Etiket Arşivleri: FE 111 GENERAL CHEMISTRY LAB SHEET

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FE 111 GENERAL CHEMISTRY LAB SHEET

Experiment: 1 Density of liquids INTRODUCTION The density, mass per unit volume, of liquids can be determined like that of solids by measuring both the mass and the volume of a given sample. The density is a characteristic property of a substance; it remains fixed unless the temperature or pressure is changed. For liquids, a relatively small change in temperature can affect the density appreciably, but a pressure change must be quite great to have a measurable effect. As an intensive property the density is independent of the quantity of material measured. In this experiment, you will practice using a balance to measure mass. In addition, you will learn how to measure volume using a graduated cylinder and then you will determine the density of a) liquid water and b) a salt solution with different concentrations. Once the density of liquid water and its change with temperature are known, you can use the information to find out what volume should be occupied by a known mass of water at a given temperature. Once a data such as concentration vs. density of the salt solution is obtained, it is possible to plot a curve what we called a calibration curve. When such a plot is obtained we can determine the density of a substance with a known concentration, or the concentration of a substance with a known density. PROCEDURE A) – Weigh an empty graduated cylinder – Add about 20 mL of distilled water in this graduated cylinder. – Record the volume to the nearest 0.1 mL. – Weigh the graduated cylinder plus water on an analytical balance and record. – Measure the temperature of the water – Calculate the density. – Compare the calculated density of the water with the values given in the Table 1. and calculate the percent error.

B) – Rinse your graduated cylinder with about 5 mL of the 4 % NaCl solution, and then pour about 20 mL of the 4 % NaCl solution into the graduated cylinder. – Determine the density of this solution as in Part A. – Determine the density of 8, 12, and 16 % NaCl solutions similarly. – Obtain a salt solution of unknown concentration of NaCl from your instructor and determine its density. – Plot a graph showing the density (on the vertical axis) vs. concentration of NaCl (horizontal axis). – Using the graph, obtain a value for the concentration of NaCl for your unknown and report it to your instructor. Table 1. Temp°C Density(g/mL) Temp°C Density (g/mL) 15 0.9979 26 0.9959 17 0.9977 28 0.9955 18 0.9975 29 0.9952 20 0.9972 31 0.9946 22 0.9968 33 0.9941 24 0.9964 35 0.9935

PRELAB QUESTIONS 1. Write the definition of the density and a) explain the effect of the temperature and pressure on the density of the liquids.How does the density of liquids changes with the temperature? b) why do we need a great pressure change to observe a change in the density of the liquids? c) if we have a gas sample instead of liquids, do we need great pressure change to have a measurable change in the density? Explain briefly. d) if we have a solid sample instead of liquid or gas, does the density of this solid change if we increase or decrease pressure too much? 2. a) Suppose that, you measure the temperature of 20 mL of the water as 30°C when it is actually 20°C. Find the minimum percent error in the calculation of the density by using the table given above. b) Suppose that you determine the mass of 41.2 mL of water as 41.052 g. at 28°C. Calculate the density from the experimental values and match it with the value given in the table. Calculate the percent error. 3. Plot the temperature vs. density values given in the table. Determine the density of the liquid water at 19°C, 27°C, 30°C and 32°C. 4. What is the calibration? Explain briefly.

FE 111 GENERAL CHEMISTRY DATA SHEET Density of liquids Experiment : 1 Date: a) Mass of empty graduated cylinder : ……………….. Volume of water in the cylinder : ……………….. Mass of graduated cylinder plus water : ……………….. Temperature of water : ………………… b) NaCl solution 4 % 8 % 12 % 16 % ……….. ……….. ……….. ……….. Mass of grd.cyl.+ known NaCl soln. ……….. ……….. ……….. ………… Volume of NaCl in cylinder Mass of grd. cyl+ unknown soln : ………… Volume of unknown soln : ………… Name of the student: Submitted to :

