Is the negative excess volume of ethanol/water the cause of my math error? I have a binary mixture, say ethanol at $90\%wt$ ethanol, $10\%wt$ water.  According to Perry's Chemical Engineers' Handbook 9th Ed., that mixture has a density of $0.81797g/cc$ (at $20^{\circ}C$).  So my $1L$ of mixture should have a mass of $817.97g$.

Now, I want to create this mixture.  I know the total mass and the mass %'s, so I just multiply them out: we find that we need $736.17g$ ethanol and $81.80g$ water.  Ok, so far so good.  However, say all of my balances are broken, so I only have graduated cylinders to work with.  Easy, just convert those masses back to volumes using their respective pure substance densities: $0.78934g/cc$ for ethanol and $0.99823g/cc$ for water.

However, doing this tells me that I should have $932.6mL$ of ethanol and $81.9mL$ of water, which adds up to a total volume of $1014.6mL$ of binary mixture, so somehow I magicked $14.6mL$ of liquid out of thin air using this conversion algorithm.

Is this due to the negative excess volume of the ethanol/water mixture upon mixing?  That's my current guess, but for the life of me I'm having a hard time figuring out the math here.  It seems like the change in volume should be much smaller: $10\%wt$ water in ethanol is about $10\%mol$ and there's about $4.5mol$ of water in the mixture, so based on this dataset it seems like there should only be a difference of about $3mL$.  Even with all the rounding I just did, it seems like this can't be the only error source.

Is there something wrong with the logic of my algorithm here?  Can anyone point me to the reason I'm seeing this numerical error whenever I run this calculation?
When you mix ethanol and water they take up less space due to the way the molecules fit together, so you will have what is called a negative volume of mixing.  In this case, the 932.6 mL of ethanol and the 81.9 mL of water will give you a 100 mL solution of 90% ethanol.