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Working with units. DO NOT GET CARELESS…EVER! Always keep track of your units I will take points off of anything you give me if you leave the units off I will stare at you when you give me an answer without units until you put them in your answer IT REALLY IS THAT IMPORTANT!!!!.

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working with units
Working with units
    • Always keep track of your units
  • I will take points off of anything you give me if you leave the units off
  • I will stare at you when you give me an answer without units until you put them in your answer
dimensional analysis
Dimensional Analysis
  • You will frequently have to convert between units during your life (Yes, your entire life!)
  • To do this, we will need to employ DIMENSIONAL ANALYSIS
  • Step 1: What do you have? What do you need?
  • Step 2: What is the conversion factor?
  • Step 3: Setup a calculation that cancels your given units and puts your target units on top.
dimensional analysis example
Dimensional Analysis: Example
  • A sample of an alloy has a density of 7.9 kg/cm3. What is the density in mg/m3?

dalton s atomic hypothesis
Dalton’s Atomic Hypothesis
  • All atoms of a particular element are identical
  • Atoms of different elements have different masses
  • A compound is a specific combination of atoms of more than one element
  • In a chemical reaction, atoms are neither created nor destroyed; they exchange partners to produce new substances

These tenets were first proposed 202 years ago and they are all true!

c compounds
C: Compounds
  • Compounds. Huh?
  • A compound is an electrically neutral substance that consists of two or more different elements with their atoms present in a definite ratio
  • Compounds: Terminology
    • Binary: Consists of only 2 elements
    • Organic: Contains Carbon and hydrogen
    • Inorganic: No Carbon
ionic and molecular compounds
Ionic and Molecular Compounds
  • Ionic Compound: Ions form compound that is electrically neutral
    • Usually formed by the reaction of a metal and a nonmetal

Na(s) + Cl(g) NaCl(s)

  • Molecular Compound: Binary molecular compounds are usually formed by the reaction of 2 nonmetals

2H2(g) + O2(g)  2H2O (l)

e moles and molar masses
E: Moles and Molar Masses
  • The Mole: The most important concept you’ve never heard about

A mole of objects contains the same

number of objects as there are carbon atoms in 12.0 g of Carbon-12

6.022x1023 somethings/mole = Avogadro’s Number

1 mole of any element always has 6.0221x1023 atoms in it
  • 1 mole of elephants always has 6.0221x1023 elephants in it.
  • In equations, you will frequently see the number of moles represented by the variable “n”
  • 1 kmol = 1000 mol
  • 1 mol = 1x10-6 mol
  • # of objects = n(NA) where NA is Avogadros’ number
molar mass
Molar Mass
  • The molar mass is the mass of one mole of material
    • Also referred to as molecular weight or formula weight for compounds
  • The molar mass of an element is the mass of one mole of its atoms
  • The molar mass of a compound is the mass of a mole of its molecules.
  • Molar mass (M) is usually given in g/mole

Mass of Sample = nM

determination of chemical formulas
Determination of Chemical Formulas
  • For hundreds of years, people knew that drinking willow bark tea would belp cure a headache
  • When scientists identified a compound form Willow bark that was biologically active, they had to determine its structure before they could synthesize it.
  • Determining the chemical formula is the first step in the structure determination process…
chemical formulas
Chemical Formulas
  • We will deal with 2 types of chemical formulas:
  • Empirical formula: Shows the relative numbers of each element present in the compound (Emp. Form. Of Hydrogen Peroxide?)
  • Molecular formula: The precise number of atoms of each element present in the compound (Mol. Form. Of Hydrogen Peroxide?)
mass percentage composition
Mass Percentage Composition
  • The mass percentage composition of a compound tells us the percentage (by mass) that a given element comprises of the compound.
  • We frequently determine the mass % composition using combustion analysis

Mass % Composition Summary:

Mass % composition is found by calculating the fraction of the total mass contributed by each element present in a compound and expressing the fraction as a percentage.

determining empirical formula from mass data
Determining Empirical Formula from Mass % Data
  • To convert the mass % composition obtained from a combustion analysis into an empirical formula, we must convert the mass % of each type of atom into the relative number of atoms.
    • To do this, assume that we have 100g of sample
    • The mass % will then be in grams

determining molecular formulas
Determining Molecular Formulas
  • Once we know the empirical formula, we need one more piece of information to determine the molecular formula: The Molar Mass
  • Say we know the empirical formula of a compound is C3H4O3.
    • All we know about this compound at this point is the ratio of the 3 elements.
    • We don’t know the exact number of each type of atom in the molecule.
    • Is the Molecular Formula C6H8O6, C12H16O12 or C18H24O18?

