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alloys / moles

You need to cast an alloy of titanium (Ti) containing 46 at.% aluminium (Al). Using the information from your periodic table, estimate how much Ti and Al you will need to put in the crucible in order to cast a 100 g ingot of this alloy. b) By mistake, you added to the crucible 32.4 g of Ti and 67.6 g of Al. Work out the composition of your ingot in at.%.

(a) Percentage composition of the two elements used:    Al = 46%        Ti = (100 - 46)% = 54% Percentage composition by mass:Al = (46% x 100)g = 46g   Ti = 54gMoles of Aluminium to be used = Mass/R.A.M                                                  = 46/27                                                  = 1.7037molMoles of Titanium to be used:Mass = (100-46) =54gR.A.M= 204Moles of Ti = 54/204                  = 0.2647mol (b) Percentage composition of the elements in the crucible:% of Titanium, If 54g = 54%therefore, 32.4g = (32.4 x 0.54) / 54                           = 0.324                           =32.4%% of Ti in the crucible = (100 - 32.4)%                                    = 67.6%That's it!!NOTE: Kindly post more questions on the same if you still have difficulty. I will be readily available to assist. Furthermore, you can hire me for tutoring so as to take you through the topic deeply into details in a very simple way of understanding the concept. Thank you.
22 July 2016
Hi Gemma,Answer to a:Moles = Mass/ RAMAssuming a percentage composition, if Al = 46%, Ti = 100-46 = 54%Therefore you have 46g of Al and 54g of Ti.Mass = 46g.RAM = 27 (found in your periodic table)Therefore, Moles of Al = Mass/RAM = 46/27 = 1.70mol.Moles of Ti:Mass of Ti = 54gRAM of Ti = 204 (from  your periodic table).Therefore, Moles = Mass/RAM = 54/204 = 0.26molAnswer to b:Ti makes up 54g of the total 100g, which equals to 54%. Therefore 32.4g = (32.4*0.54)/54 = 0.324 = 32.4% of TiTitanium = 100-32.4 = 67.6% TiIf you need any more help please don't hesitate to contact me and we can arrange a skype tutoring session. I am a female engineer and I hope that I can help you realise the doors that STEM subjects can open for you! 
27 July 2016
Hi Gemma1997, Interesting problem - interesting solutions posted too... A version that may be of interest to you follows with a few key points highlighted at the beginning as to why using percentage by mass would not get to the correct solution where an atom percentage was required.Firstly, the relative atomic mass of Ti - 47.9 not to be confused with thallium (Tl = 204.4)A key part to the solution of your problem is the appreciation that the at. % you require is not the percentage by mass which has been used in the solutions posted but the correct term is: atomic percent (or at.%), which gives the percentage of one kind of atom relative to the total number of atoms.For part a. 46% of the atoms must be aluminium, 54% of the atoms must be titanium.Using the atomic mass unit scale;46 Al atoms = 46 x 27 = 124254 Ti atoms =  54 x 47.9 = 2586.6So the 100 atoms would have a total mass on the atomic mass unit scale of 3828.6.You will find this term used in the following solution to your question: Part a. Relative atomic masses: Ti =  47.9         Al = 27   Atom Percentage at. %           54.0               46.0   Relative Mass of atoms in alloy (atomic units) Ti = 54/100 x 47.9  = 25.87Al = 46/100 x 27    = 12.42Total  = 38.29 Percentage Ti:Al by mass Ti = (25.87/38.29) x 100 Al = (12.42/38.29) x 100    Mass required to make 100g ingot Ti = 67.6g   Al = 32.4 gPart b.Relative atomic mass: Ti = 47.9                          Al = 27           Mass (g)                      Ti = 32.4                          Al = 67.6         moles                              = 32.4/47.9                       = 67.6/27moles                              = 0.676                             = 2.50 Percentage  at.           Ti = (0.676/3.18) x 100     Al = (2.50/3.18) x 100                  Percentage by moles (by atoms)                                   Ti = 21.3 %                           78.7 %Naturally, if you need any online help please let me know.Kind regards,Alastair   
Alastair B.
01 August 2016
An interesting expression of quantity.46 at.%Can I ask which exam board you are studying?
09 September 2016
Hi GemmaI advice a very simple approach, as Chemistry calculations are strongly 'maths-based' (therefore, very easy with the right method) Two variables were asked to be found (the masses of Ti and Al in the alloy), then you need just 2 equations to solve them, including the available informations, as follows :1.The total ingot mass is the sum of both single Ti and Al masses. This is the first relation2.Express the masses in terms of moles (usng atomic weights)so : 100g = n(Ti)*AW(Ti) + n(Al)*AW(Al)3. the second information/relation is the alloy composition. As you have to express one equation in 1 variable only, let's relate the compositions with their ratio :n(Ti) / n(Al) = 0,54/0,46 = 1,17so you can express , as instance, n(Ti) = 1,17 *n(Al) and replace it in the former equation.4. at the end you'll find the moles of Al = 1,20 and of Ti = 1,41; using the atomic weight you'll get tje desired masses (resp. 32,4 g and 67,6 g)5. In the second part of the problem, the masses are given and the molar % is asked. Then transform the masses in moles (using AW) and calculate the % as ratio of the single mole with the total moles sum :nTOT = n(Al) + n(Ti)nAl% = n(Al)/ nTOT          nTi% = n(Ti)/ nTOTPlease mid that with my trainings you'll learn how to face a problem, before to solve it ;)     Yours, Doc71
02 November 2016
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