1.1 The Mole Concept 1.2 Formulas

Assessment Objectives Distinguish between the terms empirical formula and molecular formula Determine the empirical formula from the percentage composition or from other experimental data Determine the molecular formula when given the empirical formula and experimental data.

Key Terms Empirical Formula Molecular Formula Formula Mass: describes the mass of the smallest repeating unit of an ionic compound. molecular mass: describes the mass of a molecule. The Percentage composition: of a compound refers to the relative mass of each element in the compound Law of definite proportions: describes that elements in a given compound are always present in the same proportions by mass.

Assessment Objectives Difference between the terms empirical formula and molecular Formula

Molecular Formula The actual number of moles of each element in 1 mol of the compound.

Empirical Formula The simplest whole number ratio of moles of each element in 1 mol of the compound. For example Hydrogen Peroxide molecular formula is H 2 O 2 However its empirical formula would be HO ‘Empirical’ means ‘by experimentation’

Empirical Vs Molecular Formula NameMolecularEmpirica l Low ratio Hydrogen peroxideH2O2H2O2 HO1:1 GlucoseC 6 H 12 O 6 CH 2 O1:2:1 Benzene C6H6C6H6 CH1:1 Ethene C2H4C2H4 CH 2 1:2 Butene C4H8C4H8 CH 2 1:2 Aniline C6H7NC6H7NC6H7NC6H7N 6:7:1 WaterH2OH2OH2OH2O2:1

Assessment Objective Determine the empirical formula from the percentage composition or from other experimental data

1. Percentage Composition In a compound, it is useful to know the how much of the mass is made up of the different elements. The formula of a compound can be used to determine the mass % of each element by using the molar relationships we have learned...

Example find the percentage composition by mass of oxygen in carbon dioxide. 1 Determine the MOLAR MASS of the compound (2 x 16.00) = g mol Determine the molar mass of the element. 2 x = 32 g mol % Comp = Molar Mass Oxygen x 100 Molar Mass total

% Comp = 32.00/44.01 x 100 = 72.71% oxygen by mass

Calculate the percentage composition of magnesium carbonate, MgCO 3. From previous slide: g g + 3(16.00 g) = g Note: It should equal 100%

PRACTICE 1. Calculate the mass percentage of nitrogen in each compound: a) N 2 O b) Sr(NO 3 ) 2 c) NH 4 NO 3 d) HNO 3 e) NH 4 f) KNO 3

2. Calculating Empirical Formula Calculate empirical formula of a compound that is 85.6% carbon and 14.4% hydrogen. First, assume that you have 100g of the substance, so you really have 85.6g of C and 14.4g of H. Second, convert it to moles using molar mass. Finally, find the lowest whole ratio between the two.

g C and 14.4g H 2. mol C = 85.6g x mol = 7.13mol C 12.01g mol H = 14.4g x mol = 14.3 mol H 1.01g 3. Lowest mole ratio: C = 7.13 /7.13 = 1 H = 14.3/ 7.13 = 2.01, so 1:2 ANSWER: CH 2

Sometimes an extra step is needed Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

Divide each value of moles by the smallest of the values. Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen: Do you see the problem here?

Multiply each number by an integer to obtain all whole numbers. Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x Empirical formula: C3H5O2C3H5O2

Find the Empirical Formula of the Following……… % hydrogen and 82.5% nitrogen % lithium and 53.7% oxygen % boron and 84.1% fluorine % chlorine and 47.48% sulfur

Assessment Objective Determine the molecular formula when given the empirical formula and experimental data.

Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = g

2. Divide the molecular mass by the mass given by the emipirical formula.

3(12.01 g) + 5(1.01) + 2(16.00) = g 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

Practice The empirical formula is CH 2 O, and its molar mass is 150g/mol. What is molecular formula?

Find empirical molar mass of CH 2 O: g/mol =30.02 g/mol Divide real molar mass by empirical molar mass: 150/30.02 = 5 Multiply subscripts of empirical formula through by ratio (5): C 5 H 10 O 5