Stoichiometry of Chemical Formulas and
Reactions
 2013, Sharmaine S. Cady
East Stroudsburg University
The law of definite proportions states that the element ratios for a given compound
remain constant. The subscripts in a chemical formula may therefore be used to define molar
ratios between the elements in the formula or between an element and the compound. For
example, the chemical formula for water is H2O. Therefore, we can write
2 mol H : 1 mol O : 1 mol H2O
or
2 mol H : 1 mol O or 1 mol O : 2 mol H (molar ratios between elements)
2 mol H : 1 mol H2O or 1 mol O : 1 mol H2O (molar ratios between elements and compound)
The molar mass of an element is its atomic mass expressed in units of g/mol. To find the molar
mass of a compound, use the Periodic Table to find the molar mass of each element and
multiply it by its subscript in the chemical formula. Subscripts in chemical formulas are exact
numbers; therefore, they do not affect the number of significant digits in a calculated value.
The sum of the element masses is the molar mass for a compound.
∑
How to calculate the molar mass of a compound
What is the molar mass of water?
Solution:
Write down the given values with their units
and what they represent.
The chemical formula for water is H2O. The
formula indicates 2 mol H and 1 mol O. The
Periodic Table gives the molar masses for H
and O.
1 mol H = 1.008 g/mol
1 mol O = 16.00 g/mol
Write down the unknown and its unit.
The molar mass of water is the unknown.
H2O = ? g/mol
Develop a strategy to solve the problem.
Use the subscripts in the chemical formula to
determine the molar mass of water:
Round off the answer to the correct number
of significant digits and give the appropriate
units.
The mass of O is only known to 2 decimal
places while H is known to 3. By the rules of
addition, the molar mass must be rounded off
to the second decimal place.
molar mass H2O = 18.02 g/mol
Check whether the answer is reasonable.
The molar mass of H is 1 g/mol and O is 16
g/mol, the sum of 2 mol H and 1 mol O is 18;
the answer is reasonable.
The relationship between the mole and mass of elements and compounds is used to solve
stoichiometry problems by dimensional analysis.
During ordinary chemical reactions, matter is neither created nor destroyed. Balanced
chemical reaction equations provide the correct ratios of reactants to products to insure there is
no loss of mass during the chemical transformation. The numbers in front of the reactant and
product formulas are known as stoichiometric coefficients and provide the molar ratios that
relate the mass of a reactant to the mass of another reactant or product. Stoichiometric
coefficients are exact numbers; therefore, they do not affect the number of significant digits in a
calculated value. The reaction between oxygen and hydrogen to form water can be used as an
example.
2
+
2
2 H2 + O2  2 H2O
Note that the stoichiometric coefficients provide for the conservation of mass among the
reactants and products. Two moles of H2 yield 4 H atoms. One mole of O2 yields 2 O atoms.
Two moles of H2O yield 4 H atoms and 2 O atoms.
The stoichiometric coefficients in front of the chemical species give the following molar ratios:
1 mol O2 : 2 mol H2 : 2 mol H2O
These reaction molar ratios are constant. Hence, they serve as an aid in solving mass
stoichiometry problems where reactants and products are involved.
How to calculate the mass of an element present
in a given mass of a compound
How many grams of carbon are in 0.987 g of carbon dioxide?
Solution:
Write down the given values with their units
and what they represent.
The chemical formula for carbon dioxide is
CO2. The formula indicates 1 mol C: 1 mol CO2.
The Periodic Table gives the molar masses for
C and O.
1 mol C = 12.01 g/mol
1 mol O = 16.00 g/mol
mass CO2 = 0.987 g
molar mass CO2 = 44.01 g/mol
Write down the unknown and its unit.
The mass of carbon is the unknown.
mass C = ? g
Develop a strategy to solve the problem.
Set up a dimensional analysis problem with the
correct molar and mass ratios to solve for g C.
Round off the answer to the correct number of
significant digits and give the appropriate
units.
The mass of CO2 has 3 significant digits while
the molar masses have 4. The mass is rounded
off to the 3 significant digits.
mass C = 0.269 g
Check whether the answer is reasonable.
The mass of CO2 given is 0.02 its molar mass
(44.01 g/mol). The mass of C should be 0.02
its molar mass (12.01 g/mol); the answer is
reasonable.
How to calculate the mass of an ion present in a
given mass of a compound
How many grams of carbonate are in 6.987 g of sodium carbonate?
Solution:
Write down the given values
with their units and what they
represent.
The chemical formula for sodium carbonate is Na2CO3. The
formula indicates 1 mol CO32- : 1 mol Na2CO3. The Periodic Table
gives the molar masses for C and O.
1 mol C = 12.01 g/mol
1 mol O = 16.00 g/mol
mass Na2CO3 = 6.987 g
molar mass Na2CO3 = 105.99 g/mol
molar mass CO32- = 60.01 g/mol
Write down the unknown and
its unit.
The mass of carbonate is the unknown.
mass CO32- = ? g
Develop a strategy to solve the
problem.
Set up a dimensional analysis problem with the correct molar and
mass ratios to solve for g CO32-.
Round off the answer to the
correct number of significant
digits and give the appropriate
units.
The mass of Na2CO3 and CO32- have 4 significant digits while the
molar mass of Na2CO3 has 5. The mass is rounded off to the 4
significant digits.
mass CO32- = 3.956 g
Check whether the answer is
reasonable.
