When 6.853 mg of sex hormone was burned in a combustion analysis,19.73 mg of CO2 and 6.391 mg of H2O were obtained. What is the empirical formula of the compound ?? Please be detailed...help me.

Answer 1

0.01973 g of CO2 has 0.000448 moles of CO2
molar mass of CO2 = 44 g/ mole
there is 1 mole of C in CO2 so moles of C in the compound =
0.000448409 moles
mass of C = 0.00538 g
0.006391 g of H2O has 0.000355 moles of H2O
molar mass of H2O = 18 g/ mole
there are 2 moles of H in H2O so moles of H in the
compound = 0.000710 moles
mass of H = 0.00072 g
mass of H + C = 0.00610 g
mass of sample = 0.006853 g
mass of O 0.000756 g
moles of O = 0.0000473 moles
molar ratio of C : H : O = 0.00045 : 0.00071 : 0.000047
smallest number 0.0000473
divide the ratio by the smallest number we get
molar ratio of C : H : O = 9.49 15.02 1
multiply by 2 to get whole numbers
molar ratio of C : H : O = 18.97099624 : 30.0429128 : 2
empirical formula = C19H30O2
empirical formula mass = 290 g
which is the same as the molecular mass
so the empirical formula is also the molecular formula
C19 H30 O2. That's what I did please help...

Answer 2

I don't agree with all of your numbers but most are close, anyway, and I don't think it will make much difference in the empirical formula. I don't know what else you want. Apparently you didn't type in the entire problem since it says nothing about the molar mass nor how you know the "missing" element is O.