Mass Relations in Chemistry Stoichiometry

1
Chapter 3

• Mass Relations in Chemistry Stoichiometry

2
Relative Atomic Mass

• Carbon-12 was assigned a mass of 12.
• Atomic mass unit (amu)
• 1/12 the mass of a carbon-12 (12C) atom
• Atomic mass
• Relative to mass of carbon-12
• Mass of atom in amus
• Same number as molar mass

3
Average Atomic Mass

• Mass shown on the periodic table
• Takes into account abundances of each isotope of
  an element
• 35Cl is 75.8 abundant 37Cl is 24.2 abundant
• Since there is more chlorine-35 in the world,
  adding the two numbers and dividing by two WONT
  WORK.
• Instead, multiply the masses by the percentages
  and add.
• (35 x 0.758) (37 x 0.242) 35.5 amu

4
Why Does This Work?

• What is the average of 12, 15, 17, and 19?
• Normally, you would add the numbers and divide by
  4.
• This gives an answer of 15.75.
• In this case, each of the numbers is given the
  same weight, 25.
• Try multiplying each by 0.25 and add the results.
• (12 x 0.25) (15 x 0.25) (17 x 0.25) (19 x
  0.25) 15.75
• If all numbers do not have the same weight, you
  can use the same process with different
  percentages.

5
The Mole

• Dozen 12 items
• Oranges, shoes, blades of grass, oxygen atoms
• Doesnt matter what item, a dozen is 12
• Mole 602,000,000,000,000,000,000,000 items
• 6.02 x 1023 (Avogadros number, the number of C
  atoms in 12 g 12C)
• Donuts, ribbons, cars, water molecules
• Can be used to measure anything, but in chemistry
  we will usually be talking about atoms and
  molecules

6
Molar Mass and Molecular Weight

• Molecular weight sum of atomic weights of atoms
  in a molecule (found on periodic table), measured
  in amu
• H2O 2(1.00) 16.00 18.00 amu
• Al2(SO4)3 2(26.98) 3(32.06) 12(16.00)
  342.14 amu
• Molar mass mass of one mole of a substance
• H2O 2(1.00) 16.00 18.00 g/mol
• Al2(SO4)3 2(26.98) 3(32.06) 12(16.00)
  342.14 g/mol
• Notice that the number is the same, but the units
  are different. Both are calculated in the same
  way.

7
Conversions
particles
6.02 x 1023
Moles
Molar mass
mass
8
Mole/Particle Conversions

• To convert from moles to particles (atoms,
  molecules, formula units), use Avogadros number
  (6.02 x 1023)
• Example How many atoms in 1.3 moles He?
• (1.3 mol He)(6.02 x 1023 atoms He/1 mol He) 7.8
  x 1023 atoms He
• Use this number to convert from particles to
  moles also
• Example How many moles NaCl are equal to 5.43 x
  1022 formula units?
• (5.43 x 1022 form units NaCl)(1 mol NaCl/6.02 x
  1023 form units NaCl) 9.02 x 10-2 mol NaCl

9
Mole/Gram Conversions

• To convert from moles to mass (grams), use the
  molar mass from the periodic table
• Example What is the mass of 2.76 mol H2O?
• (2.76 mol H2O)(18.00 g H2O/1 mol H2O) 49.7 g
  H2O
• To convert from mass to moles, use molar mass
  also
• Example How many moles Mg are equal to 68 g?
• (68 g Mg)(1 mol Mg/24.31 g Mg) 2.8 mol Mg

10
Particle/Gram Conversions

• Notice on the diagram that there was no direct
  link between particles and mass.
• Two steps are needed.
• Examples
• How many particles are in 85.2 g KCl?
• (85.2 g KCl)(1 mol KCl/74.55 g KCl)(6.02 x 1023
  form units KCl/1 mol KCl) 6.88 x 1023 form
  units KCl
• What is the mass of 5.9 x 1024 atoms Ne?
• (5.9 x 1024 atoms Ne)(1 mol Ne/6.02 x 1023 atoms
  Ne)(20.18 g Ne/1 mol Ne) 2.0 x 102 g Ne

11
Percent Composition

• Shows percent by mass of each element in a
  compound
• Can be calculated from formula
• What is the percent composition of NaNO3?
• Total molar mass of Na 22.99 g/mol
• Total molar mass of N 14.01 g/mol
• Total molar mass of O 3(16.00) 48.00 g/mol
• Total molar mass of cmpd 85.00 g/mol
• Na (22.99/85.00) x 100 27.05
• N (14.01/85.00) x 100 16.48
• O (48.00/85.00) x 100 56.17
• Sum of percentages should equal 100 .

