Chemical Composition

Chemical Composition

• Chapter 6 discusses chemical composition.

Lecture Outline

• The topics to be covered include calculating the number of atoms or molecules in a given sample of a substance and the relative amounts of elements in a compound.

How Much Sodium?

• Most materials are chemically or physically combined with other substances.
• Knowing the fraction of a sample represented by a specific material helps determine the amount of that material in the sample.
• Knowledge of composition is important for dietary considerations, steel production, and hydrogen fuel.

Counting Nails by the Pound

• A familiar analogy to counting atoms is determining how many nails are in a pound when the hardware store sells them that way.
• This is analogous to finding how many atoms are in a given mass of an element.

Counting Nails example

• A customer buys 2.60 pounds of nails where a dozen nails has a mass of 0.150 pounds.
• How many nails did the customer buy?
• Applying the conversion: 2.60 lbs x (1 dozen nails / 0.150 lbs) x (12 nails / 1 dozen) = 208 nails

Counting Atoms by Moles

• The mass of a specific number of atoms can convert the mass of an element sample to the number of atoms.
• A mole is 6.022 x 10^23 atoms.
• A mole is also known as Avogadro's number.

Chemical Packages—Moles

• A mole is the number of things equal to the number of atoms in 12 g of C-12.
• One atom of C-12 weighs exactly 12 amu.
• One mole of C-12 weighs exactly 12 g.
• In 12 g of C-12, there are 6.022 x 10^23 C-12 atoms.

Nails vs Elements

• These pictures have the same number of nails.
• The weight of one dozen nails changes for different nails.

Sulfur vs Carbon

• These pictures have the same number of atoms.
• The weight of one mole of atoms changes for different elements.
• One mole S (32.07 g)
• One mole C (12.01 g)

Converting Between Moles and Number of Atoms

• A silver ring contains 1.1 × 10^22 silver atoms.
• Given: 1.1 × 10^22 Ag atoms, Find: mol Ag, Conversion Factor: 1 mol Ag = 6.022 × 10^23 Ag atoms
• How many moles of silver are in the ring? Solution: 1.1 × 10^22 Ag atoms x (1 mol Ag / 6.022 × 10^23 Ag atoms) = 1.8 × 10^-2 mol Ag

Calculate the Number of Atoms in 2.45 mol of Copper.

• This is a practice problem in finding number of moles
• Given: 2.45 mol Cu, Find: atoms Cu, Solution: 2.45 mol Cu x (6.022x10^23 atoms / 1 mol) = 1.48×10^24 atoms Cu

Relationship Between Moles and Mass

• The molar mass is the mass of one mole of atoms.
• The molar mass of an element, in grams, is numerically equal to the element's atomic mass, in amu.
• The lighter the atom, the less a mole weighs.
• The lighter the atom, the more atoms there are in 1 g.

Mole and Mass Relationships

• This table shows pieces in 1 mole and the weight of 1 mole
• Hydrogen number of atom: 6.022 x 10^23 atoms, weight of 1 mol: 1.008 g
• Carbon number of atom: 6.022 x 10^23 atoms, weight of 1 mol: 12.01 g
• Oxygen number of atom: 6.022 x 10^23 atoms, weight of 1 mol: 16.00 g
• Sulfur number of atom: 6.022 x 10^23 atoms, weight of 1 mol: 32.06 g
• Calcium number of atom: 6.022 x 10^23 atoms, weight of 1 mol: 40.08 g
• Chlorine number of atom: 6.022 x 10^23 atoms, weight of 1 mol: 35.45 g
• Copper number of atom: 6.022 x 10^23 atoms, weight of 1 mol: 63.55 g

Calculate number of moles, example

• Calculate the number of moles of sulfur in 57.8 g of sulfur.
• Given: 57.8 g S, Find: mol S, Conversion Factor: 32.07 g S = 1 mol S, Solution: 57.8 g S X (1 mol S / 32.07 g S) = 1.80 mol S

Calculate the Moles of Carbon in Pencil Lead Example

• Calculate the Moles of Carbon in 0.0265 g of Pencil Lead
• Given: 0.0265 g C, Find: mol C, Conversion Factors: 1 mol C = 12.01 g
• Solution: 0.0265 g C× (1 mol / 12.01g) = 2.21×10^-3 mol C

