CHEM 103 Module 2 Exam (General
Chemistry) – Portage Learning|| A+
GRADE|| LATEST UPDATE 2026
2.1: MOLECULAR WEIGHT
A compound is made up of two or more elements combined in a definite ratio that is
represented by a molecular formula. Each of these elements has a certain atomic
weight, which can be found in the periodic table. The sum of the atomic weights of
the atoms in the molecular formula is called the formula weight or molecular
weight or formula mass.
Ca3(PO4)2
calcium phosphate
Molecular Weight =
3 Ca= 3 x 40.08=120.24
2 P=2 x 30.97=61.94
8 O=8 x 16.00=128.00
Total 310.18
2.2: MOLES
Chemical compounds react with one another in amounts that are based on their
molecular weights; this chemically reactive amount of compound is called a mole.
moles
= grams / molecular weight
Calculate the number of moles in 10.0 grams of each of the following compounds:
,Ca3(PO4)2 10.0 g ÷ 310.18 = 0.0322 mol (3 sig fig because of 10.0 g)
C3H5O2Cl 10.0 g ÷ 108.52 = 0.0921 mol
Al2(SO4)3 10.0 g ÷ 342.17 = 0.0292 mol
Ca3(PO4)2 0.0500 mol x 310.18 = 15.5 g
C3H5O2Cl 0.0500 mol x 108.52 = 5.43 g
Al2(SO4)3 0.0500 mol x 342.17 = 17.1 g
2.3: PERCENT COMPOSITION
he molecular formula represents the definite ratio of elements in a compound. The
weight of each element present in the compound represents a certain percentage of the
total weight of the compound. The percentage of each element present in a compound
is called the % composition of the compound.
The percentage of an element present in a compound can be calculated as shown
below:
% of an element = weight of element / molecular weight of compound x 100
Ca3(PO4)2
3 Ca = 3 x 40.08= 120.24
2 P = 2 x 30.97= 61.94
8 O = 8 x 16.00 = 128.00
, 310.18
% Ca = (120.24 ÷ 310.18) x 100 = 38.76%
%P = (61.94 ÷ 310.18) x 100 = 19.97%
%O = (128.00 ÷ 310.18) x 100 = 41.27%
2.4: EMPIRICAL FORMULA
If the formula of a compound is known, the % composition of the compound can be
determined. This process can be done in reverse: The formula of the compound can be
determined if the % of each element present in the compound is known. The formula
calculated from % composition is known as the empirical formula (or the simplest
formula). The actual molecular formula is some multiple of this simplest formula, which
is determined by knowing the molecular weight.
To determine the empirical formula:
(1) Divide each element % by its exact atomic weight to give a set of numbers.
(2) Divide the smallest of this set of decimal numbers into each of the numbers
(including itself) to yield a second set of numbers.
(3a) Round off each of the second set of decimal numbers to a whole number.
OR
(3b) If the numbers derived from the division in step 2 are recognized as exact decimal
equivalents of fractions (such as n.25 = 1/4, n.333 = 1/3, n.5 = 1/2, n.666 = 2/3, n.75 =
3/4), multiply all of the numbers derived by division by the denominator of the
recognized fraction to give whole numbers.
(4) Each whole number is the number of atoms of that element in the empirical formula.
% Composition of a compound is:
Chemistry) – Portage Learning|| A+
GRADE|| LATEST UPDATE 2026
2.1: MOLECULAR WEIGHT
A compound is made up of two or more elements combined in a definite ratio that is
represented by a molecular formula. Each of these elements has a certain atomic
weight, which can be found in the periodic table. The sum of the atomic weights of
the atoms in the molecular formula is called the formula weight or molecular
weight or formula mass.
Ca3(PO4)2
calcium phosphate
Molecular Weight =
3 Ca= 3 x 40.08=120.24
2 P=2 x 30.97=61.94
8 O=8 x 16.00=128.00
Total 310.18
2.2: MOLES
Chemical compounds react with one another in amounts that are based on their
molecular weights; this chemically reactive amount of compound is called a mole.
moles
= grams / molecular weight
Calculate the number of moles in 10.0 grams of each of the following compounds:
,Ca3(PO4)2 10.0 g ÷ 310.18 = 0.0322 mol (3 sig fig because of 10.0 g)
C3H5O2Cl 10.0 g ÷ 108.52 = 0.0921 mol
Al2(SO4)3 10.0 g ÷ 342.17 = 0.0292 mol
Ca3(PO4)2 0.0500 mol x 310.18 = 15.5 g
C3H5O2Cl 0.0500 mol x 108.52 = 5.43 g
Al2(SO4)3 0.0500 mol x 342.17 = 17.1 g
2.3: PERCENT COMPOSITION
he molecular formula represents the definite ratio of elements in a compound. The
weight of each element present in the compound represents a certain percentage of the
total weight of the compound. The percentage of each element present in a compound
is called the % composition of the compound.
The percentage of an element present in a compound can be calculated as shown
below:
% of an element = weight of element / molecular weight of compound x 100
Ca3(PO4)2
3 Ca = 3 x 40.08= 120.24
2 P = 2 x 30.97= 61.94
8 O = 8 x 16.00 = 128.00
, 310.18
% Ca = (120.24 ÷ 310.18) x 100 = 38.76%
%P = (61.94 ÷ 310.18) x 100 = 19.97%
%O = (128.00 ÷ 310.18) x 100 = 41.27%
2.4: EMPIRICAL FORMULA
If the formula of a compound is known, the % composition of the compound can be
determined. This process can be done in reverse: The formula of the compound can be
determined if the % of each element present in the compound is known. The formula
calculated from % composition is known as the empirical formula (or the simplest
formula). The actual molecular formula is some multiple of this simplest formula, which
is determined by knowing the molecular weight.
To determine the empirical formula:
(1) Divide each element % by its exact atomic weight to give a set of numbers.
(2) Divide the smallest of this set of decimal numbers into each of the numbers
(including itself) to yield a second set of numbers.
(3a) Round off each of the second set of decimal numbers to a whole number.
OR
(3b) If the numbers derived from the division in step 2 are recognized as exact decimal
equivalents of fractions (such as n.25 = 1/4, n.333 = 1/3, n.5 = 1/2, n.666 = 2/3, n.75 =
3/4), multiply all of the numbers derived by division by the denominator of the
recognized fraction to give whole numbers.
(4) Each whole number is the number of atoms of that element in the empirical formula.
% Composition of a compound is:
