Experiment #1. Math Review

Transcription

1 Experiment #1. Math Review A. Scientific notation and Significant Figures While entering a number in scientific notation in your calculator, look for the EE or the exp key on your calculator. For example using the TI3- XA, one would enter x as EE 23. Using a TI- 84 plus, the same number would be entered as nd EE 23. Using a TI- 89, the same number would be entered as EE 23 Practice entering the following numbers in your calculator a) x b) 9.11 x c) 2.3 x 10 3 Rules for Counting Significant Figures: Example # S. F. 1. All nonzero integers are significant Leading zeros are never significant Captive zeros are always significant Trailing zeros in a decimal number are significant Trailing zeros in a number with no decimal are not significant Exponential /Scientific Notation: The number of significant figures in a number written in exponential notation is easily determined as the leading and trailing zeros are removed. Examples: Decimal Exponential/Scientific notation # S. F x x x x x Rules for sig figs in mathematical operations Multiplication and division The final answer must have the same number of sig figs as the number having the least sig figs Addition and subtraction The final answer must have the same number of decimal places as the number with the least decimal places Rules for Rounding: 1. In a calculation carry all of the significant figures through to the final result, then round to the correct number of significant figure based upon the data with the smallest number of significant figures. 2. If the digit to be removed is <5, the preceding digit remains unchanged. (25.44 rounds to 25.4) 3. If the digit to be removed is 5, then the preceding digit is incremented by 1. (25.46 rounds to 25.5)

2 B. Logarithms to the base 10 (log) and Natural Logarithms (ln, to the base e) Logarithms are the opposite of exponentials. For example: Since 10 2 = 100, we can write log 10 2 = 2 or log 100 = 2 Since 10 0 = 1, then log 1=0. Similarly, e 0 = 1, therefore ln 1 = 0 e lnx = x or lne x = x Similarly, log10 x = x or 10 log x = x Look for the log and the 10 x key on your calculator to get the log and the antilog of a number. Similarly, become familiar with the ln and e x key on your calculator. Example: If log x = 2, then x = 10 2 Ex: If ln x = 5, then x = e 5 Some basic rules for logarithms are: log(a*b) = log a + Log b log a = log a log b b log a m = m log a ln (a*b) = ln a + ln b ln a = ln a ln b b ln a m = m ln a C. Graphs A graph is a visual representation of the relationship between two quantities. Usually, the x axis is the independent variable and the y axis is the dependent variable. This means that the quantity plotted on the y axis depends on the value plotted on the x axis. A graph of concentration against time implies that the concentration is on the y axis (dependent variable) and time is on the X axis (independent variable). If you are told to plot a graph of ph versus volume, then ph is plotted on the Y axis and the volume is plotted on the X axis. Chemists mostly like to deal with linear relationships and will choose y and x units such that the plot is a straight line. Consider the following graph of volume of a gas versus number of moles: From the graph, we can see that when the Volume of gas (cm 3 ) Moles of gas (n) y = 20x + 2E- 14 R² = 1 moles of the gas changes from 0.5 to 1.0, the volume of the gas changed from 10 cm 3 to 20 cm 3. The Slope of this line can be calculated Δy as: Slope = = Δx change in y change in x cm Slope = = = mol The equation of a straight line is of the form y = m x + b, where m is the slope of the line and b is called the y intercept. Consider the linear form of the Clausius- Clapeyron equation ΔHvap 1 ln PVapor = + ln β R T y = m x + b A graph of ln (P vapor ) against 1 ΔHvap, will give a straight line with a slope = T R and an intercept = ln β

3 Name CHM112 Lab Math Review Grading Rubric Criteria Points possible Points earned A. Simple Algebra (0.5 points each question) 4 B. Scientific Notation and Significant Figures (0.5 points each question) C1. Evaluating log and ln and their reciprocals (0.25 points each question) 2 2 C2. Rules of Logarithms (0.25 points each question) 1.5 C3. Solve for the unknown y in each case (0.5 points each question) 4 D1 (0.5 points each question) 1.5 D2 (0.5 points each question) 1 D3 (0.5 points each question) 1.5 D4 (0.5 points each question) 1.5 D5 (0.5 points each question) 1 Total 20 Subject to other additional penalties as per the instructor

4

5 CHM 112: Math Review A. Simple Algebra Name Solve for the unknown (y) in each of the following cases. Show your work. 2y 6 = 12 6y = = 30 5y y = y = = y y = = y y B. Scientific notation and Significant Figures Carry out the following calculations and report your answer in scientific notation with the correct sig figs a) 34 8 ( )( ) = c) ( )/ (6.022 x ) = b) 6.08 x x x 10-2 = d) ( ) / =

6 C. Logarithms C. 1. Evaluating log and ln and their reciprocals. Calculate the following quantities. 1. ln (e) = 2. log(1000) = 3. ln (e) 4 = 4. e (18200/(8.314*290)) = 5. ln (1) = 6. - log (0.050) = = 8. log (13) C. 2. Rules of Logarithms Complete the following calculations. (You can either use the rules of logarithms discussed earlier or evaluate these directly using your calculator. Try it both ways to see if you get the same answer) 1. log(20) log(2) = ln 2.5 = 2. ln(80) + ln(20) = 5. 1 log 10 = log = 6. 1 ln 40 = C3. Solve for the unknown y in each case (You can either use the Rules of logarithms discussed earlier or evaluate these directly using your calculator. Try it both ways to see if you get the same answer) 1. y log(100) = ln(5 y ) = log(36) + ln(y) = log(12 y ) = log y = log 1.25 y = y 4. ln = ln(125 y) = 4.248

7 D. Graphs D1. Complete the following table by comparing each equation to the equation of a straight line. Equation Slope Is the slope positive or negative? Intercept y = 310. x y = x y= x D2. Consider the equation y = x a. What will be the value of y when x = 4.5? b. What is the value of x when y = D3. Complete the following table by comparing each of the equations to y = m x + b Equation [A] t = [A] 0 k t What is plotted on the y axis? What is plotted on the x axis? Slope Intercept E a 1 ln k = + ln A R T 1 1 = + kt [ A] [ A] t 0

8 ΔH vap 1 D4. The linear form of the Claussius Clapeyron equation is ln P vapor = + ln β R T a) What is plotted on the X axis? b) What is plotted on the Y axis? c) What will be the slope the graph? D5. A student plotted a graph of ln (VP in torr) versus 1/T in Kelvin for dichloromethane and obtained the best fitting line as y = x a. What is the vapor pressure of dichloromethane at 75.0 C? b. What is the normal boiling point of dichloromethane? Remember that at the normal boiling point, the vapor pressure = 1 atmosphere.

