Chemical Composition Chemical Quantities

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Chemical CompositionChemical Quantities

• Chapters 910
• Pg. 221-293

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Chemical Composition

• Atomic Masses-
• Because the mass of one atom is so small- on the
  order of 10-23 gram- a special unit the atomic
  mass unit- is used to described the mass of an
  atom.
• A system has been devised for expressing the
  masses of atoms on a relative scale- that is, a
  scale based on experimental comparison with a
  standard. The scale for expressing atomic masses
  is based on the mass of carbon- 12. 1 carbon 12
  atom weighs 12 amu.
• Masses on the periodic table are weighted
  averages of all the naturally occurring isotopes
  for an element.

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• Atomic Mass the mass of an atom expressed
  relative to the mass of carbon 12
• Na
• K
• Cl
• Notice that we will be rounding to the 1/10th
  place for all calculations.
• Formula Mass / Molecular Mass the sum of the
  atomic masses for all the atoms in a compound.
• H2O
• H 2 atoms X 1.0 amu 2.0 amu
• O 1 atom X 16.0 amu 16.0 amu
• 18.0 amu
• HC2H3O2 ?

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The Mole- Three Faces

• Avogadros Number
• One mole refers to Avogadros number
  ___________________
• 1 mole of silver
• 1 mole of CO2
• Just like a dozen represents 12 of anything, and
  a gross represents 144 of anything.

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• 2. Molar Mass
• One mole of particles has a special mass
  associated with it the formula mass or molecular
  mass of the substance expressed in grams.

Substance Formula / Molecular Mass Molar Mass
Calcium (Ca) 40.1 amu 1 mole 40.1 grams 40.1 g / mol
Hydrogen Perioxide (H2O2) 34.0 amu 1 mole 34.0 grams 34.0 g / mol
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• Question Find the molar mass of the following
  substances
• Hydrogen gas
• Carbon Monoxide
• Glucose (C6H12O6)

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• 3. Molar Volume of an Ideal Gas
• Suppose the mole of particles happens to be a
  gas. In this case, the term mole has a special
  volume associated with it. If the mole of gas
  particles behaves ideally, the gas will occupy a
  volume of approximately 22.4 L at STP.
• 1 mole of He gas occupies 22.4 L
• 1 mole of hydrogen gas (H2) occupies 22.4 L
• 1 mole of carbon dioxide gas (CO2) occupies 22.4 L

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Mole Conversions

• Because the mole measures mass, number of
  particles, and volume of a gas, it is the central
  unit in converting the amount of a substance from
  one type of measurement to another.
• These conversions will require use of the Factor
  Label Method we learned at the start of the year.

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FLM

• How many apples are there in 3.75 dozen?
• 3.75 dozen X 12 apples 45 apples
• 1 dozen

Unit you want
Unit you are trying to get rid of
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Mass and Moles

• If you know the mass of a given amount of
  substance, you can calculate the number of moles
  of the substance.
• Your fact you will use in this conversion is the
  molar mass of the substance.

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• Example Suppose a laboratory procedure calls for
  0.65 moles of NaCl to be added to a solution.
• Question Suppose you produced 45.0 g of
  magnesium sulfate in the lab. How many moles of
  magnesium sulfate does this represent?

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Particles and Moles

• This conversion requires only one fact that does
  not change from atom to atom or compound to
  compound
• Avagadros Number 6.02 X 1023 stuff 1 mole
  of stuff
• Stuff could be atoms, molecules, particles,
  electron, protons, ions, .anything!

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• Example Determine the number of molecules of
  water in 2.55 moles of water.
• Question How many moles of C6H12O6 contain 7.9 X
  1024 molecules?

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Multi- Step Conversions

• Using a mole map and the FLM, link your
  conversion facts together and solve.
• Question You need 250.g of table sugar or
  sucrose (C12H22O11) to bake a cake. How many
  sucrose molecules will be in the cake? Note
  There is no direct conversion between grams and
  molecules. You must first convert grams to
  moles, and then moles to molecules.

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• Question A student fills a 1.0 L flask with
  carbon dioxide at standard temperature and
  pressure. How many molecules of gas are in the
  flask?

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Percent Composition by Mass

• Percent part per hundred
• Percentages are calculated by dividing the
  quantity under consideration by the total
  quantity involved and multiplying the result by
  100. For example, if 7 persons in 20 wear
  glasses, then the percentage of persons wearing
  glasses is

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• Similarly, we can calculate the percentage
  composition of an element in a compound by
  dividing the mass of the element by the formula
  mass of the substance and multiplying the result
  by 100.
• Question Calculate the percentage of oxygen by
  mass in CuSO4.

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Empirical Formula from Percent Composition

• Empirical Formula
• Example The empirical formula for N2O4 is NO2,
  and the empirical formula for C6H6 is CH, but the
  empirical formula for H2O is still H2O.
• Since we calculate percent composition from
  chemical formulas, we are able to reverse the
  process and determine formulas from percent
  composition. However, when we use this process,
  we are able to determine only the empirical
  formula.

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• Question Calculate the empirical formula of a
  substance that contains 50.00 sulfur and 50.00
  oxygen by mass.
• Solution Assume we have 100 grams of the
  substance. If so, then based on the percentages,
  we would have 50.00g of S and 50.00g of O. If we
  divide the given masses by the atomic masses of
  the two elements, we transform the mass into
  moles.

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• As strange as the previous formula looks, it is
  accurate because it shows the relative number of
  moles of each element. Our task is to make the
  formula look like a formula, that is, to convert
  it to one with whole numbers. Usually this is
  accomplished by dividing each element by the
  smallest number of moles.

