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Mass of Copper Oxide: 1.0495g Mass of filter paper: 0.6698g Mass of empty watch glass: 43.2063g Mass of watch glass, filter paper, and solid: 44.7007g Mass of solid copper:_____ Mass of oxygen in the copper oxide:____ A) Determine the empirical formula of the copper oxide (Show all work): B) Determine the mass percent of copper and oxygen in your sample( show all your work):

Mass of Copper Oxide: 1.0495g Mass of filter paper: 0.6698g Mass of empty watch glass: 43.2063g Mass of watch glass, filter paper, and solid: 44.7007g Mass of solid copper:_____ Mass of oxygen in the copper oxide:____ A) Determine the empirical formula of the copper oxide (Show all work): B) Determine the mass percent of copper and oxygen in your sample( show all your work):

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Mass of Copper Oxide: 1.0495g Mass of filter paper: 0.6698g Mass of empty watch glass: 43.2063g Mass of watch glass, filter paper, and solid: 44.7007g Mass of solid copper:_____ Mass of oxygen in the copper oxide:____ A) Determine the empirical formula of the copper oxide (Show all work): B) Determine the mass percent of copper and oxygen in your sample( show all your work):
LABORATORY 14 DETERMINATION OF AN EMPIRICAL FORMULA OF A COPPER OXIDE INTRODUCTION An empirical formula provides the lowest whole number ratio of the atoms found in a given molecule. For example hexene has the molecular formula C,H, the lowest whole number ratio of C to H is 1 to 2, and so the empirical formula is CH,. In ionic compounds this ratio is driven by the ionic charge on the cation and anion. For example calcium oxide has an empirical formula of CaO since the calcium ion is Ca*2 and the oxide ion is O², thus requiring a 1:1 ratio to balance the charges. Transition metals that can have more than one ionization state can have multiple ratios leading to several possible empirical formulas. Iron oxide can be made with either iron (II) or iron (III) resulting in either FeO or Fe,O. In this lab we will determine the empirical formula of copper oxide. The method used here requires a strong acid to dissolve the copper oxide. The resulting aqueous copper ion is then reduced by magnesium metal to produce a solid copper product which can be separated and weighed. This reduction is shown in Equation 14.1 for both copper (I) and copper (II). Cu*20) + Mgo → Mg", + Cu (aq) (s) or 2Cu*(a0) + Mg) → Mg* + 2Cu (Equation 14.1) Due to the color of the copper ions, it is possible to determine when the reaction is complete. Copper ions are a blue-green color. When all of the color is gone from the solution all of the copper ions have been reduced to copper solid. This reaction takes place in an environment with excess HCl, which also serves to react with any excess magnesium so that the only solid remaining during filtration is the copper. This reaction is shown in Equation 14.2. 2H*a0) + Mg → Mg²“ (aq) + H2 () (Equation 14.2) 141 Laboratory 14 Determination of an Empirical Formula of a Copper Oxide
Transcribed Image Text:LABORATORY 14 DETERMINATION OF AN EMPIRICAL FORMULA OF A COPPER OXIDE INTRODUCTION An empirical formula provides the lowest whole number ratio of the atoms found in a given molecule. For example hexene has the molecular formula C,H, the lowest whole number ratio of C to H is 1 to 2, and so the empirical formula is CH,. In ionic compounds this ratio is driven by the ionic charge on the cation and anion. For example calcium oxide has an empirical formula of CaO since the calcium ion is Ca*2 and the oxide ion is O², thus requiring a 1:1 ratio to balance the charges. Transition metals that can have more than one ionization state can have multiple ratios leading to several possible empirical formulas. Iron oxide can be made with either iron (II) or iron (III) resulting in either FeO or Fe,O. In this lab we will determine the empirical formula of copper oxide. The method used here requires a strong acid to dissolve the copper oxide. The resulting aqueous copper ion is then reduced by magnesium metal to produce a solid copper product which can be separated and weighed. This reduction is shown in Equation 14.1 for both copper (I) and copper (II). Cu*20) + Mgo → Mg", + Cu (aq) (s) or 2Cu*(a0) + Mg) → Mg* + 2Cu (Equation 14.1) Due to the color of the copper ions, it is possible to determine when the reaction is complete. Copper ions are a blue-green color. When all of the color is gone from the solution all of the copper ions have been reduced to copper solid. This reaction takes place in an environment with excess HCl, which also serves to react with any excess magnesium so that the only solid remaining during filtration is the copper. This reaction is shown in Equation 14.2. 2H*a0) + Mg → Mg²“ (aq) + H2 () (Equation 14.2) 141 Laboratory 14 Determination of an Empirical Formula of a Copper Oxide
The solid copper can then be collected by filtration, dried and weighed. To determine the empirical formula you need to know the mass of each element present. Since you were dealing with a copper oxide, the total mass of the sample comes from the mass of copper and the mass of oxygen, seen in Equation 14.3. Since you know the mass of the copper oxide and the copper, you can solve for the mass of oxygen that was present. (Equation 14.3) Mass of copper oxide = Mass of copper + Mass of oxygen Since the empirical formula is the lowest whole number ratio of moles of copper and oxygen, one can use the grams of copper and oxygen to determine the moles of copper and oxygen. These values can then be used to find the lowest ratio. EXAMPLE: Calculating the empirical formula of an Iron Oxide A 2.0 g sample of an iron oxide was determined to have 1.504 g of iron and 0.496 g of oxygen. 1 mole Fe 55.85g Fe 1.504 g Fe ( ) = 0.0269 mol Fe 1 mole Fe 16.00g Fe O = 0.031 mol O THE 0.496 g O ( Then divide all of the mole quantities by the lowest amount, in this case the 0.0269 moles. 0.0269 mol / 0.0269 mol = 1 mol Fe 0.0310 mol / 0.0269 mol = 1.15 mol O ***this most likely rounds down to 1 mol*** So there is a 1 to 1 mole ratio and the empirical formula is FeO 142 Determination of an Empirical Formula pper de Laboratory 14
Transcribed Image Text:The solid copper can then be collected by filtration, dried and weighed. To determine the empirical formula you need to know the mass of each element present. Since you were dealing with a copper oxide, the total mass of the sample comes from the mass of copper and the mass of oxygen, seen in Equation 14.3. Since you know the mass of the copper oxide and the copper, you can solve for the mass of oxygen that was present. (Equation 14.3) Mass of copper oxide = Mass of copper + Mass of oxygen Since the empirical formula is the lowest whole number ratio of moles of copper and oxygen, one can use the grams of copper and oxygen to determine the moles of copper and oxygen. These values can then be used to find the lowest ratio. EXAMPLE: Calculating the empirical formula of an Iron Oxide A 2.0 g sample of an iron oxide was determined to have 1.504 g of iron and 0.496 g of oxygen. 1 mole Fe 55.85g Fe 1.504 g Fe ( ) = 0.0269 mol Fe 1 mole Fe 16.00g Fe O = 0.031 mol O THE 0.496 g O ( Then divide all of the mole quantities by the lowest amount, in this case the 0.0269 moles. 0.0269 mol / 0.0269 mol = 1 mol Fe 0.0310 mol / 0.0269 mol = 1.15 mol O ***this most likely rounds down to 1 mol*** So there is a 1 to 1 mole ratio and the empirical formula is FeO 142 Determination of an Empirical Formula pper de Laboratory 14
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