It’s best to take the temperature of the calorimeter water before the water in the beaker so that the heated thermometer doesn’t release some of its thermal energy to the room temperature water in the calorimeter affecting the initial temperature reading. Then we transfer the metal to the calorimeter, swirl gently, and record the temperature in five second intervals for one minute and then thirty second intervals for an additional five minutes. It’s best to swirl the calorimeter and not use the thermometer to swirl the water because if the thermometer touches the metal it can increase the temperature and give a false reading of the heat the water is absorbing. We record at these time intervals because we want to make sure we have enough data points to observe the temperature change easily when viewing our graphs, that way we can see at what point our graph is linear. Then we plotted our data and repeated the experiment a second trial. In the second part of the experiment we observed a chemical reaction by mixing an acid and a base to observe the enthalpy of neutralization. First, we measured the volume and temperature of the HCl. Then we measured the volume and temperature of the NaOH. Then we quickly mixed the acid and base together in the calorimeter and recorded the temperature change every five seconds for one minute and every 30-45 seconds for an additional five minutes. We then repeated the entire experiment except we switched out our acid for HNO3. We also plotted this data on graphs as well to observe the temperature change. The procedures for these experiments were both produced by Beran, J.A. Laboratory Manual for the Principles of General Chemistry. (10th ed.); John Wiley & Sons, Inc. USA, 2015; pp 214-223. Results Table 1 displays the data necessary to find the specific heat of an unknown metal. To calculate the average specific heat of the two trials we use the equation: (specific heat = Q/m •∆T, where Q= the amount of energy or heat, m=mass in grams, and ∆T= the temperature change in °C. Specific heat is in (J/ g°C).) Discussion In this experiment, we were expected to determine the specific heat of a metal and to determine the enthalpy of neutralization for an acid-base reaction. We know that the specific heat is what causes the temperature of a gram of a substance to rise 1°C. We were expected to find the specific heat capacities of our unknown metal in our coffee cup calorimeter apparatus. We were to determine this by finding the amount of heat gained by the water in our cup, the mass of our metal, and the change in temperature. We use a calorimeter, which is an apparatus that allows us to measure the flow of energy through a system and its surroundings, to determine this because we know the specific heat capacity of water. I had hypothesized that there would not be a significant temperature change because water has a high heat capacity. This hypothesis seemed true based on my raw data and my graphs. The average specific heat capacity for my metal was 0.51(J/ g°C), we did not have enough data to determine our unknown metal but we will get this data from our instructor.

As for my chemical reactions my data for my HCl acid was that my average ∆Hn was -60.23 kj/mol, my standard deviation was 5.39, my relative standard error was 8.95%, and my percent error was -5.48%. All of my data was under 10% which means that my data was accurate and precise. Anything below 10% is considered good data. This also shows that the errors I had encountered during my experiment did not have too much of an impact on my results which is good. For my HNO3 my average ∆Hn was -54.46 kj/mol, my standard deviation was 0, my relative standard deviation was 0, and my percent error was 4.62%, these results are also below 10%, which means that these results are accurate as well. Overall, I had good data for my experiment indicating that my errors were not significant and I had good technique throughout my experiment. I then took the average of the two and got my average molar volume. My standard deviation was 2.64, anything above a 1 indicates that there was high variation in my data. This high standard deviation indicates that my technique was not precise. As for my relative standard deviation I had a 13.32%, because this percent is higher than 10%, this is also an indication of my lack of precision in this experiment. Same goes for my present error, my 11.52% error is higher than 10% indicating that my experiment not only lacked precision but accuracy as well. Precision deals with how close two or more data points are to one another, while accuracy deals with how close a measured data set is to a set standard. In terms of the results I had for my experiment I lacked both precision and accuracy. Conclusion In conclusion, my unknown sample was #6. The average molar mass of my sample was 78.24g/mol, my standard deviation of my molar mass was 15.33, and my relative standard deviation was 19.59%. Due to both my standard deviation and my relative standard deviation being high, I was neither precise not accurate when performing this experiment. Errors Errors that occurred throughout my experiment was that I reheated my sample because I thought that the condensed liquid was liquid that had not evaporated yet. This shouldn’t have had any effect on my experiment but it did waste more time. Another error that occurred was that throughout my experiment my scale began to fluctuate a lot, it could have been that my flask had not cooled enough but I tried to make sure that my sample was completely cooled before weighing. Errors that could have occurred could have been that I forgot to get the measurement of my final mass of the flask after my sample was removed. This would have affected all my data and would have prevented me from getting any real results from my experiment. A second error could have been that I did not evaporate my sample completely or that my entire sample was not removed during suctioning, this could have affected my mass readings and caused them to be too high. Experiment: Mass of a Volatile Liquid –Post Lab Questions 1. Part A.1. The mass of the flask (before the sample is placed into the flask) is measured when the outside of the flask is wet. However, in Part B.3, the outside of the flask is dried before its mass is measured. A) Will the mass of vapor in the flask be reported as too high or too low, or will it be unaffected? Explain. B) Will the molar mass of vapor in the flask be reported as too high or too low, or will it be unaffected? Explain. A) If the outside of the flask is wet when the mass is measured then the mass of the vapor in the flask will be reported as too high. This is because the additional water is

