When a 1.52 ng sample of methane burns in excess oxygen, how many grams of water and carbon dioxide are formed

I keep getting around 0.00000000003 and the answer is wrong, but I know I wrote the conversion path right..

1.52 ng x 1x10^-9 g / 1 ng x 1 mol CH4/ 16.05g x 2 mol H2O / 1 CH4 x 18.02g/ 1mol H20

Can someone find my mistake? Which I doubt there is one.

First, balance the equation

CH4 + 2O2 --> CO2 + 2H2O

Next 1.52 ng of CH4 is how many moles? Divide by 16 g/mole. This gives approximately 10^-10 moles of CH4 which produce exactly that many moles of CO2 and twice that many moles of H2O.

CO2 has a molecular mass of 44 g/mole, so you get about 4.4 x 10^-9 g of CO2
H2O has a molecular mass of 18 g/mole so you get about 3.5 x 10^-9 g of H2O

My math is being done sans calculator and you will need to report the values to three significant figures, Which will be something like 0.0000000044 for CO2

CH4 weighs 16
and CO2 weighs 44
2 H2O weighs 36


1.52 ng * 44/16 = 4.18 ng (for CO2)
1.52 ng * 36/16 = 3.42 ng (for water)

Soc dont need no stickn moles to solve these