First, we need to determine the number of moles of electrons transferred during the coulometric titration process. The relationship between charge coulombs and moles of electrons can be described by Faraday's constant F , which is approximately 96,485 C/mol.Given that 10.0 coulombs of electricity were passed through the sample, we can calculate the moles of electrons n as follows:n = charge coulombs / Faraday's constant C/mol n = 10.0 C / 96,485 C/moln 0.0001036 mol of electronsNow, we need to determine the stoichiometry of the reaction between copper and electrons. The balanced half-reaction for the reduction of copper ions to copper metal is:Cu + 2e CuFrom the stoichiometry of the reaction, we can see that 2 moles of electrons are required to reduce 1 mole of copper ions to copper metal. Therefore, we can calculate the moles of copper Cu reduced in the sample:moles of Cu = moles of electrons / 2moles of Cu 0.0001036 mol / 2moles of Cu 0.0000518 molNow, we can use the mass of the sample 0.250 g and the atomic weight of copper 63.55 g/mol to calculate the moles of copper present in the sample:moles of Cu = mass of Cu / atomic weight of Cumoles of Cu = 0.250 g / 63.55 g/molmoles of Cu 0.003934 molFinally, we can compare the moles of copper reduced 0.0000518 mol to the total moles of copper in the sample 0.003934 mol to determine the proportion of copper present:Proportion of copper = moles of Cu reduced / total moles of Cu Proportion of copper 0.0000518 mol / 0.003934 molProportion of copper 0.0132Thus, approximately 1.32% of the copper in the sample was present when 10.0 coulombs of electricity were passed through it during the coulometric titration process.