FE 111 GENERAL CHEMISTRY LAB SHEET Experiment: 2 Heat and Temperature INTRODUCTION Heat differs from temperature in that heat is a quantity of energy whereas temperature is a measure of the hotness or coldness of an object. Indeed, temperature is a physical property that determines the direction of heat flow. Heat always transfers from hot objects to cold ones. For example, if you touch an ice cube, heat flows out of your hand to the ice, cooling your hand. The specific heat capacity is the amount of heat required to raise the temperature of 1g or 1mol of substance by o 1 C. In this experiment you will investigate a) the calibration of a thermometer as a temperature-measuring device b) the determination of the specific heat capacity of a given metal Table 2-1 Boiling point of water at various pressures Pressure mm-Hg atm o Bp, C 700 0.921 97.7 The thermometer can be calibrated by 705 0.928 97.9 o immersing it first in melting ice (0 C) and than in 710 0.934 98.1 o 715 0.941 98.3 boiling water (100 C at 1 atm pressure). Since 720 0.947 98.5 the boiling point of water varies with changing barometric pressure. Table 2-1 will enable you to 725 0.954 98.7 find the t rue boiling point of water under the 730 0.961 98.9 conditions of the experiment. To determine the 735 0.967 99.1 specific heat capacity of a metal, you will heat a 740 0.974 99.3 weighed amount of metal to a known 745 0.980 99.4 temperature (that of boiling water), and then 750 0.987 99.6 place the hot metal in a known amount of water. 755 0.993 99.8 From the temperature rise of the water, you can 760 1.000 100.0 calculate the amount of heat transferred from the 765 1.007 100.2 metal to the water. This can be done directly 770 1.013 100.4 because it takes 4.2 joules (or 1 calorie) to raise o 1 g of water by 1 C. Since the density of water is almost exactly 1 g per mL. 1 g of water is very nearly 1 mL. The specific heat capacity of the metal is equal to the amount of heat liberated by the the metal divided by its temperature drop times its mass.

PROCEDURE a) Wash off some crushed ice, and place it in a small beaker. Add distilled water until the ice is nearly covered. Immerse your thermometer in the ice-water mixture as deep as possible; stir gently; record the thermometer reading when it has become constant. Place about 100 mL of distilled water in your 500 mL flask. Drop in a boiling chip to prevent bumping during boiling. Insert your thermometer into the water. Heat the water to boiling and record the thermometer reading when it has become constant. Record the barometric pressure. b) Obtain a piece of metal (about 10 g) for which the specific heat is to be found. Weigh it to the nearest 0.1 g. Tie it with a string and suspend it above boiling water in a flask as the thermometer is set in Figure. Do not let the metal get wet. (If water condenses on the metal, pull the metal out, dry, and replace.) Allow sufficient time for the metal to reach the temperature of the vapor from boiling water. Record the barometric pressure. With a graduated cylinder, measure 10.0 mL of distilled water into a test tube. Prop the test tube in the mouth of your 300 mL flask. o Measure the temperature of the water to the nearest 0.1 . Withdraw the thermometer, and hold it directly over the test tube so that any drops of water will drain back. Quickly remove the heated metal, and carefully lower it into the water in the test tube. Stir by raising and lowering the metal. Return the thermometer, and note the maximum temperature registered. PRELAB QUESTIONS 1- Define Boiling point, Normal Boiling Point, Normal Freezing point 2- What is the difference between heat capacity and specific heat capacity ?

FE 111 GENERAL CHEMISTRY DATA SHEET Heat and Temperature Experiment : 2 Date: A) Thermometer reading in ice-water mixture………………. Thermometer reading at water boiling point………………. Barometric pressure………………. B) Weight of metal ……………. Temperature of the heated metal………………. Temperature of the water in the test tube………………. Temperature of the water and metal mixture………………. Name of the student: Submitted to :

