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anonymous
 4 years ago
Suppose there are two known compounds containing generic X and Y. You have a 1.00g sample of each compound. One sample contains .34g of X and the other contains .44g of X. Identify plausible sets of formulas for the two compounds
anonymous
 4 years ago
Suppose there are two known compounds containing generic X and Y. You have a 1.00g sample of each compound. One sample contains .34g of X and the other contains .44g of X. Identify plausible sets of formulas for the two compounds

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I've started it by taking 1.00g.34=.66g Y (Sample 1) and 1.00g.44g=.56g Y (Sample 2) I've been given sets of answers such as X2Y3 and X3Y3 XY3 and XY4 and a few more. How can I determine what subscripts would work?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm not positive this is correct, but this is what I came up with: the amount of moles of each element (X1/X2 or Y1/Y2) should be related when you divide them; with this ratio, you can multiply by a number and use the numbers that are closest to whole numbers (as in the real world, measuring the weight will always be slightly off). For example, if I had 24gC and 36gC and I divided them (24/36), I get 0.666...; multiplying by 1 or 2 doesn't give a whole number, but multiplying by 3 yields 2, showing the molar ratio to be 2:3 (as you could clearly see in the above example with C as 12g/mol). So, with all of that said, it was easiest to throw the numbers into excel and have it calculate numbers and the closest one to whole numbers with viable ratios: X=7,9.059 Y=7.072,6\[X_7Y_7 and X_9Y_6\]which, reduced gives\[XY\]and\[X_3Y_2\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I figured it out after I got it wrong a couple of times. I had to find x/y for both samples. Then put (Solution 1 x/y)/ (Solution 2 x/y) After that I had to take the subscripts of the first set of values in the given answers find the ratio and put it over the x y ratio for the second formula. All in all I think it was a silly question that cost me a few points. Thank you for replying!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Can you explain what you were doing wrong?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sadly no. It was so long ago I no longer remember. I had trouble following what I wrote and don't even understand what I was saying. Sorry.
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