Calculating the Weight of Gold in a Crown: A Density Problem

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SUMMARY

The forum discussion centers on calculating the weight of gold in a crown made of an alloy of gold and copper. The crown weighs 0.358 N in air and 0.338 N when submerged in water. Participants utilized the formula D(copper)*(1-c)+D(gold)*c=D(crown) to determine the ratio of gold in the crown, ultimately arriving at an estimated weight of gold as 0.345 N. The discussion emphasizes the importance of significant figures and precision in calculations, particularly when determining the density and volume of the crown.

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Homework Statement


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.

Homework Equations


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

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?
 
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techgeek17 said:

Homework Statement


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.


Homework Equations


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


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?

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 :smile:

Why not start by determining the volume of the crown and its density?
 
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.
 
techgeek17 said:
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.

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?
 
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:
Okay. That clears up a few things :smile:

To how many significant figures should your results be reported?
 
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.
 
techgeek17 said:
I believe I need to use 2 significant digits. So that would make my density 1.8x10^4, right?

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

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.
 
gneill said:
How many significant figures are in the numbers that are given in the problem statement?

Good idea, and almost correct :smile:

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.

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
techgeek17 said:
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?

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?
 

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