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What is the empirical formula of a purified drug sample that is 74.27% C, 7.79% H, 12.99% N, and 4.95% O?

Guest May 3, 2017
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I assume I have 100g of the drug sample.

 

74.27% carbon is 74.27 g of carbon

7.79% hydrogen is 7.79g of hydrogen

etc.

Then calculate the amount in moles

74.27g of carbon = 6.184 mol

7.79g of hydrogen = 7.728 mol

12.99g of nitrogen = 0.927 mol

4.95g of oxygen = 0.309 mol

Multiply them by 1000 to become whole numbers

 

6184 units of carbon

7728 units of hydrogen

927 units of nitrogen

309 units of oxygen

The 4 have no common factors......

So the empirical formula is C6184H7728N927O309 ????

Not sure about that..... the number seems huge.

MaxWong  May 3, 2017

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