Alloy Component Calculation ``` Name: Jacey Status: student Grade: 9-12 Location: MA Country: USA Date: Summer 2013 ``` Question: Consider a piece of gold jewelry that weighs 9.30g and has a volume of 0.695cm3 . The jewelry contains only gold and silver, which have densities of 19.3 g/cm3 and 10.5 g/cm3, respectively. If the total volume of the jewelry is the sum of the volumes of the gold and silver that it contains, calculate the percentage of gold (by mass)in the jewelry. As a side note, I attempted to calculate this algebraically by letting the fraction of gold equal 'z' and the fraction of silver equal '1-z'. My equation looked something like this: z(19.3) + (1-z)(10.5) = 13.38g/1cm3. This yielded an answer of 32.7% gold in the jewelry, but the web site that my homework was off of says that the answer is wrong. I have tried this multiple times but always get the same answer. Help would be much appreciated, and it would be awesome if I could see work as well so that I can rework the problem and figure out how to do problems like this in the future. Replies: Hi, I do not want to post answers to homework questions on NEWTON, but I will try to steer you in the right direction here. Your problem is that you are assuming something different than what the question tells you to assume. You have assumed that the densities are additive, i.e., that the density due to silver plus the density due to gold adds up to the total density. However, what you are supposed to assume is that the volume due to silver plus the volume due to gold adds up to the total volume. Not the same thing... There may be easier ways to solve this, but I did it by setting up two equations and two unknowns; (1) Vg + Vs = 0.695 cm3 with Vi = mi/di (2) mg + ms = 9.30 g Since the densities (di) of gold and silver are known, we have two equations and two unknowns (mg and ms). After solving, the mass percentages are easily obtained. Good luck! best, dr. topper Hi, Jacey Thank you for your question. Although it is probably not the way that your instructor wanted you to use to solve the problem, I thought it might be helpful to try a different method. I recommend looking for different ways to figure a problem when you are stuck. I often find that what seems clear to one person may not be clear to someone else. I decided to solve the problem with a spreadsheet. I have attached both the original spreadsheet format as well as a .pdf "print". When you look at the attached results, it seems that you have calculated the percentage of gold by volume, not by mass. Please feel free to check my numbers, either with hand calculations, or your computer, or a calculator. Best regards, Robert M. Zwicker Electrical Engineer Jacey, The process you used works well if you were calculating for the ratio of *amounts* of gold to silver (something like the number of atoms of gold to that of silver). If you were to take the answer you got (.327) and then accounted for the mass of gold (196.96 - from the periodic table) and silver (107.86) you would get the mass ratio instead of the number ratio. Something like: 196.96(0.327) = 64.488 and 107.86(1-.327)=72.544 so that the total mass is: 64.488 + 72.544 = 137.033 and so the mass-percent of Au is 64.488/137.033(100) = 47.1% Alternatively - and I like this approach much better - you could solve for the mass of gold in the ring instead of the ratio of number of gold to number of silver. Something like: z g (1 cm^3 / 19.3 g) + (9.30 - z) g (1 cm^3 / 10.5 g) = 0.695 cm^3 This way, the units all match-up properly and you are actually calculating for the mass of gold in the ring (z). Note that by inverting the density unit, the mass of gold in the ring unit cancels with the mass part of the density of gold and gives just the volume of gold in the ring. Similarly, by taking z out of the mass of the ring (9.3), that gives the mass of the silver in the ring, which when multiplied by the inverse of silver density gives the volume of silver in the ring. Adding the two volumes of gold and silver in the ring gives the volume of the ring (0.695 cm^3). I like this approach much better because it takes care of units and directly gives us the mass of gold in the ring (instead of the ratio of gold to silver as in the first process). Solving for z gives 4.391g of gold, which then gives the mass-percent of gold in the ring as: 4.391/9.30(100) = 47.2%. Similar to the answer we got in the previous process. Greg (Roberto Gregorius) Canisius College Here is what I tried. Let us say Vs = volume of silver, Vg = volume of gold, Ms = mass of silver, Mg = mass of gold, V = total volume, and M = total mass. Vg + Vs = V = 0.695 cm3, Mg + Ms = M = 9.30g, Vs = V - Vg = 0.695-Vg, Ms = M - Mg = 9.30 - Ms Then, density of gold = Mg/Vg = (9.30 - Ms) / (0.695-Vs) = 19.3 g/cm3 density of silver = Ms/Vs = (9.30 - Mg) / (0.695-Vg) = 10.5 g/cm3 Thus, Mg=9.30 - 10.5 (0.695-Vg), Ms = 9.30-19.3 (0.695-Vs) We know that Vg + Vs = 0.695, so we get Vs = 0.4674 cm3 and Vg = 0.2276 cm3. Overall, Mg = 4.3923g and Ms = 4.9077g, so I got 47.2% for gold in the jewelry. Hope this works. Weonkyu Koh from Los Alamos Click here to return to the Material Science Archives

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