12
Empirical (Simplest) Formulas

• Can be determined from percent composition
• Follow these steps
• Divide the mass or percent of each element by the
  molar mass of that element
• Divide each answer by the smallest answer
• Find the smallest whole number ratio and write
  formula

13
Example

• What is the empirical formula of a compound which
  contains 26.57 K, 35.36 Cr, and 38.07 O?
• K 26.57/39.10 0.6795396
• Cr 35.36/52.00 0.68
• O 38.07/16.00 2.379375
• The smallest answer is 0.6795396.
• K 0.6795396/0.6795396 1
• Cr 0.68/0.6795396 ? 1
• O 2.379375/0.6795396 ? 3.5
• We need to keep the 113.5 ratio, but we need
  whole numbers. Multiplying each value by 2 will
  achieve this.
• K 2 Cr 2 O 7
• The empirical formula is K2Cr2O7.

14
Molecular Formulas

• Show the number of atoms of each element in a
  molecule (not just the simplest ratio like the
  empirical formula)
• Follow these steps
• Find empirical formula
• Divide molar mass given in problem by molar mass
  of your empirical formula
• Multiply each subscript of your empirical formula
  by the answer you got in step 2.

15
Example

• A compound contains 10.15 C and 89.85 Cl. If
  its molar mass is 236.7 g/mol, what is the
  molecular formula?
• C 10.15/12.01 0.8451291/0.8451291 1
• Cl 89.35/35.45 2.5204513/0.8451291 ? 3
• Empirical formula is CCl3 (molar mass 118.36
  g/mol)
• (236.7/118.36) 2
• Molecular formula is C2Cl6.

16
Chemical Reactions

• Skeletal chemical equation reaction represented
  by formulas and symbols
• H2 O2 ? H2O
• Reactants original substances (H2 and O2 in
  this case)
• Products substances formed in the reaction (H2O
  in this case)
• Physical states are often indicated in
  parentheses after the formula for a substance.
• s solid
• l liquid
• g gas
• aq aqueous (dissolved in water)
• H2 (g) O2 (g) ? H2O (l)
• Balanced equation coefficients (numbers placed
  before a neutral formula) are used to show the
  number of each substance required for the
  reaction to take place
• 2 H2 (g) O2 (g) ? 2 H2O (l)
• Review Ch 2 if you have forgotten how to write
  formulas. This is very important!

17
Balancing Chemical Equations

• Law of conservation of mass matter cannot be
  created or destroyed
• The products of a chemical reaction have to have
  the same mass as the reactants.
• The number of atoms of each element does not
  change they just get rearranged.
• Chemical formulas must be neutral.
• You cannot change the formulas of substances!
• In order to show that the number of atoms is not
  changing, you must use numbers in front of each
  formula (coefficients) to show how may molecules
  are present in the reaction.

18
Examples

• H2 O2 ? H2O
• There are 2 H atoms on each side of the arrow.
• There are 2 O atoms on the reactant side, but
  only one on the product side.
• Placing a 2 in front of H2O will balance the Os.
• H2 O2 ? 2 H2O
• There are now 2 Hs on the reactant side and 4 on
  the product side.
• Placing a 2 in front of H2 will balance the Hs.
• 2 H2 O2 ? 2 H2O

19
Examples

• KClO3 ? KCl O2
• K and Cl are already balanced.
• There are 3 Os on the reactant side and 2 on the
  product side.
• The smallest number that 2 and 3 go into is 6.
• You need 2 KClO3 to have 6 Os.
• You need 3 O2 to have 6 Os.
• 2 KClO3 ? KCl 3 O2
• K and Cl are no longer balanced. A coefficient
  of 2 will balance these.
• 2 KClO3 ? 2 KCl 3 O2

20
Balancing Tips

• Start with elements that appear in only one place
  on each side of the equation.
• If an element appears in more than one place on
  each side of the equation, and there is an odd
  number of this element on one side and an even
  number on the other multiply the substance with
  an odd number of the element by 2. You should
  then be able to work more easily toward balancing
  the equation.
• Be sure to reduce all coefficients if you can.
  (You may be able to divide all coefficients by 2,
  for example.)

21
Stoichiometry

• The math of chemical equations
• Figure out how much of one substance is needed to
  react with a certain amount of another
• Figure out how much of a product will be formed
  from a given amount of reactants
• The coefficients of the equation relate the
  amounts of each that react or are produced.
• Remember that the coefficients refer to the
  number of each substance involved and have
  nothing to do with the mass.
• If you start with mass or are trying to find
  mass, you have to do some conversions.