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Determining Molecular Formula

• The empirical formula for a compound indicates
  the simplest ratio of the atoms in the compound.
  However, it does not tell the actual numbers of
  atoms in each molecule of the compound.
• Molecular Formula

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• Question The empirical formula for glucose is
  CH2O. Its empirical formula mass is 30.0 g /
  mol. Experiments show that the molar mass of
  glucose is 180 g / mol. What is the molecular
  formula of glucose?

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Mole Problems Involving Chemical Equations

• Consider the following equation
• N2 3H2 ? 2NH3
• How do we interpret the coefficients?
• Molecules OR
• Moles of molecules
• 1 mole of N2 reacts with 3 moles of H2 to form 2
  moles of NH3

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• We can construct a mole diagram for solving mole
  problems involving equations just as we did for
  single substances. Note that the diagram
  represents simply the joining of two single mole
  maps. It does not matter whether A and B are
  both reactants, products, or one of each of a
  given chemical reaction. The important point is
  that (moles of ) A and (moles of ) B are
  connected by the coefficients the two substances
  have in the BALANCED chemical reaction.

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Gas volume of B at STP
Gas volume of A at STP
Mass of B
Mass of A
Moles of A
Moles of B
Number of Particles of A
Number of Particles of B
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• All equation problems are solved using the steps
  listed below
• Examine the chemical equation and make sure that
  it is balanced.
• Refer to the preceding diagram and prepare a
  solution map as follows
• Quantity of A ? moles of A ? moles of B ?
  Quantity of B
• given in problem from rxn from
  rxn calculate
• 3. Set up the problem, using FLM and the solution
  map
• 4. Perform the calculations on the numbers and
  units.

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Mole Problems

• Question In the equation
• N2 H2 ? NH3,
• how many moles of N2 are needed to produce 5.0
  moles of NH3?
• Tip- BALANCE EQUATION FIRST!

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Mass Mass Problems

• Question In the equation
• CH4 O2 ? CO2 H2O
• how many grams of CO2 are formed when 8.0 grams
  of CH4 reacts?
• Tip- BALANCE EQUATION FIRST

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Mass-Volume Problems

• Question In the equation
• CO (g) O2 (g) ? CO2 (g)
• how many liters of CO2 (g) at STP is produced by
  the reaction of 64.0 grams of O2 (g)?
• Tip- BALANCE EQUATION FIRST

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Volume Volume Problems

• Question In the equation
• NH3 (g) O2 (g) ? NO (g) H2O (l)
• how many liters of NH3 (g) at STP are needed to
  react with 200. liters of O2 (g) at STP?
• Tip- BALANCE EQUATION FIRST

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Problems Involving Numbers of Particles

• Question In the equation
• C2H6 O2 ? CO2 H2O
• How many molecules of C2H6 are needed to produce
  27 grams of H2O?
• Tip- BALANCE EQUATION FIRST

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Limiting Reactants and Percent Yield

• Consider the recipe for tollhouse cookies. The
  number of batches of cookies you can bake depends
  on the amount of each of the starting materials
  you have.

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• If you are in short supply of any of the
  ingredients, then the recipe must be modified in
  order to make the cookies.
• For example, say you only had 1 1/4 cups of semi
  sweet chocolate chips.
• How would that affect the rest of your recipe?
• How many cups of flour would you wind up using?

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• When the quantities of reactants are available in
  the exact ratio described by the balanced
  equation, the chemists say that the reactants are
  in stoichiometric proportions. When this is the
  case, all the reactants will take part in the
  reaction and there will be no reactants left over
  one the reaction is complete.
• More often than not, one of the reactants is
  available in limited quantities. The reactants
  are then said to be in non-stoichiometric
  proportions.

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Limiting Reactant

• The reactant that limit the amount of product
  formed in a chemical reaction. It is completely
  used up in the chemical reaction.

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• To determine the limiting reactant in an
  equation, you must complete two separate mass-
  mass problems.
• First, write the balanced equation for the
  reaction.
• Then, calculate the mass of one of the reactants
  needed based on the other reactant.
• Compare the calculated needed value to what you
  actually are given.
• If the calculated needed value is more than what
  is given, the substance is the limiting reactant.
  If not, the other reactant will by default be
  the limiting reactant.

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• Question Identify the limiting reactant when
  10.0 g of water reacts with 4.5 g of sodium to
  produce sodium hydroxide and hydrogen gas.

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• Question Identify the limiting reactant when
  12.5 L H2S at STP is bubbled through a solution
  containing 24.0 g of KOH to from K2S and H2O.

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• Once the limiting reactant is identified, it is
  easy to calculate the theoretical mass or volume
  of the products. Remember, the limiting reactant
  will be used up completely in the chemical
  reaction. The other reactant will be in excess
  meaning there will be reactant left over.

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• Question If 3.5 g of Zn and 3.5 g of S are mixed
  together and heated, what mass of ZnS will be
  produced?

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• Question What mass of barium nitride (Ba3N2) is
  produced from the reaction between 22.6 g barium
  and 4.2 g nitrogen gas?

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Percent Yield

• Expected yield- the amount of a product that
  should be produced based on calculations.
• Actual yield- the amount of product that is
  really obtained from a chemical reaction.
• Why would the expected yield be different than
  the actual yield?
• Loss of product in the chemical reaction
• Error in measuring the starting materials
• Side reactions take place between the reactants
  leading to a different product than was expected.
• Reaction did not occur because of temperature,
  pressure, etc. was not ideal for the reaction.

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• Percent yield percent of the expected yield
  that was obtained.
• Percent yield Actual yield X 100
• Expected yield
• Question Determine the percent yield for the
  reaction between 2.80 g Al(NO3)3 and excess NaOH
  if 0.966 g Al(OH)3 is recovered.

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• Question Determine the percent yield for the
  reaction between 15.0 g N2 and 15.0 g H2 if 10.5
  g NH3 is produced.