Question

Suppose a cube of aluminum which is 1.00 cm on a side accumulates a net charge of +3.00 pC.
What is the percentage of the electrons originally in the cube was removed?

Here is a list of fun facts I collected:
Atomic weight of Aluminum
26.98 g/ mol

Atoms per mole
\[6.022\times 10^{23}\]

Density
\[2.70g\cdot cm^3\]

Electric charge
\[-1.60\times 10^{-19}\;C\]
Mass of an electron
\[9.11\times 10^{-31}\;kg\]

Answer 1

\[1\; cm^3 \cdot \frac{2.70g}{cm^3}\cdot \frac{mol}{26.98\;g}\cdot \frac{6.022\times10^{23}\;atoms}{mol}\cdot \frac{13\; electrons}{atom}\cdot \frac{1.60\times10^{-19}C}{electron}\]

Answer 2

\[\small{1\; \cancel{cm^3} \cdot \frac{2.70\cancel{g}}{\cancel{cm^3}}\cdot \frac{\cancel{mol}}{26.98\;\cancel{g}}\cdot \frac{6.022\times10^{23}\;\cancel{atoms}}{\cancel{mol}}\cdot \frac{13\; \cancel{electrons}}{\cancel{atom}}\cdot \frac{1.60\times10^{-19}C}{\cancel{electron}}}\]

Answer 3

125350.4522 C

Answer 4

@phi

Answer 5

\[3pC\cdot \frac{C}{1\times10^{-12}\; pC}\cdot \frac{electron}{1.60\times10^{-19}\;C}\]

Answer 6

\[3pC\cdot \frac{C}{1\times10^{-12}\; pC}\cdot \frac{electron}{1.60\times10^{-19}\;C}=1.875\times10^{31}\; electrons\]
This must be the number of electrons that were removed...

Answer 7

(Electrons originally-electrons removed)/electrons originally *100
\[\frac{(7.8344E23)-(1.875E31)}{7.8344E23}*100=1.87E33\]
What am I doing wrong?

Answer 8

@zepdrix

Answer 9

the example attached used 2.5pC instead of 3...but it's the same idea

Answer 10

@zzr0ck3r

Answer 11

@ghazi

Answer 12

I got it.

\[\frac{accumulated\;charge\cdot Atomic\;weight}{charge\;e \cdot\left(\frac{g\; Al \;in \;1\;cm^3}{mol\; Al \; in\;1\;cm^3}\right)}\cdot 100= \]