9 Experiment #2. Lewis Structures A Lewis structure shows how the valence electrons are arranged and indicates the bonding between atoms in a molecule. We represent the elements by their symbols. The shared electron pair is shown as a line/bond between the two atoms. All the other valence electrons are shown as dots or lines around the symbol of the element. For Example: The Lewis structure for Cl 2 is Let us now see how to draw the Lewis structure for CO 2 Steps for drawing Lewis Structures Example with CO 2 1. Sum the valence electrons of all the atoms. For anions, add one electron for each negative charge. For cations, subtract one electron for each positive charge. 2. Choose the least electronegative element (other than H) as the central atom. For a molecule of type XY n, X is always the central atom. The total number of valence electrons is 16 (4 from carbon and 6 from each oxygen). Choose carbon as the central atom since it is less electronegative and also follows the XY n rule. 3. Use single bonds (lines) to connect the central atom to the surrounding atoms. 4. Subtract 2 electrons for each bond from the original total number of electrons. 5. Complete the octet for all outer atoms (other than H). Each bond to an atom will count as 2 electrons for that atom's octet. Attach the two oxygens to carbon by single bonds. O C O We subtract 4 electrons (2 for each bond) from 16 leaving 12 electrons to distribute to the remaining atoms. Complete the octet for each of the oxygen atoms which uses the last 12 electrons. (We need to add 6 electrons to each oxygen.) 6. Complete the octet for the central atom. There are no more electrons, so we cannot add more to the central atom to complete its octet. 7. If you run out of electrons before the octet for the central atom can be completed, start forming multiple bonds until the central atom has a complete octet. When you form a multiple bond, remember to remove an electron pair from the outer atom. Since carbon does not have an octet, form multiple bonds by taking the lone pairs. becomes The central atom is still electron deficient, so share another pair. becomes 8. Make sure the correct number of electrons has been used and that every atom s octet is complete. The exceptions are atoms of group 2 and 3 which can have incomplete octets and atoms in period 3 or greater which can have expanded octets. 4 bonds + 2 electron pairs equal 16 electrons. All atoms have an octet.

10 Molecular Geometry the Valence Shell Electron Pair Repulsion (VSEPR) Theory The electron groups around the central atom repel each other and therefore prefer to be as far apart from each other as possible. This is the main idea of the VSPER theory. We can apply the VSEPR theory to predict the molecular shape/geometry of a molecule. 1. Draw the Lewis structure for the molecule in question. 2. Count the total number of electron groups on the central atom. Add the number of atoms bonded to the central atom and the number of lone pairs on the central atom this is the total number of electron groups. Note that multiple bonds to one outer atom still count as one electron group. 3. The arrangement of the electron groups is determined by minimizing the repulsions between them. 4. Remember that lone pairs require more space than bonding pairs. Therefore, choose an arrangement that gives lone pairs as much room as possible. Electron Groups 2 0 Lone Pairs 1 Lone Pair 2 Lone Pairs 3 Lone Pairs 4 Lone Pairs linear (180 o ) 3 trigonal planar (120 o ) bent tetrahedral (109 o ) trigonal pyramid bent 4 trigonal bipyramid (120 o, 90 o ) seesaw T- shaped linear 5 octahedral(90 o ) square pyramid square planar T- shaped linear 6

11 Polarity of Molecules A covalent bond is polar if there is a difference in electronegativity between the bonded atoms. A molecule like HCl has a polar covalent bond since there is a difference in electronegativity between the two atoms (the difference is greater than 0.4 and less than 1.8). Thus HCl possesses a permanent dipole moment because the molecule has a distinct negative end and a distinct positive end. However, just because there is a polar bond present in a molecule does not necessarily mean that the molecule is polar. If all the dipoles in the molecule cancel each other out, then the molecule will be non- polar. For example, CO 2 has two polar bonds, but they point in opposite directions and cancel each other out. For simple molecules, polarity can usually be determined by looking for lone pairs on a central atom or non- identical outer atoms. Both of these things cause uneven electron distribution and generally lead to polar molecules. Polar molecules bond dipoles do not cancel. Have a lone pair on the central atom AND/OR different outer atoms. Non- polar molecules bond dipoles cancel. Have no lone pairs on the central atom AND have all outer atoms the same. Condensed Structural Formulas Organic (carbon- based) molecules can contain chains of several central atoms. Condensed formulas are often used to indicate how the atoms are connected. To draw a Lewis structure from a condensed structure, connect any elements that are not hydrogens or halogens in a chain in the order written. Connect the hydrogens and halogens with single bonds to the atom they are written next to. Count valence electrons and add lone pairs and double/triple bonds to complete octets as needed. Molecular Formula Condensed Formula Lewis structure C 3 H 8 O CH 3 CH 2 OCH 3 H H C H H C H H O C H H C 3 H 8 O CH 3 CH 2 CH 2 OH H H H C C H H H C H O H F H F C C C N C 3 H 2 NF 3 CF 3 CH 2 CN F H

12

13 Name CHM112 Lab Lewis Structures Grading Rubric Criteria Points possible Points earned Question 1 (1.33 points each question) 8 Question 2 (1 point each question) 4 Question 3 (1 point each question) 2 Question 4 (1 point each question) 4 Question 5 (1 points each question) 2 Total 20 Subject to additional penalties at the discretion of the instructor.

14

15 CHM 112: Lewis Structures Lab Name 1. Fill in the table below. CH 2 Cl 2 Lewis Structure (redraw to reflect geometry, show net dipole direction if polar) Polar? Y/N Geometry CBr 4 CBr 2 O NBr 3 CH 3 CH 2 OH at C? at O? C 5 H 10 O (connect C s & O in a 6- membered ring. H s are attached to C s ) at C? at O?