FE 111 GENERAL CHEMISTRY LAB SHEET Experiment 3 Charles’s Law INTRODUCTION Charles’s law gives the relation between the volume and temperature of a gas at constant pressure. It states that the volume of a given mass of gas is directly proportional to the absolute temperature when pressure is constant. Mathematically we may state this in either of two ways: V = kT (P constant) or V /T = V /T (P constant) 1 1 2 2 In this experiment the student will heat the air inside a flask to the boiling temperature of water. Then he will cool the flask by placing it in a cold water bath. As the air inside the flask cools, its volume will decrease, and water will be drawn into the flask. The volume of this water is the difference between the initial and final air volumes. One correction must be made. The water drawn into the flask has a vapor pressure (about 20 mm). Therefore, the air in the flask at the end of the experiment will be mixed with water vapor. The pressure of air: Pair = Patm — PH 2O Now we can use Boyle’s law to calculate the volume the air would occupy if P = P . Let’s call this volume V . Then air atm 2 Patm V2 = (Patm – P 2 ) ( V1 – V 2 ) H O H O where V 2 is the volume of water drawn into the flask and V is the volume of the H O 1 flask. Now it should be possible to compare V /T (the volume and temperature of the 1 1 air at the boiling temperature of water) with V /T (where T is room temperature). 2 2 2 Since in both cases P = Patm, if Charles’s law is obeyed, we will find that V /T = V /T 1 1 2 2 PROCEDURE Take a clean 250 mL flask and make sure it contains no moisture by passing it back and forth through a (luminous) bunsen burner flame, until it is completely dry. (You have to hold the flask in your hand and heat the area that contains moisture). Take a one-holed rubber stopper that fits the flask and insert a 7-8 cm length of fire- polished glass tubing so that one end is even with the bottom of the stopper. Fit the stopper firmly into the flask and mark with a pencil where it comes in contact with the flask. Place the flask in a 1000 mL beaker and set up apparatus as shown below. The flask should be almost completely immersed in water.

Gently boil the water for about 10 minutes. Record the temperature of the boiling water (T 1) and the atmospheric pressure (Patm). Turn the burner off and cover the open end of the glass tube with your finger. Loosen the clamp and, keeping your finger over the glass tube, turn the flask upside down and immerse it in a room temperature water bath. Remove your finger and keep the flask completely submerged for 10 minutes. At the end of this time, raise the inverted flask until the water levels inside and outside are the same. Cover the end of the glass tube with your finger, remove the flask from the bath, and restore it to an upright position. Record the temperature of the water bath (T ). Pour the water from the flask into a 2 graduated cylinder and record its volume ( V 2 ). H O Fill the flask with water and place the rubber stopper in its original position (using the pencil mark). The glass tube should be completely filled with water, and no air bubbles should be trapped around the bottom of the stopper. Dry the outside of the stopper, then remove the stopper and carefully drain the contents of the glass tube into the flask. Pour the water from the flask into a graduated cylinder and record its volume (V ). 1 From the appendix of your textbook obtain the vapor pressure of water ( PH 2O ) at room temperature T . 2 CALCULATIONS Remember to convert T and T to the absolute temperature scale. Now calculate 1 2 V , as shown in the introductory discussion. Calculate and compare V /T and V /T . 2 1 1 2 2

QUESTIONS 1- Assuming your values of V , T , and T were correct, what is the theoretical value 1 1 2 of V ? (Calculate from Charles’s law). Calculate your percent error. 2 % error = difference between V (theoretically) and V (experimentally) x 100 2 2 V (theoretically) 2 2- How would the experimental results be affected if the flask contained moisture at the beginning of the experiment ? 3- Why was it necessary to match the water levels inside and outside the cooled flask before removing it from the water bath ?

FE 111 GENERAL CHEMISTRY DATA SHEET Charles’s Law Experiment : 3 Date: Temperature of the Boiling water (T ) : ……………… 1 Atmospheric Pressure (Patm ) : ………………. Temperature of Water Bath (T ) : ………………. 2 Volume of Water in the Flask ( V 2 ) : ………………. H O Vapor Pressure of Water at T ( P ) : ………………. 2 2 H O Volume of the Flask (V ) : ………………. 1 Name of the student: Submitted to :

