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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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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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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Sean Dowling Julia Phaltankar Mrs. Oakes Chemistry w/ Algebra 10/G February 18‚ 2015 Determining Empirical Formula Lab Introduction: One can find an empirical formula by taking a sample of a compound and dividing the number of moles of one element in the compound by the number of moles of another element in the compound to form a small wholenumber formula. For example‚ in a sample of a made up compound of oxygen and lead‚ one mole of lead has a molar mass of 207.2 g/mole‚ and oxygen
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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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Lab of Determining an Empirical Formula B2 Honors 12/18/13 Abstract In this lab‚ to help better understand the concepts of gram atomic masses and empirical formulas‚ we found the gram atomic masses and empirical formula of a binary compound. The two compounds should form a definite whole number ratio by mass. This ratio will also help determine the subscripts of the elements in the empirical formula. Errors in this experiment can stem from measuring wrong or doing equations incorrectly.
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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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A) January 18‚ 2011B) Empirical Formula C) The purpose is to determine the empirical formula of a metallic oxide. D) Pre Lab Questions: After heating the metal‚ the crucible and contents should mass less than it did before it was heated. This is because heating the crucible may rid of other residue that was left in it; bringing it a to a constant mass. A yellow flame will deposit soot on the crucible. This would be a problem because the soot left on the crucible would vary from our constant
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Pre-Lab Discussion An empirical formula is a formula for a chemical compound found by direct laboratory examination. Laboratory procedures allow the chemist to find the simplest whole number ratio of elements within the compound. In order to find the true molecular formula‚ the chemist also needs to know the compound’s molecular mass. The general procedure is to use laboratory techniques to determine the mass of each element in the compound. In this lab‚ we will react a known quantity of magnesium
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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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