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Chemical Equations
Molar quantities
Balancing equations
Objectives
Perform calculations using moles,
Avogadro’s number and molar masses
Determine percent composition from formula
Determine empirical formula from percent
composition
Determine molecular formula
Write and balance chemical equations
Counting crows:
How many H atoms are in the molecule
C6H4(CH3)(C2H5)?
1. 12
2. 4
3. 7
4. 9
How many H atoms are in the molecule
C6H3(CH3)2(C2H5)?
1. 5
2. 8
3. 11
4. 14
How many O atoms in
CuSO4.5H2O?
1. 4
2. 5
3. 9
4. 14
Molecules or moles
The numbers (coefficients) in chemical equation
can refer to molecules
But for practical applications, we need a more
useful number: we cannot count molecules
The Mole
The mole is a unit of quantity used in
chemistry to measure the number of atoms
or molecules
DEFINITION:
The number of atoms in exactly 12 g of 12C
A mole of anything always has the same
number of particles: atoms, molecules or
potatoes – 6.02 x 1023 – Avogadro’s number
Mole conversions
Particle – mole conversions
23 1No particles = Moles (mol) x 6.02 x 10 ( )
mol
23
No particles Moles (mol) =
16.02 x 10 ( )
mol
Atomic and molecular weights
Two scales:
Atomic mass unit scale
The mass of an individual atom or molecule in
atomic mass units (amu)
Molar mass scale
The mass of a mole of atoms or molecules in
grams
Confusing?...
The Good News
The mass of a single atom or molecule in amu has
same numerical value as molar mass in grams
The atomic mass of carbon is 12 amu
The molar mass of carbon is 12 g/mol
The same is true for molecules and compounds
The formula mass of H2O is 18 amu (1+1+16)
The molar mass of H2O is 18 g/mol
Calculations with molar mass
Moles =
How many moles are in 13.88 g of lithium if
the atomic mass of Li is 6.94 amu?
2.00
massmolar
mass
How many H atoms in 1 mol of CH4? 1 mol = 6.02 x 1023 particles
How many moles of O2 in 64 g of oxygen?
Atomic mass of O = 16 AMU
Calculate the formula mass of Na2SO4 in AMU.
Use atomic mass Na = 23 AMU, O = 16 AMU, S = 32 AMU
Significance of formula unit
Ionic compounds do not contain molecules. Simplest
formula is the formula unit
Covalent compounds, the molecular formula is the
formula unit
Percent composition and empirical
formula
Chemical analysis
gives the mass % of
each element in the
compound
Molar masses give the
number of moles
Obtain mole ratios
Determine empirical
formula
Determining percent composition
Percent composition is obtained from the actual masses.
Example: Sample contained 0.4205 g of C and 0.0795 g of H.
Total mass = 0.5000 g (0.4205 + 0.0795)
Therefore: in 100 g there are:
(84.10 %)
(15.90 %)
Percent composition: 84.10 % C, 15.90 % H
100 g0.4205 = 84.10 g C
.5000 gx
100 g0.0795 = 15.90 g H
.5000 gx
Percent composition from formula
What is percent composition of C5H10O2? 1 mol C5H10O2 contains 5 mol C, 10 mol H and 2 mol O
atoms
Mass of each element
Total mass = 102.13 g
12.01 g C5 mol C = 60.05 g C
1 mol C1.008 g H
10 mol H = 10.08 g H1 mol H
16.00 g O2 mol O = 32.00 g O
1 mol O
Convert masses into percents
Percent composition:
58.8 % C, 9.870 % H, 31.33 % OCheck result:
58.80 % C + 9.870 % H + 31.33 % O = 100.00%
5 10 2
60.05 g C% C = x100 = 58.80 % C
102.13 g C H O
5 10 2
10.08 g H% H = x100 = 9.870 % H
102.13 g C H O
5 10 2
32.00 g O% O = x100 = 31.33 % O
102.13 g C H O
Empirical formula from percent
composition: 84.1 % C, 15.9 % H
1. Convert percents into moles84.10 g of C ≡ 7.00 mol C
15.9 g of H ≡ 15.8 mol H
2. Determine mole ratioMole ratio H:C =
Simplest formula (decimal form): C1H2.26
Make smallest integers by multiplying
C4H9
May require rounding. Errors in real data cause problems
Do percent composition and empirical formula exercises
84.10 g C
12.00 g/mol
15.9 g H
1.008 g/mol
15.8 mol H2.26 :1
7.00 mol C
What is percent C content of C2H6? Molar mass C = 12 g/mol; molar mass H = 1 g/mol
Empirical formula with more than two elements
Percent composition of vitamin C is: 40.9 % C, 4.58 % H, 54.5 % O
1. Convert into moles
2. Determine mole ratios
3. Find lowest whole numbers
Rounding or not: the role of chemical
intuition
Formulae are always written with integers
Experimental ratios are always fractions
Two choices:
Round to nearest whole number
Multiply top and bottom to find ratio of whole
numbers with same value
Choice depends on the type of substance
Hydrocarbons: A case for not
rounding
There are millions of different hydrocarbons
What if H:C is 2.20 rather than 2.26? An error of only 3 %
Formula becomes C5H11 rather than C4H9
What if H:C is 2.30 rather than 2.26? An error of only 2 %
Formula becomes C3H7
All formulae are reasonable
So how do I know what the composition is?