g molarity
G: Molarity
  • Hands down one of the most important concepts you need to master is you are going to stay in the sciences. Period.
  • The Molar Concentration, c, of a solute in solution is the number of moles of solute divided by the volume of the solution (in liters).
    • Also referred to as Molarity

The symbol M is used to denote the molarity of the solution

1M NaCl = 1 mole NaCl per liter of H2O

g4 dilutions
G4: Dilutions
  • Frequently in the laboratory, you will need to make dilutions from a stock solution.
  • This involves taking a volume from the stock and bringing it to a new volume with solvent.
  • In order to perform these dilutions, we can use the following equation:

c1V1 = c2V2

Where: c1 = Stock concentration

V1 = Volume removed from stock

c2 = Target conc of new sol’n

V2 = Volume of new solution

law of conservation of matter
Law of Conservation of Matter
  • “Matter can neither be created nor destroyed” – Antoine Lavoisier, 1774

If a complete chemical reaction has occurred, all of the

reactant atoms must be present in the product(s)

law of conservation of matter1
Law of Conservation of Matter



  • Stoichiometry coefficients are necessary to balance the equation so that the Law of Conservation of Matter is not violated
    • 6 molecules of Cl2 react with 1 molecule of P4
    • 3 molecules of Cl2 react with 2 molecules of Fe
balancing chemical reactions
Balancing Chemical Reactions
  • Let’s look at Oxide Formation
  • Metals/Nonmetals may react with oxygen to form an oxide with the formula MxOy
  • Example 1: Iron reacts with oxygen to give Iron (III) Oxide

Fe (s) + O2 (g) → Fe2O3 (s)

how do we solve it
How do we solve it?

Fe (s) + O2 (g) → Fe2O3 (s)

  • Step 1: Look at the product. There are 3 atoms of oxygen in the product, but we start with an even number of oxygen atoms.
    • Let’s convert the # of oxygens in the product to an even number

Result: Fe (s) + O2 (g) → 2Fe2O3 (s)

how do we solve it1
How do we Solve It?

Fe (s) + O2 (g) → 2Fe2O3 (s)

  • Then, balance the reactant side and make sure the number/type of atoms on each side balance.

Balanced Equation: 4Fe (s) + 3O2 (g) → 2Fe2O3 (s)

how do we solve it2
How do we Solve It?
  • Example 2: Sulfur and oxygen react to form sulfur dioxide.

S (s) + O2 (g) → SO2 (g)

  • Step 1: Look at the reaction. We lucked out!

Balanced Equation: S (s) + O2 (g) → SO2 (g)

how do we solve it3
How do we Solve It?
  • Example 3: Phosphorus (P4) reacts with oxygen to give tetraphosphorus decaoxide.

P4 (s) + O2 (g) → P4O10 (s)

  • Step 1: Look at the reaction. The phosphorus atoms are balanced, so let’s balance the oxygens.

Balanced Equation: P4 (s) + 5O2 (g) → P4O10 (g)

how do we solve it4
How do we Solve It?
  • Example 4: Combustion of Octane (C8H18).

C8H18 (l) + O2 (g) → CO2 (g) + H2O (g)

  • Step 1: Look at the reaction. Then:
    • Balance the Carbons

C8H18 (l) + O2 (g) → 8CO2 (g) + H2O (g)

how do we solve it5
How do we Solve It?

C8H18 (l) + O2 (g) → 8CO2 (g) + H2O (g)

  • Step 2: Balance the Hydrogens

C8H18 (l) + O2 (g) → 8CO2 (g) + 9H2O (g)

  • Step 3: Balance the Oxygens
    • Problem! Odd number of oxygen atoms
      • 12.5 Oxygens on reactant side
    • Solution: Double EVERY coefficient (even those with a value of ‘1’)
how do we solve it6
How do we Solve It?

C8H18 (l) + 12.5O2 (g) → 8CO2 (g) + 9H2O (g)

  • Step 3 (cont’d): Balance the Oxygens

2C8H18 (l) + 25O2 (g) → 16CO2 (g) + 18H2O (g)

  • Step 4: Make sure everything checks out