12.5 mg C,H,N.
Convert mg CO2 to mg C.
22.3 CO2 x (1 mol/molar mass CO2) x (12 g C/1 mol) = about 0.5 mg C (but you should do it more accurately.
Do the same for H2O to hydrogen.
15.2 mg H2O x ( mol H2O/molar mass H2O)x (2 g H2/1mol H2) = about 1.7 mg H.
12.5 mg total - mg CO2 - mg H = mg N = about 0.34 mg.
Then convert mg to mols (I would use millimols).
0.5/12 = millimols C
1.7/1 = mmoles H
0.34/14 = mmoles N.
Now find the ratio of these elements to each other in small whole numbers. The easy way to do that is to divide the smallest number by itelf then divide the other two numbers by the same small number. Round to whole numbers BUI don't round too much; i.e., 5.9 can be called 6.0 but 1.5 can't be rounded to 1 or 2.
• A compound is known to contain only C, H, and N.
• On combustion of a sample of 12.5 mg, 22.3 mg of CO2 and 15.2 mg of H2O were produced.
•Use this data to find the empirical formula for the compound. (Please show working out)
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