# Homework Help: Charge on a gold bar

1. Jan 16, 2005

### Purduenub03

I'm sure im missing something simple here but...

Suppose a 1.72 g nugget of pure gold has zero net charge. What would be its net charge after it has 1% of its electrons removed?

So i know that an electron is 9.109 x 10^-31 kg, and has a charge of -1.602 x 10^-19C. However, how do i know the split of electrons and protons in the gold bar?

I'd assume im trying to get to mass of electrons / mass of electron, that times the charge of an electron divided by 100???

2. Jan 16, 2005

### Gokul43201

Staff Emeritus
How many electrons in a Gold atom ? How many atoms in 1.72 g ?

3. Jan 16, 2005

### Purduenub03

79 electrons in a gold atom

1.72/196.96654 = .008732 atoms

so .008732 * .79 * 1.602 x 10 ^-19 ?

4. Jan 16, 2005

### dextercioby

Okay,you're on the wrong track...

197g gold----------------------------->6.023 10^{23} atoms

1.72g gold----------------------------> x atoms

1 atom gold---------------------------> 79 electrons
x atoms gold---------------------------> y electrons

What do you get for "y"??

Daniel.

5. Jan 16, 2005

### Purduenub03

(1.71 / 197) * 6.023 X 10 ^23 * 79 = 4.154 x 10 ^ 23

AH!

and that divided by 100 x the charge is the answer

thank you

Ryan

Last edited: Jan 16, 2005
6. Jan 16, 2005

### Gokul43201

Staff Emeritus
Not so fast !

You've only found the number of electrons removed. Multiply this number by the charge on an electron to get the net positive charge.