# Answer to Question #57180 in General Chemistry for julie apedo

Question #57180
1.The number of copper ions in 3 lumps of copper is 2.8*10^22. How many electrons are present
in the sample.How much do they contribute to their mass.
(Cu=29, mass of Electron=9.11*10^-28g)
1
Expert's answer
2016-01-04T04:45:22-0500
1. If we really assume that lumps of copper consist of ions, than:
a) Total mass of copper in 3 pieces:
m = Number of ions * Mass of 1 ion = 2.8*10^22 * 105.49*10(-23) = 29.54 g
b) There are different ions of copper - I, II, III, IV. Cu(II) is the most wide-spread, so we may suppose that those lumps of copper consist of this ion. Every Cu(II) ion have 29-2=27 electrons.
c) Total number of electrons:
m(e) = 27 * N(ions) = 27 * 2.8*10^22 = 75.6*10^22 g
d) Total mass of electrons:
m(e) = N(e) * M(e) = 75.6*10^22 * 9.11*10^(-28) = 0.69*10^(-3) g
e) The contribution of electrons to the total mass:
% = m(e) / m = 0.69*10^(-3) / 29.54 = 2.33*10^(-5) = 0.00233%

2. If we assume that lumps of copper consist of atoms, than:
a) Total mass of copper in 3 pieces:
m = Number of ions * Mass of 1 ion = 2.8*10^22 * 105.49*10(-23) = 29.54 g
b) Total number of electrons:
m(e) = 29 * N(ions) = 29 * 2.8*10^22 = 81.2*10^22 g
c) Total mass of electrons:
m(e) = N(e) * M(e) = 81.2*10^22 * 9.11*10^(-28) = 0.74*10^(-3) g
d) The contribution of electrons to the total mass:
% = m(e) / m = 0.74*10^(-3) / 29.54 = 2.5*10^(-5) = 0.0025 %

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