Portage Chemistry 103 Module 2 Exam
Study 54 correct solutions.
Portage Chemistry 103 Module 2 Exam
Study 54 correct solutions.
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. - ANSWER 2.1: MOLECULAR WEIGHT
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 - ANSWER Ca3(PO4)2
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. - ANSWER 2.2: MOLES
= grams / molecular weight - ANSWER moles
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
,Portage Chemistry 103 Module 2 Exam
Study 54 correct solutions.
C3H5O2Cl 0.0500 mol x 108.52 = 5.43 g
Al2(SO4)3 0.0500 mol x 342.17 = 17.1 g - ANSWER Calculate the number of moles in 10.0 grams of each
of the following compounds:
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. -
ANSWER 2.3: PERCENT COMPOSITION
% of an element = weight of element / molecular weight of compound x 100 - ANSWER The percentage
of an element present in a compound can be calculated as shown below:
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% - ANSWER Ca3(PO4)2
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. - ANSWER 2.4: EMPIRICAL
FORMULA
(1) Divide each element % by its exact atomic weight to give a set of numbers.
, Portage Chemistry 103 Module 2 Exam
Study 54 correct solutions.
(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. - ANSWER To
determine the empirical formula:
32.37% Na; 32.37% Na ÷ 22.99 = 1.408
22.58%S; 22.58% S ÷ 32.07 = 0.704 (smallest number of the set)
45.05% O; 45.05% O ÷ 16.00 = 2.816
0.704 is the smallest of this set of numbers, so it is divided into each of the set of numbers.
Na = 1.408 ÷ 0.704 = 2 Na
S = 0.704 ÷ 0.704 = 1 S
O = 2.816 ÷ 0.704 = 4 O
Na2SO4 - ANSWER % Composition of a compound is:
When certain chemical materials are added to one another, they undergo a chemical reaction in which
the atoms of the materials separate from one another and recombine in a new way to form new
materials. This chemical reaction can be described by a chemical reaction equation in which the
reactants (starting materials) are written on the left side of the equation and the products (final
Study 54 correct solutions.
Portage Chemistry 103 Module 2 Exam
Study 54 correct solutions.
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. - ANSWER 2.1: MOLECULAR WEIGHT
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 - ANSWER Ca3(PO4)2
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. - ANSWER 2.2: MOLES
= grams / molecular weight - ANSWER moles
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
,Portage Chemistry 103 Module 2 Exam
Study 54 correct solutions.
C3H5O2Cl 0.0500 mol x 108.52 = 5.43 g
Al2(SO4)3 0.0500 mol x 342.17 = 17.1 g - ANSWER Calculate the number of moles in 10.0 grams of each
of the following compounds:
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. -
ANSWER 2.3: PERCENT COMPOSITION
% of an element = weight of element / molecular weight of compound x 100 - ANSWER The percentage
of an element present in a compound can be calculated as shown below:
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% - ANSWER Ca3(PO4)2
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. - ANSWER 2.4: EMPIRICAL
FORMULA
(1) Divide each element % by its exact atomic weight to give a set of numbers.
, Portage Chemistry 103 Module 2 Exam
Study 54 correct solutions.
(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. - ANSWER To
determine the empirical formula:
32.37% Na; 32.37% Na ÷ 22.99 = 1.408
22.58%S; 22.58% S ÷ 32.07 = 0.704 (smallest number of the set)
45.05% O; 45.05% O ÷ 16.00 = 2.816
0.704 is the smallest of this set of numbers, so it is divided into each of the set of numbers.
Na = 1.408 ÷ 0.704 = 2 Na
S = 0.704 ÷ 0.704 = 1 S
O = 2.816 ÷ 0.704 = 4 O
Na2SO4 - ANSWER % Composition of a compound is:
When certain chemical materials are added to one another, they undergo a chemical reaction in which
the atoms of the materials separate from one another and recombine in a new way to form new
materials. This chemical reaction can be described by a chemical reaction equation in which the
reactants (starting materials) are written on the left side of the equation and the products (final