16 2. Draw the Lewis Structures of each of the following acids: (These structures, like most acids, have each H attached to one of the oxygens. (A) H 2 SO 4 (B) HIO 2 (C) H 3 PO 3 (D) CF 3 CO 2 H (Both O atoms are attached to second C) 3. Draw the Lewis Structure for these anions. Each anion is a conjugate base of an acid. 2 (E) HCO 3 (H attached to one of the O s) (F) SO 3 4. Weak bases frequently contain N: (nitrogen with a lone pair of electrons). Draw the Lewis Structure for each of the following weak bases. (A) NH 3 (B) CH 3 CONH 2 (Connectivity is C C N ) O (C) CH 3 NH 2 (D) C 5 H 5 N (six membered ring made of C and N, Hs attached to Cs) 5. Draw the Lewis structures of the following cations. Each cation is the conjugate acid of a weak base: + + (A) NH 4 (B) CH 3 NH 2 CH 3

17 Experiment #3. Graphing with Excel Study the "Graphing with Excel" instructions that have been provided. Additional help with learning to use Excel can be found on several web sites, including menu.html. The "help" function of Excel also can be useful to answer specific questions. Note that Excel can eliminate the need to make manual calculations for example, in this exercise, instead of using a calculator to find the natural log (ln) values, simply type the following in the cell where the ln data is to be put: =ln(x1), where x is the column and 1 is the row of the cell containing the data you want to take the natural log of. Similarly, to obtain the reciprocal of the temperature data, simply type the following in the cell where you want the reciprocal: =1/x1, where x is the column and 1 is the row of the cell containing the temperature data you want to take the reciprocal of. You can use the cut and paste functions to copy formulas such as this into the other cells in the column so you don't have to retype the formula each time Excel will automatically enter the corresponding VP or T data from the same row. If you take some time to practice, Excel can become a very powerful tool for you. Relationship of Vapor Pressure to Temperature: Developing the Clausius- Clapeyron Equation 1. Type in the temperature and vapor pressure data for Toluene (columns 1 and 2) into an Excel spreadsheet as seen below. Label the top cell in each column and include that cell when you plot the data. T ( o C) VP (Torr) T in K 1/T in K VP (Torr) ln (Vapor Pressure) (A) Make column 3 by converting the temperature in column 1 into Kelvin. (B) Make column 4 by taking the inverse of the temperature in Kelvin (1/T). (C) Copy column 2 into column 5. (D) Make column 6 by taking the natural logarithm (ln) of the VP in Torr. (column 5) 3. You will be making a series of XY scatter plots. Temperature in some form will be the x axis and Vapor Pressure in some form will be the y axis. The goal is to find a mathematical relationship between the x and y data. In chemistry we like straight lines. If the plot yields a straight line then the equation: y = mx + b can be used. To generate a graph in excel, select the columns you want to plot by dragging the cursor while holding the control button (or command button on Macs). Excel will automatically plot the left- hand column on the x- axis and the right hand column on the y- axis. In this exercise, be sure that your selected temperature data is on the left side of your VP data, otherwise your graph will be inverted. This is why we repasted the VP data after the temperature data in our table.

18 Once the appropriate data are highlighted, choose the type of graph you want to generate in Excel. In this class we will always graph using the Marked Scatter Plot option in Excel. Do not choose plot options that have lines. When we want lines on our Excel graphs we will add them using the trendline feature. You are not finished! The graph automatically generated by Excel is never correctly formatted for submission in a science course. Below are features of a good graph. Be sure any graph you turn in for this class adheres to these guidelines. Each version of excel has different ways to change graph formatting. If your version is different from the one in the lab video on the CHM112 website, you may need to use your help button to learn how to make these changes. o o o o o o o o Give the graph a descriptive title. Don t just repeat the information in the axes titles. Pressure vs Temperature for what? Look in the instructions above to find what substance this data is for. Your title should describe what the experiment was about. Label axes with the type of unit and the unit itself. For example: Temperature ( o C). Axes should be at the bottom and left of the graph not in the middle of the data. Remove the legend if it provides no information. Most Excel versions default to having legends for every graph. If you are only plotting one set of data per graph, the legend contributes nothing get rid of it. Change the start and end points of each axis so that the data takes up the majority of the graph. Excel usually defaults to having both axes start at zero. This often leads to the graph having a lot of white space, which we do not want. Change the minimum value of each axis so that it starts just below your lowest data point. Add a trendline when appropriate. If you want to find a mathematical relationship in the data, be sure to include a trendline and to check the box to display the equation on the chart. You can also choose to display the R 2 value. The closer the value of R 2 is to 1, the better the fit. You will add trendlines to all three graphs in the exercise. Add more gridlines when appropriate. In this exercise we are only concerned with the trendline, but in some labs you will need to estimate a value off the graph after printing. If this is the case, be sure to add enough gridlines so that you can accurately estimate your value. If you need to read a value to the nearest 0.1 ml, but your x- axis only has gridlines every 5 ml, your estimate will be poor. Print your graphs at a reasonable size. Usually about ½ page is appropriate. Also print the spreadsheet when asked. With the data in your tables, generate and print the following graphs as well as your spreadsheet data and attach them to this lab. (A) Plot VP vs T in K. Add trendlines (both linear and exponential) and display the equations for both trendlines on the chart. (B) Plot VP vs 1/T in K. Add trendlines (both linear and exponential) and display the equations on the chart. (C) Plot ln(vp) vs 1/T in K. Add a linear trendline and display the equation on the chart.

19 Name CHM112 Lab Graphing with Excel Grading Rubric Criteria Points possible Points earned Graphs correctly plotted and adhere to all guidelines (including descriptive title, properly formatted axes, trendline equations displayed) Graph 1 2 Graph 2 2 Graph 3 2 Question 1 (1 point each part) 1 Question 2 (1 point each part) 6 Question 3 1 Question 4 1 Question 5 (1 point each part) 5 Total 20 Subject to additional penalties at the discretion of the instructor.

