Download CH1710-Lecture #5-Chemical Formulas and Equations.key

Chapter 3
Calculations Related to
Chemical Formulas
Formula Mass
The mass of an individual molecule or formula unit
Also known as molecular mass or molecular weight
Sum of the masses of the atoms in a single molecule or formula
unit
mass of 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
mass of 1 formula unit of MgCl2
= 2(35.45 amu Cl) + 24.30 amu Mg = 95.20 amu
Molar Mass
Molar Mass of Compounds
The relative masses of molecules can be calculated
from atomic masses.
Formula Mass = 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
1 mole of H2O contains 2 moles of H and 1 mole of O.
molar mass = 1 mole H2O
= 2(1.01 g H) + 16.00 g O = 18.02 g
so the Molar Mass of H2O is 18.02 g/mole
1. How many moles are in 50.0 g of PbO2?
(Pb, 207.2 g/mol; O,16.00 g/mol)
g
PbO2
mol
Pb = 1 x 207.2
= 207.200000
O = 2 x 16.00 = 32.000000
PbO2 = 239.20g/mol
PbO2
239.2 g PbO2
1 mol PbO2
1 mol PbO2
239.2 g PbO2
1 mol PbO2
50.0 g PbO2 x 239.2 g PbO = 0.20903
mol PbO2
0.209
2
2. Find the number of CO2 molecules in 10.8 g of CO2.
1 mol CO2 = 6.022 x 1023 molecules
g
CO2
mol
CO2
C = 1 x 12.011 = 12.011000
O = 2 x 16.00 = 32.000000
CO2 = 44.01g/mol
10.8 g CO2 x
molec CO2
44.01 g CO2
1 mol CO2
1 mol CO2
44.01 g CO2
1 mol CO2 x 6.022 x 1023 molecules =
44.01 g CO2
1 mol CO2
23 molecules CO2
23
1.4778
x
10
1.48 x 10
3. What is the mass of 4.78 x 1024 NO2 molecules?
1 mol = 6.022 x 1023 ,1 mol NO2 = 46.01 g
molec NO2
6.022 x 1023 molec NO2
1 mol NO2
4.78 x
1024 molec
mol
NO2
1 mol NO2
6.022 x 1023 molec NO2
NO2
x
g
46.01 g NO2
1 mol NO2
1 mol NO2
6.022 x 1023 molec NO2
x
NO2
1 mol NO2
46.01 g NO2
46.01 g NO2
1 mol NO2
= 365.207
NO
365 ggNO
22
Percent Composition
Percent Composition
Percentage of each element in a compound
by mass
Can be determined from
1. the formula of the compound
2. the experimental mass analysis of the compound
4. Find the mass percent of Cl in C2Cl4F2
Mass Percent as a Conversion Factor
5. If NaCl is 39% sodium, find the mass of table salt containing 2.4 g of Na.
g
Na
g NaCl
100 g NaCl
39 g Na
100
g
NaCl
2.4 g Na x
39 g Na
39 g Na
100 g NaCl
=
6.1538
6.2 g NaCl
Empirical Formulas
Empirical Formula
Simplest, whole-number ratio of the
atoms of elements in a compound
Can be determined from elemental analysis
Finding an Empirical Formula from %
Composition
1. Convert the percentages to grams
2. Convert grams to moles
3. Write a pseudoformula using moles as subscripts
4. Divide all by smallest number of moles
5. Multiply all mole ratios by number to make all
whole numbers
Example:
Find the empirical formula of
aspirin with the given mass
percent composition
Given: C = 60.00%
H = 4.48%
O = 35.53%
Therefore, in 100 g of aspirin there are 60.00 g C,
4.48 g H, and 35.53 g O
However, ratios of mass are not useful in
determining empirical formulas !!
We must convert to a ratio of moles !!
Methodology for determining empirical formula:
gC
mol C
gH
gO
mol H
mol O
pseudoformula
CxHyOz
Manipulate
subscripts to obtain
whole-number ratio
empirical formula
CxHyOz
Calculate the moles of each element
Write a pseudoformula
C4.996H4.44O2.220
C4.996H4.44O2.220
Find the mole ratio
÷ 2.220
C2.25H2.00O1.00
Multiply subscripts by factor to give whole number
(x 4)
C9H8O4
6. Determine the empirical formula of stannous
fluoride, which contains 75.7% Sn (118.70 g/mol)
and the rest fluorine (19.00 g/mol)
Given: 75.7% Sn, (100 – 75.3) = 24.3% F
g
Sn
mol Sn
F
mol F
g
pseudo
formula
empirical
formula
6. Determine the empirical formula of stannous
fluoride, which contains 75.7% Sn (118.70 g/mol)
and the rest fluorine (19.00 g/mol)
÷0.6377
Element
Ratio in Grams
Molar Mass
Ratio in Moles
Ratio in Moles
Sn
75.7g
X
1mol
118.7g
0.6377
1
F
24.3g
X
1mol
19.00g
1.279
2.005
SnF2
7. Determine the empirical formula of magnetite, which
contains 72.4% Fe (55.85) and the rest oxygen (16.00)
Given: 72.4% Fe, (100 – 72.4) = 27.6% O
g
Fe
mol Fe
g
O
mol O
pseudo
formula
empirical
formula
7. Determine the empirical formula of magnetite, which contains
72.4% Fe (55.85) and the rest oxygen (16.00)
÷1.296
Element
Fe
O
Ratio in Grams
72.4g
27.6g
Molar Mass
x3
Ratio in Moles Ratio in Moles Ratio in Moles
X
1mol
55.85g
1.296
1
3
X
1mol
16.00g
1.725
1.33
4
Fe3O4
Chapter 3
Molecular Formulas
Molecular Formulas
The molecular formula is a multiple
of the empirical formula.