The mass of Na2CO3 given is 0.07 its molar mass (105.99 g/mol).
The mass of CO32- should be 0.07 its molar mass (60.01 g/mol);
the answer is reasonable.
How to calculate the mass of a reactant needed
for given mass of another reactant
If 3.00 g of oxygen are used for the following reaction, how many grams of hydrogen are
needed?
2 H2 + O2
 2 H2O
Solution:
Write down the given values with their
units and what they represent.
The stoichiometric molar ratio between H2 and O2 is 2
mol H2 : 1 mol O2. The Periodic Table gives the molar
masses for H and O.
1 mol H = 1.008 g/mol
1 mol O = 16.00 g/mol
mass O2 = 3.00 g
molar mass O2 = 32.00 g/mol
molar mass H2 = 2.016 g/mol
Write down the unknown and its unit.
The mass of H2 needed for the reaction is the unknown.
mass H2 = ? g
Develop a strategy to solve the problem.
Set up a dimensional analysis problem with the correct
molar and mass ratios to solve for g H2.
Round off the answer to the correct
number of significant digits and give the
appropriate units.
The mass of O2 has 3 significant digits, while the molar
masses of H2 and O2 have 4. The mass is rounded off to
3 significant digits.
mass CO32- = 0.378 g
Check whether the answer is reasonable.
The mass of O2 given is 0.1 its molar mass (32.00
g/mol). The mass of H2 should be 0.2 its molar mass
(2.016 g/mol); the answer is reasonable.
How to calculate theoretical yield based on the
limiting reagent
If a student uses 0.536 g of H2 and 2.48 g of O2, how many grams of water can theoretically be
formed from the reaction?
2 H2 + O2
 2 H2O
Solution:
Write down the given values with their
units and what they represent.
The stoichiometric molar ratio between H2 and O2 is 2
mol H2 : 1 mol O2. The Periodic Table gives the molar
masses for H and O.
1 mol H = 1.008 g/mol
1 mol O = 16.00 g/mol
mass H2 = 0.536 g
molar mass H2 = 2.016 g/mol
mass O2 = 2.48 g
molar mass O2 = 32.00 g/mol
molar mass H2O = 18.02 g/mol
Write down the unknown and its unit.
The maximum theoretical yield of H2O produced is the
unknown.
mass H2O = ? g
Develop a strategy to solve the problem.
Set up a dimensional analysis problem with the correct
molar and mass ratios to solve for g H2O formed from
the given mass of H2. Then set up a dimensional
analysis problem with the correct molar and mass
ratios to solve for g H2O formed from the given mass of
O2. If the two masses differ, then the reagent with the
lowest yield is the limiting reagent. The mass of water
produced by this reagent is the theoretical yield.
2.79 g H2O is the lower of the two yields. O2 is the
limiting reagent, and the theoretical yield is 2.79 g H2O.
Round off the answer to the correct
number of significant digits and give the
appropriate units.
The mass of O2 has 3 significant digits while the molar
masses of H2O and O2 have 4. The mass is rounded off
to 3 significant digits.
mass H2O = 2.79 g
Check whether the answer is reasonable.
The mass of O2 given is 0.08 its molar mass (32.00
g/mol). The mass of H2O should be 0.16 its molar
mass (18.02 g/mol); the answer is reasonable.
How to use gravimetric analysis to determine
m/m% of a component in a mixture
A student weighed and dissolved 4.019 g of Miracle-Gro Shake 'n Feed. The water-soluble
phosphate was precipitated with MgSO4 to give the following precipitate. The precipitate mass
after isolation and drying was found to have a mass of 1.309 g. Calculate the m/m% P2O5 in the
fertilizer sample.
PO43-(aq) + Mg2+ (aq) + NH3 (aq) + 7 H2O (l)  MgNH4PO4·6H2O(s) + OH-(aq)
Solution:
Write down the given values
with their units and what they
represent.
The Periodic Table gives the molar masses for Mg, N, H, P, and O.
1 mol Mg = 24.31 g/mol
1 mol N = 14.01 g/mol
1 mol H = 1.008 g/mol
1 mol P = 30.97 g/mol
1 mol O = 16.00 g/mol
mass MgNH4PO4·6H2O = 1.039 g
molar mass MgNH4PO4·6H2O = 245.42 g/mol
molar mass P2O5 = 141.94 g/mol
mass fertilizer = 4.019 g
Write down the unknown and
its unit.
The m/m% of P2O5 in the fertilizer sample is the unknown.
m/m% P2O5 = ? %
Develop a strategy to solve
the problem.
The mass of P2O5 is directly related to the moles of P in the precipitate. From
the chemical formulas, there is a 2:1 molar ratio between the moles of P in
P2O5 and the moles of P in MgNH4PO4·6H2O (ppt). Set up a dimensional
analysis problem with the correct masses and molar ratios to find g P2O5.
Round off the answer to the
correct number of significant
digits and give the appropriate
units.
The precipitate and fertilizer masses have 4 significant digits while the molar
masses have 5. The m/m% is rounded off to 4 significant digits.
Check whether the answer is
reasonable.
The mass of P2O5 is 0.1 the fertilizer mass; the answer is reasonable.
m/m% P2O5 = 9.419 %