Matter and Measurement

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Chemical Formulas
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• MOLECULAR FORMULAS representation of molecules
  in terms of their constituent atoms - H2, H2O

DIATOMIC - two atoms example H2, N2, CO (carbon
monoxide) TRIATOMIC - three atoms example CO2
(carbon dioxide), O3(ozone) POLYATOMIC - more
than three atoms examples NH3 (ammonia), CH4
(methane), C2H6O2 (ethylene glycol)
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• What limits the boundary of a molecule?
• Atoms in molecules are held together by strong
  interactions called CHEMICAL BONDS
• Interactions between neutral atoms in a molecule
  (e.g. H2, H2O) is called a COVALENT bond, forming
  covalent compounds

Interactions between charged elements (IONS)
result in a different kind of chemical bond,
called an IONIC BOND
Compounds formed via interactions between ionic
(charged) elements are called ionic compounds.
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In a crystal of NaCl a Na ion interacts with the
neighboring Cl- ion, hence a NaCl molecule does
not exist.
However, if one counted the number of sodium ions
and the number of Cl- ions in a given volume we
would find that the ratio of Na Cl- is 11.
A crystal of sodium chloride is electrically
neutral and its CHEMICAL FORMULA can be expressed
as NaCl.
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Chemical formulas indicate the constituent
elements in a compound (covalent or ionic) The
term molecular formula refers specifically to
covalent compounds. How are chemical formulas
determined?
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• ELEMENTAL ANALYSIS
• Determination of the relative amounts of the
  elements in a compound
• Determines the EMPIRICAL FORMULA which is the
  simplest possible formula and indicates the
  relative amounts of constituent elements
• For example, the molecular formula of hydrogen
  peroxide is H2O2 its empirical formula is HO.
• The empirical and the chemical formula can be the
  same, for example, H2O.
• For ionic compounds, the empirical formula is the
  same as the chemical formula (NaCl)

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• Problem
• An oxide of nitrogen is analyzed and found to
  contain 25.9 N and 74.1 O. What is the
  empirical formula of the compound?
• In 100.0 g of compound
• moles of N 25.9 g N / (14.01 g/mol) 1.85
  mol
• moles of O 74.1 g O / (16.00 g/mol) 4.63
  mol
• Ratio of O N 2.51
• Must be whole numbers N2O5

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Combustion Analysis - elemental analysis of
organic compounds. Organic compounds contain
predominantly C and H. Minor elements include O,
N, S, Cl, P, etc. In combustion analysis the
organic compound is burnt in the presence of
oxygen (O2) to form carbon dioxide (CO2), water
(H2O) and other gases (nitrogen (N2), sulfur
dioxide (SO2))
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A sample of a compound that is known to contain
only carbon, hydrogen, and oxygen is combusted,
and the CO2 and H2O produced are trapped and
weighed. The original sample weighed 8.38 g and
yielded 16.0 g CO2 and 9.5 g H2O. What is the
empirical formula?
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Need to determine the number of moles of C and H
in the compound, and then determine the mass of O
in the compound. Moles of C (16.0 g CO2 / 44.0
g/mol CO2) x (1 mol C/1 mol CO2) 0.364 mol
C Moles of H (9.80 g H2O / 18.0 g/mol H2O) x
(2 mol H / 1 mol H2O) 1.09 mol H
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Mass of C in compound 0.364 mol C x 12.0 g/mol
C 4.37 g C Mass of H in compound 1.09
mol H x 1.01 g/ mol H 1.10 g H Mass of
O in compound 8.38 g - 4.37 g - 1.10 g 2.91 g
O Moles of O in compound (2.91 g O / 16.0 g/mol
O) 0.182 mol O Mole ratio of C H O is
0.364 1.09 0.182 or 261 Hence empirical
formula is C2H6O
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• Determining Chemical Formulas
• The empirical formula tells you the simplest
  ratio of the individual elements in the compound.
  
• For an ionic compound this information is enough.
  
• For a molecular compound this may not be enough
  since the empirical formula may not be the
  molecular formula.
• Knowledge of the MOLAR MASS of the compound and
  its empirical formula, allows the molecular
  formula to be determined.

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• Elemental analysis of a sugar shows that it
  consists of 40.0 carbon (C), 6.7 hydrogen (H),
  53.3 oxygen (O). The molar mass of the compound
  was found to be 180.0 g/mol. What is the
  molecular formula of the compound?

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moles of C in 100.0 g of compound (mass of C
g)/(atomic mass g/mol) 40.0/12.01 3.33 mol C
in 100.0 g of compound moles of H in 100.0 g of
compound (mass of H g)/(atomic mass g/mol)
6.7/1.01 6.7 mol H in 100.0 g of
compound moles of O in 100.0 g of compound
(mass of O g)/(atomic mass g/mol) 53.3/16.00
3.33 mol O in 100.0 g of compound Ratio of CHO
121 hence empirical formula is CH2O
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• Molar mass of empirical formula 12.01 2(1.01)
  16.00 30.0 g/mol
• Ratio of molar mass of compound molar mass of
  empirical formula
• 180.0/30.0 6.0
• molecular formula is (CH2O) 6 or C6H12O6

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Chemical Reactions Equations

• In a chemical reaction elements and/or compounds
  collide, interact and react to form new
  compounds.
• The compounds that come together are called the
  REACTANTS and the new compounds formed are the
  PRODUCTS.

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• An important aspect of a chemical reaction is
  that MASS IS ALWAYS CONSERVED - i.e. the total
  mass of the reactants must equal the total mass
  of the products.
• To ensure that mass is conserved, we have to keep
  track of the number of atoms of each element in
  the reactants and number of atoms of each element
  in the products

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• Writing Chemical Equations

Mass has not been conserved - equation is not
BALANCED
Need to make sure that the number of atoms of a
given element is the same on either side.
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• H2 O2 ? H2O

H is balanced, but O is not. To balance O,
multiply H2O by 2
H2 O2 ? 2 H2O
Now O is balanced but not H (4 Hs on right, 2 on
left) Multiply H2 by 2
2H2 O2 ? 2 H2O
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2H2 O2 ? 2 H2O

• Now the equation is balanced.
• However, for a complete chemical equation the
  state of all reactants and products must be
  included

2H2(g) O2 (g) ? 2 H2O (g)
g - gas l - liquid s- solid aq - denotes a
solution of a solute in water (solvent)
Note the subscripts in the molecular formula
must never be changed in balancing equations.
Changing the subscript corresponds to a different
molecular formula, hence a different molecule,
with completely different properties
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• Rules to balance chemical equations
• Consider the reaction of methane, CH4, burning in
  air to produce CO2 and H2O (combustion reaction)
• CH4 O2 ? CO2 H2O
• It is usually best to first balance those
  elements that occur in the fewest chemical
  formulas on each side of the equation.
• So in this example balance C and H first

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• C is already balanced. However 4 Hs on the left
  and 2 on the right
• Hence,
• CH4 O2 ? CO2 2H2O
• Now balance O
• CH4 2O2 ? CO2 2H2O
• Now specify the state of each reactant and
  product
• CH4(g) 2O2(g) ? CO2(g) 2H2O(g)

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Ag2S(s) KCN(aq) O2(g) H2O(l) ?
KAg(CN)2(aq) S(s) KOH(aq) Sometimes it
is hard to balance a chemical reaction by
inspection Can the use algebraic expressions
to balance equations
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a Ag2S(s) b KCN(aq) c O2(g) d H2O(l) ? 1
KAg(CN)2(aq) e S(s) f KOH(aq) Element Reacta
nt Product Ag 2 a 1 S a e K b 1
f C b 2 N b 2 O 2c d f H 2d f
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2 a 1 a 0.5 b 2 a e e 0.5 b 1
f f 1 2d f d 0.5 2c d f c
0.25 Divide all by 0.25 a 2 b 8 c 1 d
2 e 2 f 4 Hence 2 Ag2S(s) 8 KCN(aq)
O2(g) 2 H2O(l) ? 4 KAg(CN)2(aq) 2 S(s) 4
KOH(aq)
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Chemical Stoichiometry