22
Mole/Mole Stoichiometry

• 2 Na Cl2 ? 2 NaCl
• This is the simplest calculation since only one
  step is involved. Just use coefficients.
• How many moles of Cl2 are needed to produce 1.3
  mol NaCl?
• (1.3 mol NaCl)(1 mol Cl2/2 mol NaCl) 0.65 mol
  Cl2
• This is the key step in stoichiometry. You can
  only switch from one substance to another through
  the number of moles. All other steps are just
  mole-gram conversion like you were doing at the
  beginning of the chapter.

Coefficients of 1 are not shown in the equation.
23
Mole/Gram Stoichiometry

• 2 Na Cl2 ? 2 NaCl
• How many moles of sodium are needed to produce
  6.7 g NaCl?
• Since youre starting with grams, first convert
  this to moles. Then convert to moles of the
  other substance.
• (6.7 g NaCl)(1 mol NaCl/58.44 g NaCl)(2 mol Na/2
  mol NaCl) 0.11 mol Na
• How many grams of Cl2 are needed to react with
  0.954 mol Na?
• This time youre starting with moles, so first
  find moles of the other substance then convert to
  grams.
• (0.954 mol Na)(1 mol Cl2/2 mol Na)(70.90 g Cl2/1
  mol Cl2) 33.8 g Cl2

24
Gram/Gram Stoichiometry

• 2 Na Cl2 ? 2 NaCl
• How many grams of sodium chloride can be produced
  from 35 g Cl2?
• (35 g Cl2)(1 mol Cl2/70.90 g Cl2)(2 mol NaCl/1
  mol Cl2)
• (58.44 g NaCl/1 mol NaCl) 58 g NaCl

25
Stoichiometry Summary
A 2 B ? 3 C 4 D
coefficients
mol
mol
Molar mass of B
Molar mass of C
g
g
To determine the mass of C which can be produced
from a certain mass of B, you must first convert
the g B to mol B using molar mass. You can then
use coefficients to figure out mol C. Last,
convert mol C to g C using molar mass. You can
only switch substances through moles!
26
Limiting Reagents

• Until now, we have been determining the amount of
  one substance in the equation given an amount of
  another.
• We have assumed that there was enough of all
  other substances to complete the reaction.
• Usually, there will actually be excess of one (or
  more) reactant.
• The other, the limiting reagent, will run out
  before the excess reactant is completely used up.
• Figure out how much product can be formed from
  each reactant.
• The reactant that makes less product is the
  limiting reagent.

27
Limiting Reagents

• 2 Na Cl2 ? 2 NaCl
• How much sodium chloride can be produced produced
  if 57.8 g Na and 62.4 g Cl2 are combined?
• (57.8 g Na)(1 mol Na/22.99 g Na)(2 mol NaCl/2 mol
  Na)(58.44 g NaCl/1 mol NaCl) 147 g NaCl can be
  made from the available Na if there is enough Cl2
• (62.4 g Cl2)(1 mol Cl2/70.90 g Cl2)(2 mol NaCl/1
  mol Cl2)(58.44 g NaCl/1 mol NaCl) 103 g NaCl
  can be made from the available Cl2 if there is
  enough Na
• Cl2 is the limiting reagent since it makes less
  of the product. Only 103 g NaCl can be made
  before running out of this reactant.

28
Percent Yield

• Percent yield (actual yield/theoretical yield)
  x 100
• Theoretical yield is the amount of product
  calculated based on the amount of reactants
  available (just like weve been doing).
• Experiments never work perfectly. The actual
  yield is the amount we really got when we did the
  experiment. (This information would be given if
  you were doing a homework problem.)

29
Percent Yield

• Lets add this step to the previous problem
• Suppose 57.8 g Na and 62.4 g Cl2 are combined.
  If the actual yield is 99.2 g NaCl, what is the
  percent yield?
• First figure out which reactant makes less. This
  is the theoretical yield, the maximum amount of
  product you can get.
• Sometimes you will be given only the amount of
  one reactant, and you must assume that you have
  excess of the other.
• (57.8 g Na)(1 mol Na/22.99 g Na)(2 mol NaCl/2 mol
  Na)(58.44 g NaCl/1 mol NaCl) 147 g NaCl can be
  made from the available Na if there is enough Cl2
• (62.4 g Cl2)(1 mol Cl2/70.90 g Cl2)(2 mol NaCl/1
  mol Cl2)(58.44 g NaCl/1 mol NaCl) 103 g NaCl
  can be made from the available Cl2 if there is
  enough Na
• Calculate the percent yield (99.2 g NaCl/103 g
  NaCl) x 100 96.3