To determine the molecular formula you need to know the
empirical formula and the molar mass of the compound.
Find the molecular formula of butanedione if its empirical
formula is C2H3O and its molar mass (MM) is 86.03 g/mol.
Factor of 2
Molar Mass (emp. form.)
= 2 x (12.01 gC/molC) + 3 x (1.008 gH/molH) + 1 x (16.00 gO/molO) = 43.04
Molecular formula = C2H3O
x 2 = C4H6O2
g/mol
Practice – Benzopyrene has a molar mass of 252 g and an
empirical formula of C5H3. What is its molecular formula?
(C = 12.01, H=1.01)
C5 = 5(12.01 g) = 60.05 g
H3 = 3(1.01 g) = 3.03 g
C 5H 3
= 63.08 g
252
?
Molecular formula = {C5H3} x 4 = C20H12
Combustion Analysis
Combustion Analysis
(Generally used for organic compounds containing C, H, O)
A known mass of compound is burned in oxygen and the
masses of the products formed (CO2 and H2O) are determined.
By knowing the masses of the products and composition of
constituent elements in the product, the original amount of
constituent elements can be determined.
It is assumed that all of the carbon in the original
sample is converted to carbon dioxide and all of the
hydrogen in the sample is converted to water.
Combustion Analysis
Example of Combustion Analysis
Combustion of a 0.8233 g sample of a compound containing only
carbon, hydrogen, and oxygen produced the following:
This came from C.
This came from H.
CO2 = 2.445 g
H2O = 0.6003 g
Determine the empirical formula of the compound.
mol
CO2, H2O
g
CO2, H2O
g
C, H
mol
C,H,O
g
C, H
mol
C, H
g
O
pseudoformula
mol
O
empirical
formula
1 mole H = 1.008 g H
1 mole C = 12.01 g C
1 mole O = 16.00 g O
1 mole CO2 = 44.01 g CO2
1 mole H2O = 18.02 g H2O
1 mole C
1 mole CO2
2 mole H
1 mole H2O
molar masses of elements
molar masses of compounds
ratios of compounds formed in
combustion
In the original sample
1 mole H = 1.008 g H
1 mole C = 12.01 g C
1 mole O = 16.00 g O
molar masses of elements
In the original
sample
In the original sample
In the original sample
In the original sample
Pseudo formula
C0.05556H0.06662O0.00556
÷ 0.00556
Empirical
formula
8. Combustion of 0.844 g of caproic acid produced
0.784 g of H2O and 1.92 g of CO2.
If the molar mass of caproic acid is 116.2 g/mol,
what is the molecular formula of caproic acid?
In the original sample
C
H
O
g
0.524 0.0877 0.232
moles 0.0436 0.0870 0.0145
In the original sample
Molecular formula = {C3H6O} x 2 = C6H12O2
Chapter 3
Chemical Reactions and
Equations
Chemical Reactions
Reactions involve rearrangement and exchange
of atoms to produce new pure substances.
Reactants
Products
Chemical Equations
Shorthand way of describing a reaction
Provides information about the reaction
1. formulas of reactants and products
2. states of reactants and products
3. relative numbers of reactant and product molecules
Combustion of Methane
Methane gas reacts with oxygen gas to produce
carbon dioxide gas and gaseous water.
CH4(g) + O2(g) ➜ CO2(g) + H2O(g)
This equation reads
“1 molecule of CH4 gas combines with 1 molecule of O2
gas to make 1 molecule of CO2 gas and 1 molecule of
H2O gas.”
+
+
What about conservation of mass ??
+
1C + 4H
+
+ 2O
X
1C + 2O
+2H + O
Combustion of Methane, Balanced
To show the reaction obeys the Law of Conservation of Mass, the
equation must be balanced.
CH4(g) + O2(g) ➜ CO2(g) + H2O(g)
CH4(g) + O2(g) ➜ CO2(g) + 2 H2O(g)
CH4(g) + 2 O2(g) ➜ CO2(g) + 2 H2O(g)
+
+
“1 molecule of CH4 gas combines with 2 molecules of
O2 gas to make 1 molecule of CO2 gas and 2 molecules
of H2O gas.”
