# How do you do mass to mass stoichiometry problems?

Jun 22, 2016

I'm assuming you mean reacting masses - for example:
"Calculate the amount of $C {O}_{2}$ formed when 58g of $C a C {O}_{3}$ decomposes."

So first you need to write a balanced equation:
$C a C {O}_{3} \to C a O + C {O}_{2}$

Then you need to divide the mass you are given by its molar mass, to find the number of moles that react, so..
Molar mass $C a C {O}_{3}$ = 40 + 12 + (3 x 16) = 100
Moles of $C a C {O}_{3}$ reacting = 58g / 100 = 0.58

Then you need to multiply the number of moles of $C a C {O}_{3}$ by the molar mass of the thing you have to find ($C {O}_{2}$) so...
Molar mass $C {O}_{2}$ = 12 + (1 x 16) = 44
Mass $C {O}_{2}$ = 0.58 x 44 = 25.5g.

If you have different stoichiometric ratios, it is simplest to multiply the molar mass by the coefficient first, so if you had a question, "Calculate the mass of $F {e}_{2} {O}_{3}$ formed when 10.0g of Oxygen reacts with excess iron"...

First write that balanced equation:
$4 F e + 3 {O}_{2} \to 2 F {e}_{2} {O}_{3}$

Then divide the moles of ${O}_{2}$ given by the molar mass x 3, so...
Molar mass ${O}_{2}$ = 2 x 16 = 32
Moles = 10 / (32 x 3) = 0.10417

Then multiply this number by the mass of $2 F {e}_{2} {O}_{3}$:
Molar mass $F {e}_{2} {O}_{3}$ = (2 x 56) + (3 x 16) = 160
Mass $F {e}_{2} {O}_{3}$ formed = (160 x 2) x 0.10417 = 33.3g