0.482% of the electrons are removed from a sphere of iron (Z=26,A=56),resulting in a net charge of 1.00 C on the sphere.What is the mass of the iron sphere?
A)4.64 mg
B)10.0 mg
C)3.19 mg
D)3.99 mg
E)6.20 mg

Question

Quizplus · Accepted Answer

The net charge of 1.00 C implies a certain number of electrons have been removed. Since each electron has a charge of $1.6 \times 10^{-19}$ C, the number of electrons removed ($N_e$) can be calculated as $N_e = \frac{1.00\,C}{1.6 \times 10^{-19}\,C/e^-} = 6.25 \times 10^{18}$ electrons. Given that 0.482% of the electrons are removed, the total number of electrons originally present in the sphere ($N_{total}$) is $N_{total} = \frac{N_e}{0.00482} = 1.297 \times 10^{21}$ electrons. Since iron (Fe) has an atomic number (Z) of 26, it has 26 electrons per atom in its neutral state. Thus, the number of iron atoms in the sphere is $N_{atoms} = \frac{N_{total}}{Z} = \frac{1.297 \times 10^{21}}{26} = 4.989 \times 10^{19}$ atoms. The molar mass of iron is approximately 56 g/mol, and Avogadro's number is $6.022 \times 10^{23}$ atoms/mol, so the mass of the iron sphere can be calculated using the formula $mass = \frac{N_{atoms} \times Molar\,Mass}{Avogadro's\,Number}$. Substituting the values, we get $mass = \frac{4.989 \times 10^{19} \times 56}{6.022 \times 10^{23}} = 4.64 \times 10^{-3}$ g, or 4.64 mg.

0482% of the Electrons Are Removed from a Sphere of Iron