Okay so i am doing an experiment called the Analysis of Copper in Brass and i am completely lost on my post lab. i need to calculate (a) Cu concentraion (b) mass of copper in Brass (c) percent copper in Brass (d) the average percent Cu in brass.Here's my data.I have 0.712g of Brass with an absorption of 1.4 and I have 0.723g of Brass with an absorption of 1.1.I have an equation C= (A-b)/m and i've been playing around with it but i'm really not confident i'm getting the right Cu concentration.... Any help would be appreciated. Easy best answer :))
OK, assumptions first - you dissolved the brass in nitric acid (or similar), you diluted it down and added some kind of color-enhancing reagent - possibly just ammonia; you measured the blue color of the solution by UV-vis. 0.712 g of brass gave you a solution whose absorbance was 1.4 (etc.). (This also works if you burned the acid solution in a flame AA experiment.) Your equation relates the concentration to the absorbance. The problem is, you don't know how much concentration gives you a certain amount of absorbance. You probably have additional data: the absorbance of a couple copper solutions whose concentrations you know. (Might be in your notes, or the manual, under standards.) If you plot the absorbance of these solutions (NOT your brass solutions!) as Y, and the concentration as X, you should get a fairly straight line. A spreadsheet will tell you the equation for that line: A = m * C + b. Now you rearrange that to solve for C (getting the form in the problem); the m and b values will give you the concentration from the absorption. That's the concentration of copper in the solution. To figure out how much copper is in the brass, take the concentration of copper and the total volume of solution you made, to get the total amount of copper in the brass. (Could be grams or moles, depending on the concentration units. You want it in grams, so convert if needed.) Divide by the sample mass to get the percent copper. Example: Slope m = 0.56, b = 0.031. C = (1.4 - 0.031) / 0.56 = 2.44 g/L Cu, in 0.050 L solution; this is 0.122 g of copper in 0.712 g brass, so 17.1% copper. (Which would be really low for most brasses.)
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A very simple experiment is to obtain wires of the same thickness (or thin bars of the same thickness) of the metals you are investigating. put them on a tripod radiating out from cental point. At a fixed length say 12 cm on each rod attach a matchstick (or cocktail stick) with a small piece of candle wax. Heat strongly with roaring bunsen, and time for each stick to fall. Independant metal used dependant time for sticl to fall controlled Heat, amount of wax used GL
