Stoichiometry: gram-gram calculations tutorial
(to download this document, click on “Resources” after going to
www.myteacherpages/webpages/Dbarnes/, then click on Stoichiometry
problem-formation of ammonia. This tutorial is for use with Ammonia
Stoichiometry gram-gram quiz-basic skills and Stoichiometry gram-gram
regular quiz-Ammonia)
Consider the reaction for the commercial preparation of ammonia:
N2 + 3H2 2NH3
How many grams of hydrogen are needed to prepare 200.0 grams of
ammonia (NH3)?
The question of grams of hydrogen is placed on the right, because that is
what we are looking for. Elemental hydrogen is H2, so that is what we write
on the right.
Note: Both oxygen and nitrogen gases are diatomic as found in nature,
meaning that we write oxygen as “O2” and nitrogen as “N2”.
What we are given goes on the left. In this case, 200.0 grams of ammonia
(NH3) is given, so this is what we write on the left.
When we go from grams to grams, we have three bridges/blanks, so check
out the three blanks with multiplication signs:
200.0 g NH3 _________ ________ __________ = g H2
given info left blank middle blank right blank what we are
or bridge or bridge or bridge looking for
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Dr. Barnes’ Stoichiometry gram-gram tutorial
The leftmost blank belongs to the chemical on the left, in this case, ammonia
(NH3). This blank is reserved to tell how many grams in one mole of
ammonia. To do this we must determine the formula mass of ammonia. To
do this, I write the formula of ammonia, tell how many moles of each
element and then look up the atomic mass of each element on a periodic
chart. The number of moles of each element is multiplied by its atomic
mass. The total masses of each element are then added to find the formula
mass of that element.
Warning! DANGER! DO NOT use coefficients in this step. Coefficients
are only used in the middle blank, which we will discuss later!
Periodic Chart (condensed)
1.01
H
1
6.94 12.01 14.01 15.99 18.99
Li C N O F
3 6 7 8 9
22.99 24.31 26.98 32.06 35.45
Na Mg Al S Cl
11 12 13 16 7
39.10 40.08 55.85
K Ca Fe
19 20 26
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Dr. Barnes’ Stoichiometry gram-gram tutorial
200.0 g NH3 _________ ________ __________ = g H2
Determination of formula mass:
NH3 H2
N = 1 x 14.01 = 14.01 H = 2 x 1.01 = 2.02
H = 3 x 1.01 = 3.03 ___________________
____________________ 2.02 g/mol
17.04 g/mol
Next, the formula masses are used in the equation:
200.0 g NH3 1 mol NH3 _________ 2.02 grams H2 = g H2
17.04 g NH3 1 mol H2
Note that the molar mass of NH3 is placed on the bottom in the left blank.
On the leftmost blank, “1 mole of NH3” is written on top.
Notice on the rightmost blank that the grams statement is placed on top and
the 1 mole of H2 is placed on the bottom. This top and bottom pattern of
placement is how we always do the gram-gram stoichiometry problems.
On the right, 2.02 grams of H2 is placed on top and 1 mole of H2 is placed on
bottom. Notice that the 2.02 grams of H2 carries the same unit as what we
want in the answer: grams of H2.
Please remember that the left most blank belongs to the chemical on the left
and the rightmost blank belongs to the chemical on the right.
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Dr. Barnes’ Stoichiometry gram-gram tutorial
Next, we must complete the middle blank, the blank that we use to allow for
the coefficients in the balanced equation.
200.0 g NH3 1 mol NH3 ___________ 2.02 grams H2 = g H2
17.04 g NH3 1 mol H2
left blank middle blank right blank
formula molar formula
mass coefficient mass
N2 + 3H2 2NH3
In the middle blank, the top belongs to the right chemical, in this case H2. In
the middle blank, the bottom belongs to the left chemical, in this case NH3.
So, to complete the top blank, find the number in front of H2 in the balanced
equation (3H2). The number is “3” so write 3 moles of H2 in the top blank.
200.0 g NH3 1 mol NH3 3 moles of H2 2.02 grams H2 = g H2
17.04 g NH3 1 mol H2
We will now complete the bottom blank of the middle “bridge”. The
chemical on the left is NH3 so we will put the molar coefficient from the
balanced equation in this bottom of the middle blank or bridge. When we
look at the balanced equation:
N2 + 3H2 2NH3
We notice that NH3 has a “2” in front of it (2NH3). We therefore will write
2 moles of NH3 in the bottom blank of the middle bridge.
200 g NH3 1 mol NH3 3 mol H2 2.02 grams H2 = g H2
17.04 g NH3 2 mol NH3 1 mol H2
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Dr. Barnes’ Stoichiometry gram-gram tutorial
Notice that g NH3 in the given information cancels with the bottom g NH3 in
the left blank.
Notice that the mol NH3 on the top of the left blank cancels with the mol
NH3 on the bottom of the middle blank or bridge.
Notice that the mol H2 on the top of the middle bridge or blank cancels with
the mol H2 on the bottom of the right bridge or blank.
Use your writing tool and cancel these units in the equation below.
200.0 g NH3 1 mol NH3 3 mol H2 2.02 grams H2 = g H2
17.04 g NH3 2 mol NH3 1 mol H2
Lastly we have to plug the numbers into a calculator and let it crank out an
answer:
Turn your calculator “on”. Hit “clear” (usually the C button) several times.
Punch the 200.0 into your calculator, multiply by 3 and then multiply by
2.02. Hit the equals key. You should get “1212”. Press the “divided by”
button, enter 17.04, hit the “equals” button and you should see
“71.12676056”(or something like it). Press the “divided by” button, enter 2,
hit the “equals” button and you should see “35.56338028” (or something
like it).
We will round this to four significant figures, only retaining the same
amount of significant figures as we see in the original number (200.0).
In the end, we will retain the number 35.56 as our answer.
200.0 g NH3 1 mol NH3 3 mol H2 2.02 grams H2 = 35.56 g H2
17.04 g NH3 2 mol NH3 1 mol H2
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Dr. Barnes’ Stoichiometry gram-gram tutorial
This answer, resulting from a long problem, tells us that it will take 35.56
grams of H2 to make 200.0 grams of NH3.
Stoichiometry is a way to figure how much of this stuff we have to react
with that stuff to get a product. It’s kinda like a recipe for cooking.
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Dr. Barnes’ Stoichiometry gram-gram tutorial