Converting Between Grams and Number of Atoms

• Find count of aluminum atoms in an aluminum can with a mass of 16.2 g?
• Given: 16.2 g Al, Find: Al atoms, Conversion Factor: 26.98 g Al = 1 mol Al and 6.022 × 10^23 = 1 mol
• Solution: 16.2 g Al x (1 mol Al / 26.98 g Al) x (6.022 × 10^23 Al atoms / 1 mol Al) = 3.62 x 10^23 Al atoms

Copper Atoms in a Penny Example

• Solving for Cu atoms in a penny with a mass of 3.10 g
• Given data: 3.10 g Cu, 1 mol Cu = 63.55 g, 1 mol = 6.022 x 10^23
• Solution: 3.10 g Cu x (1 mol Cu / 63.55 g Cu) x (6.022x10^23 atoms / 1 mol)
• Answer: = 2.94×10^22 atoms Cu

Counting Molecules by the Gram

• The relative weights of molecules can be calculated from atomic weights.
• Formula mass of 1 molecule of H₂O = 2(1.01 amu H) + 16.00 amu O = 18.02 amu.
• 1 mole of H₂O contains 2 moles of H and 1 mole of O
• Molar mass 1 mole H₂O = 2(1.01 g H) + 16.00 g O = 18.02 g.

Grams and Moles for Compounds Example

• Calculate the mass (in grams) of 1.75 mol of water.
• Given: 1.75 H2O, Find: g H2O, Conversion Factor: H₂O molar mass = 2(1.01) + 1(16.00) = 18.02 g/mol
• Solution: 1.75 mol H2O × (18.02 g H2O/ 1 mol H2O) = 31.5 g H2O

Pennny Given Lead Moles Example

• Calculate the Moles of Carbon in 0.0265 g of Pencil Lead
• Given: 0.0265 g C - Find: mol C, Relationship: 1 mol C = 12.01 g
• Solution: 0.0265 g C× (1 mol / 12.01g)= 2.21×10^-3 mol C

Practice Formula Units

• How Many Formula Units Are in 50.0 g of PbO2? (PbO2 = 239.2)
• Solution Map: g PbO2 -> mol PbO2 -> units PbO2
• Relationships: 1 mol PbO2 = 239.2 g,1 mol = 6.022 x 1023
• Solution: 50.0 g PbO2 x (1 mol PbO2 / 239.2 g PbO2) x (6.022×1023 units / 1 mol) = 1.26×1023 units PbO2

Finding Mass with Molecular Factor

• Find the mass of 4.78 X 10^24 NO2 molecules.
• Given: 4.78 × 10^24 NO2 molecules, Find: g NO2, Conversion Factor: 6.022 × 10^23 = 1 mol, NO2 molar mass = 46.01 g/mol
• Solution: 4.78 × 10^24 NO₂ molecules x (1 mol NO2 / 6.022 × 10^23 NO2 molecules) x (46.01 g NO2 / 1 mol NO2) = 365 g NO2

Chemical Formulas

• Chemical Formulas act as conversion factors
• 1 spider = 8 legs.
• 1 chair = 4 legs.
• 1 H₂O molecule = 2 H atoms = 1 O atom.

Counting Parts

• Counts of parts can be used to count the number of whole units,
• If we know how many parts are in the whole unit.
• For example, if all desks have 4 legs and there are 30 desks in the room, there will be 120 legs (4 x 30).
• For example, if every H₂O molecule has 2 H atoms, in 100 H₂O molecules, there are 200 H atoms. Or In 1 mole of H₂O molecules, there are 2 moles of H atoms.

