### Video Transcript

A copper atom has 29 protons and a
neutral ball of copper has one electron per proton. Copper has an atomic mass of 63.5
unified atomic mass units. A unified atomic mass unit is
equivalent to 1.66 times 10 to the negative 24th grams. How many electrons would be
contained in a neutral copper ball with a mass of 50.0 grams? A copper ball with a mass of 50.0
grams has a net charge of positive 2.00 microcoulombs. What percent of the electrons in
the copper ball have been removed?

Weβll start off on this first
question solving for the number of electrons that are contained in a copper ball of
this given mass. We can call this number of
electrons π sub π. And that will be equal to the
number of copper atoms multiplied by 29, since there are 29 electrons as well as
protons per copper atom.

So if π sub π is equal to 29
times the number of atoms, π sub π, the next question is what is π sub π. Weβre told that each copper atom
has an atomic mass of 63.5 unified atomic mass units. And weβre given the mass of an
atomic mass unit in grams. So we can say that the number of
copper atoms total is equal to the overall mass of our sample divided by the number
of atomic mass units per atom multiplied by the mass of each πππ’. And weβre told what the overall
mass of our copper sample is. Itβs 50.0 grams.

When we enter in this value, we see
that the units of grams cancel out, as do the units of πππ’. Weβre left with a unitless number,
which is good because we want to solve for a number of electrons. When we enter this expression on
our calculator, to three significant figures, we find a result of 1.38 times 10 to
the 25th. Thatβs the number of electrons
contained in this copper ball.

Now letβs move on to the second
part of the question. Part two of our question asks, βA
copper ball with a mass of 50.0 grams has a net charge of positive 2.00
microcoulombs. What percent of the electrons in
the copper ball have been removed?β

In part one, we solved for the
number of electrons in a neutral copper ball with the same mass, 50.0 grams. Now we imagine a copper ball with
that mass, but a net charge of positive 2.00 microcoulombs.

If this is the same ball of copper
from part one, that means that some electrons have been removed. And we want to solve for the
percent of the total number, π sub π, that have been. Weβll call that percent of
electrons that have been removed capital π sub π. And we can say that π sub π is
equal to the number of electrons removed β weβll call that π sub ππ β divided by
the total number of original electrons, π sub π, all multiplied by 100
percent.

Since weβve already solved for the
total number of electrons, π sub π, in part one, that means all we need to do is
solve for the number of electrons that have been removed in order to create this
positive net charge.

We can say that the number of
electrons removed is equal to the net charge on this copper ball divided by the
magnitude of the charge of a single electron. We can recall that the charge on a
single electron is equal to negative 1.60 times 10 to the negative 19th
coulombs. So we plug that value in for our
charge on an electron and then take the magnitude of it, turning it into a positive
number.

Before we calculate this fraction
to solve for the number of electrons removed, notice weβve converted the net charge
on this copper ball, 2.00 microcoulombs, to its expression in scientific notation,
2.00 times 10 to the negative sixth coulombs. With all that said, we then
calculate this fraction and we find itβs equal to 1.25 times 10 to the 13th. Thatβs the number of electrons that
had to be removed from the copper ball in order to create a net charge of positive
2.00 microcoulombs.

Now that weβve solved for both the
number of electrons total originally and the number of electrons removed, weβre
ready to plug in and solve for π sub π, the percent of electrons that have been
removed. When we enter the exact values for
π sub π and π sub ππ and calculate this whole expression, we find a result of
9.09 times 10 to the negative 11th percent. Thatβs the percent of electrons
that have been removed from the copper ball.