Additional knowledge about the substance is helpful: melting point, boiling point, molar mass
Inorganic compounds: Rounding
makes sense
Inorganic compounds tend to have few
compositions
Iron forms three oxides: FeO, Fe3O4 and
Fe2O3
Experimental formula FeO1.75 would indicate
Fe2O3 not FeO2
Practice empirical formula problem A compound contains 62.1 % C, 5.21 % H, 12.1 % N and
20.7 % O. What is the empirical formula?
Empirical and molecular formula
Percent composition gives the empirical(simplest) formula. It says nothing about the molecular formula.
Molecular formula describes number of atoms in the molecule
May be much larger than the empirical formula in the case of molecular covalent compounds
For ionic compounds empirical formula = “molecular” formula
Elements and compounds can have molecular
formula different from simplest formula
Substance Empirical
formula
Molecular
formula
Substance Empirical
formula
Molecular
formula
Sulphur S S8Phosphorous P P4
Benzene CH C6H6Acetylene CH C2H2
Ethylene CH2 C2H4Cyclohexane CH2 C6H12
Determination of molecular formula
Require:
1. Empirical formula from percent composition analysis
2. Molar mass from some other source
Number of empirical formula units in molecule:
There are n (AaBbCc) in molecule:
Molecular formula is AnaBnbCnc
Molar mass
Empirical formula massn
Molecular formula of vitamin C
Empirical formula of vitamin C is C3H4O3
Molar mass vitamin C is 176.12 g/mol
Mass of empirical formula = 88.06 g/mol
(3 x 12.01 + 4 x 1.008 + 3 x 16.00)
Number of formula units per molecule =
Molecular formula = 2(C3H4O3) = C6H8O6
Molar mass vitamin C 176.122
Empirical formula mass vitamin C 88.06n
Ionic solids do not have molecular
formulas
Infinite lattices like ionic solids and
covalently bonded lattices are not molecular.
The formula used for an ionic compound is
the same as the empirical formula – with
one or two exceptions Hg2Cl2 rather than HgCl
Structural formula provides more
information
The molecular formula indicates the number of atoms in the molecule
The structural formula indicates how those atoms are arranged
C2H6O is the molecular formula for ethanol and
dimethyl ether
Structural formula for ethanol is CH3CH2OH
Structural formula for ether is CH3OCH3
In a recipe we would need to use the structural
formula to identify the correct reagent
In some cases the mole contents of the
compound are obtained indirectly
Analysis of the
hydrocarbon is performed
by combustion.
Mole ratios of the elements
are derived from the mole
ratios of the combustion
products CO2 + H2O
(1 mol H2O ≡ 2 mol H)
(1 mol CO2 ≡ 1 mol C)
The molecular or empirical formula can
be used to determine the percent
composition
Formula of aspirin is C9H8O4
Molar mass is 180 g
Mass of C = 108 g
Mass of H = 8 g
Mass of O = 64 g
% C = 108/180 x 100 %
% H = 8/180 x 100 %
% O = 64/180 x 100 %
The chemical equation
aA + bB = cC + dD
The Law of Conservation of Matter states that matter is neither created nor destroyed
All the atoms on the left must be the same as those on the right
Reactant
sideProduct
side
coefficient ELEMENT or
COMPOUND
Chemical book-keeping
The key to writing correct equations is to ask
the question, “Have I gained or lost any
atoms?”
Another thing is to put down the correct
formula for each reactant or product
Formulas cannot be changed in order to
balance the equation
In the reaction of hydrogen with oxygen to produce
water, the reactants are the elements H2 and O2,
and the product is H2O
Count the atoms: 4 H and 2 O 4 H and 2 O
The big number
multiplies every
atom after it
The subscript
only multiplies
the atom before
it
Molecular representation of the
same reaction
Balance the equations
CH4 + O2 = CO2 + H2O CH4 + 2O2 = CO2 + 2H2O
C3H8 + O2 = CO2 + H2O C3H8 + 5O2 = 3CO2 + 4H2O
N2 + H2 = NH3
N2 + 3H2 = 2NH3
Do balancing equation exercises
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