Transcript Formula Mass

Formula Mass
 the mass of an individual molecule or formula unit
 also known as molecular mass or molecular weight
(MW)
 sum of the masses of the atoms in a single molecule or
formula unit
 whole = sum of the parts!
mass of 1 molecule of H2O
= 2(1.01 amu H) + 16.00 amu O = 18.02 amu
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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
 since 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
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Example
 Example – Find the number of CO2 molecules in
10.8 g of dry ice
 How many molecules are in 50.0 g of PbO2?
(PbO2 = 239.2)
Percent Composition
Percentage of each element in a compound



1.
2.

By mass
Can be determined from
the formula of the compound
the experimental mass analysis of the compound
The percentages may not always total to 100% due to
rounding
part
Percentage 
 100%
whole
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Example
 Calculate the mass percent composition of calcium chloride
 Calculate the mass percent of oxygen in acetic acid
(C2H4O2)
Mass Percent as a
Conversion Factor
 the mass percent tells you the mass of a constituent
element in 100 g of the compound
 the fact that CCl2F2 is 58.64% Cl by mass means that 100 g
of CCl2F2 contains 58.64 g Cl
 this can be used as a conversion factor
 100 g CCl2F2 : 58.64 g Cl
58.64 g Cl
g CCl2 F2 
 g Cl
100 g CCl2 F2
100 g CCl2 F2
g Cl 
 g CCl2 F2
58.64 g Cl
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Example
 Find the mass of table salt containing 2.4 g of Na
 Benzaldehyde is 79.2% carbon. What mass of
benzaldehyde contains 19.8 g of C?
Conversion Factors in Chemical
Formulas
 chemical formulas have inherent in them relationships
between numbers of atoms and molecules
 or moles of atoms and molecules
 these relationships can be used to convert between amounts
of constituent elements and molecules
 like percent composition
 E. g Find the mass of hydrogen in 1.00 gal of water
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Grams to Grams conversion
 Grams A  Moles A  Moles B  Grams B
 How many grams of sodium are in 6.2 g of NaCl? (Na
= 22.99; Cl = 35.45)
Empirical Formula
 simplest, whole-number ratio of the atoms of elements in a
compound
 can be determined from elemental analysis
 masses of elements formed when decompose or react
compound

