# The weight of gold in a crown

1. May 15, 2012

### techgeek17

1. The problem statement, all variables and given/known data
A jeweler has made a crown of an alloy of gold and copper. The crown weighs 0.358 N when measured in air and 0.338 N when submerged completely in water. The density of water is 1.00x10^3 kg/m^3, the density of copper is 8.92X10^3 kg/m^3, and the density of gold is 1.93x10^4 kg/m^3. Calculate the weight of the gold in the crown.

2. Relevant equations
Unknown. None are given and can't find any in my textbook.

3. The attempt at a solution
None.

I can't seem to find any information about how exactly to do this problem in my textbook or online. Help please?

2. May 15, 2012

### Staff: Mentor

If you can't find a specific example or a recipe for how to do the problem, you may have to resort to devising your own method

Why not start by determining the volume of the crown and its density?

3. May 16, 2012

### techgeek17

OK, I attempted to find the volume and density. I got:
mass(in air) as 36.5g
appearant mass(in water) as 34.5g
volume = 2.1x10^6 L(?)
density = 1.83 x 10^4 kg/l

I am assuming now that I need to find a percentage or something of the sort for the gold in the crown.

4. May 16, 2012

### Staff: Mentor

Your mass looks fine, although "apparent mass in water" doesn't make physical sense; stick with the apparent weight.

Your volume looks a bit odd; that's a LOT of liters! How did you arrive at it? Can you demonstrate your calculation?

5. May 16, 2012

### techgeek17

Whoops! I didn't catch that before I posted. The volume I calculated is supposed to be 2.0x10^-6 m^3 (got confused with the unit), not 10^6. and I wrote the apparent mass title based off the equation I used:

m(in air) - m(apparent/in water) = d(water)*v , solving for v

also density unit should be kg/m^3

Last edited: May 16, 2012
6. May 17, 2012

### Staff: Mentor

Okay. That clears up a few things

To how many significant figures should your results be reported?

7. May 17, 2012

### techgeek17

I believe I need to use 2 significant digits. So that would make my density 1.8x10^4, right?

Also, I did some calculations and managed to get what looks like a possibly correct answer. Here's what I did:

Formula: D(copper)*(c-1)+D(gold)*c=D(crown), where c is the ratio of gold to crown
Solved for c - c=[D(crown)+D(copper)]/[D(copper)+D(gold)]
Substituted values - c=[1.8*10^4+8.92*10^3]/[8.92*10^3+1.93*10^4]
And solved - c=.965

then i took weight of crown times c and got 3.52*10^-2 kg or .345 N

I am unsure is this is right or not though, since the amount of gold is so high, but it is my best bet at the moment.

8. May 17, 2012

### Staff: Mentor

How many significant figures are in the numbers that are given in the problem statement?
Good idea, and almost correct

The fractions of the whole should sum to the whole, so your terms c and c-1 should add up to 1. That is, they should be c and 1-c.

9. May 17, 2012

### techgeek17

Significant Digits = 3

And the formula should be D(copper)*(1-c)+D(gold)*c=D(crown), right?

If so, I got 3.52*10^-2 kg or .345 N

How am I doing?

10. May 17, 2012

### Staff: Mentor

Your approach looks fine now. I'm a bit concerned with the final value though, it's in the right ballpark but seems a tad high to me. Could it be rounding or truncation error creeping in from intermediate values? Did you keep sufficient extra digits of precision in your intermediate steps?