Calorimetry Lab Report

The purpose of this lab is to figure out the mass percentage of copper in a penny. Furthermore, by doing this lab we will practice using a spectrophotometer and review the names of equipment such as volumetric glassware, pipets, and volumetric flasks.

In order to measure the heats of reactions, add the reactants into the calorimeter and measure the difference between the initial and final temperature. The temperature difference helps us calculate the heat released or absorbed by the reaction. The equation for calorimetry is q=mc(ΔT). ΔT is the temperature change, m is the mass, c is the specific heat capacity of the solution, and q is the heat transfer. Given that the experiment is operated under constant pressure in the lab, the temperature change is due to the enthalpy of the reaction, therefore the heat of the reaction can be calculated.

Washing of the copper is necessary in this experiment to separate the iron from the copper and make sure the iron is not counted in the mass of the copper.

For q, you found how much heat was gained by the water so you know that same amount of heat was lost by the metal. Therefore, qmetal = -qwater. The mass of the sample was recorded from the baggie. The temperature change

Weight 30 dry pre-82 pennies which get 89.77g, using 30ml initial volume measuring the volume of 30 pennies, record the data 10.0ml. Using equation Density= Mass/Volume, get the density of the pre-82 pennies is 8.98g/ml. Then calculate the error%=0.10%, and the deviation%=1.29%.

The first part of this lab’s objective was to find the calorimeter constant using DI water. We accomplished this by first checking the temperature and then adding 20 mL of cold DI water into our calorimeter. Next we collected 20 more mL of DI water in a 50 mL beaker and placed it onto a burner in order to heat it. We removed the beaker once the temperature of the water reached 60º celsius. After we removed the water, we poured it into the calorimeter with the cold water and took the temperature. The temperature ended up being 37º celsius. From this information we were able to calculate for qhot, qcold, qcal and Ccal. To be as accurate as possible we conducted this same test three more times and used the averages from all Ccal calculations as the final Ccal.

The purpose of this experiment was to find, compare, and contrast the mass, volume, and densities of copper and zinc pennies. Information about the pennies was acquired before the start of the lab. The pre-lab research stated that any pennies that were minted before 1982 are composed of pure copper, while any pennies after 1982 are made out of zinc. The densities of copper -- 8.96 grams per centimeters cubed -- and zinc -- 7.13 grams per centimeters cubed -- were also information that was acquired before starting the experiment. With knowing the density of the two metals, this hypothesis was formed: If the mass and volume of pennies made out of zinc and copper are both measured, then the copper pennies will have a greater mass, because copper has a greater density than zinc. This hypothesis was formed because density equals mass over volume and if the mass is greater in an object that has the same volume, then the density of the object will be greater than the other.

The objective of this lab was first to convert the mass of a compound to the number of moles and number of molecules and then determine the concentration of salt and its component. The first thing we did was get the mass of an empty container by using a scale and it came out to be 16.87g. Next thing we did was pick a substance which in this case it was Potassium Chloride and placed it on the scale to get a total mass of 31.20g. The container the Potassium Chloride was in only had a mass of 16.87g which means that the mass of the substance was 14.33g. To convert the mass to the number of moles we took the amount of the substance 14.33g and divided it by the mass of Potassium Chloride 74.55g and figured out that the number of moles was 0.192.

The percent mass of each component of the mixture was 23% iron, 61% sand, and 8.7% salt.

The boiling point elevation constant for water that was experimentally determined in this analysis was 0.4396 °C/m, which was derived from the slope of the trend line in Figure 2. This is slightly lower than the constant provided in lecture of 0.51 °C/m. This could be due to further evaporation of water from the solutions tested via refractive index after the boiling temperature was recorded.

If there is an additional unmeasured amount of water in the Erlenmeyer flask, then this would reduce the concentration of the HCl, and therefore reduce the molarity. The volume of the amount of HCl solution added would increase, yet the concentration of the HCl would remain the same, which would ultimately result in the molarity of the HCl being lower than in reality.

In this experiment, a mixture of unknown #3 was used. That mixture had acid, base, and neutral. We added solvent to the unknown. It is important to know the density of the solvent in order to determine which is the aqueous layer and which is the organic layer. If the solvent that has more density than water, so the organic layer will be the lower layer, while if the solvent has lower density than water, the organic layer will be the upper layer. This will make an error if the determination of the layers was wrong after added the strong acid or the strong base. We added 5% HCl to the mixture in order to separate the base in the aqueous layer and form its salt. Same thing, we add 5% NaOH to the mixture in order to separate the acid and form its salt. In order to recover the base, we add 10% NaOH to the HCl extraction. The result will be salt with a base. Same thing for the acid, in order to recovered it, we added 10% HCl. The reaction will give us salt with an acid. For the neutral, we added sodium sulfate as a drying reagent in order to dry water and separate the neutral part as pure.

1. In the human blood, there is the bicarbonate buffer system. CO2 is released from cellular respiration and then taken up by red blood cells. Next, it is changed to carbonic acid which dissociates to form bicarbonate and H+ ions.

Finally, we measured the mass of the remaining copper in the test tube. This lead to the discovery that the mass of the copper in relation to our original 1.08 grams of copper oxide was 0.96 grams. From this data, we could find that the measurement of oxygen in our amount of copper oxide was 0.12 grams. With these significant weights, we were then able to convert those results to the amount of moles each element had in contribution to our total quantity of copper oxide. Specifically, we found that there was 0.015 moles of copper in our copper oxide, and 0.0075 moles of oxygen. So, that made the calculated ratio between moles of copper and moles of oxygen to be 2.0, or a 2:1 ratio of copper to oxygen. Thus, the formula of the red copper oxide can be expressed as . The lab worked the way it did primarily because of the chemical characteristics of copper oxide. Since the oxygen could separate from the copper when presented with a heat source, it allowed for the amount of each element in our quantity of copper oxide to be measured and then calculated into a ratio, which determined the formula for the

To achieve this, the final value from each thermocouple was set to be equal to the warm water bath temperature (370C), and the initial reading was set equal to the ice water bath temperature. Thus, for each thermocouple an equation was obtained using the two points to convert voltage readings to temperature. An example of the calibration for one of the thermocouples is shown in Appendix II.