Determine the mass of water (in g) produced from the complete combustion of 7.60×10-3 g of C8H14O with excess oxygen.

Heres my solution but I doubt I did it right:

C8H14O + 11O2 >> 8CO2 + 7 H2O

with excess oxygen means to add oxygen on left side

Moles C8H14O = 7.60×10^-3 g / 126.1969 g/mol = 6.022×10^-5

Moles H2O = 7 x 6.022×10^-5 = 0.00042154

Mass CO2 = 0.00042154 mol x 18.02 g/mol = 0.0075961508 g

sorry that says Mass CO2 but it should say, Mass H2O, lol