20

21 Graphing with Excel: Post Lab and Analysis Name: 1. Consider Graphs (A) and (B) 2. a) Which trendline was a better fit, the linear or the exponential? b) What was the equation for the best fitting trendline for each graph? Equation for Graph A: Equation for Graph B: 3. Consider Graph (C) a) Did you obtain a linear relationship for the data plotted? b) What was the equation of the best fitting trendline? c) What is the form of the Clausius- Clapeyron equation that we use for graphing? (the form that can be written as y = mx + b). d) Based the equation for the best fit line (the equation for the trendline in C), calculate the enthalpy of vaporization (ΔH vap ) for Toluene in kj/mol. (R = J/mol K.) Show your calculations and all units. e) Is the sign of the ΔH vap that you calculated positive or negative? Does this fit with what you know about heat of vaporization? Explain your answer (your explanation must be chemistry based and not mathematical). f) Look up ΔH vap for Toluene in a reliable source (like the CRC handbook of Chemistry and Physics or the NIST Chemistry webbook ( ) and NOT Wikepedia). (Include proper units) ΔH vap = Source from where you obtained the data

22 4. Using the equation for the best fit line from Graph C, calculate the vapor pressure of Toluene at 45.0 o C. Show your work. 5. Using the equation for the best fit line from Graph C, calculate the normal boiling point of Toluene in C. Recall that normal boiling point means that the pressure is 760 torr. Show your work in detail. 6. Ethanol and dimethyl ether have the same molecular formula (C 2 H 6 O) and molar mass (approx 46 g/mol). a) Look up ΔH vap for both these compounds in a reliable source (like the CRC handbook of Chemistry and Physics or the NIST Chemistry webbook ( and NOT Wikepedia). List the values below with the correct units. ΔH vap for ethanol = ΔH vap for dimethyl ether = Source from where you obtained the data b) Which of the two compounds has the larger ΔH vap? c) Draw the Lewis structure for both compounds. d) list the IMFs present in each e) Use your answers in c) & d) to explain why one of the compounds has a larger ΔH vap.

23 Experiment #4. Molar Mass by Freezing Point Depression Introduction When a nonvolatile solute is dissolved in a solvent, the freezing point of the solution is lowered. This process is called Freezing Point Depression and is a colligative property of solutions. Colligative properties depend only on the amount and not the nature of the solute dissolved in the solvent. Sprinkling salt on icy roads to melt the ice is an example of Freezing Point Depression. One way to undestand why the freezing point of a mixture is lower than that of the pure solvent is to consider the equilibria of freezing and melting. In a pure solvent at its freezing point, there is some liquid and some solid, and they are both freezing and melting at the same time, and at the same speed. If it s warmed, melting speeds up, and the solid melts. If it s cooled, freezing speeds up, and the liquid freezes. However, when another substance is dissolved in the liquid, it is (in most cases) largely excluded from the freezing solid because it doesn t fit into the solvent s solid crystal structure. Now, when molecules approach the solid to freeze, some of them aren t solvent anymore, and these solute molecules mostly bounce off without freezing. This slows the freezing process, but has little or no effect on the melting speed, and we must cool the mixture more than usual to freeze the solvent. Thus, the freezing point is lower. Each solvent responds differently to the addition of a solute. This response is given by K f, the molal freezing point depression constant. It is important that molality is used as the unit for concentration in temperature studies, as mass (used in molality calculations) is temperature independent while volume (used in molarity calculations) is not. The equation for freezing point depression is: ΔT f = imk f where: ΔT f = the absolute decrease in temperature. This is always a positive value. i = the van t Hoff factor i = moles of solute particles in solution moles of solute dissolved. (For a nonpolar organic solute such as our unkowns used in this experiment, i is always 1, because nonpolar organic substances do not dissociate into multiple particles.) m = molality of the solution = (moles of solute)/(mass of solvent in kg) K f = molal freezing point depression constant (this constant is solvent dependent) K f (naphthalene) = 6.80 C/molal The aim of this experiment is to determine the molar mass and identity of an unknown compound by measuring the depression in freezing point. In this experiment, we will first measure the freezing point of pure naphthalene, then measure the freezing point again after adding an unknown solute. From the decrease in freezing point of the solution versus pure naphthalene, the molar mass of the unknown will be determined. Equipment: analytical balance ( measure to g) wide (fat) 6" test tube 600 ml beaker wide mouth gas bottle 2 ring stands test tube clamp hot plate Venier with thermometer probe Chemicals: Naphthalene Possible Unknowns: p- dichlorobenzene, benzophenone, benzil

24 Procedure Determining the Freezing Point of Naphthalene 1. Using a plastic weighing boat, weigh between 4.5 and 5.5 g of naphthalene to the nearest g. Record this mass on the data sheet and place the naphthalene into a dry fat 6" test tube. 2. Place a hot plate on the base of the stand. Place a 600 ml (or a similar size) beaker on the hot plate. Add ~ 300 ml of water to the beaker and suspend the test tube in the beaker using a test tube clamp. Check that the water level in the beaker is above the level of the mixture in the test tube. 3. Now place the wire stirrer and a thermometer in the test tube, so that the loop of the stirrer surrounds the thermometer. Heat the water until the naphthalene melts (heat to at least 95 C). Do not stir until the naphthalene melts. 4. Measure the temperature of the melted naphthalene. This is the temperature at time zero. Start the stopwatch. Quickly remove the test tube from the hot water. Place the test tube into a wide mouth bottle. (Be sure to keep the water hot, since you will need it again.) 5. Gently stir the naphthalene with the wire stirrer and record the temperature (to the nearest 0.2 C) of the liquid every 30 seconds. Stir gently as you take your readings (be sure not to poke the bottom of the test tube, as it can break easily). Continue to take readings until about half the liquid has frozen/solidified. The naphthalene inside the test tube will be like slush. Determining the Freezing Point of a Mixture 6. Using a plastic boat, weigh between 0.5 and 0.6 g of the unknown compound to the nearest g. Your instructor may assign you the unknown. Follow your professor s instructions. Record the mass and number or letter of the unknown on the data sheet and place the unknown in the test tube containing your original sample of naphthalene. Do not stir at this time. 7. Using the same sample, now with the unknown in it: melt and solidify the mixture, recording a second set of data with the same procedure as you used for the pure naphthalene. 8. When you have completed the experiment, cafefully follow the Cleanup procedure as outlined below: A. Warm the test tube in the hot water bath until all of the mixture has melted and all you have is a liquid in the test tube. B. Remove the thermometer and stirrer and wipe them clean using a paper towel wetted with acetone. DO NOT THROW THIS PAPER TOWEL IN THE TRASH. Place the paper towel in plexiglass contaminated paper bin in the waste hood. C. Pour the molten naphthalene mixture into the specified waste container. (lift the red lid before pouring) DO NOT POUR THE LIQUID INTO THE SINK!!! D. Rinse the test tube with small amounts (2 ml) of acetone, pouring this waste solution into the appropriate waste container until the test tube is clean. Have we mentioned that you will not pour anything from this lab down the sink?!?

