Empirical formula from Combustion Analysis (©2002, François G. Amar, All rights reserved)
Exercise 5.5 Gillespie (similar to BLB 3.92 and BLB pp. 86)
0.8874 g phenol are burned to produce 2.491 g CO2 & 0.510 g H2O
Determine the empirical formula of phenol (it contains C, H, O)
Combustion equation: CnHmOq + b O2 ---> c CO2 + d H2O
It appears that we are missing a lot of stoichiometric information, however we do know that:
1) all the C in the
2.491 g of CO2 came from the
phenol
[so the coefficient c = the subscript
n,
c=n]. The total grams of C are:
g C = mass fraction C in CO2 x mass of CO2
12.01
g C/mol CO2
= _____________________
x 2.491 g CO2
44.01
g CO2/mol
CO2
g C = 0.6798 g
2) all the H in
0.510 g of H2O came from the phenol
[so the coefficient d = one-half the
subscript m, d=m/2]. The total grams of
H are:
g H = mass fraction H in H2O x mass of
H2O
2.016
g H/mol H2O
= _____________________
x 0.510 g H2O
18.016 g H2O/mol H2O
g H = 0.0571 g
3) the amount of O
in the phenol can be determined by difference
[note that b=c+(d-q)/2]
g O = g phenol - g C - g H
= 0.8874 g - 0.6798 g - 0.0571 g
g O = 0.1505 g
Now that we have the no. of grams of each
element, we can determine
moles, mole ratios, and look for small integer
coefficients in the
table as follows
element grams ---> moles ---> mole ratios C 0.6798 (/12.01) 0.05660 (/0.00941) 6.01 H 0.0571 (/1.008) 0.05665 (/0.00941) 6.02 O 0.1505 (/16.00) 0.00941 (/0.00941) 1 Empirical formula is C6H6O. There is no need to look for multiples since all the mole ratios are already very close to integers. We can now rewrite the combustion reaction:
C6H6O + 7 O2 ---> 6 CO2 + 3 H2O .
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