Chemical Accounting:

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Chapter 5 Chemical Accounting Mass and Volume
Relationships
Outlines 1. Introduction 2. Chemical equation 3.
Volume relationship 4. Mole, formula mass, molar
mass, molar volume, 5. Mole-mass relationship in
reactions 6. Chemical accounting involving
solutions.
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1. Introduction

• Chemists often need to do some calculations, e.g.
  
• To determine how much of a particular product
  will be produced when certain quantities of the
  reactants are consumed.
• e.g. how many grams of CO2 are produced in the
  combustion of one liter of gasoline?
• To determine how much of a particular starting
  material will be needed to produce certain
  quantities of products.
• To prepare a solution of acid or base with a
  certain concentration.

• You may also encounter the terms such as mole,
  molecular mass, formula mass, molar mass,
  molecular weight, molarity, percent by volume,
  percent by weight. What are they?

• Purpose of this chapter (i) to introduce these
  terms, and (ii) to describe common chemical
  calculations.

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2. Chemical Equations
Chemical equations. Equations using the symbols
and formula to represent the elements and
compounds involved in a chemical reaction. E.g.
C O2 ? CO2
Substances on left Reactants or Starting
materials
Substances on right Products
and ? reacts to produce
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Reading a Chemical Reaction
The physical states of matter of a species in an
equation may also be specified. e.g.

• C(s) O2(g) ? CO2(g)
• (s) solid
• (l) liquid
• (g) gas
• (aq) aqueous or in water

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Balancing Chemical Equations

• A balanced equation has the same number of each
  atom on both sides of the equation.

Coefficients (whole numbers) are used to balance
a chemical equation. e.g.
2 H2 O2 ? 2 H2O
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Suggestions for Balancing Chemical Equations

• If an element occurs once on each side, balance
  it first
• Balance any reactants or products that exist as
  free elements last
• DO NOT change subscripts
• DO NOT add/delete products or reactants

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Example. Balance the following chemical equation
Un-balanced
Balanced
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Exercise 1. Balance the following reaction.

• Aluminum(s) oxygen(g) aluminum oxide(s)
• Al(s) O2(g) Al2O3(s)

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Exercise 2. Balance the following reaction.

• C2H5OH(l) O2(g) CO2(g) H2O(g)

Balance _______ first, balance ___(element) last.
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• Exercise 3. Balance the following equation
• NH3(g) O2(g) NO(g) H2O(g)

Balance _______ first, balance ____ (element)
last.
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2 NH3(g) O2(g) 2 NO(g) 3 H2O(g)

• Balance O last
• We want to make the O on the left equal _,
  therefore we will multiply it by _
• 2 NH3(g) _ O2(g) 2 NO(g) 3 H2O(g)

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3. Volume Relationships

• Law of combining volumes when all measurements
  are made at same temperature and pressure,
  volumes of gaseous reactants and products are in
  small whole-number ratio. in 1809, Joseph
  Louis Gay-Lussac (1778?1850)

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• Avogadros hypothesis equal volumes of gases at
  constant pressure and temperature have the same
  number of molecules
• in 1811, Amedeo Avogadro (1776?1856)

The volume ratio of gaseous reactants/product can
be determined by the coefficients of a balanced
equation
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4. Mole, molar mass, molar volume, formula mass
Mole

• An atom or typical molecule are too small to see
  or weigh.
• Chemist count atoms or molecules by mole.
• A mole (mol) is defined as the amount of a
  substance that contains 6.02 x 1023 particles.
• Abbreviated as mol
• Easier to weigh out moles instead of individual
  molecules
• 1 mol H2O has how many moles of O? H?

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Avogadros number

• Avogadros number 6.02 x 1023
• the number of atoms in a 12-g sample of
  carbon-12.

How big is Avogadros number?
1 mole of chemistry textbooks would cover the
surface of the earth to a depth of 300 km. If you
won 1 mole of dollars when you were born and
spent a billion dollars per second, 99.999 would
still be left at 90 years old.
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A Mole of Particles Contains 6.02 x 1023
particles

• 6.02 x 1023 C atoms
• 6.02 x 1023 H2O molecules
• 6.02 x 1023 NaCl molecules
• (technically, ionics are compounds not molecules
  so they are called formula units)
• 6.02 x 1023 Na ions and
• 6.02 x 1023 Cl ions

1 mole C 1 mole H2O 1 mole NaCl
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exercise

• 1. The number of atoms in 0.500 mole of Al is
• a) 500 Al atoms
• b) 6.02 x 1023 Al atoms
• c) 3.01 x 1023 Al atoms
• 2. The Number of moles of S in 1.8 x 1024 S atoms
  is
• a) 1.0 mole S atoms
• b) 3.0 mole S atoms
• c) 1.1 x 1048 mole S atoms

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Formula Masses (or molecular mass)

