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Matter and Measurement

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Title: Matter and Measurement


1
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2
Chemical Formulas
3
  • 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)
4
  • 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.
5
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.
6
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?
7
  • 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)

8
  • 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

9
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))
10
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?
11
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
12
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
13
  • 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.

14
  • 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?

15
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
16
  • 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

17
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.

18
  • 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

19
  • 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.
20
  • 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
21
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
22
  • 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

23
  • 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)

24
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
25
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
26
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)
27
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.

28
  • 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)

29
  • 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
30
  • 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

31
  • 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
32
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.
33
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
34
  • 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.

35
  • 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
36
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

37
  • 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.

38
  • 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

39
  • 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

40
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.

41
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.)
42
  • 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
43
  • 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

44
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.
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