Step 1: Determine the mass of C and H produced from the reaction
So the mass of C from 8.49 g CO2 is
= 2.3169 g C
So the mass of H from 2.14 g H2O is
= 0.2395 g H
Step 2: Determine the mass percent of C, H and N
Recall that mass percent is given by:
= 65.4492 % C
= 6.7655 % H
= 8.4681 % N
Step 3: Determine the mass and moles C, H, O and N from their mass percent
%C + %H + %O + %N = 100%
%O = 100% – (% C + % H + %N)
%O = 100% - (65.4492 %C + 6.7655% H + 8.4681% N)
%O = 19.3172 % O
Assuming we have 100 g of the compound, this means we have 65.4492 g C, 6.7655 g H, 8.4681 g N and 19.3172 g O.
= 5.4496 mol C
= 6.6585 mol H
= 0.6044 mol N
= 1.2073 mol O
Step 4: Determine the lowest whole number ratio of C, H, O and N to get the empirical formula
= 9 C
= 11 H
= 1 N
= 2 O
The empirical formula of benzocaine is C9H11NO2
Step 5: Determine the molecular formula
Researchers obtained the following data from experiments to find the molecular formula of benzocaine, a local anesthetic, which contains only carbon, hydrogen, nitrogen, and oxygen. Complete combustion of a 3.54-g sample of benzocaine with excess O2 forms 8.49 g CO2 and 2.14 g H2 O. Another 2.35 g sample contains 0.199 g of N. The molar mass of benzocaine is 165 g/mol.
Find the molecular formula of benzocaine.
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