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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