Purpose
The purpose of this experiment is to use the freezing point depression method to determine the molar mass of ethylene glycol (antifreeze) by analyzing the effect of antifreeze on the freezing points of varying solutions of water and antifreeze.
Data Table
Freezing point of Water
Freezing point of Solution 1 Freezing point of Solution 2 |
-1 degrees Celsius
-4.5 degrees Celsius -7 degrees Celsius |
Calculations
Freezing point depression of Solutions 1 and 2.
freezing point of solution - freezing point distilled water= change in temperature (freezing point depression)
Solution 1: -4.5 degrees Celsius- -1 degrees Celsius= -3.5 degrees Celsius
Solution 2: -7 degrees Celsius- -1 degrees Celsius = -6.5 degrees Celsius
Solution 1: -4.5 degrees Celsius- -1 degrees Celsius= -3.5 degrees Celsius
Solution 2: -7 degrees Celsius- -1 degrees Celsius = -6.5 degrees Celsius
Molality of Solutions 1 and 2.
Molality of Solution 1: 1.882 mol/kg
Molality of Solution 2: 3.226 mol/kg
Grams of antifreeze (solute) per 100 grams of solvent (water).
Molar Mass of antifreeze, calculated from both solutions.
Conclusion
After completing the experiment and the calculations, it was determined that the addition of antifreeze to water decreases the freezing point, thus causing the solution of water and antifreeze to undergo the phase change from liquid to solid at a lower temperature. For solution 1, the freezing point depression was calculated to be -3.5 degrees Celsius, and -6 degrees Celsius for Solution 2. This decrease in the freezing point as a result of adding antifreeze to water demonstrates the purpose of using antifreeze in cooling systems of automobiles, for it reduces the risk of the liquids in the system from freezing during the winter months.
Discussion of Theory
The addition of antifreeze is a very useful technique to reduce the freezing point of solutions, such as those found in the cooling systems of automobiles. This technique is effective due to the colligative property that is freezing point depression. A colligative property is one that is dependent solely on the number of solute particles present in a solution, not the identity of said particles. Examples of colligative properties include boiling point elevation, vapor pressure lowering, osmotic pressure, and freezing point depression. Freezing point depression is the lowering of the freezing point of the solvent by a nonvolatile solute. When a nonvolatile solute such as antifreeze is added to a solvent such as water, the number of particles in the solution is increased. The number of particles a compound splits into is called its van't Hoff factor (i factor). For ionic compounds, the i factor is an integer greater than one, because ionic compounds that are soluble in water will split into ions. The van't Hoff factor for biocompounds (those containing carbon, hydrogen or oxygen) it is always one, because the covalent bonds that hold these molecules together do not allow for the compound to split into particles. For acids and bases, however, the van't Hoff factor is not a whole number because there is partial dissociation of particles. Strong acids dissociate almost completely, but there is part that does not. Conversely, weak acids barely dissociate at all, but a small portion of H+ ions will dissociate when dissolved in water. For example, in strong acids like H2SO4, most of the acid splits into H+ and HSO4- ions, but since the resulting HSO4- is a weak acid, it dissociated a little as well. This slight dissociation of the weak acid HSO4- is represented with a + on the van't Hoff factor.
The i factor is used when calculating the freezing point depression of a solution, by multiplying the i value by the molal freezing point depression constant of the solvent and the molality of the solute in the solution. Therefore, because of the direct relationship, when the van't Hoff factor is increased by the addition of particles in the solution, the freezing point depression increases as well. The freezing point depression increases as a result of the freezing point decreasing, because the freezing point depression is the difference in the new and standard freezing points. So to decrease the temperature at which water freezes, additional particles should be added to increase the van't Hoff factor and consequently the freezing point. Adding antifreeze, composed of mostly ethylene glycol, increases the i factor of the solution. When the solution of water and antifreeze is exposed to low temperatures, the particles of antifreeze that dissociated in the water prevent the water molecules from compressing into a solid (the process of freezing). Since these extremely low temperatures are not commonly encountered in nature, the solution of water and antifreeze will not turn into a solid nearly as quickly nor as often as pure water. The addition of antifreeze to water decreases the point at which the solution will freeze but will not prevent it from freezing all together, as the name suggests.
The i factor is used when calculating the freezing point depression of a solution, by multiplying the i value by the molal freezing point depression constant of the solvent and the molality of the solute in the solution. Therefore, because of the direct relationship, when the van't Hoff factor is increased by the addition of particles in the solution, the freezing point depression increases as well. The freezing point depression increases as a result of the freezing point decreasing, because the freezing point depression is the difference in the new and standard freezing points. So to decrease the temperature at which water freezes, additional particles should be added to increase the van't Hoff factor and consequently the freezing point. Adding antifreeze, composed of mostly ethylene glycol, increases the i factor of the solution. When the solution of water and antifreeze is exposed to low temperatures, the particles of antifreeze that dissociated in the water prevent the water molecules from compressing into a solid (the process of freezing). Since these extremely low temperatures are not commonly encountered in nature, the solution of water and antifreeze will not turn into a solid nearly as quickly nor as often as pure water. The addition of antifreeze to water decreases the point at which the solution will freeze but will not prevent it from freezing all together, as the name suggests.