Mole Relationships in Chemical Formulas

• Atoms and molecules are counted in moles.
• The number of moles of a constituent element if we know the number of moles of the compound is calculable.
• Examples: 1 mol NaCl -> 1 mol Na, 1 mol Cl or 1 mol H2O -> 2 mol H, 1 mol O or 1 mol C6H12O6 ->6 mol C, 12 mol H, 6 mol O

Moles in CaCO Example

• Determine the number of moles of O in 1.7 mol of CaCO3.
• Given: 1.7 mol CaCO3, Find: mol O, Conversion Factor: 3 mol O = 1 mol CaCO3
• Solution: 1.7 mol CaCO3 X (3 mol O / 1 mol CaCO3)= 5.1 mol O

Determine Find Mass Carvone Example

• Find the mass of carbon in 55.4 g of carvone.
• Carvone (C10H14O) is the main component of spearmint oil and is in chewing gum, liqueurs, soaps, and perfumes.
• Given: 55.4 g C10H14, Find: g C, Conversion Factors: Molar mass = 150.2 g/mol, 1 mole C10H14O = 150.2 g C10H14O, 1 mole C10H14O = 10 mol C, 1 mole C = 12.01 g C
• Solution: 55.4 g C10H14O x (1 mol C10H14O/ 150.2 g C10H14O) x (10 mol C / 1 mol C10H14O) x (12.01 g C / 1 mol C )= 44.3 g C

Finding Mass of Sodium Examples

• Find the Mass of Sodium in 6.2 g of NaCl
• Given: 6.2 g NaCl - Find: g Na - Relationships Example: 1 mol NaCl = 58.44 g, 1 mol Na = 22.99 g, Solution Map: g NaCl ->mol NaCl ->mol Na ->g Na
• Solution: 6.2 g NaClx (1 mol NaCl / 58.44 g) x (1 mol Na / 1 mol NaCl) x (22.99 g Na / 1 mol Na) = 2.4 g Na

Mass Percent Conversion.

• The mass percent shows the mass of a constituent element in 100 g of the compound.
• Fact: NaCl @ 39% Na by mass means that 100 g of NaCl contains 39 g Na.
• Conversion examples: g NaCl x (39 g Na/ 100 g NaCl) = g Na or g Na x (100 g NaCl/ 39 g Na) = g NaCl

Sodium Chloride Example

• How many grams of sodium chloride can be consumed still be within FDA guidelines of less than 2.4 g of sodium per day?
• Given: 2.4 g Na, Find: g NaCl, Conversion Factor: 39 g Na = 100 g NaCl.
• Solution: 2.4 g Na X (100 g NaCl / 39 g Na) = 6.2 g NaCl.

Mass Percent Composition

• It can be determined by mass
• The formula of the chemical can help determien it.
• The experimental mass analysis of the compound.
• The percentages may not always total to 100% due to rounding.
• Percentage = (mass of element X in 1 mol) / (mass of 1 mol of the compound) ×100%
• Percentage = (part / whole) ×100%

Finding Cl Example

• Find the Mass Percent of Cl in C2Cl4F2.
• Given C2Cl4F2 and you need to find % Cl by mass
• Then you need the molar mass C2Cl4F2 and divide that by chlorine molecular mass.

Percent Composition CalcL2 Example

• Determine the Percent Composition of CaCl₂ (Ca = 40.08, Cl = 35.45)
• Mass % Ca = (molar mass Ca) / (molar mass CaCl2) ×100%
• Mass % Cl = (2x molar mass Cl) / (molar mass CaCl2) ×100%

Whole Number Atoms

• The empirical formula, is the simplest, whole-number ratio of atoms in a molecule.
• Can be determined from percent composition or combining masses.
• The molecular formula is a multiple of the empirical formula.

Empirical Formulas

• The molecular formula for Hydrogen Peroxide = H2O2 and the Empirical formula = HO
• Molecular formula for Benzene= C6H6 and Empirical formula = CH
• Molecular formula for Glucose = C6H12O6 and Empirical formula = CH2O

greatest common Formula

• A way for determining the empirical formula.
• Find the greatest common factor(GCF)
• 20 factors = (10 x 2), (5 x 4) & 12 factors = (6 x 2), (4 x 3), gives a GCF = 4
• Take C20H12 = (C5H3)4 giving a Empirical formula of = C5H3

Finding an Empirical Formula

• Convert percentages to grams
• Convert grams to moles Use molar mass of each element.
• Write a pseudoformula using moles as subscripts
• Divide all by smallest number of moles.
• Multiply all mole ratios by number to make all whole numbers, if necessary.

Computing Empricial formula.

• Calculate the empirical formula of a compound containing 24.5 g of nitrogen and 70.0 g of oxygen.

