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.

• 1 Mole = 6.022 x 1023

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.

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

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.

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

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 %

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

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)

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

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

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

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

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.

5 ml of methanol 1 cm3

1 mlx 5 cm3 of methanol=

Stoichiometry

• Ratios can be used to convert one number into another.

• Example: 1 ml / 1 cm3 or 1cm3 / 1 ml

• Possible question: How many cm3 are there in 5 ml of methanol?

5 ml of methanol 1 cm3

1 mlx 5 cm3 of methanol=

Stoichiometry

• How many ml are in 0.125 liters of water?

0.125 l of water 1000 ml 1 liter

x 125 ml of water=

Stoichiometry

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

2.54 grams of ethanol 1.50 grams1 liter

x 3.81grams of ethanol=