25 9. Plot a graph of temperature versus time for each data set. Guidelines for graphing: Use a scatter plot without lines in excel. Do not add a trend line. You will draw two lines by hand after printing. Give the graph a descriptive title. For example Pressure versus Temperature of Gas in a Weather Balloon, not just Pressure vs Temperature. Label the axes with the type of unit (ex: pressure, temperature, time) and the unit itself (ex: atm, o C, s) Number the axes so that the data takes up most of the graph. For most axes, it is not appropriate to start at zero. Insert gridlines at least every 1 o C otherwise it will be difficult to interpret the temperature at the line intersection of your hand drawn lines. Print your graphs so that each one takes up at least a half a page. You will not be able to accurately interpret the freezing point on tiny graphs. These plots are cooling curves they show how temperature falls over time as the sample loses energy to the surroundings. During phase changes, the temperature will plateau. The phase change from liquid to solid is exothermic. As this phase change occurs, all energy loss goes to this transformation, not to changing the temperature. In an idealized sample, the temperature will not drop again until all the liquid has been converted to solid. Example 1 shows what this would look like, and this is also how cooling curves in your textbook will look. The freezing point of the sample is the temperature of the plateau. However, in lab, some naphthalene evaporates as the experiment procedes, so we do got get idealized graphs like Example 1, we get graphs that look like Example 2. The temperature stays relatively steady during the phase change, but it drops off at a slow rate. Determining the freezing point of a sample. Determine which data points represent pure liquid cooling (the steeper part of the curve). Using a straight edge, draw an estimated best fit line for those data points. The rest of the data points (with the shallow slope) represent the liquid to solid phase change. Using a straight edge, draw an estimated best fit line for these data points. The intersection of the two lines is your freezing point. Estimate its value to the nearest 0.2 o C

26

27 Name CHM112 Lab Molar Mass by Freezing Point Depression Grading Rubric Criteria Points possible Points earned Lab Performance Printed lab handout and rubric was brought to lab 3 Safety and proper waste disposal procedures observed 2 Followed procedure correctly without depending too much on instructor or lab partner 3 Work space and glassware was cleaned up 1 Lab Report Calculations for molar mass shown clearly with units. 2 Graph 1 & 2 (titled, axis labeled, gridlines every 1 o, data takes up majority of graph, two lines hand drawn on each, freezing point clearly labeled.) 4 Question 1 1 Question 2 3 Question 3 1 Total 20 Subject to additional penalties as per the instructor.

28

29 Molar Mass by FPD: Data Sheet Name Unknown letter Mass of unknown compound Mass of naphthalene Time Temp: Napthalene Run Temp: Mixture Run Graph your results. Attach both graphs to the lab report. Follow the graphing guidelines in the instructions. Freezing point of naphthalene (from graph 1) Freezing point of mixture (from graph 2) Change in Freezing point DT: Calculation for Molar mass (Show your work in detail with units

30 1. Look up the formulas and molar masses of all the possible unknowns (p- dichloro benzene, Benzophenone, Benzil) from a reliable source (CRC handbook or the NIST web book) and list them below. Based on your results, what is your unknown? Explain clearly the reason for your choice. 2. Find the structures of naphthalene and your unknown (based on your answer in 1). a) Draw the Lewis structures for each. Include all bonds and lone pairs. b) State which intermolecular forces are present in naphthalene and your unknown. c) Explain why your unknown dissolves in naphthalene based on your answer to b). 3. Compare the molar mass that you obtained from your experiment to the correct molar mass for your unknown. Calculate the percent error. percent error = actual value- experimental value actual value 100%

31 Experiment #5. Iodine Clock Reaction Part 1 Introduction In this experiment you will determine the Rate Law for the following oxidation- reduction reaction: 2 H + (aq) + 2 I (aq) + H 2 O 2 (aq) I 2 (aq) + 2 H 2 O (l) (1) The rate or speed of the reaction is dependent on the concentrations of iodide ion (I - ) and hydrogen peroxide, H 2 O 2. (The spectator ions are left off the reaction.) Therefore, we can write the Rate Law (concentration dependence) for the reaction as: Rate = k [I ] x [H 2 O 2 ] y (2) Where: x is the order of the reaction in I -, y is the order of the reaction in H 2 O 2, and k is the rate constant. The temperature dependence of the rate is seen in k that is, there is a separate value of k for each temperature at which the reaction takes place. The temperature must therefore be held constant to accurately calculate x, y and k. Since the Rate Law is empirical, we have to go to the lab to make measurements that will enable these values to be calculated. The rate will be measured for the reaction near time = 0, so that few products been formed and there will be no reverse reaction. The concentrations of iodide and hydrogen peroxide will be varied and the rates compared to find each order (i.e., the exponents x and y). This is the Method of Initial Rates and it will be used to find x, y and k. As with a lot of kinetics, the concentration of reactants or products at any instant is difficult to measure directly, so in this lab the rate will be determined indirectly. We have a very handy test for the presence of one of the products, iodine (I 2 ), namely starch. Starch reacts with iodine to form a blue/black colored complex. Unfortunately as soon as any iodine is produced it will react to make the complex and the solution will turn blue/black instantaneously. Thus, using starch as an indicator by itself would not be of much help. It confirms that some amount of I 2 is being formed, but it tells us nothing about what we are trying to measure - the rate (how long it takes to produce a given quantity of I 2.) To get around this problem we will introduce a side reaction that will remove the initial I 2 that is produced by our main reaction. This will prevent the solution from turning black long enough so that we can make some time measurements. We will use the following side reaction: I 2 (aq) + 2 S 2 O 3 2 (aq) 2 I (aq) + S 4 O 6 2 (aq) (3) S 2 O 3 2, thiosulfate ion, reacts with I 2 which prevents the solution from turning blue/black. How will this help? Since we have carefully measured the amount of thiosulfate (a small amount that will run out fairly quickly), we know exactly how much iodine it will take to react with this thiosulfate. Once the small amount thiosulfate has completely reacted, I 2 will start to build up in the solution. As soon as the thiosulfate runs out, I 2 will react with the starch and the solution will turn blue/black. By putting in this "time delay", we can now calculate the rate at which I 2 is being formed. The rate of reaction is equal to the change in concentration divided by the change in time. The change in time will be the time it takes for the solution to turn dark. We will calculate the change in concentration in I 2 based on the amount of thiosulfate added. Using the known volume and molarity, we can calcuate moles of thiosulfate (S 2 O 3 2 ). Based on stoichiometry, we can calculate the moles of I 2 : according to equation (3), 2 moles of thiosulfate react with every 1 mole of I 2, this gives us the change in moles, however for the rate formula we need change in concentration. Divide the moles of I 2 reacted by the total volume to find the change in molarity. Rate = [ I 2 ] / t (4)

