Okay, I'm trying to figure out how to make 10% sodium methoxide. I know 10g solute in 100g solution. However, it's the form of w/v which I despise. Grr. And I'm probably making this a lot harder than it really is. But anyhow, here's the gest of it. I want a gallon of methanol as the solution, so how much NaOH do I need to add? 0.1 gallons right? Now to convert this to metric (as is the standard). There are roughly 3,785 mL in a gallon and the density of MeOH is 0.7918g/mL, so in terms of weight I have ~2997g per gallon. How much of this needs to be NaOH to make 10% sodium methoxide, provided that sodium methoxide is a reaction between NaOH and MeOH in a 1:1 ratio---forming water and sodium methoxide. 10% of 2997g is 299.7g, however, I can't help but think that is a lot of NaOH for a 10% solution of sodium methoxide. If the sodium is reacting with the methanol (with methanol left in excess), shouldn't the NaOH be reduced by half? Otherwise the 10% NaOH that was added is going to react with 10% of the methanol and form I guess 10% sodium methoxide. Hmmm. But what about the water? Say I take it out, how much NaOH do I need to add then? Oh I have got to be making this harder than it really is. But it's late and doing weight/volume percentage equations should be tackled after dusk.

Any help is greatly appreciated. Thank you!