Chapter 3: Mass Relations in Chemistry Stoichiometry

1
Chapter 3 Mass Relations in Chemistry
Stoichiometry

• Chapter Outline
• 3.1 Atomic Masses 3.4 Mass Relations
• 3.2 The Mole in Reactions
• 3.3 Mass Relations in
• Chemical Formula

2
Overview

• Chemistry is quantitative atoms differ not only
  in atomic number but in mass
• Chemical formulas reveal not only the atom ratios
  in which elements are present but also the mass
  ratios
• Stoichiometry is the study of mass relations in
  chemistry

3
Atomic Masses

• The relative masses of atoms of different
  elements are expressed in terms of their atomic
  masses (atomic weights).
• The atomic mass indicates how heavy, on average,
  one atom of an element is compared to an atom of
  another element.

4
Relative Scale

• In order to set up a relative scale of atomic
  masses, it is necessary to establish a standard
  value for one element.
• Modern scale of atomic masses is based on the
  most common isotope of carbon, carbon-12. This
  isotope is assigned exactly the mass of 12.
• Mass of C-12 atom 12 amu (exactly)

5
Atomic Masses

• Most modern periodic tables show the masses of
  the elements to 4 decimal places.
• e.g. He 4.003 amu which, on average, is 1/3
  that of a C-12 atom.
• In general, ratio of the atomic masses of two
  elements is equal to the ratio of the masses of
  the individual atoms of those elements.

6
Atomic Masses and Isotopic Abundances

• Relative masses of individual atoms are
  determined using a mass spectrometer.
• To determine the atomic mass of elements with 1
  or more isotopes, it is necessary to know both
  the masses of the isotopes as well as their
  abundance in nature.
• Isotopic abundances can also be determined by
  mass spectrometry.

7
Mass Spectrum of Chlorine

• Atomic Mass Abundance
• Cl-35 34.97amu 75.53
• Cl-37 36.97amu 24.47
• From this data, we can calculate the atomic mass
  of chlorine using the following equation

8
Calculation of Atomic Mass of Chlorine

• Atomic mass of Y (atomic mass Y1) x
  Y1/100 (atomic mass Y2) x Y2/100 where
  Y1, Y2 are isotopes of element Y.
• Atomic Mass of Cl 34.97amu x (75.53/100)
  36.97 x (24.47/100) 35.46
• If atomic mass of an element is known and if it
  has only two stable isotopes, their abundances
  can be calculated as shown.

9
Masses of Individual Atoms

• It is usually sufficient to consider the relative
  masses of different atoms.
• What if we want to calculate the mass in grams of
  individual atoms?
• e.g. A He atom (4.003amu) is 4 times as heavy
  as a H (1.008amu) atom. It follows that a sample
  containing 100 atoms of He will weigh 4 times as
  much as a sample containing 100 atoms of H.

10
Avogadros Number

• It also follows that a sample of Helium weighing
  4 grams has the same number of atoms of a sample
  of H weighing 1 gram.
• No. of He atoms in 4.003 g He No. of H atoms in
  1.008 g H.
• A sample of any element with a mass in grams
  equal to its atomic mass contains the same number
  of atoms, NA, regardless of its identity.

11
Avogadros Number (Contd)

• NA is equal to Avogadros number and has a value
  of 6.022 x 1023.
• Suppose the entire population of the world was
  assigned to counting the atoms in 4.003 g of He.
  If each person counted one atom per second and
  worked a 48 hour week, the task would take more
  than ten million years!

12
Importance of Avogadros Number

• Avogadros number represents the number of atoms
  of an element in a sample whose mass in grams is
  numerically equal to the atomic mass of the
  element.
• Using this value, it is possible to calculate the
  mass of an individual atom or determine the
  number of atoms in a weighed sample of any
  element.