32 Equipment Three 125 or 250 ml Erlenmeyer flasks Three 100 or 150 ml beakers 10 ml and 5 ml Pipettes Thermometer Stop watch or other time keeper One bin of chemicals per group that will contain: M KI M NaCl M Na 2 S 2 O M H 2 O M H 2 SO 4 1% starch solution Disposal: All mixtures Spill/Disposal B1 (down the sink) Procedure: 1. Before coming to lab, complete the initial concentration table below showing all work. Initial concentrations are not the concentration of the solutions you start with (i.e. not 0.050M for I - and H 2 O 2 ). Initial concentrations are the concentration of species in the final total volume of contents of the Erlenmeyer plus beaker for each run (see tables in steps 4 & 6). Determine initial concentrations using the formula M 1 V 1 = M 2 V 2. M 1 and V 1 are the molarity and volume of the solutions added, V2 is the total volume of the Erlenmeyer plus beaker. Copy these initial values onto the table in the data sheet. Run [ I ] (initial) [H 2 O 2 ] (initial) 1 (Flask 1 + beaker 1) (15.0 ml)(0.050m) = (42.5 ml)(m 2 ) Solving for M 2, we get [ I ] (initial)= M 2 (Flask 2 + beaker 2) 3 (Flask 3 + beaker 3) 2. Clean and mostly dry three Erlenmeyer flasks. Label them 1, 2 and Obtain a bin of chemicals for your group. Use only these chemicals for all of your runs. Use fresh pipettes for each solution. Rinse pipette twice with the solution that you will be measuring and keep this prepared pipette with the corresponding solution. 4. Add the amounts of the solutions below to prepare each flask. The chemicals must be added in the order listed (top to bottom). Flask #1 Flask #2 Flask # M KI 15.0 ml 15.0 ml 7.5 ml 1% Starch 5.0 ml 5.0 ml 5.0 ml M Na 2 S 2 O ml 2.5 ml 2.5 ml 1 M H 2 SO ml 5.0 ml 5.0 ml M NaCl ml Note: The NaCl solution is added so that the ionic strength of and volume of each solution is the same.

33 5. Rinse and mostly dry 3 beakers and label as 1, 2 and Prepare the following solutions in clean beakers: Beaker #1 Beaker #2 Beaker # M H 2 O ml 7.5 ml 15.0 ml Deionized H 2 O ml 0 Note: deionized water are added so that the total volume of each run is the same. (42.5 ml) 7. Get your timer ready. Add the contents of beaker 1 to flask Start the stopwatch as soon as you mix the solutions. Swirl the flask to mix and note the time it takes for the color to change. This is Run 1. Record the temperature of the mixture. 9. Add the contents of beaker 2 to flask 2. Repeat procedure in step 8. This is Run Add the contents of beaker 3 to flask 3. Repeat procedure in step 8. This is Run 3. The shade of color you observe should be the same in each of the three runs. If it is not, consult your instructor. Disposal: All contents of the reaction flasks may be disposed of into the sink. Calculations: Review the method of initial rates in your text. You will need the values for the initial concentrations of each reactant as well as the rate. Hypothetical data for reaction: A + 2B C + 2D Run [A] 0 [B] 0 Initial [C]/ t (Molarity/sec) M M M M M M 32.0 Rate = k[a] x [B] y. We need to calculate x, y and k To get x, we have to hold [B] constant and just see how [A] affects the rate. Take the ratio of Run 3/Run 1 Run 3 = Initial [C]/ t = k[a] x [B] y = 32.0 M/sec = k(0.300m) x (0.150M) y Run 1 Initial [C]/ t k[a] x [B] y 8.0 M/sec k(0.150m) x (0.150M) y Simplifying: 4.0 = (0.150M) x / (0.150M) x or 4.0 = (0.300/0.150) x or 4.0 = 2 x Or x = 2 Do the same for B (holding [A] constant) and discover that y = 1. You can calculate k as well. Using run 1: 8.00 M/sec = k (0.150M) 1 (0.150M) 1 and solving for k, we get k = 356 M - 1 s - 1. k can be calculated using any of the three runs from Part I For your data: Find x and y as explained above. Calculate k for each run and average the three values at the end. If your values for x and y do not turn out to be whole numbers, round to the nearest whole number. Write the Rate Law using your values for x, y and k: Rate = k [I ] x [H 2 O 2 ] y

34

35 Name CHM112 Lab Iodine Clock Reaction Part 1 Grading Rubric Criteria Points possible Points earned Lab Performance Printed lab handout and rubric was brought to lab 3 Initial concentrations completed before coming to lab. 2 Safety and proper waste disposal procedures observed 2 Followed procedure correctly without depending too much on instructor or lab partner 3 Work space and glassware was cleaned up 1 Lab Report Calculations for change in molarity (work shown) 1 Rate law calculations 2 Rate constant calculations 2 Question Question 2 1 Question Question 4 1 Total 20 Subject to additional penalties at the discretion of the instructor.

36 Iodine Clock Reaction Part 1: Data Sheet Name Show your calculations for moles of I 2 reacted: (find through moles of S 2 O 3 2- ) Show your calculation for molarity of I 2 reacted [ I 2 ]: (divide moles reacted by total reaction volume) Run Temp( o C) Initial [ I - ] M Initial [H 2 O 2 ] M ReactionTime (in s) t Reaction Rate (in Ms - 1 ) = [ I 2 ] / t Show calculations for x and y: (Include appropriate units, attach a separate sheet if necessary.) x = y = (round to the nearest whole number) Rate Law : Show calculations for k: k (run 1) = k (run 2) = k (run 3) = Average k =

37 Iodine Clock Reaction Part 1: PostLab Name 1. What is the overall order of the reaction in this lab? 2. Use your Rate Law and average k to calculate the reaction rate when [I - ] = M and [H 2 O 2 ] = M. 3. Consider the reaction A + B Products. This reaction was experimentally determined to be first order in A and second order in B. a. Write down the rate law for this reaction. b. What would be the unit of rate constant k for the above reaction? c. If the concentration of A is doubled while keeping the concentration of B constant, by how much will the rate of the reaction change? 4. The decomposition of a certain chemical X is found to be second order with resepect to [X] and have a rate constant of 5.5 x 10-2 M - 1 s - 1. If the initial concentration of X is 2.54 M, what will be the concentration after 89 seconds?

