Molar mass Essays

uniquely identifying molar mass as well as individual properties. In this lab, the purpose was to use the properties that three unknown alkali metal carbonate powers have, to identify the molar mass. For the three unknown substances, they all were white powders of about the same texture, and they all reacted when added to hydrochloric acid. Based on these properties, it is impossible to distinguish which one is which due to the similarities, so it is necessary to solve for the molar mass since that is unique

Introduction A way to determine the molar mass of an unknown substance is to use other properties of that substance and solve for desired information. In this experiment, a colligative property, like the freezing point of an aqueous solution of the unknown substance, was used to find the molar mass of the substance. With the molar mass discovered, the identity of the substance was found. Material and Methods First, a Vernier temperature probe was attached to a plastic rod using rubber bands. A cork

Determination of molar mass of an element and a compound. Introduction: Aim: To determine the molar mass of an element, copper, and a compound, barium sulfate. Background Information: In this experiment, the limiting reagent was the copper oxide. The limiting reagent is the reactant that is completely used in a reaction, and thus determines when the reaction stops. Copper oxide is a pitch black solid at room temperature. Zinc reacts with acid, whereas copper does not react. The molar mass formula is:

Determination of the molar mass of magnesium Aim To determine the molar mass of magnesium with the ideal gas equations and Dalton’s law about partial pressures. Materials and chemicals gas measuring tube 50.0cm3 graduated cylinder 1000.0 cm graduated cylinder 10.0cm3 cork with a hole copper wire thermometer barometer magnesium ribbon HCL 35% (concentrated) Background theory Dalton’s Law of Partial states that the total pressure of a gas is equal to the sum of pressure of each individual

characterize a volatile liquid is determination of its molar mass. Dumas method also known as the vapor density method uses the vapor density of the unknown volatile liquid in determining its molar mass. The major assumptions of these methods are the substance behaves ideally. The molar mass of a volatile liquid can be obtained by measuring the temperature, pressure, mass, and volume in a gaseous state. The equation used to determine the molar mass is derived from the Ideal Gas Law equation. The objective

Determination of the molar mass of a chosen compound/element Fran Jurinec 1.M Introduction Molar mass is a physical property of a chemical element or substance which shows the mass per amount of substance. My task is to determine the molar mass of a product substance from one of the following equations: a. Zn(s) + 2HCl (aq) → ZnCl2 (aq) + H2 (g) b. CaCO3 (s) + 2HCl(aq) → CaCl2 (aq) + CO2 (g) + H2O(l) c. Na2SO3 (aq) + 2 HCl (aq) → 2NaCl (aq) + S (s) + SO2 (g) For my experiment,

experiment was to find the molar mass of the butane using the gas produced by the lighter. The lighter produces the butane gas and the volume of gas released and the mass change of the lighter are used to calculate the mole and the molar mass of the butane. To calculate the molar mass, the Ideal gas law, PV=nRT is used. The theoretical value of the C4H10 is 58.12g/mol. However, the experimental value that I could get from the calculation was 51.46g/mol for the molar mass of butane. This indicates

.1-amEr (c) Gram molecular mass of HNO3 = Mass of 1-molecule of HNO3x NA = 63 amu x NA = 63 gm X NA = 63 gram NA Solved Example-4:Find out the mass of carbon -12 that would contain 1.0 x1019carbon-12 atoms. Solution : Mass of 6.022 x1023 carbon-12 atoms = 12 g Mass of 1.0 x1019 carbon-12 atoms = 12x1x10 19 g 6.022 x1023 = 1.99 x10-4g Solved Example-5: How many molecules are present in 100 g sample of NH3? Solution : Molar mass of NH3 = (14 + 3) g mol-1 = 17 g mol-1 17 g sample of NH3 contains 6

81%. For the iron ion the mean percentage was 12.20% and the average percent for oxalate was 57.20%. The total percentage of the mean adds up to 97.28%. The empirical formula using the means provided created the formula K2Fe(C2O4)3*2H2O. The total molar mass in this equations is 434.05 grams. The percent yield of iron salt using the actual yield of 8.71 grams with a theoretical yield of 17.8 grams is 49%. Discussion: The percent yield calculated in this experiment was 49% that indicates errors occurred

