Chapter 6: Standard Review Worksheet
1. 1 amu = 1.66 _ 10–24 g. For example, the average atomic mass of sodium is 22.99 amu,
which represents the average mass of all the sodium atoms in the world (including all the
various isotopes and their relative abundances). So that we will be able to use the mass of a
sample of sodium to count the number of atoms of sodium present in the sample, we
consider that every sodium atom in a sample has exactly the same mass (the average
atomic mass). The average atomic mass of an element is typically not a whole number of
amu’s because of the presence of the different isotopes of the element, each with its own
relative abundance. Since the relative abundance of an element can be any number, when
the weighted average atomic mass of the element is calculated, the average is unlikely to be
a whole number.
2. On a microscopic basis, one mole of a substance represents Avogadro’s number
(6.022 _ 1023) of individual units (atoms or molecules) of the substance. On a macroscopic
basis, one mole of a substance represents the amount of substance present when the molar
mass of the substance in grams is taken (for example, 12.01 g of carbon will be one mole of
carbon).
3. The molar mass of a compound is the mass in grams of one mole of the compound (6.022 _
1023 molecules of the compound) and is calculated by summing the average atomic masses
of all the atoms present in a molecule of the compound. For example, a molecule of the
compound H3PO4 contains three hydrogen atoms, one phosphorus atom, and four oxygen
atoms. The molar mass is obtained by adding up the average atomic masses of these atoms:
Molar mass H3PO4 = 3(1.008 g) + 1(30.97 g) + 4(16.00 g) = 97.99 g.
4. When a new compound is prepared (the formula is not known), the percent composition must
be determined on an experimental basis. An elemental analysis must be done of a sample of
the new compound to see what mass of each element is present in the sample. For example, if
a 1.000-g sample of a hydrocarbon is analyzed, and it is found that the sample contains
0.7487 g of C, then the percent by mass of carbon present in the compound is
0.7487 g
C
1.000-g sam
p
100% = 74.87% C
Since we can use the formula of a known compound to calculate the percent composition
by mass of the compound, it is not surprising that we can go in the opposite
direction—from experimentally determined percent compositions for an unknown
compound, we can calculate the formula of the compound.
Answer Key
