Honors Chemistry In Class Notes
Indications of a Chemical Reaction
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


 What do we look for to determine if a chemical reaction is occurring? o Evolution of heat and light
 The burning of wood or coal
 The burning of magnesium
 This is not always an indication of a chemical reaction.
 Phosphorescence or fluorescence produce light without a chemical reaction.
 Compression of gases produce heat without a chemical reaction. o Production of a gas
 Gas bubbles when two things are mixed
 Zinc and acid
 Baking soda and vinegar o Formation of a precipitate
 A solid formed when mixing two things
 Zinc sulfate and barium nitrate o Color change
 Rusting causes iron to go from silvery to red o Formation of water
 Burning hydrocarbons produce carbon dioxide and water.
 Combining acids and bases form water.
 The formation of water does not necessarily indicate a chemical reaction.
 Water condenses on a cold glass.
1

The Mole: A Measurement of Matter 2
 Chemical formulas tell us about the number of atoms in a compound.
 In general, there are two kinds of chemical formulas. o In molecular formulas, the total number of atoms in the compound is used. o In empirical formulas, the lowest whole number ratio of atoms in the compound is used.
 In molecular formulas, the total number of atoms in the compound is used. o For example, benzene has 6 carbon atoms and 6 hydrogen atoms in each molecule.
 Therefore, its molecular formula is C
6
H
6
. o For example, acetic acid has 2 carbon atoms, 4 hydrogen atoms, and 2 oxygen atoms in each molecule.
 Therefore, its molecular formula is C
2
H
4
O
2
. o For example, propane gas has 3 carbon atoms and 8 hydrogen atoms in each molecule.
 Therefore, its molecular formula is C
3
H
8
.
 In empirical formulas, the lowest whole number ratio of atoms in the compound are used. o For example, benzene has 6 carbon atoms and 6 hydrogen atoms in each molecule.
 Therefore, its molecular formula is C
6
H
6
and its empirical formula is
CH (we divide all subscripts by 6).

Honors Chemistry In Class Notes
Indications of a Chemical Reaction
3 o For example, acetic acid has 2 carbon atoms, 6 hydrogen atoms, and 2 oxygen atoms in each molecule.
 Therefore, its molecular formula is C
2
H
4
O
2
. and its empirical formula is CH
2
O (we divide all subscripts by 2). o For example, propane gas has 3 carbon atoms and 8 hydrogen atoms in each molecule.
 Therefore, its molecular formula is C
3
H
8
and its empirical formula is also C
3
H
8
(there is no common divisor for all subscripts).
Percent Composition
 If we know the mass of a compound and the mass of one or more of the elements in the compound, then we can find the percent composition of those atoms in the compound. mass of atoms
 100 %
 For example, o 1.716 g of C, 0.577 g of H, and 2.286 g of O combine together to form 4.579 g of a compound.
Sample Problem 10.9
When a 13.60 g sample of a compound containing only magnesium and oxygen is decomposed, 5.40 g of oxygen is obtained. What is the percent composition of this compound?
Known: mass of compound = 13.60 g mass of O = 5.40 g mass of Mg = 13.60 g – 5.40 g = 8.20 g
Unknown: % Mg = ?%
% O = ?%

The Mole: A Measurement of Matter
 If we know the chemical formula of a compound, then we can find the percent composition of each of the atoms in the compound.
 We use the molar mass of the compound and the average atomic masses of the atoms in the compound. atomic mass of atoms molar mass of compound
 100 %
 For example: o the percent composition of benzene, C
6
H
6
, is: o the percent composition of acetic acid, C
2
H
3
O
2
, is:
Sample Problem 10.10
Propane (C
3
H
8
), the fuel commonly used in gas grills, is one of the compounds obtained from petroleum. Calculate the percent composition of propane
Known: molar mass of C
3
H
8
= 44.0 g mass of C = 3  12.01 g/mol = 36.03 g/mol
Unknown: % C = ?%
% H = ?% mass of H = 8  1.01 g/mol = 8.08 g/mol
4

Honors Chemistry In Class Notes
Indications of a Chemical Reaction
Practice Problems: find the percent composition of …
1.
C
6
H
14
2.
NaCl
3.
KNO
3
4.
CuSO
4
5.
FeCO
3
Empirical Formulas
 In empirical formulas, the lowest whole number ratios of atoms in a compound is used. o Benzene, C
6
H
6
, has an empirical formula of CH.
 A ratio of 6/6 = 1/1. o Acetic acid, C
2
H
4
O
2
, has an empirical formula of CH
2
O.
 A ratio of 2/4/2 = 1/2/1 o Propane, C
3
H
8
, has an empirical formula of C
3
H
8
.
 A ratio of 3/8 =3/8.
 If we know the percent composition of a compound, then we can find the empirical formula of the compound. o First, we assume that we have 100 g of the compound and find the mass of each atom in the compound. o Second, we find the number of mols of each atom. o Third, we find the lowest whole number ratio of mols.
5

