Fraction of Electrons Removed

[mysticbms]

Homework Statement

A 50.0 g ball of copper has a net charge of 2.00µC. What fraction of the copper’s electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.)

Homework Equations

1.00 C× (1proton)/(1.60×10^−19C)=6.25×10^18 protons
C = coloumb, the SI unit of charge

The Attempt at a Solution

I don't know where to begin. I haven't taken Chemistry in a long time so I'm pretty stumped.

 

[SammyS]

Homework Statement

A 50.0 g ball of copper has a net charge of 2.00µC. What fraction of the copper’s electrons has been removed? (Each copper atom has 29 protons, and copper has an atomic mass of 63.5.)

Homework Equations

1.00 C× (1proton)/(1.60×10^−19C)=6.25×10^18 protons
C = coloumb, the SI unit of charge

The Attempt at a Solution

I don't know where to begin. I haven't taken Chemistry in a long time so I'm pretty stumped.

How many copper atoms are there in 63.5 grams of copper?

 

[gneill]

The atomic mass tells you the number of grams making up a mole of a substance. A mole is a specific number of particles (look up Avogadro's Number). So start by determining how many atoms comprise the 50.0 g ball of copper.EDIT: Beaten to the punch by SammyS!

 

[mysticbms]

Got it thanks.

 

[Nickie]

I would approach this problem by first converting the given charge of 2.00µC into coulombs, the SI unit of charge. This can be done by using the conversion factor provided in the homework equations, where 1.00 C is equal to 6.25×10^18 protons. This gives us a charge of (2.00×10^-6 C) × (6.25×10^18 protons/1 C) = 1.25×10^13 protons.

Next, I would calculate the number of electrons present in the 50.0 g ball of copper by using the atomic mass of copper (63.5 g/mol) and Avogadro's number (6.02×10^23 particles/mol). This gives us (50.0 g)/(63.5 g/mol) × (6.02×10^23 particles/mol) = 4.76×10^23 electrons.

Finally, I would calculate the fraction of electrons removed by dividing the number of protons removed (1.25×10^13 protons) by the total number of electrons in the copper ball (4.76×10^23 electrons). This gives us a fraction of 2.63×10^-11, meaning that only a very small fraction (0.0000000000263%) of the copper's electrons have been removed.

Overall, this problem requires knowledge of basic chemistry concepts such as atomic mass and Avogadro's number, as well as the ability to convert between units and use conversion factors.