38

39 Experiment #6. Iodine Clock Reaction Part 2 Introduction In this experiment you will determine the Rate Law for the following oxidation- reduction reaction: 2 H + (aq) + 2 I (aq) + H 2 O 2 (aq) I 2 (aq) + 2 H 2 O (l) (1) The rate or speed of the reaction is dependent on the concentrations of iodide ion (I - ) and hydrogen peroxide, H 2 O 2. (The spectator ions are left off the reaction.) Therefore, we can write the Rate Law (concentration dependence) for the reaction as: Rate = k [I ] 1 [H 2 O 2 ] 1 (2) The temperature dependence of the rate is seen in k that is, there is a separate value of k for each temperature at which the reaction takes place. As with a lot of kinetics, the concentration of reactants or products at any instant is difficult to measure directly, so in this lab the rate will be determined indirectly. We have a very handy test for the presence of one of the products, iodine (I 2 ), namely starch. Starch reacts with iodine to form a blue/black colored complex. Unfortunately as soon as any iodine is produced it will react to make the complex and the solution will turn blue/black instantaneously. Thus, using starch as an indicator by itself would not be of much help. It confirms that some amount of I 2 is being formed, but it tells us nothing about what we are trying to measure - the rate (how long it takes to produce a given quantity of I 2.) To get around this problem we will introduce a side reaction that will remove the initial I 2 that is produced by our main reaction. This will prevent the solution from turning black long enough so that we can make some time measurements. We will use the following side reaction: I 2 (aq) + 2 S 2 O 3 2 (aq) 2 I (aq) + S 4 O 6 2 (aq) (3) S 2 O 3 2, thiosulfate ion, reacts with I 2 which prevents the solution from turning blue/black. How will this help? Since we have carefully measured the amount of thiosulfate (a small amount that will run out fairly quickly), we know exactly how much iodine it will take to react with this thiosulfate. Once the small amount thiosulfate has completely reacted, I 2 will start to build up in the solution. As soon as the thiosulfate runs out, I 2 will react with the starch and the solution will turn blue/black. By putting in this "time delay", we can now calculate the rate at which I 2 is being formed. The rate of reaction is equal to the change in concentration divided by the change in time. The change in time will be the time it takes for the solution to turn dark. We will calculate the change in concentration in I 2 based on the amount of thiosulfate added. Using the known volume and molarity, we can calcuate moles of thiosulfate (S 2 O 3 2 ). Based on stoichiometry, we can calculate the moles of I 2 : according to equation (3), 2 moles of thiosulfate react with every 1 mole of I 2, This gives us the change in moles, however for the rate formula we need change in concentration. Divide the moles of I 2 reacted by the total volume to find the change in molarity. Rate = [ I 2 ] / t (4)

40 Equipment Three 125 or 250 ml Erlenmeyer flasks Three 100 or 150 ml beakers 10 ml and 5 ml Pipettes Thermometer Stop watch or other time keeper Hot water bath Ice bath One bin of chemicals per group that will contain: M KI M Na 2 S 2 O M H 2 O M H 2 SO 4 (Spill : B1) 1% starch solution Disposal: All mixtures Spill/Disposal B1 (down the sink) Procedure: 11. Before coming to lab, complete the initial concentration table below showing all work. Determine initial concentrations using the formula M 1 V 1 = M 2 V 2. Note that since we are changing temperature, concentration will be kept constant. Copy these values onto the table in the data sheet. Run [ I ] (initial) [H 2 O 2 ] (initial) 1 (Flask 1 + beaker 1) 2 (Flask 2 + beaker 2) 3 (Flask 3 + beaker 3) 12. Clean and mostly dry three Erlenmeyer flasks. Label them 1, 2 and Obtain a bin of chemicals for your group. Use only these chemicals for all of your runs. Use fresh pipettes for each solution. Rinse pipette twice with the solution that you will be measuring and keep this prepared pipette with the corresponding solution. 14. Add the amounts of the solutions below to prepare each flask. The chemicals must be added in the order listed (top to bottom). Flask #1 Flask #2 Flask # M KI 15.0 ml 15.0 ml 15.0 ml 1% Starch 5.0 ml 5.0 ml 5.0 ml M Na 2 S 2 O ml 2.5 ml 2.5 ml 1 M H 2 SO ml 5.0 ml 5.0 ml

41 15. Rinse and mostly dry 3 beakers and label as 1, 2 and Prepare the following solutions in clean beakers: Beaker #1 Beaker #2 Beaker # M H 2 O ml 15.0 ml 15.0 ml 17. Get your timer ready. Add the contents of beaker 1 to flask 1. Start the stopwatch as soon as you mix the solutions. Swirl the flask to mix and note the time it takes for the color to change. This is Run 1. Record the temperature of the mixture. 18. Place Flask #2 and beaker #2 in an ice water bath. Let both of them cool until the temperature of both solutions is about 10 C lower than the room temperature. Record the exact temperature, mix the contents of the two flasks and note the time it takes for the color change. 19. Place Flasks #3 and beaker #3 in a hot water bath. Let both flasks warm up until the temperature of the two solutions is about 10 C higher than the room temperature. Record the exact temperature, mix the contents and note the time it takes for the color change. Disposal: All contents of the reaction flasks may be disposed of into the sink. Calculations: The Arrhenius Equation will be used to determine the Temperature dependence. Using the rate law expression determined part 1 (equation 2), calculate k for each temperature. Using Excel, plot a graph of ln k versus 1/T. If the reaction exhibits Arrhenius behavior, a straight line will be obtained where: y = m x + b ln k = (Ea/R)(1/T) + ln A where Ea is the activation energy of the reaction, T is the temperature in Kelvin, k is the rate constant at this temperature, and R is the ideal gas constant (8.314 J/mol K). add best fit line and display the equation. Label axes and give the graph a descriptive title edit axes such that the data takes up most of the graph use the slope of the best fit line to calculate activation energy Remember to attach the graph and spreadsheet to your lab report.

