Stoichiometry Cheat Sheet
1 mole = molar mass of the substance
1 mole = 6.02 x 1023 particles
1 mole = 22.4L of a gas at STP
Amount given
x
1 mol given
x
(y) mol sought
x
d) Molar mass
e) 6.02 x 1023 particles
f) 22.4L
= ANSWER
a) Molar mass
b) 6.02 x 1023 particles
c) 22.4L
(x) mol given
1 mol sought
a) – used when amount given is a mass (g)
b) – used when amount given is particles (molecules, formula units, atoms)
c) – used when amount given is liters of gas at STP (L)
d) – used when answer is sought in mass (g)
e) – used when answer is sought in particles (molecules, formula units, atoms)
f) – used when answer is sought in liters of gas at STP (L)
(x) = coefficient of given substance from balanced equation
(y) = coeeficient of sought substance from balanced equation
Example: (mass to mass problem)
Given the equation Al(NO3)3 + NaOH Al(OH)3 + NaNO3, find the mass of NaNO3 formed if 275g of NaOH reacts.
1. Given substance = NaOH
2. Sought substance = NaNO3
3. Balance equation = Al(NO3)3 + 3NaOH Al(OH)3 + 3NaNO3
4. Start with given amount (275g NaOH) and solve for grams of NaNO3.
275g NaOH
x
1 mol NaOH
x
3 mol NaNO3
x
84.994 g NaNO3
= 584g NaNO3
39.997g NaOH
3 mol NaOH
1 mol NaNO3
Molar mass of NaNO3 because question
asks for “mass”
Molar mass of NaOH because question
gives info “275g of NaOH”
“3” mol of NaOH because coefficient of
NaOH in balanced equation is “3”
“3” mol of NaNO3 because coefficient
of NaNO3 in balanced equation is “3”