Compute Empirical formula (2)

• A laboratory analysis of aspirin determined the following mass percent composition: C 60.00%, Η 4.48%, Ο 35.53%
• 1. First, turn each into a 100-g sample: C is 60.00 g C, Η is 4.48 g H & Ο is 35.53 g O.
• 2. After this, calculate the moles of each element using molar mass
• 3. Then, you divid by 2.221 - ISOI &

Titanium reacts Formula Oxide

• a 3.24 g sample of titanium reacts with oxygen to form 5.40 g of the metal oxide. What is the formula of the oxide?

Formulas: Stannous Fluoride

• Determine the Empirical Formula of Stannous Fluoride, which Contains 75.7% Sn (118.70) and the Rest Fluorine (19.00).
• Given: 75.7% Sn, (100 – 75.3) = 24.3% F in 100 g stannous fluoride there are 75.7 g Sn and 24.3 g F.

Hemitite Contains Oxyen

• Determine the Empirical Formula of Hematite, which Contains 72.4% Fe (55.85) and the Rest Oxygen (16.00). Find FexOy then 72.4% Fe, (100 – 72.4) = 27.6% O with 100 g of hematite having 72.4 g Fe and 27.6 g O.

Molicules Share Common Formulae

• Molecular Formula: Glyceraldehyde is C3H6O3, Erythrose is C4H8O4, Arabinose is C5H10O5 and Glucose is C6H12O6
• Empirical formula for all four is CH2O with masses respectively of 90, 120, 150 and 180.

Formulas: compounds for moleculars

• The molecular formula is a multiple of the empirical formula.
• This is dertermined with Molar mass real formula = Factor used to multiply subscripts and Molar massempirical formula

Mothballs formula

• Formula as follows-Naphthalene is a compound containing carbon and hydrogen that is often used in mothballs.
• With its empirical formula = C5H4 and it's molar mass of -128.16 g/mol you can find the molecular formula.

Benzopyrene Formula

• There's more details on determining compounds, Molar Mass has Molicular Formulas and this is found with a solution map

Nicotine Formula

• Determine the Molecular Formula of Nicotine, which has a Molecular Mass of 162 g and is composed of { 74.0% C, and 8.7% H with remaining 17.3% being N

Oxidation and Reduction

• Redox: reactions are also called redox reactions.
• All redox reactions involve the transfer of electrons from one atom to another.
• In combustion reactions, O2 combines with all the elements in another reactant to make the products.

Oyxgen in redox

• Reactions in which O₂ is gained or lost are redox reactions. When an element attaches to an oxygen during the course of a reaction it is generally being oxidized.
• When an element loses an attachment to oxygen during the course of a reaction, it is generally being reduced. For example: In 2 Fe2O3(s) → 4 Fe(s) + 3 O2(g) Fe is being reduced.
• One definition of redox is the gain or loss of O, but it is not the best.

Redox Reacting

• Reaction involves a metal reacting with a nonmetal while converting a free element into ions.

Another Redox Formula

• In order to convert a free element into an ion, the atoms must gain or lose electrons.
• Of course, if one atom loses electrons, another must accept them.
• Atoms that lose electrons are being oxidized, gaining electrons mean they are reduced.

Identfifying formula element

• Identify the Element Being Oxidized and the Element Being Reduced. 2 C + O2 → 2 CO then C is oxidized.
• Identify the Element Being Oxidized and the Element Being Reduced. Mg + Cl2→ MgCl2; Mg is oxidized

Oxidation Formula element

• Oxidation and reduction must occur simultaneously and must take electrons or lose electrons.
• The reactant that reduces an element in another reactant is called reducing agent and will contain the element that is isoxidized.
• The reactant that oxidizes an element in another reactant is called oxidizing agent and will contain the element thatis reduced.

oxidizig element element

• Examples C + O2 → 2 CO: Oxygen is the oxidizign agent.

oxidation equation state

• Need a method for determining how the electrons are transferred with Reactions that are not metal + nonmetal, or do not involve O2.
• Chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reaction.
• Although they look like them, oxidation states are not ion charges!. Oxidation states are imaginary charges assigned based on a set of rules where ion charges are real.

