ipl-logo

Molar Volume Lab

1390 Words6 Pages

Determining the Molar Volume of a Gas Lab Report The purpose of the experiment is to determine the molar volume of hydrogen gas at standard pressure and temperature. Excess hydrochloric acid was placed in a eudiometer and decanted with deionized water. A piece of magnesium ribbon was trapped in a copper wire cage in the eudiometer in order to keep it in place as the more dense hydrochloric acid diffused downward in the inverted eudiometer. The eudiometer was placed in a water bath and the magnesium ribbon and hydrochloric acid reacted to form hydrogen gas. The rubber stopper was removed in order to balance the water levels in the eudiometer and the bath. The process was repeated. Chemical theories and concepts used in this experiment includes …show more content…

Vapor pressure (H2O): 17.444 torr = 17.444 mmHg Correction: (263 mm * 1.00 mmHg) / 13.6 mmHg = 19.3 mmHg Corrected Pressure of H2: 758 mmHg - 19.3 mmHg = Pressure of H2 + 17.444 mmHg Pressure of H2 = 721 mmHg Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) (1.60 cm * 0.0200 g) / 1 cm = 0.0320 g (0.0320 g * 1 mol * 1 mol H2) / (24.3050 g * 1 mol Mg) = 0.00132 mol H2 Volume at STP: (760 mmHg)(V1)(292.9 K) = (721 mmHg)(.0314 L)(273 K) V1 = 0.0278 L Molar Volume: 0.0278 L / .00132 mol H2 = 21.1 L / mol Trial 2 Pressure: (101.1 kPa * 760 mmHg) / 101.3 kPa = 758 mmHg Vapor pressure (H2O): 17.444 torr = 17.444 mmHg Correction: (268 mm * 1.00 mmHg) / 13.6 mmHg = 19.7 mmHg Corrected Pressure of H2: 758 mmHg - 19.7 mmHg = Pressure of H2 + 17.444 mmHg Pressure of H2 = 721 mmHg Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) (1.50 cm * 0.0200 g) / 1 cm = 0.0300 g (0.0300 g * 1 mol * 1 mol H2) / (24.3050 g * 1 mol Mg) = 0.00123 mol H2 Volume at STP: (760 mmHg)(V1)(292.9 K) = (721 mmHg)(.0302 L)(273 K) V1 = 0.0267 L Molar Volume: 0.0267 L / .00123 mol H2 = 21.7 L / mol Average Molar Volume: (21.1 L/mol + 21.7 L/mol + 21.9 L/mol + 20.4 L/mol + 21.0 L/mol + 22.7 L/mol) / 6 = 21.5 L/mol Percent difference: | 21.5 L/mol - 22.4 L/mol | / 24.4 L/mol = 0.0369 * 100% = 3.69% Results Table

Show More
Related

  • Eastmamol/540 Week 4

    563 Words  | 3 Pages

    Prelab week 1 Calculations Preparation of 1.5μmol/L mixed low-level standard dilution 150μmol/L × V1=1.5μmol/L × 10ml V1=(1.5μmol/L×10ml)/(150μmol/L)=0.1ml Conversion of milliliters to microliters (0.1ml×1000)μL= 100μL Preparation of 3μmol/L mixed low-level standard dilution 150μmol/L × V1=3μmol/L × 10ml V1=(3μmol/L×10ml)/(150μmol/L)=0.2ml Conversion of milliliters to microliters (0.2ml×1000)μL= 200μL Preparation of 3μmol/L mixed low-level standard dilution 150μmol/L × V1=7.5μmol/L × 10ml V1=(7.5μmol/L×10ml)/(150μmol/L)=0.5ml Conversion of milliliters to microliters (0.5ml×1000)μL= 500μL Preparation of the blank samples The volumetric flask will be filled to the mark with 150μmole/L of stock solution to act as blank (reference). Additional two blanks will

    Read More

  • Unknown 6 Essay

    680 Words  | 3 Pages

    Characteristic Property- Test 2- Density Materials: Triple Beam balance, distilled water, graduated cylinder, unknown 6 Procedure: first we found the mass of the empty graduated cylinder and then its mass with the now distilled unknown. After subtracting the mass of the graduated cylinder, we were able to find the volume. For every 1mL=1cm³ so there we had the volume found with the graduated cylinder. We divided the mass by the volume in order to get the density Data: We found that the density of our unknown was 0.76 g/cm3.

    Read More

  • Unknown Mixture Lab Report

    302 Words  | 2 Pages

    A two-hundred and fifty milliliter(ml) flask was zeroed out on the scale. Fifty milliliters of HCl was put into the flask and weighed. The unknown mixture was poured into the HCl while

    Read More

  • Hydrochlorimetry Lab Essay

    1866 Words  | 8 Pages

    The purpose of this lab is to observe the reaction between hydrochloric acid and magnesium metal. When the substances are reacted over water, the products produced are a salt in aqueous solution and a gas. While the salt remains in the water as part of a solution, the gas produced will float to the top. Though water vapor pressure will affect the pressure of the gas in the eudiometer, it is possible to apply Dalton’s law of partial pursues to find the dry pressure of the gas. When the dry pressure is determined, the volume of the gas at STP can then be determined and what the experimental volume of one mole of the gas would be at STP.

