Video: Percentage of Electrons Removed from a Positively Charged Object

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 Γ— 10⁻²⁴ g. How many electrons would be contained in a neutral copper ball with a mass of 50.0 g? A copper ball with a mass of 50.0 g has a net charge of +2.00 πœ‡C. What percent of the electrons in the copper ball have been removed?

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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?

We’ll 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 𝑁 sub 𝑒. 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 𝑁 sub 𝑒 is equal to 29 times the number of atoms, 𝑁 sub π‘Ž, the next question is what is 𝑁 sub π‘Ž. We’re told that each copper atom has an atomic mass of 63.5 unified atomic mass units. And we’re 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 π‘Žπ‘šπ‘’. And we’re told what the overall mass of our copper sample is. It’s 50.0 grams.

When we enter in this value, we see that the units of grams cancel out, as do the units of π‘Žπ‘šπ‘’. We’re 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. That’s the number of electrons contained in this copper ball.

Now let’s move on to the second part of the question. Part two of our question asks, β€œA 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?”

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, 𝑁 sub 𝑒, that have been. We’ll call that percent of electrons that have been removed capital 𝑃 sub π‘Ÿ. And we can say that 𝑃 sub π‘Ÿ is equal to the number of electrons removed β€” we’ll call that 𝑁 sub π‘’π‘Ÿ β€” divided by the total number of original electrons, 𝑁 sub 𝑒, all multiplied by 100 percent.

Since we’ve already solved for the total number of electrons, 𝑁 sub 𝑒, 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 we’ve 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 it’s equal to 1.25 times 10 to the 13th. That’s 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 we’ve solved for both the number of electrons total originally and the number of electrons removed, we’re ready to plug in and solve for 𝑃 sub π‘Ÿ, the percent of electrons that have been removed. When we enter the exact values for 𝑁 sub 𝑒 and 𝑁 sub π‘’π‘Ÿ and calculate this whole expression, we find a result of 9.09 times 10 to the negative 11th percent. That’s the percent of electrons that have been removed from the copper ball.

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