Symbols Used in Equations
Symbols used to indicate state after chemical:
(g) = gas; (l) = liquid; (s) = solid
(aq) = aqueous = dissolved in water
Energy symbols used above the arrow for
conditions for reactions:
Δ = heat
hν = light
shock = mechanical
elec = electrical
Steps in Balancing Equations
1. In compounds balance elements other than H and O.
a. Balance elements which occur only once on each side of
the equation.
b. Start with the elements which occur the most.
c. Balance polyatomic ions which do not change in the
reaction.
2. Be prepared to rebalance if something changes!!!
3. Balance H.
4. Balance O.
5. Balance elements which appear in their “elemental” forms.
1. When aluminum metal reacts with air, it produces a white,
powdery compound, aluminum oxide.
aluminum(s) + oxygen(g) ➜ aluminum oxide(s)
Al(s) + O2(g) ➜ Al2O3(s)
2 Al(s) + O2(g) ➜ Al2O3(s)
2 Al(s) + 3 O2(g) ➜ 2 Al2O3(s)
4 Al(s) + 3 O2(g) ➜ 2 Al2O3(s)
2. Solid phosphorous (P4) reacts with hydrogen gas to
produce phosphorous trihydride.
P4 (s)
+
H2 (g)
---------->
PH3 (g)
P4 (s)
+
H2 (g)
----------> 4 PH3 (g)
P4 (s)
+ 6 H2 (g)
----------> 4 PH3 (g)
3. Solid potassium chlorate decomposes to produce
oxygen gas and potassium chloride.
KClO3 (s) ------------------->
O2 (g) + KCl (s)
2 KClO3 (s) ------------------->
3 O2 (g) + KCl (s)
2 KClO3 (s) ------------------->
3 O2 (g) + 2 KCl (s)
4. Aqueous sulfuric acid reacts with solid sodium cyanide
to produce aqueous sodium sulfate and hydrogen cyanide gas.
H2SO4 (aq) +
NaCN (s) ------> Na2SO4 (aq) + HCN(g)
H2SO4 (aq) + 2 NaCN (s) ------> Na2SO4 (aq) + HCN(g)
H2SO4 (aq) + 2 NaCN (s) ------> Na2SO4 (aq) + 2 HCN(g)
5. Aqueous potassium phosphate reacts with aqueous calcium nitrate
to produce solid calcium phosphate and aqueous potassium nitrate.
K3PO4 (aq) + Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + KNO3 (aq)
K3PO4 (aq) + 3 Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + KNO3 (aq)
K3PO4 (aq) + 3 Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + 6 KNO3 (aq)
2 K3PO4 (aq) + 3 Ca(NO3)2 (aq) ----------> Ca3(PO4)2 (s) + 6 KNO3 (aq)
Organic Chemistry
Classifying Compounds
Organic vs. Inorganic
In the18th century, compounds from living things
were called organic; compounds from the nonliving
environment were called inorganic.
Organic compounds easily decomposed and could
not be made in the 18th-century lab.
Inorganic compounds were very difficult to
decompose, but could be synthesized.
Modern Classification of Compounds
Organic vs. Inorganic
Today we commonly make organic compounds
in the lab and find them all around us.
Organic compounds are mainly made of C and
H, sometimes with O, N, P, S, halogens, and
trace amounts of other elements.
The main element that is the focus of organic
chemistry is carbon.
Write a balanced equation for the
combustion of butane, C4H10.
C4H10 (g) +
O2 (g) ➝
CO2 (g)
+
H2O (g)
C4H10 (g) +
O2 (g) ➝ 4 CO2 (g)
+
H2O (g)
C4H10 (g) +
O2 (g) ➝ 4 CO2 (g)
+ 5 H2O (g)
C4H10 (g) + 13/2 O2 (g) ➝ 4 CO2 (g)
+ 5 H2O (g)
2 C4H10 (g) + 13 O2 (g) ➝ 8 CO2 (g) + 10 H2O (g)
Carbon Bonding in Organic Compounds
Carbon atoms bond almost exclusively
covalently in organic compounds.
When C bonds, it forms four covalent bonds.
Carbon is unique in that it can form limitless
chains of C atoms, both straight and branched,
and rings of C atoms.
Classifying Organic Compounds
There are two main categories of organic compounds,
hydrocarbons and functionalized
hydrocarbons.
All chemical
compounds
Inorganic
Organic
Hydrocarbons
Functionalized
Hydrocarbons
Hydrocarbons-Contain only Carbon and Hydrogen
Name
Molecular formula
Structure
Use
methane
CH4
Natural gas
propane
C3H8
LP gas
n-butane
C4H10
Butane lighters
n-pentane
C5H12
Gasolines
ethene
C2H4
Polymers
ethyne
C2H2
Welding
Families of Organic Compounds
C
O
Alkanes
C
C
Aldehydes
H
O
C
C
C
C
C
C
Alkenes
C
Ketones
C
O
C
O
H
Carboxylic
Acids
O
C
Aromatics
C
C
C
Alkynes
C
C
O
H
C
C
O
Alcohols
Esters
Amines
N
O
C
O
C
Ethers
C
N
Amides