13
Problem

• Selenium (Se) is used an additive to make colored
  glass. Using Avogadros number, calculate
• a. the mass of a selenium atom
• b.the number of selenium atoms in a 1.000g
  sample of the element

14
Problem (Contd)

• Mass of a Selenium Atom
• The atomic mass of selenium is 78.96 amu (from
  periodic table). Therefore,
• 6.022 x 1023 Se atoms 78.96 g Se (conversion
  factor)
• Mass 1 Se atom x 78.96g Se/6.022 x 1023 Se
  atoms
• Mass of Se atom 1.311 x 10-22g

15
Problem (Contd)

• No. of Se atoms
• 1.000g x 6.022 x 1023 Se atoms/78.96g
• 7.627 x 1021 Se atoms
• Reality Check!

16
The Mole

• The quantity represented by avogadros number is
  called the mole.
• A mole represents 6.022 x 1023 objects.
• One mole of pennies is enough to pay all of the
  expenses of the U.S. For the next billion years.

17
Molar Mass

• A mole represents not only a specific number of
  particles but also a definite mass of a substance
  as represented by its formula (O, O2, H2O, NaCl,
  etc).
• The molar mass, M, in grams per mole, is
  numerically equal to the sum of the masses (in
  amu) of the atoms in the formula.

18
Examples of Molar Mass

• Formula Sum of Atomic Masses Molar Mass
• O 16.00amu 16.00g/mole
• O2 2(16.00amu) 32.00amu 32.00g/mole
• H2O 2(1.008amu) 16.00amu 18.02g/mole
• 18.02amu
• NaCl 22.99amu 35.45amu 58.44g/mole
• 58.44amu

19
Mole-Gram Conversion

• Conversion from moles to grams and vice versa
• m M x n
• m mass in grams
• M molar mass (g/mol)
• n amount in moles

20
Examples 1

• Calculate the number of moles of Calcium
  Carbonate in a stick of chalk containing 14.8g of
  calcium carbonate.
• Strategy Find the molar mass of calcium
  carbonate and use it to convert 14.8g to moles.
  Must first come up with chemical formula.

21
Example 1 (Contd)

• Solution The formula of calcium carbonate is
  CaCO3, so the molar mass is
• M 40.08 12.01 3(16.00)g/mol
• 100.09 g/mol
• n 14.8g CaCO3 x (1 mole of CaCO3/100.09g CaCO3)
• 0.148 moles CaCO3

22
Example 2

• Acetylsalicyclic acid, C9H8O4, is the active
  ingredient of aspirin. What is the mass in grams
  of 0.287 moles.
• Strategy Find the molar mass of C9H8O4 and use
  it to obtain the conversion factor required.

23
Example 2 ( Contd)

• Solution M 9(12.01) 8(1.008)
  4(16.00)g/mol 180.15g/mole
• Mass C9H8O4
• 0.287mol C9H8O4 x 180.15g C9H8O4/1mol
• 51.7g C9H8O4

24
Percent Composition from Formula

• The percent composition of a compound is
  specified by citing the mass percents of the
  elements present.
• e.g. 100g sample of H2O, there is 11.19g of H
  and 88.81g of O.
• 11.19g H/100.00g x 100 11.19 H
• 88.81g O/100.00g x 100 88.81 O

25
Example 3

• Sodium Hydrogen Carbonate, commonly called
  bicarbonate of soda, is used in many commercial
  products (antacids). It has the chemical formula
  NaHCO3. What are the mass percents of Na, H, C,
  and O.
• Strategy Find the mass in grams of each element
  in 1 mole of NaHCO3. Then find elem.t mass
  elem./total mass x 100

26
Example 3 (contd)

• n x M m
• Na 1 mole x 22.99g/mol 22.99g
• H 1 mole x 1.008g/mol 1.008g
• C 1 mole x 12.01g/mol 12.01g
• O 3 mole x 16.00g/mol
  48.00g
• 84.01g NaHCO3

