Hello Dr. Bodwin. I've been working on some old exams for studying and on the exam 1a from fall 2011, the last question is asking about the empirical formula, and I was just wondering when solving for each part, where do the last two numbers come from?

For example,

C -> (71.98 g)/(12.011 g/mol) = 5.993 mols

**-> 4.5 -> 9**

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Let me pull up the whole problem from the exam key:

{from: http://www.drbodwin.com/teaching/exams/c150fe1ak.pdf}

When determining empirical formulas from percent composition data, the first step is to assume that you have 100g of sample. You can assume any amount of sample you like, but 100g simplifies things a little because if I have 100g of a sample and I know that 71.98% of that sample is carbon, then there must be 71.98g of carbon in the sample. Using the percentages given in the problem, we know that the "100g" sample contains 71.98g of carbon, 6.711g of hydrogen, and 21.31g of oxygen. Grams are great, but we want to count the number of different atoms, so we need to convert grams to moles... That's the first step in the calculation that is shown.

Once we have moles, we know the relative amounts of each element present in the sample, and we can write a balanced chemical formula:

C

Hmm, that doesn't look quite right... But at this point, we have the correct _{5.993}H_{6.658}O_{1.332}*relationship*between the moles of each element, so we can force those relationships to be whole numbers by dividing all of them by the smallest one. Essentially, we're saying "what if 1.332 actually represents 1 oxygen atom?" Dividing them all gives the forumla:

C

Still not perfect, but it's a lot more "normal" looking than the first formula. Again, since we know that the _{4.5}H_{5}O_{1}*relationship*between moles is correct here, we can multiple all the subscripts by something that gives us a nice, round, whole number ratio. If we double everything, we get:

C

And now we have a good empirical formula for this elemental analysis._{9}H_{10}O_{2}Other questions? Let me know.

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