42

43 Name CHM112 Lab Iodine Clock Reaction Part 2 Grading Rubric Criteria Points possible Points earned Lab Performance Printed lab handout and rubric was brought to lab 3 Initial concentrations completed before coming to lab. 2 Safety and proper waste disposal procedures observed 2 Followed procedure correctly without depending too much on instructor or lab partner 3 Work space and glassware was cleaned up 1 Lab Report Calculations for reaction rate and rate constant 3 Graph & spreadsheet attached (trendline displayed, follows guidelines in lab instructions) 3 Calculations for activation energy 1 Question 1 1 Question 2 1 Total 20 Subject to additional penalties at the discretion of the instructor.

44

45 Iodine Clock Reaction Part 2: PostLab Name Moles I 2 reacted: Molarity of I 2 reacted [ I 2 ] : (same as values from Part 1) Run Temp ( o C) Initial [I - ] M Initial [H 2 O 2 ] M Reaction Time (secs) Reaction Rate (Ms - 1 )= [ I 2 ] / t Cold Room temp Hot In the space below show work for the rate constants at each temperature and copy the values into the table below. Plot a graph of ln (k) vs 1/T in Excel, insert a trendline, and display the equation on the chart. Attach your graph and spreadsheet with the lab report. Run Temperature (in C) Temperature ( in Kelvin) 1/T in Kelvin Rate constant k (include proper units) ln (k) Cold Room temp Hot Trendline equation: Based on your graph, what is the activation energy of the reaction in kj/mol? Show your calculations below:

46 1. If the actual value for the activation energy is 53 kj/mole, using your calculated value for E a, calculate the percent error. 2. Using your Arrhenius Plot and the best fit equation calculate the value for your rate constant at 85.0 C.

47 Experiment #7. Determination of an Equilibrium Constant Introduction It is frequently assumed that reactions go to completion, that all of the reactants are converted into products. Most chemical reactions do not go to completion because they are equilibrium systems where the reaction proceeds in both directions. As the reactants are used up, the rate of the forward reaction decreases. Conversely, as the concentrations of the products increase, the rate of the reverse reaction increases. Eventually, the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of the reactants and the products stay constant. The system has reached a state of dynamic equilibrium. At equilibrium, both the forward and reverse reactions are occuring, but no net change is observed. Consider the general reaction: aa + bb cc + dd ( 1 ) where a,b,c and d are the stoiochiometric coefficicents. Experimental evidence shows that the ratio of products to reactants (with each product and reactant expressed as a molar concentration and raised to its stiochiometric coefficient) is a constant for a reaction that has reached equilibrium. This constant, which is different for each chemical reaction, is known as the equilibrium constant and is designated with the letter K. There is a separate value of K for each temperature at which the reaction occurs. Thus, at equilibrium, the equilibrium constant K is equal to: [ ] c C [ D ] K = a [ A] [ B] d b ( 2 ) where the brackets [ ] imply molarity and the exponents are the stoichiometric coeffients of the balanced chemical equation. The equilibrium constant measures the extent to which a chemical reaction occurs. The larger the value for K, the greater the tendency for the reaction to go to completion is and the more products will be formed relative to the reactants. In this experiment you will determine the equilibrium constant for the following reaction: (Spectator ions are not shown.) Fe 3+ (aq) + SCN - (aq) FeSCN 2+ (aq) ( 3 ) 2+ [ FeSCN ] K = 3+ [ Fe ][ SCN ] ( 4 ) Solutions of Fe 3+ and SCN - will be mixed and will react to form some FeSCN 2+. The initial amounts of Fe 3+ and SCN - can be calculated. The equilibrium concentration of FeSCN 2+ will be found using its spectroscopic properties how much light it absorbs at a specific wavelength. FeSCN 2+ is a blood red complex that absorbs the blue- green wavelengths of visible light. Its absorbance is directly proportional to its concentration. The absorbance (a measure of the amount of light absorbed) will be measured by a spectrophotometer. Solutions to be measured are placed in cuvettes; these are square tubes have minimal absorbance in the wavelength range of the spectrophotometer.

48 A cuvette for measuring light absorption. It is usually made of quartz or plastic. On some, two of the sides are frosted. For a solution placed in a 1 cm cuvette, the absorbance, A, is equal to the extinction coefficient, ε (epsilon), times the molar concentration, C. The value of ε is can be determined experimentally for each substance from solutions of known concentration at a particualr wavelength. ε varies with the wavelength of light. In this experiment we will be measuring absorbtion at 470 nm. A = ε C ( 5 ) We will use a best- fit line equation to find the concentration from the absorbance measurement for each sample. Using an equilibrium (ICE) chart, the equilibrium concentrations of Fe 3+ and HSCN are then calculated. Finally, the equilibrium concentrations are put into equation ( 4 ) to find the equilibrium constant, K. Note: All of the solutions are made in 1.0M HNO 3 (aq), so be cautious and wear gloves. Equipment 4 small beakers 3 test tubes stirring rod 5 cuvettes Venier equipment and colorimeter thermometer Chemicals M KSCN(aq), potassium thiocyanate (source of SCN - ) M Fe(NO 3 ) 3 (aq), iron (III) nitrate dissolved in 1.0M HNO 3 (aq). (Source of Fe 3+ ) M Fe(NO 3 ) 3 (aq), iron (III) nitrate dissolved in 1.0M HNO 3 (aq). (Source of Fe 3+ ) Spill/Disposal The contents of all test tubes, cuvettes, and beakers may be disposed of in the sink. Flush with a large volume of water. Prelab Calculations Note: you MUST have these calculations done before coming to lab. For all 4 of the solutions you will prepare in Step 3 of the procedure, calculate [FeSCN 2+ ]. For these solutions, assume that moles of SCN - = moles of FeSCN 2+ (a large excess of Fe 3+ makes the SCN - react completely). Use M 1 V 1 = M 2 V 2. V 1 should be the volume of SCN - in that beaker, and V 2 should be the total volume of everything in that beaker. Beaker [FeSCN 2+ ] *You will need to enter these concentrations into the Labquest in Step 3, so have these ready when you come to lab.*