    Read More

  • From 5.0 Ml, What Does Your Data Show Happens To The Pressure?

    205 Words  | 1 Pages

    When the volume was 5.8 ml the pressure was 192.52 kPa but when the volume double (10.8 ml), the pressure was 133.38 kPa. 2. If the volume is halved from 20.0 mL to 10.0 mL, what does your data show happens to the pressure? Show the pressure values in your answer.

    Read More

  • Magnesium Sulfate Heptahydrate Lab Report

    478 Words  | 2 Pages

    The purpose of this lab was to determine the percent water in magnesium sulfate heptahydrate, or Epsom salt. The experimental percent water is determined to be 42.06% in both trials, making the average also 42.06%. To determine this percent water a heating and cooling procedure was used. First, the vials were cleaned of impurities using the lab oven and were not touched after this point.

    Read More

  • Mole Mass Lab Report

    372 Words  | 2 Pages

    I. Purpose: To experimentally determine the mass and the mole content of a measured sample. II. Materials: The materials used in this experiment a 50-mL beaker, 12 samples, a balance and paper towels. III.

    Read More

  • Mat 540 Final Paper

    801 Words  | 4 Pages

    = 10^-3 M = 1,000 mL Here C1,C2; are the first and second concentrations of solution V1 and V2 ; are the required and current volumes. The impeller turned on and DDA, and tap water left to be mixed properly with water for 2 minutes. Approximately 150 grams of quartz added into the solution.

    Read More

  • 1. How Did The Final Naoh Molarities Compare For Each Trial?

    700 Words  | 3 Pages

    Conclusion: Based on the results of molarity from Trials 1, 2, and 3, it is concluded that our experimental for each trial is .410M NaOH, .410M NaOH, and .450M NaOH. The actual molarity of the NaOH concentration used was found to be 1.5M NaOH. The percent error of the results resulted in 72%. The large error may have occurred due to over titration of the NaOH, as the color of the solution in the flask was a darker pink in comparison for the needed faint pink. Discussion of Theory:

    Read More

  • Aluminum Synthesis Lab Report

    1073 Words  | 5 Pages

    Once the material was acquired, 1.0094 grams of Aluminum were weighed and then transferred to a 250mL beaker. The 250mL beaker continued to remain in use for the next few steps. 1.4M KOH solution was added to the Aluminum sample that was previously obtained. For gas to escape the lab, there was a fume

    Read More

  • Density Of Water Lab Report

    376 Words  | 2 Pages

    Moving forward, to find the volume of the metal object, the graduated cylinder was filled with water. The initial volume in the graduated cylinder read 21.0 mL. The metal object was then placed in the graduated cylinder carefully, as to not splash water. In the graduated cylinder of 21.0 mL, the metal object raised the water to 24.1 mL, thus the object was 3.1 mL for measurement 1. In a graduated cylinder of 21.0 mL, the metal object raised the water to 24.0 mL, thus the metal object was 3.0 mL for measurement 2. In a graduated cylinder of 21.0 mL, the metal object raised the water level to 23.9 mL, thus the object was 2.9 mL for measurement 3.

    Read More

  • Gas Law Lab

    596 Words  | 3 Pages

    The goal of the experiment is to determine the molar mass of an unknown liquid using the Ideal Gas Law and once it is found the empirical and molecular formula could be determine. The theory behind this experiment is the idea that if the pressure, volume, number of moles and temperature are all constant we could easily use the Ideal gas law equation to find the molar mass of a gas. obtain this information we would need to heat up and unknown liquid in a closed flask in boiling water and once the liquid has evaporated the volume and mass is calculated. The values obtained once completing this experiment is 103 g/mol as the molar mass and CH2 for the empirical formula and C7H14 for the molecular formula.

    Read More

  • Acid Base Neutralization Lab Report

    1001 Words  | 5 Pages

    Procedure A. Preparation of NaOH solution The molarity of a solution is the ratio of the number of solutes dissolved in a liter of solution. To figure out the needed mass (in grams) of NaOH pellets to be dissolved in a 0.25 L of water, remember that a mole is equivalent to the quotient of mass over the molar mass of the substance. This was used to rearrange the base formula and to derive the mathematical equation of mass in terms of molarity. mass (g) =

    Read More

  • Electrolysis In Chemistry Ia

    2800 Words  | 12 Pages

    so when you multiply the 2.9 by 2 it gives you 5.8 cm3 which was my result. As the voltage increases the volume of hydrogen and oxygen increases. The standard deviation when I used 9 volts for hydrogen it was 0.7 + 0.23= 0.93 and 0.7-0.23= 0.47 so the range of values is between 0.47 cm3 to 0.93 cm3. When I used 9 volts for oxygen it was 0.3+0.12=0.42 and 0.3-0.12=0.18 so the range of values is between 0.18 cm3 to 0.42 cm3.

    Read More

  • Benzoic Acid Lab Report

    767 Words  | 4 Pages

    That caused a new initial reading of NaOH on the burette (see Table1 & 2). The drops were caused because the burette was not tightened enough at the bottom to avoid it from being hard to release the basic solution for titrating the acid. The volume of the acid used for each titration was 25ml. The volume of the solution was then calculated by subtracting the initial volume from the final volume. We then calculated the average volume at each temperature.

    Read More

More about Molar Volume Lab

Open Document