27
Example 3 (contd)

• Mass Na 22.99g/84.01g x 100 27.36
• Mass H 1.008g/84.01g x 100 1.200
• Mass C 12.01g/84.01g x 100 14.30
• Mass O 48.00g/84.01g x 100 57.14
• 27.36 1.200 14.30 57.14 100

28
Conclusions

• In 1 mole of NaHCO3, there is 1 mole of Na
  (22.99g), 1 mole of H (1.008g), 1 mole of C
  (12.01g) and 3 moles of O (48.00g).
• In general, the subscripts in a chemical formula
  represent not only the atom ratio in which the
  different elements are combined but also the mole
  ratio

29
Example 4

• An iron-containing mineral responsible for the
  red color of soils in the southeast U.S. is
  limonite, which has the formula Fe2O3.3/2 H2O.
  What mass of iron in grams can be obtained from a
  metric ton (103kg 106g) of limonite.
• Molar Mass 117 3(16.00) 3/2(18.02)g/mol
  186.7 g/mol.

30
Example 4 (Contd)

• In 1 mole of limonite there is
• 2 mole Fe x 55.85g/1 mole Fe 111.7g Fe
• Fe 111.7g/186.7g x 100 59.83
• In 1 metric ton
• mass Fe 1.000 x 106g limonite x 59.83g/100.0g
  5.983 x 105g Fe

31
Simplest Formula from Chemical Analysis

• The simplest formula is the simplest whole number
  ratio of the atoms present.
• In ionic compounds, the simplest formula is
  usually the only one that can be written (NaCl,
  CaCl2).
• In covalent compounds, the actual formula is a
  whole-number ratio of the simplest formula

32
Example 5

• A 25.00g sample of an orange solid contains 6.64g
  of potassium, 8.84g of chromium and 9.52g of
  oxygen. Find the simplest formula.
• Strategy (1) convert mass in grams to moles.
  (2) Calculate the mole ratios. (3) Equate the
  mole ratios to the atom ratio, which gives you
  the simplest formula.

33
Example 5 (Contd)

• nK 6.64g K x 1mol K/39.10g K 0.170mol K
• nCr 8.84g Cr x 1mol Cr/52.00g Cr 0.170mol Cr
• nO 9.52g O x 1mol O/16.00g O 0.595mol O
• Find the mole ratios by dividing by the smallest
  number, 0.170
• 0.170mol Cr/0.170mol K 1.00mol Cr/1mol K
• 0.595mol O/0.170mol K 3.50mol O/1mol K

34
Example 5 (Contd)

• The mole ratio is 1mol K 1mol Cr 3.50mol O
• The mole ratio is the same as the atom ratio and
  must be a whole number. We, therefore, multiple
  the mole ratio by the smallest whole number
  possible to give all whole numbers
• Multiply by 2
• 2K 2 Cr 7 O
• Simplest formula is K2Cr2O7.

35
Simplest formula

• If you are given the mass percents in a problem,
  assume a 100g sample and then calculate the
  number of moles
• e.g. In a compound the percentages of the
  elements are 26.6 K, 35.4 Cr, and 38.0 O. If
  we assume a 100g sample, then there is 26.6g of
  K, 35.4g of Cr, and 38.0g of O. Then follow the
  previous example.

36
Example 6

• The compound ethanol contains the elements C, H,
  and O. When a sample of ethanol is burned in
  air, it is found that
• 5.00g ethanol ? 9.55g CO2 5.87g H2O
• What is the simplest formula for ethanol?

37
Example 6 (Contd)

• Strategy (1) Calculate the masses of C, H, and O
  in the 5.00g sample. All of the carbon is
  converted to CO2.
• 1 mole C(12.01g) ? 1 mole CO2 (44.01g)
• To find the mass of C in CO2, use following
  conversion factor
• 12.01g C/44.01g CO2

38
Example 6 (Contd)

• All of the hydrogen is converted to H2O. There
  are 2 moles of H in water, therefore, the
  conversion factor is
• 2.016g H/18.02g H2O
• We can not determine the mass of oxygen by the
  same method because some of it comes from the
  air. Find the mass of O by difference.