• Reactants are consumed and products are formed in
  definite proportions.
• These proportions are given by the coefficients
  in the balanced equations for chemical reactions
• The calculation of the quantities of reactants
  and products is called STOICHIOMETRY.
• Stoichiometry is the use of chemical equations to
  calculate quantities of substances that take part
  in chemical reactions.

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• To do a stoichiometric calculation, the chemical
  equation for the calculation must be balanced.
• Equations are read in terms of moles of reactants
  and product
• 2H2(g) O2 (g) ? 2H2O(g)
• 2 moles of H2 (g) reacts with 1 mole of O2 (g) to
  form 2 moles of H2O(g)
• Or
• 2No molecules of H2 (g) No molecules of O2 (g)
  ? 2No molecules of H2O(g)

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• Problem
• Consider the reaction of 100 g of H2(g) with
  sufficient O2 (g) to produce the stoichiometric
  quantity of H2O(g). (stoichiometric quantities
  are the exact amounts of reactants and products
  predicted by balanced equations). Calculate the
  mass of H2O formed.
• Need a balanced equation for this reaction
• 2H2(g) O2 (g) ? 2H2O(g)

To find the mass of water formed, need to find
the number of moles of H2 that reacted
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• 2 moles of H2 reacts with 1 mole of O2 to form 2
  moles of H2O
• gt 49.5 moles of H2 will form 49.5 moles of H2O
• Hence, the mass of H2O(g) formed
• (49.5 moles H2O) x (18.02 g/mol) 892 g H2O

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• Problem Diethyl ether (C4H10O) combusts in air,
  reacting with O2 to form H2O and CO2. How many
  grams of CO2 would be produced if 350.0 mL of
  diethyl ether were combusted in an unlimited
  amount of oxygen? The density of diethyl ether
  is 0.713 g/mL.

First write a balanced equation ?C4H10O(l)
?O2(g) ? ?H2O(l) ?CO2(g) Then convert the
volume of diethyl ether to mass Next, determine
the number of moles of diethyl ether
reacted Determine the number of moles of CO2
formed based on the stoichiometry of the
reaction Finally determine the mass of CO2
produced Answer 592 g CO2
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For reactions occurring in solution, instead of
using moles as a unit of quantity, define
moles/liter MOLARITY Molarity is defined as the
number of moles of a solute per liter of
solution. Note the solute can be a solid,
liquid, or gas dissolved in a solvent.
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Problem For the reaction 2 PbO2(s) 4 HNO3(aq)
? 2 PbNO3(aq) 2 H2O (l) O2(g) What volume of
7.91M solution of nitric acid, HNO3, is just
sufficient to react with 15.9 g of lead dioxide,
PbO2? Answer 0.0168 L
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• Volume Relationship of Gases in Chemical
  Reactions
• Gay Lussacs law of combining volumes states
  that the volumes of gaseous reactants and
  products stand in ratios of simple integers, as
  long as those volumes are measured at the SAME
  temperature and pressure.
• These integers are the same as the integers used
  to balance the chemical equation.
• A balanced chemical equation therefore provides a
  relationship between the volumes of gases
  reacting.

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• Problem
• Gaseous H2 and N2 react according to the equation
• 3H2 (g) N2 (g) ? 2 NH3(g)
• If in a given reaction 5.00 L of NH3 are formed,
  what volume of H2 reacted, assuming that the
  temperature and pressure before and after the
  reaction are the same.