• Average mass of a formula unit relative to that
  of a carbon-12 atom.
• It is simply the sum of the atomic masses for all
  atoms in a formula.
• It is expressed in atomic mass unit (u)

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Molar Mass (Molecular Weight)

• Mass of 1 mol of a substance
• Numerically equivalent to formula mass
• Unit gram/mole (g/mol)
• e.g. Formula mass of SO2 64.1 u
• Molar mass of SO2 64.1 g/mol
• Can be used to convert between moles and mass

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Exercise 1. How many grams of N2 are in 0.400 mol
N2?
Atomic mass N 14.0 u Na 23.0 u
or
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Exercise 2. Calculate the number of moles of NaN3
in a 10.0 g sample of the solid.
Atomic mass N 14.0 u Na 23.0 u
Molar mass of NaN3
Mole of NaN3
or
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Molar Volume

• Volume occupied by 1 mol of gas
• At standard temperature and pressure (STP, 1 atm
  pressure and 0 oC)
• 1 mole of gas has a volume of 22.4 L

Exercise. Calculate the density of (a) nitrogen
and (b) oxygen at STP.
Atomic mass N 14.0 u O 16.0
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5. Mole and Mass Relationships in reactions

• The balanced chemical reaction gives the
  following information

CO2
C
O2

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Example
molar mass NO 14.0 16.0 30 g/mol O2
16.0x2 32.0 g/mol NO2 14 16x2 46 g/mol
Atomic mass N 14.0 u O 16.0
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Mole and Mass Relationships in reactions

• In general

x mole of A reacts with y mole of B X mole of A
reacts with B to give z mole of C and q mole of D.
Question 5 g of A can reacts with ? g of B? 5 g
of A reacts enough B to give ? g of C or D?
Answer Depend on molar mass of A, B, C, D.
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Mole and Mass Relationships in reactions

• Steps in a Stoichiometric Calculation
• Write and balance the chemical equation for the
  reaction.
• Determine molar masses of substances involved in
  the calculation.
• Use the coefficients of the balanced equation to
  convert the moles of the given substance to the
  moles of the desired substance.
• Use the molar mass to convert the moles of the
  desired substance to grams of the desired
  substance.

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Exercise 1.
Atomic mass C 12.0 u O 16.0 u H, 1.0 u
C O2 ? CO2
Balanced equation
Molar mass C, 12.0 g/mol O2, 2x16.0 32.0
g/mol
? mol C (in 10 g) 10.0/12.0 0.833 mol
? mol O2 required 1 x 0.833 0.833 mol
? g O2 required mol O2x molar mass of O2
0.833 x 32.0 26.7 g
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Exercise 2.

• Acetylene (C2H2) burns in pure oxygen with a
  very hot flame. The products of this reaction are
  carbon dioxide and water. How much oxygen is
  required to react with 52.0 g of acetylene?
• a) 32.0 g b) 52.0 g c) 160.0 g d) 240.0 g

Atomic mass C 12.0 u O 16.0 u H, 1.0 u
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Exercise 3.

• Calcium metal reacts with water to form calcium
  hydroxide and hydrogen gas. How many grams of
  hydrogen are formed when 0.50 g of calcium are
  added to water?
• a) 0.025 g b) 0.050 g c) 0.10 g d) 0.50 g

Atomic mass Ca 40.1 u O 16.0 u H, 1.0 u
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6. Chemical accounting involving solutions

• Solution Homogeneous mixture of two or more
  substances
• Solute what is being dissolved
• Solvent what is doing the dissolving
• Aqueous solutions water is solvent

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Solubility

• Soluble an appreciable quantity dissolves
• Insoluble very little, if any, quantity
  dissolves
• Dilute solution little solute in a lot of the
  solvent (for example water)
• Concentrated solution lots of solute in the
  solvent

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Measurement of Concentration

• Molarity (M) the number of moles of solute per
  liter of solution.

Exercise 1
Atomic mass Na 23.0 u Cl 16.0 u H, 1.0 u
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Exercise 2. How many grams of NaCl is required to
prepare 0.500 L of 0.15 M NaCl?
Atomic mass Na 23.0 u Cl 35.4 u
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Percent Concentration

• Percent Concentration
• Percent by volume
  x 100

Percent by mass
x 100
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Percent Concentration

• If both solute and solvent are liquids, often use
  percent by volume

Exercise. 120 mL of oil dissolved in enough
gasoline to make 4.0 L of fuel. What is the
percent by volume of the oil in the mixture?
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• Percent by mass is commonly used for commercial
  solutions
• 35.0 HCl means 35.0 g HCl for every 100.0 g of
  solution

Exercise. What is the percent by mass of a
solution of 25.0 g of NaCl dissolved in 475 g
(475 mL) of water?
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• Note the differences between percent by mass,
  percent by volume, and molarity
• 10 HCl solution is considerably different from
  10 M HCl
• Require different amounts of HCl