FE 111 GENERAL CHEMISTRY LAB SHEET Experimet: 4 Chemical Equilibrium INTRODUCTION Chemical reactions occur so as to approach a state of chemical equilibrium. The equilibrium state can be characterized by specifying its equilibrium constant, i.e., by indicating the numerical value of the equilibrium constant expression. In this experiment you will determine the value of the equilibrium constant for the reaction 3+ – between ferric ion, Fe and isothiocyanate ion, SCN . 3+ – 2+ Fe + SCN ══ FeSCN for which the equilibrium condition is 2+ [FeSCN ] ———————- = K 3+ – [Fe ][SCN ] To find the value of K, it is necessary to determine the concentration of each of 3+ – 2+ the species Fe , SCN , and FeSCN in the system at equilibrium. This will be done 2+ colorimetrically, taking advantage of the fact that FeSCN is the only highly colored species in the solution. The color intensity of a solution depends on the concentration of the colored species and on the depth of solution viewed. Thus 2 cm of a 0.1 M solution of a colored species appears to have the same color intensity as 1 cm of a 0.2 M solution. Consequently, if the depths of two solutions of unequal concentrations are chosen so that the solutions appear equally colored, then the ratio of concentrations is simply the inverse of the ratio of the two depths ( h M = h M ). It should be noted that this 1 1 2 2 procedure permits only a comparison between concentrations. It does not give an absolute value of either one of the concentrations. To know absolute values it is necessary to compare with a standard of known concentration. 2+ For color determination of FeSCN concentration, you must have a standard 2+ solution in which the concentration of FeSCN is known. Such a solution can be – prepared by starting with a small known concentration of SCN and adding such a 3+ – 2+ large excess of Fe that essentially all the SCN is converted to FeSCN . Then, 2+ the concentration of FeSCN may be calculated from the equation:

PROCEDURE – Obtain five clean test tubes. Rinse with distilled water and let drain. – Give numbers to each test tube. – Add 5 mL of 0.0020 M NaSCN solution to each. – To the first test tube, which will serve as your standard, add 5 mL of 0.20 M Fe(NO ) . 3 3 – For the remaining four tubes proceed as described below. – Add 10 mL of 0.20 M Fe(NO ) to your graduated cylinder. 3 3 – Add 15 mL distilled water to make the volume exactly 25 mL. Stir this solution thoroughly. – Pour 5 mL of the this solution into the second test tube. – Pour out the half of the remaining 20 mL of the solution in graduated cylinder so that the volume will be exactly 10 mL. – Make the volume again 25 mL with distilled water and mix thoroughly. – Pour 5 mL of this solution into the third test tube. – Pour out half of the solution in the graduated cylinder until exactly 10 mL remain. – Add 15 mL of distilled water and mix thoroughly. – Pour 5 mL of this solution into the fourth test tube. – Again pour out half of the solution in the graduated cylinder until exactly 10 mL solution remain. – Add 15 mL of distilled water and mix thoroughly. – Pour 5 mL of this solution into the fifth test tube. 2+ The next step is to determine the relative FeSCN concentration in each test tube. To do this you will compare the color intensity in test tube 1 with that in each of other four test tubes. – Take test tube 1 and test tube 2. – Hold them side by side. – Wrap a piece of white paper around them.

– Look down through the solution towards the white background that your table makes. – If color intensities appear identical record this fact. – If not, pour out some of the standard in the test tube 1 until the colour intensities are the same. – Measure the heights of solutions in the two tubes being compared. – Repeat this comparison for the remaining three test tubes and record the heihts of both the standard solution and solution being compared. RESULTS AND CALCULATIONS 3+ – 1- Calculate the concentration of Fe and SCN in each of the five tubes, assuming 2 3+ – that no FeSCN + had been formed as [ Fe ] and [ SCN ] . ( Use M V = M V ). o o 1 1 2 2 – 2+ 2- Assume that all the SCN in the test tube 1 was converted to FeSCN . Then 2+ 2+ calculate the concentration of FeSCN , [ FeSCN ], in each of the other test tubes from (h ) M = (h ) M 1 1 2 2 3+ – 3- Calculate [ Fe ] and [ SCN ] from 3+ 3+ 2+ – – 2+ [ Fe ] = [ Fe ] – [ FeSCN ] and [ SCN ] = [ SCN ] – [ FeSCN ] o o 4- Determine the K value separately for each tube by using the equilibrium constant expression. PRELAB QUESTIONS 1- Define the following terms. Equilibrium Constant, Le Chatelier’s Principle, Chemical Equilibrium, Homogeneous Equilibria and Heterogeneous Equilibria 2- Explain the effect of concentration, temperature, pressure and catalyst according to Le Chatelier’s principle on chemical equilibrium. 3- Write a brief procedure for the experiment in your own words. QUESTIONS 1- What are possible sources of error in this experiment ? 2- Why are the values of K determined for test tubes 3, 4, and 5 probably more reliable than that determined for tube 2 ?