Claim: Electrons with higher energy will be positioned further away from the nucleus than electrons with lower energy. Evidence and Analysis: In this lab, eight solutions were tested to show how the energy of light given off by an atom describes the location of an electron in the atom. When certain solutions were placed in the flame from a bunsen burner, the flame turned different colours. Lithium made the flame change to a violet/teal colour, barium and copper made the flame become green, and

As mentioned earlier, the molar mass of benzaldehyde is 106.121g/mol, and the molar mass of benzoic acid is 122.12g/mol. With that in mind, to calculate the maximum possible mass of benzoic acid, which is the product of the experiment, Equation #2 must be used. Referencing from Table 1.1, the observed benzaldehyde mass is 5.02g, thus the plugging that into Equation #2, the calculation would be, (122.12 g⁄mol C_7 H_6 O_2)/(106.121 g⁄mol C_7 H_6 O)×5.02g, equaling to, 5.78g±0.01g of benzoic acid.

10 February 2017 Title: Experiment 2: Determining The Molar Mass of a Volatile Liquid Introduction: In this lab experiment, the ideal gas law will be used to find the average molecular mass of the unknown liquid as a gas. This will be done by heating the liquid until it is in its gaseous state and then once it is there it will be determined what the average molecular mass is. The purpose of this lab experiment is to determine the molar mass of a volatile gas by using the ideal gas law. Procedure:

the molar mass of elements to identify unknown substances by weighing the mass of the substance and calculating the grams per mole. The main objective of this experiment is to determine the mass similarities between the given and unknown substances. The equation used in this experiment is: Molar Mass = Mass (g) / Mol Where the molar mass of the unknown substance, the mass of the unknown substance, Mol is the number of moles in the unknown substance. These results will be compared to the molar mass

Purpose: The purpose of this experiment was to determine the molar mass of unknown #43 using the derived freezing point depression. To obtain the freezing point depression, t-butyl alcohol was placed in a cold-water bath and frozen solid for a total of two runs. Then, unknown #43 was dissolved in t-butyl alcohol and placed in a cold-water bath until frozen solid. This process was repeated for a total of three runs, with the first two runs containing half of the unknown, and the last run containing

It justifies the idea that increasing the concentration of sucrose does in fact speed up the rate of osmosis, an therefore increase the mass of the Visking tube. This happens due to the water molecules moving from a high water concentration to a low water concentration. Another reason why this hypothesis is correct is because sucrose particles are too large to go through the membrane therefore

Copper Cycle Lab Report Ameerah Alajmi Abstract: A specific amount of Copper will undergo several chemical reactions and then recovered as a solid copper. A and percent recovery will be calculated and sources of loss or gain will be determined. The percent recovery for this experiment was 20.46%. Introduction: The purpose of this experiment is to demonstrate the different types of chemical reactions, those including Copper. There are different types of chemical reactions. A double displacement reaction

Determining the Identity of an Unknown Diprotic Acid Through Titration Kevin S. Burton; Madison Gallegos April 2, 2018 Abstract There is more than a single way to determine the molar mass of a compound or element. Titration is one such way to determine molar mass by reacting an unknown compound or element with a known compound or element. In this experiment an unknown diprotic acid was combined with a known base, in this particular case, NaOH. As it was known to be diprotic, the unknown acid

I. Title: Mass and Mole Relationships in a Chemical Reaction II. Background: Percent yield is the ratio of actual yield to theoretical yield. Amount in percent of one product formed in chemical reaction. Actual yield is the information found is experiments or is given. It is also the real amount. Theoretical yield is found through a mathematical equation. The amount produced is another way of identifying theoretical yield. The limiting reactant is the reactant that is completely used in the reaction

initial mass of the solid sodium thiosulfate pentahydrate was 0.21 grams. Once the ten milliliters of water was added, the mass of the sodium thiosulfate solution as 9.70 grams(Table 1). In Table 2, the initial mass of the weight buret in trial one was 15.23 grams, the final mass of the buret was 14.14 grams. The second trial’s initial buret mass was also 14.14 grams, because no solution was used in between trials; however, the final mass was 13.04 grams.

The central purpose of this experiment was to determine the experimental empirical formula of an oxide of magnesium by performing a synthesis reaction. It was hypothesized that the formula that was derived from the recorded data would be identical to the theoretical empirical formula. After performing calculations with the data that had been collected within the duration of the experiment, it was deduced that the empirical formula of the product generated by the synthesis reaction was Mg5O6. Since