39
Example 6 (Contd)

• Mass O mass of sample (mass of C mass of H)
• Mass C 9.55g CO2 x 12.01gC/44.01g CO2 2.61g
• Mass H 5.87g H2O x 2.016g H/18.02g H2O 0.657g
• Mass O 5.00g (2.61g C 0.657g H) 1.73g O

40
Example 6 (Contd)

• (2) Find the Number of Moles
• nC 2.61g C x 1mol C/12.01g C 0.217mol
• nH 0.657g H x 1mol H/1.008g H 0.652mol
• nO 1.73g O x 1mol O/16.00g O 0.108mol
• (3) Find the mole ratios and the simplest
  formula
• 0.217 mol C/0.108mol O 2.01mol C/1mol O
• 0.652mol H/0.108mol O 6.04mol H/1mol O

41
Example 6 (Contd)

• Round off to whole numbers
• C2.01H6.04O C2H6O

42
Mass Relations In Reactions

• starting materials (reactants) ? products
• Chemical reactions are represented by chemical
  equations, which identify reactants and products.
• (i) reactants always appear to the left of the
  arrow
• (ii) products always appear to the right of the
  arrow
• In a balanced equation, there are the same number
  atoms of a given element on both sides of the
  arrow.

43
Writing and Balancing Chemical Equations

• Any calculation involving a reaction must be
  based on the balanced equation for that reaction.
• You cannot write an equation unless you
  understand what happens in the reaction that it
  represents.

44
Rocket Fuel

• 1. Write a skeleton equation in which the
  formulas of the reactants appear to the left of
  the arrow and the products to the right.
• N2H4 N2O4 ? N2 H2O

45
Balancing Equations (Contd)

• Indicate the physical state of each reactant and
  product, after the formula, by writing
• (g) for a gaseous substance
• (l) for a pure liquid
• (s) for a solid
• (aq) for an ion or molecule in water
  solution
• N2H4 (l) N2O4 (l) ? N2 (g) H2O (g)

46
Balancing Equations (Contd)

• Balance the Equation. This is accomplished by
  checking to see if the number of atoms of each
  element is equal on both sides of the equation.
  If not, multiply by the appropriate coefficient.

47
Balancing Equations (Contd)

• N2H4 (l) N2O4 (l) ? N2 (g) H2O (g)
• Balance the number of O by multiplying the right
  side by 4 to give 8 O on each side.
• N2H4 N2O4 ? N2 4H2O
• There are now 8 H on the right side. Multiply
  the left side by 2 to give 8H on each side.
• 2N2H4 N2O4 ? N2 4H2O

48
Balancing Equations (Contd)

• There are now 6 N on the left side. Multiply the
  right side by 2 to give 6 N on each side
• 2N2H4 N2O4 ? 3N2 4H2O
• Check to see that both sides of the equation are
  balanced.

49
Hints for Balancing Equations

• Equations are balanced by adjusting coefficients
  in front of formulas, never by changing
  subscripts within formulas.
• In balancing an equation, it is best to start
  with an element that appears in only on species
  on each side of the equation.
• In principle, there are an infinite number of
  balanced equations that can be written for any
  reaction. The one with the simplest whole-number
  coefficients is preferred.

50
Mass Relations From Equations

• The coefficients of balanced equations represent
  the number of moles of reactants and products.
• 2N2H4 N2O4 ? 3N2 4H2O
• A balanced equation remains valid if each
  coefficient is multiplied by the same number,
  including NA.

