PERCENTAGE COMPOSITION WORKSHEET 1. Calculate the COMPLETE percentage composition (by mass) of the following: (a) iron (III) oxide (Fe = 70.0%‚ O = 30.0%) (b) barium phosphate (Ba = 68.4%‚ P = 10.3%‚ O= 21.3%) 2. What is the percentage of sodium (by mass) in sodium phosphate? (42.1%) 3. For the hydrate sodium sulfate decahydrate‚ calculate the following: (a) the percent of sodium (by mass) in the hydrate (14.3%) (b) percent of TOTAL oxygen (by mass) in
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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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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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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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Finding the formula of magnesium oxide Aims When magnesium is heated in air‚ it reacts with oxygen. During this oxidation reaction‚ magnesium oxide is produced. This increases the mass. If we know the mass of magnesium at the start‚ and the mass of magnesium oxide produced at the end‚ we can work out the mass of oxygen which has been combined with the magnesium. We can use these masses to work out the formula of magnesium oxide. Eye protection must be worn Apparatus Crucible and lid
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Percent Copper and Formula of a Copper Compound Sofia Economides Date Preformed: 3/19/2012 Date Submitted: 3/26/2012 Lab Partners: Anita Smith‚ Taylor Bell‚ Derek Bourgeois 1. Purpose a. To determine the percent of copper and the formula weight of a copper compound. 2. Introduction a. Magnesium is a very active metal which can reduce copper (II) ions in a weight sample of a compound to metallic copper. The reduction of copper ions in a compound to a copper metal
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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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CHEMISTRY EXPERIMENTAL REPORT SALLY LI INTRODUCTION: In chemistry‚ compounds can be distinguished by using the empirical formula. The formula provides the simplest positive integer ratio of elements in a compound. The empirical formula is largely useful in determining the ratio of elements within ionic compounds where the structure is of a non-directional nature of bonding where any ion at any time could be surrounded by
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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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