 # Stoichiometry By Ellis Benjamin. Definitions I Compounds - is a pure substance that is composed of two or more elements Molecules – is a combination of.

## Presentation on theme: "Stoichiometry By Ellis Benjamin. Definitions I Compounds - is a pure substance that is composed of two or more elements Molecules – is a combination of."— Presentation transcript:

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 10 23

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 C 2 H 6 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 K 2 Cr 2 O 7 ? 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 = C 9 H 12 O 9

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.

Stoichiometry Ratios can be used to convert one number into another. Example: 1 ml / 1 cm 3 or 1cm 3 / 1 ml Possible question: How many cm 3 are there in 5 ml of methanol?

Stoichiometry How many ml are in 0.125 liters of water?

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

Download ppt "Stoichiometry By Ellis Benjamin. Definitions I Compounds - is a pure substance that is composed of two or more elements Molecules – is a combination of."

Similar presentations