Chapter 3 – Mass Relationships in Chemical Reactions
3.1 –Atomic Mass
• Atomic mass is the mass of an atom, measured in atomic mass units (amu).
o One atomic mass unit (amu) is a mass equal to 1/12 the mass of a carbon-12 atom.
• Once the atomic mass of carbon-12 was set to 12 amu, all other elements were given amu
relative to carbon-12.
o For example, through experiments, it was determined that the mass of an H atom
is 8.400% the mass of a carbon-12 atom (1.008 amu).
Average Atomic Mass
• Most of the elements in nature have more than one isotope.
• This means that when calculating the atomic mass of an element, we must find the
average atomic mass of the different isotopes that exist in nature.
o C is 98.90% abundant (12.00 amu) and C is 1.10% abundant (13.00335amu).
Average Atomic Massof NaturalCarbon=(0.9890)(12.00000amu)+(0.0110)(13.00335amu)
• This value is slightly over 12.00 amu since a carbon-13 isotope also occurs in nature, but
in small amounts.
• In the periodic table, average atomic masses appear for all the elements. However, for
simplicity, we refer to them as atomic mass.
3.2 –Avogadro’s Number and theAtomic Mass of an Element
• The mole is the amount of a substance that contains as many elementary entities (atoms,
molecules, ions, and formula units) as there are atoms in 12 grams of carbon-12.
o The number of atoms in 12 g of carbon is known as Avogadro’s number (N ).
N A 6.022 × 10 23
• Molar mass (M) is the mass per mole of units (atoms, molecules....) of a substance (in
• The molar mass of an element is the same as the atomic mass of the substance. 1amu=1 g
1amu=1.66×10 −2grams 1g=6.02×10 amu3
Look at examples 3.2, 3.3, and 3.4
3.3 – Molecular Mass
• The molecular mass is the sum of the atomic masses of all the atoms in a molecule.
o It is measured in amu (Not g/mol. Only molar mass is measured in g/mol).
Molecular Mass of H O = 2 (atomic mass of H) + 1(atomic mass of O)
= 2(1.008 amu) + 1(16.00 amu)
= 18.02 amu
• For ionic compounds such as NaCl and MgO, the term formula mass is used instead of
molecular mass, since these are formula units.
Formula Mass of NaCl = (atomic mass of Na ) + 1(atomic mass of Cl )-
= (22.99 amu) + (35.45 amu)
= 58.44 amu
3.4 – The Mass Spectrometer
• In a mass spectrometer, a gaseous sample of an element is bombarded into a stream of
high energy electrons.
• When the gaseous atoms/molecules collide with these electrons, they each lose an
electron, and become a positively charged ion.
• These ions pass through oppositely charged plates and accelerate as they pass through
• After leaving the plates, they are deflected by a magnet and begin to curve. The amount
of curve depends on the charge-to-mass ratio of each of the ions. • Since the charges of the ions are all the same, their differences in mass would cause
different deflections. Higher masses would deflect less, whereas lower masses would
• The atoms/molecules that had similar masses would strike the detecting screen around the
same location (each ion would create a small current as it strikes the screen). Using this
info, the mass of the different isotopes and their abundance could be calculated.
• The first mass, spectrometer, discovered in the 1920s by F. W.Aston, provided evidence
that Neon isotopes do exist (neon-20 and neon-22). Neon-21 was discovered later using a
more accurate mass spectrometer.
3.5 – Percent Composition of Compounds
• Percent composition by mass is the percentage (by mass) of each element present in a
PercentComposition()= element(100 )
( M compound)
• n represents the amount of moles of the element for every mole of the
compound (subscript to element)
• In the formula, Mcompoundn also be replaced with empirical formula
o This is because when the empirical formula is used, both the n and
M compoundlues would decrease (by the same factor), leading to the
same % composition
Refer to Example 3.8 – 3.10 When we are given the % composition and have to determine the empirical formula, assume 100
g of sample, and find the mass of each element in the sample. Then determine moles, mole ratios
of the elements and then the empirical formula
3.6 – Experimental Determination of Empirical Formulas
Consider a reaction in which ethanol is burned in the presence of Oto form H O and CO .
2 (g) 2 2
Since no C or H atoms are present in the2O , all the C and H atoms will be in the ethanol.
Although O atoms are presen