Experiment 9 Empirical Formula of Zinc Iodide Objectives Upon completion of this experiment‚ students should have learned: 1. The law of conservation of mass. 2. How to calculate an empirical formula. 3. The concept of limiting reagents. Introduction Synthesis and the determination of empirical formulas are two extremely important parts of chemistry. In this experiment‚ you will synthesize zinc iodide and determine its empirical formula. The molecular formula gives the actual
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Empirical Formula Lab Class: Chemistry 1405 Fall 2013 Aim: The aim of this Lab Exercise is to use the mass of a chemical and use that mass to find the amount of moles of the final product you can get using the empirical formula. Introduction: The empirical formula of a compound is the simplest whole-number ratio of the elements in the compound‚ which as you will discover‚ is a ratio of the moles of those elements. “Empirical” also means “experimentally determined”. In this experiment
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Chemistry pre- IB Empirical Formula Observations/Qualitative Data: I have used my sense to observe that the magnesium is a solid that is bendable‚ is very light and its color is silver. After being put in a Crucible covered by a lid‚ under a Bunsen burner for a few minutes‚ it has lit up and turned red. After the experiment was over‚ the magnesium was turned into an ash/powdery state and its color became white/grey. Data collection and Processing (DCP): Quantitative Data: Weight in grams
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Empirical Formula of a Compound * Purpose: To determine the empirical formula of Magnesium Chloride. * Data 1. Mass of evaporating dish = 45.08g 2. Mass of evaporating dish and Magnesium = 45.17g 3. Mass of Magnesium: { 2 } – { 1 } = 0.09 4. Mass of evaporating dish and Magnesium Chloride First weighing = 45.48g (After heating and cooling) second weighing = 45.49g 5. Mass of Magnesium Chloride: { 4} – { 1 } = 0.41g
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What is the empirical formula of a compound that contains 75% Ag and 25% Cl by mass? AGCL Calculate the approximate number of molecules in a drop of water with a mass of 0.10 g. 3 x 1021 molecules What is the percentage composition of CaSO4? 29.44% Ca‚ 23.55% S‚ 47.01% O What mass of calcium bromide is needed to prepare 150.0 mL of a 3.50 M solution? (Assume that the molecular weight of CaBr2 is 200.618 g/mol) 105 g Nitrous oxide (N2O)‚ or laughing gas‚ is commonly used as an anesthetic
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Empirical and Molecular Formula | Key Concepts * Empirical Formula of a compound shows the ratio of elements present in a compound. * Molecular Formula of a compound shows how many atoms of each element are present in a molecule of the compound. * The empirical formula mass of a compound refers to the sum of the atomic masses of the elements present in the empirical formula. * The Molecular Mass (formula mass‚ formula weight or molecular weight) of a compound is a multiple of the empirical
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nitrogen gas?(2.8 mol) EMPIRICAL FORMULA WORKSHEET 1. What is the empirical formula for a compound which contains 0.0134 g of iron‚ 0.00769 g of sulfur and 0.0115 g of oxygen? (FeSO3) 2. Find the empirical formula for a compound which contains 32.8% chromium and 67.2% chlorine. (CrCl3) 3. NAME the compound which contains 0.463 g Tl (#81)‚ 0.0544 g of carbon‚ 0.00685 g of hydrogen and 0.0725 g of oxygen by finding its empirical formula. (TlC2H3O2)
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Steven Leung 9/19/06 Lab Report The Empirical Formula of a Copper Oxide Purpose: To convert an unknown copper oxide to copper (Cu) metal using natural gas to provide a reducing environment as shown below: Cu O (s) + CH (g) ¨ Cu (s) + Co (g) + H O (g) From the mass difference between the unknown copper oxide and the Cu metal generated at the completion of the reaction and the molar mass of Cu and oxygen‚ the empirical formula of the original copper oxide can be calculated. Materials:
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structure is derived. Firstly it is important to determine the percentage composition of elements to work out the empirical formula. The empirical formula was found to be C10H12O and the mass of the unknown was 148.09 m/z which when calculating the molecular weight of the empirical formula it did equal 148.09 g mol -1. This means that the empirical formula is also the molecular formula. As 12.01x10 carbons +1.008x12 hydrogens + 16= 148.09 09 g mol -1. From this knowledge the unknown molecule must
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Name: J.T Empirical Formula of Magnesium Oxide: Lab Report The objective of the experiment is to determine the empirical formula of Magnesium Oxide through a procedure of heating magnesium ribbon to react with oxygen to form a magnesium oxide compound with the correct ratio of atoms within each element; 1:1. Equipment: REFER TO EXPERIMENT SHEET Method: REFER TO EXPERIMENT SHEET Results: Object | Mass (g) | Crucible + Lid | 38.23 | Crucible + Lid + Magnesium | 38.57 | Crucible
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