From the balanced reaction above, assuming the
same temperature and pressure before and after
the reaction, 2 volumes
of NH3 are formed from 3 volumes of H2 Hence,
5.00 L of NH3 are formed from (3/2)x5.00 L 7.5
L of H2
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Limiting Reactants and Product Yields

• If everything went perfectly in the reaction
  Fe(s) S (s) ? FeS (s)
• 55.8 g (1 mole) of iron will react with 32.1 g (1
  mole) of S to form 87.9 g (1 mole) of FeS.
• If stoichiometric amounts of reactants were used,
  and assuming that there are no competing factors
  to limit the amount of products being formed,
  then a stoichiometric amount of product is formed

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• If we were to mix reactants together in
  non-stoichiometric amounts then what determines
  the amount of product formed?
• The reactant which is first used up determines
  the amount of product formed.
• This reactants is called the LIMITING REACTANT.
• The other reactants are in EXCESS when the
  reaction stops.
• If additional amounts of the limiting reactant is
  added the reaction starts again.

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• Problem
• Calcium carbonate CaCO3(s) is decomposed by
  HCl(aq) to give CaCl2 (aq), CO2(g) and H2O(l).
  If 10.0g of CaCO3 are treated with 10.0 g of HCl,
  how many grams of CO2 are generated?
• First write a balanced equation for the reaction
• CaCO3 (s) 2HCl(aq) ? CaCl2(aq) H2O(l)
  CO2(g)
• 1 mole of CaCO3 reacts with 2 moles of HCl to
  form 1 mole of CO2
• Moles of CaCO3 (10.0 g)/(100.0g/mol) 0.100
  mole CaCO3
• Moles of HCl (10.0 g)/(36.5 g/mol) 0.274 mol
  HCl

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• Hence 0.1 mole of CaCO3 reacts with 0.2 moles of
  HCl. Since the amount of HCl present is greater
  that 0.2, HCl is in excess and CaCO3 is the
  limiting reactant.
• Hence, 0.1 mole of CO2 formed.
• Mass of CO2 formed 0.1 mol x 44.01 g/mol 4.40
  g CO2

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Product Yields

• In the previous calculation 4.40 g is the amount
  of CO2 we would expect to be formed
• 4.40 g CO2 is the CALCULATED or the THEORETICAL
  PRODUCT YIELD.
• This assumes that the reaction goes to
  completion, and that there are no competing
  factors that may reduce the amount of CO2 formed.
  
• The measured amount of product formed is called
  the ACTUAL YIELD which is often smaller than the
  theoretical yield.

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The larger the yield the more cost effective is
the process and hence a more likely candidate for
industrial scale processes (other factors are
also important the nature of the by-products -
are they environmentally safe -, the cost of the
starting materials, etc.)
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• Problem
• The principle step in the recovery of elemental
  iron (Fe) from iron oxide (FeO) is a process
  known as reduction. In this reaction hydrogen
  (H2) at elevated temperatures is used as the
  reducing agent. Calculate the percent yield of
  Fe if 1.00 g of H2 reacts with 30.0 g of FeO and
  produces 19.5 g of Fe. The reaction between FeO
  and H2 is
• FeO(s) H2 (g) ? Fe(s) H2O(g)

First we need to determine the limiting
reactant Moles of H2 in reaction mixture 1.00
g / (2.02g/mol) 0.500 mol Moles of FeO
reaction mixture 30.0 g / (71.85 g/mol)
0.418 mol
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• From the stoichiometry of the equation FeO and H2
  react in a 11 ratio.
• Hence, FeO is the limiting reactant
• The number of moles of Fe that should be formed
  0.418 moles
• Mass of Fe that should be formed
• 0.418 mole x 55.85g.mol 23.3 g of Fe.
• The actual yield of FeO is 19.5 g

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Ideally for an industrial process, the yield
should be large. Performing stoichiometric
calculations, determining yields, are important
in analyzing the success of a chemical
reaction. Sometime, a reaction performed on a
lab-scale may have good product yields but when
the reaction is scaled up to an industrial
process, the yield may be lower.