MOLES, FORMULAS, YIELDS, CONCENTRATIONS AND OTHER COMMON
CHEMICAL CALCULATIONS
Atomic mass unit: is defined as 1/12th the mass of a 12C atom, this is equivalent
to
.

Atoms range in mass between from about 1-300 amu.
o It is difficult to measure masses this small.
o Amu is not a very practical unit for determining the number of atoms in a sample.
The SI unit for quantity is the mole (mol); a mole is the amount of a substance that contains the
same number of particles in 12 g or 12C.
e.g.
1 atom of 12C weighs 12 amu.
1 mole of 12C atoms weighs 12 grams.
One mole of an atom weighs an equivalent number of grams as the mass of the atom in amu
e.g.
1 atom of Na weighs (on average) 22.99 amu
1 mole of sodium atoms weighs 22.99 grams
Molar mass (M): mass in grams of one mole of a substance, has units g/mol.
The relationship between mass, number of moles and molar mass is:
Where: n = number of moles of a substance
m = mass of substance in grams
M = molar mass of substance in g/mol
Avogadro’s number (NA): The number of particles in one mole of a substance:
Hence:
1 mole of H2O contains
water molecules
1 mole of CO2 contains
carbon dioxide molecules
1 mole of Ag contains
silver atoms
0.5 moles of students contains
3 moles of rabbits contains
students
rabbits
The molar mass of a compound is determined by summing the molar masses of the atoms it is
composed of.
e.g.
molar mass of C = 12.01 g/mol
molar mass of O = 16.00 g/mol
molar mass of O2 =
molar mass of CO2 =
Mass percent from formulas: the mass percent of an element in a compound can be calculated
from its formula using the following relationship:
Empirical formula: with our knowledge of the concept of moles we can define the empirical
formula of a compound in the following way:
“The empirical formula of a compound is the simplest whole number ratio of moles of each type
of atom in the compound”
Determining the empirical formula of a compound can be achieved as follows:
a. Calculate the number of moles of each atom type.
b. Divide the number of moles of each atom type by the smallest number.
c. Multiply each of these numbers by the smallest integer possible to make a
whole number of each atom type.
e.g. analysis of a sample of compounds revels it to contain 4.34g of Na, 4.53g of O and 1.13g
of C. How can we determine the empirical formula of the compound?
Calculate the number of moles of each atom type:
Divide the number of moles of each atom type by the smallest number.
Multiply each of these numbers by the smallest integer possible to make them all whole numbers.
Multiply by 1. The mole ratio of the elements in the compound is Na:O:C = 2:3:1
Thus, the empirical formula is:
Na2CO3
Molecular formulas: we can define the molecular formula as the number of moles of each atom
type in a molecule of the compound. If we can determine the empirical formula and we know the
molar mass of a compound then by multiplying by a whole number we can determine the
molecular formula that has the correct molar mass.
i.e. Ethanol (M = 46.068 g/mol) is found to contain 51.14% C, 13.13% H and 34.73% O by
mass. How can we determine the molecular formula?
Calculate the mass of each atom type in a sample:
In a 100g sample of ethanol we would have 51.14g of C, 13.13g of H and 34.73g of O.
Calculate the number of moles of each atom type:
Divide the number of moles of each atom type by the smallest number:
Empirical formula is C2H6O
Multiplying empirical formula by a whole number to obtain molecular formula with correct
molar mass:
Empirical formula mass =
Molar mass of ethanol = 46.068 g/mol (given)
Whole number multiple =
Hence, molecular formula of ethanol is C2H6O.
Actual yield: the yield obtained experimentally.
Theoretical yield: the yield obtained if the entire amount of limiting reagent is converted to
products.
%Yield: the %yield of a chemical reaction is calculated by dividing the observed or
experimental yield by the maximum possible, or theoretical yield and multiplying by 100%.
The yield of a multistep reaction: is calculated by multiplying the fractional yields for each
step in the reaction sequence together and then multiplying by 100%. Where the fractional yield
is given by:
Consider the following:
Na metal is reacted with water to produce H2 and NaOH according to the equation:
Na(s) + H2O(l)  NaOH(aq) + 0.5H2(g)
Fractional yield = 0.8
The NaOH may then be reacted with HCl(aq) as described below:
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
Fractional yield = 0.9
The NaCl(aq) produced can then be reacted with AgNO3(aq) to produce AgCl(s):
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
Fractional yield = 0.7
The overall %yield for the formation of AgCl using this sequence of reactions is:
Solution: formed when a relative small amount of one substance, the solute, is dissolved in a
much larger amount of another substance, the solvent.
Concentration: the amount of solute dissolved in a given volume of solution.
Molarity: the number of moles of solute per liter of solution. Molarity is the most common way
for expressing the concentration of a solution used in chemistry and has units of molL-1.
We can dilute a more concentrated solution to get a solution with a lower concentration. To
calculate the concentration of the new solution we use the dilution equation below:
CinitialVinitial = CfinalVfinal
Where:
Cinitial = initial concentration
Cfinal = final concentration
Vinitial= volume of original solution
and Vfinal = volume of final solution
We may use whatever units we wish in this equation but they must be the same for both the final
and initial solution.
Consider the following: What would be the concentration of a sulfuric acid, H2SO4(aq), solution
that was prepared by diluting 5.00 mL of 2.00molL-1 H2SO4(aq) solution with enough deionized
water to give 100.0 mL of solution.
V1 = 5.00 mL
C1 = 2.00 molL-1
V2 = 100.00 mL
C2 = ?
V1C1 = V2C2
(V1C1)/V2 = C2
C2 = (5.00 mL x 2.00 molL-1)/100.00 mL = 0.1 molL-1
The concentration of the solution formed would 0.1 molL-1.