The Mole: A Measurement of Matter
 For example, we have a compound with 52.2% C, 13.1% H, and 34.7% O. o First, we assume that we have 100 g of the compound and find the mass of each atom in the compound.
 52.2 g of C
 13.1 g of H
 34.7 g of O o Next, we we find the number of mols of each atom.
 n
C
= m
C
/M
C
= 52.2 g/12.0 g/mol = 4.35 mol C
 n
H
= m
H
/M
H
= 13.1 g/1.01 g/mol = 13.0 mol H
 n
O
= m
O
/M
O
= 34.7 g/16.0 g/mol = 2.17 mol O o Finally, we find the lowest whole number ratio of mols.
 n
C
/n
O
= 4.35 mol/2.17 mol = 2/1
 n
H
/n
O
= 13.0 mol/2.17 mol = 6/1 o This means that there is a ratio of 2:6:1 for C:H:O
 The empirical formula is C
2
H
6
O
 For example, we have a compound with 44.9% K, 18.4% S, and 36.7% O. o First, we assume that we have 100 g of the compound and find the mass of each atom in the compound.
 44.9 g of K
 18.4 g of S
 36.7 g of O o Next, we we find the number of mols of each atom.
 n
K
= m
K
/M
K
= 44.9 g/39.1 g/mol = 1.15 mol K
 n
S
= m
S
/M
S
= 18.4 g/32.1 g/mol = 0.573 mol S
 n
O
= m
O
/M
O
= 36.7 g/16.0 g/mol = 2.29 mol O o Finally, we find the lowest whole number ratio of mols.
 n
K
/n
S
= 1.15 mol/0.573 mol = 2/1
 n
O
/n
S
= 2.29 mol/0.573 mol = 4/1 o This means that there is a ratio of 2:1:4 for K:S:O
 The empirical formula is K
2
SO
4
6

Honors Chemistry In Class Notes
Indications of a Chemical Reaction
Sample Problem 10.11
A compound is analyzed and found to contain25.9% nitrogen and 74.1% oxygen. What is the empirical formula of the compound?
 First, we assume that we have 100 g of the compound and find the mass of each atom in the compound.
 Next, we find the number of mols of each atom.
 Finally, we find the lowest whole number ratio of mols.
Practice problems: find the empirical formulas of compounds that have the following percent compositions.
1.
11.21% H and 88.79% O
7
2.
92.24% C and 7.76% H
3.
39.99% C, 6.73% H, and 53.28% O
4.
24.27% C, 4.08%H, and 71.65% Cl
5.
39.96% N, 14.40% H, and 45.64% O

The Mole: A Measurement of Matter 8
Molecular Formulas
 In molecular formulas, the total number of atoms in the compound is used. o Benzene has a molecular formula of C
6
H
6
. o Acetic acid has a molecular formula of C
2
H
4
O
2
. o Propane has a molecular formula of C
3
H
8
.
 If we know the empirical formula of a compound, then we can find the molecular formula of the compound, if we know the molar mass of the compound. o First, we determine the empirical mass of the compound. o Second, we divide the molar mass by the empirical mass to get our multiplier. o Third, we multiply the subscripts of the empirical formula by the multiplier to get the molecular formula.
 For example, we have a compound with a molecular formula of CH
2
O and a molar mass of 180.16 g/mol. What is the molecular formula of the compound? o First, we determine the empirical mass of the compound.
EM = (1×12.01) + (2×1.01) + (1×16.00) = 30.03 o Next, we divide the molar mass by the empirical mass to find the multiplier.
M /EM = (180.16)/(30.03) = 5.999333999 ≈ 6 o Finally, we multiply the subscripts of the empirical formula by the multiplier to get the molecular formula.
CH
2
O ⇒ C
6
H
12
O
6
Sample Problem 10.12
Calculate the molecular formula of a compound whose molar mass is 60.0 g/mol and empirical formula is CH
4
N.

Honors Chemistry In Class Notes
Indications of a Chemical Reaction
Practice problems: find the molecular formulas of compounds that have the following empirical formulas and molar masses.
1.
CH
2
O; M = 60.0 g/mol
9
2.
CH
2
; M = 42.1 g/mol
3.
NaCO
2
; M = 134.0 g/mol
4.
CH
2
Cl; M = 98.96 g/mol
5.
NH
5
O; M = 35.06 g/mol