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taken into account for the initial mass but will evaporate and is not part of the mass of the vapor. B) If the mass of the flask is recorded with the additional water on the outside of the flask then the molar mass will be recorded as too high. The moles are not affected by a larger mass, but because the mass is higher and it is divided by the number of moles of vapor then this will cause the molar mass to be higher as a result. Part A.1. From the time the mass of the flask is first measured in Part A.1 until the time it is finally measured in Part B.3, it is handled a number of times with oily fingers. Does this lack of proper technique result in the molar mass of the vapor in the flask being reported as too high or too low or as unaffected? Explain. If the flask is constantly held with oily fingers then the oil on the flask will add an additional mass, therefore the molar mass will be recorded at too high. The oil that gets on the flask remains for the final reading, because the mass of the oil is unaccounted for then the larger mass with result in a larger molar mass. Part A. 2. The aluminum foil is pierced several times with large pencil-size holes instead of pin size. A) How will this oversight in the procedure affect the mass of vapor measured in Part B.3, too low, too high, or unaffected? Explain. B) Will the reported molar mass of the liquid be reported too low, too high, or unaffected? Explain. A) If the foil is pierced several times with large pencil holes then the mass of the vapor will be too low. If the holes are too large then the liquid will bubble out when it is heated. This will result in lower mass of vapor because some of it will be lost as liquid through the holes. This is why pin-sized holes are important, smaller holes will prevent the liquid from escaping. B) If there are large holes in the foil then more of the liquid will be lost as vapor, therefore a lower mass of vapor results in a lower molar mass of the liquid. So, by the holes being too large this results in a lower reported molar mass than there should be. Part B.2. The flask is completely filled with vapor only when it is removed from the hot water bath in Part B.3. However, when the flask cools, some of the vapor condenses in the flask. As a result of this observation, will the reported molar mass of the liquid be too high, too low, or unaffected? Explain. If there is only vapor when it is removed from the bath and the vapor then condenses this will have no effect on the molar mass. This is because the mass in the flask is still from the liquid itself. The only change that occurred was that the liquid in the flask underwent a physical change. I made this mistake because I thought that I had not heated the flask efficiently when really it was just the vapor that condensed. No additional liquid is being added, so the molar mass is not affected. Part B.2 Supposed the thermometer is miscalibrated to read 0.3°C higher than actual. Does this error in calibration result in the molar mass of the vapor in the flask being reported as too high, too low, or as unaffected? Explain. If the measurement of the thermometer is higher than it should be then the molar mass of the vapor will be higher than it should be. When we calculate the moles of the vapor we will get a result of a lower molar mass because we are dividing by a small mole value, leading to a larger molar mass than what is actually present.

6. Part C.1. If the volume of the flask is assumed to be 125ml instead of the measured volume, would the calculated molar mass of the unknown liquid be too high, too low, or unaffected by this experimental error? Explain. If we used a different volume from the measured volume we will alter the result of the molar mass. If the volume is 125ml instead of the measured volume or less then this would result in a lower molar mass, this is because a larger volume results in a higher mole value for the vapor. 7. Part C.2. The pressure reading from the barometer is recorded higher than it actually is. How does this affect the reported molar mass of the liquid: too high, too low, or unaffected? Explain. If the barometer is reading higher than it is then this will result in a molar mass being reported as too low. If there is a higher pressure reading then there will be a higher mole value than what there actually is. When dividing the mass by this higher mole value we get a molar mass of the vapor being higher than it should be....