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Stoichiometry

Stoichiometry. By Ellis Benjamin. Definitions I. Compounds - is a pure substance that is composed of two or more elements Molecules – is a combination of two or more atoms held together by covalent bonds Molecular Weight – is the sum of atomic weights (amu) of the atoms in a molecule.

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Stoichiometry

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  1. Stoichiometry By Ellis Benjamin

  2. Definitions I • Compounds - is a pure substance that is composed of two or more elements • Molecules – is a combination of two or more atoms held together by covalent bonds • Molecular Weight – is the sum of atomic weights (amu) of the atoms in a molecule. • 1 Mole = 6.022 x 1023

  3. Definitions II • Molar Weight – is the mass of 1 mole of a compound (equals the molar mass and molecular weight) • Equivalent – is used to define equal amounts of atoms and molecules used in a reaction • Stoichiometry – is the conversion of one amount to another using equal values.

  4. Important Equations • Moles = Grams / Molecular Weight (eq. 1) • % Comp = (Element Mass in Formula / Formula Weight) * 100 (eq. 2) Empirical and Molecular Formula • 1) # moles of atom = Percent = grams / Atomic Weight of Atom (eq.3) • 2) Divide by lowest # moles of atoms then multiply to get the whole # • 3) Divide the molecular weight given by this number to get the molecular formula • Stoichiometry – calculation using equal values

  5. Let look at Equation 1 • Moles = Grams / Molecular Weight • What this means is that if we are given a certain amount of Grams of a given substance (molecule, atom, or compound). And it is possible for us to calculate the Molecular Weight. Then we can calculate the Moles of substance.

  6. Calculate Equation 1 • If we have 5 grams of C2H6 how many moles would we have? • 1) Identify what the question is asking for. Moles • 2) What knows do we have. Grams and Molecular Weight • 3) Calculate Molecular Weight. C=12*2 and H = 6*1 so (12*2) + (6*1) = 30 g/mol • 4) 0.167 Moles = 5 grams / 30 g/mol

  7. Let look at Equation 2 % Comp = (Element Mass / Formula Weight) * 100 • What is the percent composition of Chromium in K2Cr2O7? • Molecular Weight = (2*39) + (2*52) + (7*16) • Molecular Weight = 294 g/mol • Elemental Mass of Cr = (2*52) • Elemental Mass of Cr = 104 g/mol • % Comp = (104 (g/mol) / 294 (g/mol)) * 100 • % Comp = 35.4 %

  8. Let’s Look at Equation 3 • What are the empirical and molecular formula of a compound that contains 40.9% carbon, 4.58% hydrogen, 54.52% oxygen and has a molecular weight of 264 g/mol? • Step 1. Make percent equal grams • 40.9% = 40.9 g • 4.58% = 4.58 g • 54.52% = 54.52 g

  9. Equation 3: Step 2 • Divide the grams by the atomic weight of the atom. • 3.41 mol of C = 40.9 g / 12 (g/mol) • 4.58 mol of H = 4.58 g / 1 (g/mol) • 3.41 mol of O = 54.52 g / 16 (g/mol)

  10. Equation 3: Step 2 • Divide by the lowest number of moles. • 3.41 mol of C / 3.41 = 1 • 4.58 mol of H / 3.41 = 1.33 • 3.41 mol of O / 3.41 = 1.0

  11. Equation 3: Step 2 • Make the all values equal the lowest whole number. • 1 mol of C * 3 = 3 • 1.33 mol of H * 3 = 4 • 1.0 mol of O * 3 = 3

  12. Equation 3: Step 3 • Calculate the Molecular Formula of the given empirical formula • 1 mol of C * 3 = 3 * 12 • 1.33 mol of H * 3 = 4 * 1 • 1.0 mol of O * 3 = 3 * 16 • Empirical Formula = 88 g/mol

  13. Equation 3: Step 3 • Divide the given molecular weight by the empirical formula. • Given Molecular Weight = 264 g/mol • Empirical Formula = 88 g/mol • Molecular Percent = 264 g/mol / 88 g/mol • Molecular Percent = 3 • Molecular Formula = C9H12O9

  14. Stoichiometry • Ratios can be used to convert one number into another. • The easiest way to convert is to place the initial number on the bottom of the conversion and the desired number on the top.

  15. Stoichiometry • Ratios can be used to convert one number into another. • Example: 1 ml / 1 cm3or 1cm3 / 1 ml • Possible question: How many cm3 are there in 5 ml of methanol?

  16. Stoichiometry • How many ml are in 0.125 liters of water?

  17. Stoichiometry • How many grams of Methanol are found in 2.54 liters (density 1.50 g/l)?

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