Transcript Chapter 3 Sections 3.5-3.6 Powerpoint

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•There are two common ways to describe the composition
of a compound: the numbers of its constituents atoms
(chemical formula) or by percentages (by mass) of its
elements (percent composition).
•The mass percents of the elements in a compound can be
determined by comparing the mass of each element present
in 1 mole of the compound to the total mass of 1 mole of
the compound.
Mass percent of element
mass of element in 1 mol of compound
=
x 100%
mass of 1 mol of compound
Example:
Find the percent composition of the compound N2O5.
Solution: First determine the mass of each element in the
compound.
We know that 1 mol of N2O5 must contain 2 mol N and 5
mol O. The number of grams of N and O are found as
follows:
14.01 g
Mass of N = 2 mol x
= 28.02 g N
1 mol
16.00 g
Mass of O = 5 mol x
= 80.00 g O
1 mol
Total mass of compound =
28.02 g N + 80.00 g O = 108.02 g N2O5
The mass percent of N and O in N2O5 can be computed by
comparing the mass of N and O in 1 mole of N2O5 to the total
mass of 1 mole of N2O5.
28.02 g
Mass percent of N =
x 100% = 25.94% N
108.02 g
80.00 g
Mass percent of O =
x 100% = 74.06% O
108.02 g
Notice the percents add up to 100.00%; this provides a check
that the calculations are correct.
Example:
A sample of a liquid with a mass of 8.657 g was decomposed
into its elements and gave 5.217 g of carbon, 0.9620 g of
hydrogen, and 2.478 g of oxygen. What is the percent
composition of this compound?
Solution:
We can apply an equation for the percent by mass of an
element to each element. The “mass of whole sample” here
is 8.657 g so we can take each element in turn and perform
the calculations.
mass of element
% by mass of element =
x 100%
mass of whole sample
For C:
5.217 g
x 100% = 60.26% C
8.657 g
For H:
0.9620 g
x 100% = 11.11% H
8.657 g
For O:
2.478 g
x 100% = 28.62% O
8.657 g
The percentages must add up to 100%, allowing for small
differences caused by rounding.
Sum of percentages: 99.99%
•When a new compound is prepared, one of the first items
of interest is the formula of the compound.
•For example, the compound that forms when phosphorus
burns in oxygen consists of molecules with the formula
P4O10.
•When a formula gives the composition of one molecule, it
is called a molecular formula.
•Molecular formula: the exact formula of a molecule,
giving the types of atoms and the number of each type.
•Notice that in the formula P4O10 that both the subscripts 4
and 10 are divisible by 2, so the smallest numbers that tell
us the ratio of P to O are 2 and 5.
•In a simpler kind of formula, the empirical formula, the
subscripts are the smallest whole numbers that describe the
ratios of the atoms in the substance.
•Empirical formula: the simplest whole number ratio of
atoms in a compound.
•It is quite common for the empirical and molecular
formulas to be different; some examples are shown below.
•Example: Calculating an Empirical Formula from Percent
Composition
A white powder used in paints, enamels, and ceramics has
the following percent composition: Ba, 69.58%, C, 6.090%,
and O, 24.32%. What is its empirical formula?
Solution:
Since mass percentage gives the number of grams of a
particular element per 100 grams of compound, base the
calculation on 100 grams of compound. Each percent will
then represent the mass in grams of that element.
If we have 100 g of the above compound, based on the
percentages we would have 69.58 g of Ba, 6.090 g of C, and
24.32 g of O.
Determine the number of moles of each element present in
100 grams of compound using the atomic masses of the
elements present.
Ba:
C:
O:
1 mol Ba
69.58 g Ba x
= 0.5067 mol Ba
137.33 g Ba
1 mol C
6.090 g C x
= 0.5071 mol C
12.01 g C
1 mol O
24.32 g O x
= 1.520 mol O
16.00 g O
Divide each value of the number of moles by the smallest of
the values. If each resulting number is a whole number
(after appropriate rounding), these numbers represent the
subscripts of the elements in the empirical formula.
If the numbers obtained in the previous process are not
whole numbers, multiply each number by an integer so that
the results are all whole numbers.
0.5067
Ba :
=1
The coefficients are acceptably
0.5067
close to whole numbers, so the
0.5070
C:
= 1.001 = 1 empirical formula is
0.5067
BaCO3.
1.520
O:
= 3.000 = 3
0.5067
Example: One compound of mercury with a molar mass of
519 contains 77.26% Hg, 9.25% C, and 1.17% H (with the
balance being O). Calculate the empirical and molecular
formulas, arranging the symbols in the order HgCHO.
Solution:
First obtain the empirical formula.
Assume a 100 g sample therefore, we have 77.26 g Hg, 9.25
g C, 1.17 g H, and 12.32 g O.
The amount of oxygen was determined by subtracting the
total amounts of the other three elements from the total
assumed mass of 100 g.
Convert each of these masses into moles.
1 mol Hg
Hg : 77.26 g Hg x
= 0.3852 mol Hg
200.59 g Hg
1 mol C
C : 9.25 g C x
= 0.770 mol C
12.01 g C
1 mol H
H : 1.17 g H x
= 1.16 mol H
1.01 g H
1 mol O
O : 12.32 g O x
= 0.7700 mol O
16.00 g O
Divide each number of moles by the smallest number of
moles.
0.3852
Hg :
= 1.000 = 1 Hg
Empirical Formula:
0.3852
HgC2H3O2
0.770
C:
= 2.00 = 2 C
0.3852
1.16
H:
= 3.01 = 3 H
0.3852
0.7700
O:
= 1.999 = 2 O
0.3852
Compute the mass corresponding to the empirical formula.
(1 Hg x 200.59 amu) + (2 C x 12.01 amu) + (3 H x 1.01 amu)
+ (2 O x 16.00 amu) = 259.64 g/mol
Calculate the ratio:
Molar mass
519 g / mol
=
= 2.00
Empirical formula mass 259.64 g / mol
The integer from the previous step represents the number
of empirical formula units in one molecule. When the
empirical formula subscripts are multiplied by this integer,
the molecular formula results.
Molecular formula = 2(HgC2H3O2) = Hg2C4H6O4