 # 5- Determining the Formula of a Compound: The formula of a compound gives the atoms (or mole) ratio. It can be represented as: ❶empirical formula of a.

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5- Determining the Formula of a Compound: The formula of a compound gives the atoms (or mole) ratio. It can be represented as: ❶empirical formula of a chemical compound is the simplest whole number ratio of atoms in a compound (C n H n O n..) ❷molecular formula is the exact formula of a molecule, giving the types of atoms and the exact number of each type. (C n H n O n ….) n ❸structural formula of a chemical compound is a graphical representation of the molecular structure showing how the atoms are arranged

Determination of empirical formula Calcuate mass in grams of each element in 100 g of compound Calculate number of moles of each element in 100 g of compound Determine the molar ratio ( whole number if not multiply by integer to get the whole number) Obtain the empirical formula

Determination of molecular formula Compute the empirical formula mass Calculate the ratio

Example : Determine the empirical and molecular formulas for a compound that gives the following mass: 71.65 % Cl24.27 % C 4.07% H Molar mass = 98.96 g/mol Answer: Determination of empirical formula Assume we have a 100.00 g sample 1- Calcuate mass in grams of each element in 100 g of compound m Cl = 71.65 g,m C = 24.27 g andm H = 4.07 g 2- Calculate number of moles of each element in 100 g of compound n Cl = (71.65 g)/(35.45 g mol -1 ) = 2.021 mol Cl

n C = (24.27 g)/(12.01 g mol -1 ) = 2.021 mol C n H = (4.07 g)/(1.008 g mol -1 ) = 4.04 mol H 3 -Determine the molar ratio ( whole number if not multiply by integer to get the whole number) n Cl : n C : n H 2.021 : 2.021 : 4.04 1 : 1 : 2 Obtain the empirical formula Empirical formula = Cl C H 2 Determination of molecular formula 1-Compute the empirical formula mass Empirical mass = (1 x 35.45) + (1 x 12.01) + (2 x 1.008) = 49.48 g 2- Calculate the ratio = (98.96 g)/(49.48 g) =2 Molecular formula = (ClCH 2 ) 2 = Cl 2 C 2 H 4

6-Chemical Equations: A process in which one or more substances is changed into one or more new substances is a chemical reaction Note: In chemical reaction atoms are neither created nor destroyed. The chemical equation gives two important information: 1- The type (or nature) and the relative numbers (stoichiometric coefficients) of each of the reactants and the products. 2- The physical states of the reactants and products

7 7.Writing and balancing Chemical equations Chemical reactions are represented by chemical equations, which identify reactants and products. Formulas of reactants appear on the left side of the equation and those of products are written on the right. In a balanced chemical equation, there are the same number of atoms of a given element on both sides. You can’t write an equation unless you know what happens in the reaction that it represents. So all the reactants and products should be identified.

8 Balancing Chemical Equation: C 2 H 6 + O 2 CO 2 + H 2 O C 2 H 6 + O 2 CO 2 + H 2 O start with C or H but not O 2 carbon on left 1 carbon on right multiply CO 2 by 2 C 2 H 6 + O 2 2CO 2 + H 2 O 6 hydrogen on left 2 hydrogen on right multiply H 2 O by 3  Start by balancing those elements that appear in only one reactant and one product or the most complicated one.  Write the correct formula(s) for the reactants and the product(s) Ethane reacts with oxygen to form carbon dioxide and water

Balance the following chemical equation. Example 1: CH 4 + O 2  CO 2 + H 2 O answer CH 4 + 2 O 2  CO 2 + 2 H 2 O

Home work: Balance the following equations: 1- 2- 3-

8-Stoichiometric Calculations: Amounts of Reactants and Products: The coefficients in chemical equations represent number of molecules or (moles), not masses of molecules. Example: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g) What mass of oxygen will react with 96.1 grams of propane?

Home work: What mass of carbon dioxide is produced when 96.1 grams of propane is combusted with oxygen? Balance chemical equation C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O (g)

9- Calculations Involving a Limiting Reactant: The reaction between reactants occurs in Stoichiometric quantities: CH 4 (g) + H 2 O (g)  3H 2 (g) + CO (g) To calculate the mass of water required to react exactly with all amount of methane. That is, how much water will just consume all the amount of methane, leaving no methane or water remaining?

Example : If a sample containing 18.1 g of NH 3 is reacted with 90.4 g of CuO, which is the limiting reactant? How many grams of N 2 will be produced?

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