combustion analysis
 percent composition
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Finding an Empirical Formula
1)
convert the percentages to grams
a)
b)
2)
assume you start with 100 g of the compound
skip if already grams
convert grams to moles
use molar mass of each element
a)
3)
4)
write a pseudoformula using moles as subscripts
divide all by smallest number of moles
if result is within 0.1 of whole number, round to whole
number
a)
5)
multiply all mole ratios by number to make all whole
numbers
a)
b)
if ratio ?.5, multiply all by 2; if ratio ?.33 or ?.67, multiply all
by 3; if ratio 0.25 or 0.75, multiply all by 4; etc.
skip if already whole numbers
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Example 3.17
 Laboratory analysis of aspirin determined the following
mass percent composition. Find the empirical formula.
C = 60.00%
H = 4.48%
O = 35.53%
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Example
 Determine the empirical formula of stannous fluoride,
which contains 75.7% Sn (118.70) and the rest fluorine
(19.00)
 Determine the empirical formula of hematite, which
contains 72.4% Fe (55.85) and the rest oxygen (16.00)
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
Molar Mass molecular formula
 multiplyin g factor, n
Empirical Formula Mass
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Example
 Benzopyrene has a molar mass of
252 g/mol and an empirical formula of C5H3. What is
its molecular formula? (C = 12.01, H=1.01)
 A compound has the emperical formular CH and a
molar mass of 78.11 g/mol. Find its molecular formula
Combustion Analysis
 a common technique for analyzing compounds is to burn
a known mass of compound and weigh the amounts of
product made
 generally used for organic compounds containing C, H, O
 by knowing the mass of the product and composition of
constituent element in the product, the original amount
of constituent element can be determined
 all the original C forms CO2, the original H forms H2O, the
original mass of O is found by subtraction
 once the masses of all the constituent elements in the
original compound have been determined, the empirical
formula can be found
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Combustion Analysis
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Combustion Analysis
 Unknown formula: CxHyOx (Oxygen can be replaced with
other nonmetal)
 gCO2  moles CO2  moles C  gC
 gH2O  moles H2O  moles H  gH
 g O = g sample – (g H + g C)
 gO  moles O
 Follow steps in determine the empirical formula and
molecular formula
Example 3.20
 Combustion of a 0.8233 g sample of a compound containing only
carbon, hydrogen, and oxygen produced the following:
CO2 = 2.445 g
H2O = 0.6003 g
Determine the empirical formula of the compound
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Example
 Upon combustion, a compound containing only carbon
and hydrogen produced 1.60g CO2 and 0.819g H2O.
Find the empirical formula
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Chemical Reactions
 Reactions involve chemical changes in matter resulting in
new substances
 Reactions involve rearrangement and exchange of atoms to
produce new molecules
 Elements are not transmuted during a reaction
Reactants 
Products
21
Chemical Equations
 Shorthand way of describing a reaction
 Provides information about the reaction
 Formulas of reactants and products
 States of reactants and products
 Relative numbers of reactant and product molecules that are
required
 Can be used to determine weights of reactants used and
products that can be made
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Combustion of Methane
 methane gas burns to produce carbon dioxide gas
and gaseous water
 whenever something burns it combines with O2(g)
CH4(g) + O2(g)  CO2(g) + H2O(g)
O
H
H
C
H
H
+
O
O
C
+
O
H
H
O
1C+4H
+
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2O
1C+2O +2H+O
1C+2H+3O
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Combustion of Methane
Balanced
 to show the reaction obeys the Law of Conservation
of Mass, it must be balanced
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
H
H
C
H
H
O
+
O
+
O
O
C
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+
H
H
+
O
O
1C + 4H + 4O
O
O
H
H
1C + 4H + 4O
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Chemical Equations
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
 CH4 and O2 are the reactants, and CO2 and H2O are
the products
 the (g) after the formulas tells us the state of the
chemical
 the number in front of each substance tells us the
numbers of those molecules in the reaction
 called the coefficients
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Chemical Equations
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
 this equation is balanced, meaning that there are equal
numbers of atoms of each element on the reactant and
product sides
 to obtain the number of atoms of an element, multiply the
subscript by the coefficient
1C1
4H4
4O2+2
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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 decomposition
reactions
 D = heat
 hn = light
 shock = mechanical
 elec = electrical
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Example
 Write a balanced equation for the combustion of
butane, C4H10
Practice
when aluminum metal reacts with air, it produces a white,
powdery compound aluminum oxide
 reacting with air means reacting with O2
aluminum(s) + oxygen(g) aluminum oxide(s)
Al(s) + O2(g)  Al2O3(s)
Acetic acid reacts with the metal aluminum to make aqueous
aluminum acetate and gaseous hydrogen
 acids are always aqueous
 metals are solid except for mercury
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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 18th century lab
 inorganic compounds very difficult to decompose, but able
to be synthesized
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Modern Classifying 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, and trace amounts of other
elements
 the main element that is the focus of organic chemistry is
carbon
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Carbon Bonding
 carbon atoms bond almost exclusively covalently
 compounds with ionic bonding C are generally inorganic
 when C bonds, it forms 4 covalent bonds
 4 single bonds, 2 double bonds, 1 triple + 1 single, etc.
 carbon is unique in that it can form limitless chains of C atoms,
both straight and branched, and rings of C atoms
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Carbon Bonding
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Classifying Organic
Compounds
 there are two main categories of
organic compounds, hydrocarbons
and functionalized hydrocarbons
 hydrocarbons contain only C and H
 most fuels are mixtures of
hydrocarbons
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Classifying Hydrocarbons
 hydrocarbons containing only single bonds are called
alkanes
 hydrocarbons containing one or more C=C are called
alkenes
 hydrocarbons containing one or more CC are called
alkynes
 hydrocarbons containing C6 “benzene” ring are called
aromatic
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Naming
Straight
Chain
Hydrocarbons
 consists of a base name to indicate the number of carbons in
the chain, with a suffix to indicate the class and position of
multiple bonds
 suffix –ane for alkane, –ene for alkene, –yne for alkyne
Base Name
methethpropbutpent-
No. of C
1
2
3
4
5
Base Name
hexheptoctnondec-
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No. of C
6
7
8
9
10
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Functionalized Hydrocarbons
 functional groups are non-carbon groups that are on the
molecule
 substitute one or more functional groups replacing H’s on
the hydrocarbon chain
 generally, the chemical reactions of the compound are
determined by the kinds of functional groups on the
molecule
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Functional Groups
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