Rules:Assignning Equation state

• They must be in the order of priority as follows: first the Free elements have an oxidation state = 0 such as Na +2 and Cl2 + 0 and so Monoatomic ions have an oxidation state equal to their charge such as Na +1 and Cl = -1 in NaCl(s) and the sum must Equal 0.

state of atoms - Equation state

• The sum of the oxidation states of all the atoms in a polyatomic ion equals the charge on the ion: for expample N = +5 and O = -2 in NO3

Equation state I metals

• Group I metals have an oxidation state of +1 in all their compounds and Group II metals have an oxidation state of +2 in all their compounds.

Non metal equation state

• Nonmetals have oxidation states which are higher in prioritiy and for an element to be considered they Must be higher on the table.

Element following formula

• Assign an Oxidation State to Each Element in the Following: F2, Mg2+, KC1 ,502 , PO43-, BaO2.

Oxidation Reduction state

• Oxidation occurs when an atom's oxidation state increases during a reaction and Reduction occurs when an atom's oxidation state decreases during a reaction.

Oxidation and Reduction Reactants

• In an oxidation reaction the amount of substance used and made in a chemical reaction is related to the amounts of all the other substances in the reaction due to the law if conservations.
• The study of the numerical relationship between chemical quantities in a chemical reaction is stoichiometry.

Balancing Equations

• The number of pancakes you can make depends on the amount of the ingredients you use.
• A mathematical expression would be: 1 cup flour = 2 eggs = ½ tsp baking powder = 5 pancakes

Pancake Formulas

• What the egg coefficient? If you want to make more or less that 5 pancakes eggs amount must be right.
• This is because assuming you have enough flour and baking powder 8 eggs ×(5 pancakes /2 eggs) = 20 pancakes

Chemical Molocules

• The balanced equation is the "recipe" for a chemical reaction as seen with 3 H2(g) + N2(g) → 2 NH3(g) that has 3 molecules of H₂ reacting -with 1 molecule of N₂ and makes 2 molecules of NH3 or: -3 molecules H₂ = 1 molecule N2 = 2 molecules NH3 then when we use Moles, and count molecules it's: 3 moles H₂ = 1 mole N2 = 2 moles NH3

Many moles?

• How Many Moles of NaCl Result from the Complete Reaction of 3.4 Mol of Cl2 With 2 Na -2 Na(s) + Cl2(g) → 2 NaCl Given quantity, we apply a solution map.

Mass of Glucose water.

• Practice with the following how many moles of water are made in the combustion of 0.10 moles of glucose? using the equation is C6H12O6 + 6 O2 → 6 CO2 + 6 H₂O.

Mass to Mass

• We want to have a relationship between the mass and number of moles of a chemical.
• 1 mole = Molar Mass in grams. And thanks to the balanced chemical equation we can go back and forth.

Photo synthesis with plants (mass of glucose)

• Plants in photo synthesis make mass conversions to determine to following reaction -6 CO2(g) + 6 H2O(l) ---- sunlight ---> 6 O2(g) + C6H12O6(aq) This is to demonstrate that grams of glucose cane be synthesized from 58.5 g of CO₂ under normal conditions.

Mass of kilgrams with pollution.

• Acid rain the forms reacts with the chemicals -4 NO2(g) + O2(g) + 2 H2O(l) ==4 HNO3(aq) and assuming there is more than enough O₂ and H₂O, is to show the kilograms using 1.5 × 103 kg of kilgrams with pollution

Determining Empirical Hemite with Oxide

• Practice finding the Molecular Formula or Hematite, which Contains in 72.4% Fe (55.85) and the rest Oxygen (16.00).

Limiting Reactants?

• The Limiting Reagent (or Reactant) is used find out what will limit quantities of -the products but the Excess quantity is a reactant that has insufficient quantity. With limited, Theoretical Yield would result if all the limited reacts reacts in that form. The Extant of the reaction determines if it would be present In a limit.

Limited Reaction Formula

• We can use the limited reactant to determine why it important.
• How do know what reaction has limited reactant then if one is been given masses then u start thinking about the mole number then can have limit.

Limited reaction step

• For the balance equation of Finding what is a limited. This is done by setting and using the ratio the formula must use the equation between required substance for number of mole requirement or balanced numbers.

Molor

limiting reactant and theoretical step from Al

• Find a chemical equation of and with an alluminum equation this is determined to see If it has limiting Reactant and Theoretical Yield from Initial Moles of Reactants