Answer
$\Delta H^o_f : C_4H_{10}O = -279.45kJ/mol$
Work Step by Step
As we found in 3.78a:
$C_4H_{10}O(l) + 6O_2(g) --> 4CO_2(g) + 5H_2O(l)$
$\Delta H^o : -2723.7 kJ$
$\Delta H^o_f : O_2(g) = 0 kJ/mol$
$\Delta H^o_f : CO_2(g) = -393.5 kJ/mol$
$\Delta H^o_f : H_2O(l) = -285.83 kJ/mol$
$\Delta H^o_f : C_4H_{10}O(l) = x$
*These values are given in the book
$-2723.7 = [ 4*(-393.5) + 5*(-285.83) ] - [(x) + 6*(0) ]$
$-2723.7 = [-1574 + (-1429.15)] - x$
$-2723.7 = -3003.15 - x$
$279.45 = -x$
$x = -279.45$
