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18Karat Gold Problem 
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#1
Jan3105, 03:47 AM

P: 3

I just need a little help with this question..how should I finish a problem like this?
In a sample of 18karat gold, 75 percent of the total mass is pure gold, density_gold = 19.3 g/cm^3, while the rest is typically 16 percent silver density_silver = 10.5 g/cm^3 and 9 percent copper density_copper = 8.9 g/cm^3. If the density of pure gold is , while the densitites of silver and copper are respectively and , what is the overall density (density_18k) of this alloy of 18karat gold? Express your answer in grams per cubic centimeter to three significant figures. I first thought of the formula D = M/V and went from there. I got the mass of pure gold in 1 gram of alloy, which was just .75. Then I got the volume of it, which was 3.89 *10^2. But when I did mass/volume, I just got 19.3. They are asking for the overall density of the 18 karat gold sample, so I just added all three densities and my answer was wrong. So, I'm stuck b/c I think I was just going in circles for this problem. And can any of you tell me why it didn't work when I added them all up b/c it didn't make sense. 


#2
Jan3105, 05:23 AM

P: 29

100g of the alloy would contain 75g of gold (75/19.3 cc in volume), 16g of silver( 16/10.5 cc) & 9g of copper( 9/8.9 cc).Hence, the density would be equal to
100/( 75/19.3 +...) g/cc. which, I think, will be about 15g/cc. Regards, Einstone. 


#3
Jan3105, 11:40 AM

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PF Gold
P: 39,687

Adding the three densities won't work! You wind up with a density greater than the density of each part! Suppose you had pure gold. Would you say that since you can think of it as one part with density 19.3 and another with density 19.3 that the whole sample has density 38.4?? You need to use an average.
Since you don't have equal amounts, you need to use a "weighted" average: 0.75*19.3 + 0.16*10.5+ 0.09*8.9= 16.9 g/cm^2 


#4
Nov2011, 11:29 PM

P: 30

18Karat Gold Problem
Einstone is correct, it's the mass divided by volume. You take an arbitrary mass (100g, let's say), and see how much volume it would occupy (based on the given densities and percentages). You CANNOT just do a "weighted" average. This would mean you mixed three metals of equal volumes, and then multiplied each of their densities by their respective "proportions."
100g/(75/19.3+...) =/= (75*0.19.3+...) The left side is a density, the right side doesn't give you anything useful. Hollumber 


#5
Nov2011, 11:39 PM

HW Helper
P: 2,318

Now you need the volume of .16 g of Silver [ 16% of 1g is silver] then the volume of 0.9 g of copper. That will give you the total volume of the 1 g. You can then do M/V to find the final density. NOTE: that always assumes that when these volumes are added together, they simply accumulate. For example if you put 50ml of water with 50ml of methylated spirits, the final volume is LESS than 100ml. Hopefully the same thing doesn't happen with gold, silver and copper 


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