Sources of Error
This laboratory experiment contained various sources of error that impacted the data collected and subsequently the calculations derived thereafter. A major source of error in this experiment was the use of tap water instead of distilled water. Distilled water contains no contaminants or minerals, while tap water has many. The presence of the minerals and other materials in the tap water used increased the van't Hoff factor (i factor) of the water, and potentially lowered the freezing point slightly. This would cause the temperature recorded for the freezing points of the two solutions to be a little lower than the actual, and therefore the calculations of the freezing point depressions for each solution to be slightly skewed as well. A second source of error occurred when the temperature of the water was suppose to be measured the instant ice crystals appeared in the bottom of the test tube. The temperature of the tap water was not measured the moment the crystals were present, resulting in a slightly lower temperature recorded for the water. In addition, when solution 1 was placed in the ice bath and stirred until slush ice was present in the test tube, an ice chip was suppose to be added to prevent the solution from supercooling. This ice chip was not added, resulting in the temperature measured to be slightly cooler than it actually was, due to potential supercooling. Another source of error was the amount of salt added to the beaker filled with ice in step one. This amount was not constant throughout the experiment, and may have affected the temperatures of the water and solutions. A final source of error that occurred during this experiment was the presence of water in the test tubes before the solutions were poured into them. This water, present as a result of the test tubes being cleaned, could have affected the freezing points of the solutions and therefore the measured temperatures and calculated freezing point depression values were not truly accurate.
Critical Thinking: Analysis and Conclusions
1. The molar mass of permanent antifreeze, almost 100% ethylene glycol, is 62 grams/mole. This is determined by adding the molar masses of carbon, hydrogen and oxygen, multiplying by the subscripts where applicable to determine the number of molecules present in the compound.
2. The percent error for Solution 1 was -14.775%. The formula used to calculate percent error is: experimental-theoretical x100
Calculations: 53.135-62.00 x100= -14.775% theoretical
62.00
The percent error for Solution 2 was -0.00645%.
Calculations: 61.996-62.00 x100= -0.00645%.
62.00
*The negative percent errors mean that the calculated molar mass was lower than the actual molar mass.
3. The major sources of error in this investigation came from errors in measurements and the lack of distilled water. When the amounts of antifreeze were measured, they were done so using a triple beam balance instead of a digital scale. The triple beam balances are less accurate than digital scales, for they require more interpretation from the experimenter to gain the reading. This caused the measurements recorded for the masses of antifreeze used in each solution to be not exact. This can be reduced by using more accurate balances and improving techniques for measuring and interpreting readings from the balances. The second major source of error was the use of tap water in place of distilled water. Tap water contains minerals and other substances that increase its density. This increased density causes the molality and other calculations to be slightly increased than they would be if distilled water was used. This can be remedied by using distilled water instead of tap water, or adjusting the calculations for the higher density of tap water.
2. The percent error for Solution 1 was -14.775%. The formula used to calculate percent error is: experimental-theoretical x100
Calculations: 53.135-62.00 x100= -14.775% theoretical
62.00
The percent error for Solution 2 was -0.00645%.
Calculations: 61.996-62.00 x100= -0.00645%.
62.00
*The negative percent errors mean that the calculated molar mass was lower than the actual molar mass.
3. The major sources of error in this investigation came from errors in measurements and the lack of distilled water. When the amounts of antifreeze were measured, they were done so using a triple beam balance instead of a digital scale. The triple beam balances are less accurate than digital scales, for they require more interpretation from the experimenter to gain the reading. This caused the measurements recorded for the masses of antifreeze used in each solution to be not exact. This can be reduced by using more accurate balances and improving techniques for measuring and interpreting readings from the balances. The second major source of error was the use of tap water in place of distilled water. Tap water contains minerals and other substances that increase its density. This increased density causes the molality and other calculations to be slightly increased than they would be if distilled water was used. This can be remedied by using distilled water instead of tap water, or adjusting the calculations for the higher density of tap water.
Critical Thinking: Application
1. The freezing point depression method could not be used for substances not soluble in water because freezing point depression is a colligative property. Colligative properties only exist in substances that are soluble in water because they are measured by the number of particles or ions a substance breaks into, called the van't Hoff factor. If a substance is not soluble in water, it will not break down into any particles, giving it an i factor of zero. This results in a value of zero for the freezing point depression, for the van't Hoff factor is multiplied by the molal freezing point depression constant and the molality. A substance must have an i factor greater than one to have a freezing point depression, which means it must be soluble in water.
2. The freezing point depression of water would increase due to a 1m solution of NH4PO3. This is because the ionic substance breaks into more ions, increasing the i factor and therefore the freezing point depression, because it is calculated by multiplying the molality by the van't Hoff factor and the constant.
3. It is assumed that the density of distilled water is exactly 1 gram per 1 milliliter of water. This assumption is made because distilled water theoretically contains no excess minerals or other substances, making its density exactly that of pure water.
4. This method of molar mass determination is practical for other substances soluble in water, as long as a different molal freezing point depression constant is used for each substance. Since any substance that dissolves in water has a van't Hoff factor greater than 1, the freezing point depression can be calculated, as well as the molality. From the molality and the number of grams of solute, the molar mass can be determined.
2. The freezing point depression of water would increase due to a 1m solution of NH4PO3. This is because the ionic substance breaks into more ions, increasing the i factor and therefore the freezing point depression, because it is calculated by multiplying the molality by the van't Hoff factor and the constant.
3. It is assumed that the density of distilled water is exactly 1 gram per 1 milliliter of water. This assumption is made because distilled water theoretically contains no excess minerals or other substances, making its density exactly that of pure water.
4. This method of molar mass determination is practical for other substances soluble in water, as long as a different molal freezing point depression constant is used for each substance. Since any substance that dissolves in water has a van't Hoff factor greater than 1, the freezing point depression can be calculated, as well as the molality. From the molality and the number of grams of solute, the molar mass can be determined.