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I had a doubt in part b) of the following question.

a) When 0.527 g of platinum is heated in fluorine, 0.732 g of a dark red volatile solid forms. What is the empirical formula of this product?

b) Suppose fluorine was a limiting reagent, such that 0.155 g of platinum remains after all the fluorine has been consumed. Calculate the percent yield of this reaction. How many grams of product would be produced?

I've given part a) also because it was told that we should use data from a).

This is my attempt:

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The answer for b) is given as 70.6% and 0.517 g, but I am unable to find where I am wrong. Is my logic correct? (I've double checked that my calculations at least are correct.)

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    $\begingroup$ Please format your question to include the information contained in the photo in a typeset format. That will enable others to find the information contained therein and make a more valuable contribution to the site. $\endgroup$ – Todd Minehardt Nov 5 '15 at 3:45
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Well the problem doesn't say anything about how to get the % yield. If you start with 0.527 grams of platinum, but have 0.155 grams left, then 0.372 grams reacted.

$\frac{0.372}{0.527} * 100$% = $70.6$%

As for the yield

$\frac{0.372}{0.527} * 0.732 = 0.517$ grams

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  • $\begingroup$ OK, but where did I go wrong? $\endgroup$ – Leponzo Nov 5 '15 at 3:06
  • $\begingroup$ To start look at your yield calculation. Yes $0.732g - 0.155g = 0.577g$ but what makes you think that you can subtract $0.155 g$ $\ce{Pt}$ from $0.732 g$ $\ce{PTF4}$ ?!? // It like an orange weighs 8 ounces and is 3 inches in diameter. Therefore it has a measurement of 5 (8-3). $\endgroup$ – MaxW Nov 5 '15 at 3:45
  • $\begingroup$ Yeah, I think the actual yield is the main issue. I used a reaction table and the law of mass conservation to arrive at that. $\endgroup$ – Leponzo Nov 5 '15 at 3:53
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    $\begingroup$ For the complete reaction you do use 0.00540 moles of fluorine gas ($\ce{F2}$) as a reactant, but you only get 0.00270 moles of platinum tetrafluoride ($\ce{PtF4}$) which does have a molecular weight of 271.1 g/mol. // How could you get 1.46 g of product for the 100% conversion when the problem tells you that you got 0.732 g of $\ce{PtF4}$ ?!? $\endgroup$ – MaxW Nov 5 '15 at 3:54
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    $\begingroup$ Of course. If you start with 0.527 grams of platinum, but have 0.155 grams left, then 0.372 grams of platinum reacted. From that you can figure out how many moles, then use molecular weight of platinum tetrafluoride to get grams of product. $\endgroup$ – MaxW Nov 5 '15 at 15:49

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