51
Mass Relations (Contd)

• 2N2H4 N2O4 ? 3N2 4H2O
• Coefficients can be used for conversion factors
• 2 mol N2H4/3 mol N2
• 3 mol N2/1 mol N2O4

52
Examples

• How many moles of hydrazine is required to
  produce 1.80mol of nitrogen
• nhydrazine 1.80 mol N2 x 2 mol N2H4/3 mol N2
  1.20 mol N2H4
• These conversions will be used to relate moles of
  one substance to grams of another and grams of
  one substance to grams of another

53
Example 7

• Determine (a) mass in grams of NH3, formed when
  1,34mol of N2 reacts. (b) the mass in grams of N2
  required to form 1.00kg of NH3 and (c) the mass
  in grams of H2 required to react with 6.00g of
  N2.
• Strategy In each case, you will use the mole
  ratios given by the coefficients of the balanced
  equations to relate moles of one substance to
  moles of another

54
Example 7 (Contd)

• (a) nnitrogen ? nammonia ? mass of NH3
• (b) Mass of NH3 ? nammonia ? nnitrogen ? mass of
  N2
• (c) Mass of N2 ? nnitrogen ? nhydrogen ? mass of
  H2

55
Balanced Equation

• N2 3H2 ? 2NH3
• Mass of NH3 1.34mol N2 x
• (2mol NH3/1mol N2) x 17.03g NH3/1mol NH3
  45.6g NH3.
• (b) Mass of N2 1.00 x 103g NH3 x
• 1mol NH3/17.03g NH3 x 1mol N2/2mol NH3 x 28.02g
  N2/1 mol N2 823g N2

56
Example 7 (Contd)

• (c) Mass of H2 6.00g of N2 x 1mol N2/28.02g N2
  x 3mol H2/1mol N2 x 2.016g H2/1mol H2 1.30g H2

57
Limiting Reactant and Theoretical Yield

• 2Sb (s) 3I2 (s) ? 2SbI3 (s)
• Coefficients show the 2 moles of Sb (243.6g)
  reacts with exactly 3 moles of I2 (761.4g) to
  form 2 moles of SbI3 (1005.0g)
• Another way of stating this is that the maximum
  amount of SbI3 that can be obtained under these
  conditions if the reaction goes to completion and
  no product is lost is 1005.0g.
• This quantity is referred to as the theoretical
  yield.

58
Limiting Reagent (Contd)

• In the lab, reactants are not usually mixed in
  exactly the required ratio. An excess of one
  reagent, usually the cheaper one, is used to
  ensure all of the more expensive reagent reacts.
• e.q. 3mol Sb reacted with 3 mol I2
• After the reaction, 1 mole of Sb remains.

59
Limiting Reagent (Contd)

• The limiting reagent is the reagent that is
  present in the least amount (in moles).
• The amount of product formed is determined
  (limited) by the limiting reagent.
• Often, you will be given the amounts of two
  reactants (in grams) and asked to determine the
  limiting reagent and to calculate the theoretical
  yield.

60
Hints for Determining the Limiting Reagent

• Calculate the amount of product that would be
  formed if the first reactant were completely
  consumed.
• Repeat this calculation for the second reagent.
• Choose the smaller of the two amounts calculated
  in (1) and (2). This is the theoretical yield.
  The reactant that produces the smaller amount is
  the limiting reagent.

61
Example 8

• 2Sb(s) 3I2(s) ? 2SbI3(s)
• Determine the limiting reagent and the
  theoretical yield when
• 1.20mol of Sb and 2.40mol of I2 are mixed.
• 1.20g of Sb and 2.40g of I2 are mixed. What mass
  of excess reagent is left when the reaction is
  complete.

62
Example 8 (Contd)

• 1. nSbI3 from Sb 1.20mol x 2mol SbI3/2 mol Sb
  1.20 mol SbI3
• 2. nSbI3 from I2 2.40mol x 2mol SbI3/3mol I2
  1.60mol SbI3
• 1.20moles is the smaller number so it is the
  theoretical yield and thus Sb is the limiting
  reagent.