Mass stoichiometry: using the U-diagram
Recall yesterday’s question: Methane reacts with sulfur to produce carbon disulfide and
hydrogen sulfide. What mass of CH4 is required if 4.09g of hydrogen sulfide is produced ?
 we are not given any moles, but instead mass.
This is one way of breaking up this type of question into steps:
balanced chemical equation
known in g
unknown* in g
use
Mknown
known in mol
use
Munknown
use
unknown* in mol
mole ratio
*what was unknown in the beginning, what you are looking for
You can do this for any of the mole ratios you can form for your equation.
Example 4 - see next page !
Example 4
Methane, CH4(g), reacts with sulfur, S8(s) to produce carbon disulfide, CS2(l), and hydrogen
sulfide, H2S(g). Carbon disulfide is often used in the production of cellophane. What mass of
CH4 is required if 4.09g of hydrogen sulfide is produced ?
2CH4(g) + S8(s)
mole ratio
2
M [ g/ mol] 16.05
:
 2CS2(l) + 4H2S(g)
1
:
2
:
4
36.11
0.909g CH4
4.09g H2S
MCH4 = 16.05 g/mol
MH2S = 36.11g/mol
nCH4 = 0.05663mol CH4 x 16.05 gCH4/ mol CH4 use M
use M
nCH4 = 0.90895g CH4
4.09g H2S x mol H2S
36.11g H2S
mH2S = 0.1133 mol H2S
0.05663mol C
use mole ratio
nCH4
=
0.1133mol H2S
Answer: 0.909 g CH4 are required.
mH2S =
nCH4
=
nCH4
=
0.1133mol H2S
2mol CH4
4mol H2S
2mol CH4
4mol H2S
x 0.1133mol H2S
0.05663mol CH4