Stoichiometry

so beginning today, we will begin to put together all the things we have learned in the last couple months so that they work together. anything from mole conversions to balancing equations will be involved and it sounds like pretty interesting stuff!

mole to mole calculations

• coefficients in balanced equations respresent the number of moles in each part
• they can also be used as conversion factors

example:

if .15 mol of methane reacts with oxygen, how many moles of water are produced?

first we find the balanced equation:

CH4 + 2O2 -> CO2 + 2H2O

then we find out how many moles of water are needed for the equation by using the formula:

(what you're given) X (what you need)/(what you know)

so... (.15 moles of methane) X (2 moles of water)/(1 mole of methane) = .30 moles of H2O

so by using the quantity we are given, we can find out any other of the factors in the equation

we can also take it one step farther by multiplying the mole ratio we found by the molar mass of the chemical we are trying to find the mass of

for example we can take the .30 moles of water and multiply that by its molar mass (18.02 g/mol) and we find that 5.4 g of water has been used in the equation.

so thats all the notes for today!