CH 3 Mass Relations
in Chemistry;
Stoichiometry
Atomic Mass
Indicates how heavy an element is
compared to another element.
Units AMU---Atomic Mass Unit
Defined as 1/12 of the mass of a C-12 atom
Atomic Mass from isotope composition
Isotopic Abundance: the natural of an isotope.
Isotope Atomic Mass Percent Contribution to atomic
abundance mass
Ne-20 20.00 90.92 =
Ne-21 21.00 00.26 =
Ne-22 22.00 08.82 =
Mass of individual atoms
Mass of 1 atom = molar mass/ NA
(Avogadro's #)
Reacquaint yourself with the mole wheel.
The Mole
1 mol= 6.022 x 1023 items
1mol H = 6.022 x 1023 H atoms = 1.008g
1mol Cl= 6.022 x 1023 Cl atoms = 35.45g
1mol Cl2= 6.022 x 1023 Cl2 molecules = 70.90g
Molar mass (MM)
Molar mass is numerically equal to the
sum of the atomic masses.
Mass % from formula
% composition of K2CrO4
Use part / whole, assume you have 1 mole of
compound. (the math is easier)
1mol = K2CrO4 194.20 g
%K = 78.90/194.20 =
%Cr =
%O =
Simplest / empirical formula
Simplest whole number ratio of atoms
present in a compound.
Simplest (empirical) formula
from % composition
Steps:
1. Find the mass of each element in the
sample compound, assume 100g total.
2. Find the numbers of moles of each
compound.
3. Divide each by the smallest # of moles and
look for obvious ratios.
Use the following K= 26.6%, Cr= 35.4%, O = 38.0%
Simplest formula (empirical)
from analytical data
An organic sample containing only C, H, O atoms weighs
1.000g
Burning the sample gives 1.466g CO2,
0.6001g H2O
Find the simplest formula…
All the carbon from the sample is “locked up” in CO2
The portion of CO2 that is carbon can be determined by
the mass ratio in the formula (12.01/44.01)
This multiply this by the mass of CO2 and you find grams
of carbon in the sample.
Yield of product in a reaction
Ordinarily, reactants are not present in the
exact ratio required for reaction.
Usually 1 is in excess; some left when reaction
is over.
1 is limiting; completely consumed to give the
theoretical yield of product.
Calculating theoretical yield
1. Calculate the yield expected if the first
reactant is limiting.
2. Repeat the calculation for the second
reactant
3. The theoretical yield is the SMALLER
of these two quantities. The reactant that
gave the smaller theoretical yield is the
limiting reactant.
2 Ag (s) + I2 (s) 2 AgI (s)
Calculate the theoretical yield of AgI and
determine the limiting reactant.
There is 1.00g Ag, and 1.00g I2.
% Yield
Suppose the actual yield is 1.50g AgI,
what was the % yield?
Actual/Theoretical (x100) = % Yield