CHEM1111

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Chemistry

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Apr 3, 2024

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CHEM 1111 v.0916 General Chemistry I Laboratory Page 1 Laboratory 6. Empirical and Molecular Formulas Tommie Villanueva Why? An empirical formula is the simplest ratio of elements in a compound. It is important to be aware of what empirical formulas are and how they are used since they are used widely in chemical analysis and representations. A molecular formula is used to represent the exact formula of covalent molecules like the organic molecules discussed previously. The relationship between the two types is that in analyzing chemical compounds an empirical formula is used to find the molecular formula. Problem Statement: How can the empirical formula of a compound be experimentally found? Learning Objectives: ● To prepare a metal oxide from a metal ● To analyze a molecule and determine its formula ● To determine the empirical formula from the ratio of the reactants ● To observe and record data ● To make inferences from observations ● To critically analyze sources of error Resources ● Read the section(s) in your textbook about empirical and molecular formulas. ● Safety glasses ● Graduated cylinder ● Steel wool ● Short plastic beaker ● Two plastic pipette bulbs ● Two 3 inch pieces of stiff steel wire ● Two wire leads with alligator clips ● 9 V battery clip ● 9 V battery ● Epsom salts ● Distilled water ● Thermometer Information - Empirical Formulas An empirical formula is the lowest whole number ratio of elements in a compound. There are two primary uses for empirical formulas. One use is the representation of the formulas of ionic and macromolecular compounds. In these compounds there are not distinct molecules, there are large number of repeating units with the same formula like NaCl or SiO 2 (sand). The other use is the chemical analysis of covalent molecules, by a method called elemental analysis , in which the ratio of elements are determined in one experiment and the molecular

CHEM 1111 v.0916 General Chemistry I Laboratory Page 2 weight is determined in another. Model - Empirical Formulas Compound Formula H 2 O 2 H 2 O C 2 H 4 O 2 C 2 H 6 O C 6 H 14 CaCl 2 Fe 2 O 3 N 2 O 4 N 2 O Empirical Formula HO H 2 O CH 2 O C 2 H 6 O CH 7 CaCl 2 Fe 2 O 3 NO 2 N 2 O Model - Determining an Empirical Formula A particular compound of sulfur and oxygen is found to have a 1:1 mass ratio, what must be its empirical formula? The steps an expert problem solver might use to solve this problem are: What assumptions fit the data? There are 100 grams of the compound. 1% of 100 is 1, making calculations easier. How many moles of each element are present in 100g? 50.0 g S/ 32.07 g/mol S = 1.56 mol S 50.0 g O/ 15.99 g/mol O = 3.12 mol O Moles are a count, grams are an amount. What is the ratio of the number of each element? 1.56:3.12 = 1:2 1.56 S to 3.12 O or 1 S per 2 O What is its empirical formula? S 1 O 2 or properly SO 2 Subscripts of 1 are not shown. The 4 steps to solve this problem are: STEP 1 = Assume 100g. STEP 2 = Find the moles of each element. STEP 3 = Find the ratio by dividing the moles by the smallest amount of moles. STEP 4 = Write the Empirical Formula. SAFETY: i) Safety glasses must be worn at all times in lab.

CHEM 1111 v.0916 General Chemistry I Laboratory Page 3 ii) Any chemical contact with your skin or eyes needs to be immediately and throughly washed. iii) Acetone should be used in a well ventilated area and over exposure to its vapor can be harmful. a) Gloves should be worn. b) Don't rub your eyes. c) Acetone vapors are flammable. d) Acetone can damage plastics, varnished and painted sufaces. iv) Do not use a microwave to dry these samples, metal placed in a microwave can cause sparking that can damage the oven and/or ignite flammable vapors. Procedure You will study the product produced by leaving vinegar-wetted steel wool to air over an extended period. i) Determine and record the mass of the steel wool left over from a previous lab in Table 1. If not all of the steel wool has been converted try to separate as much of the product from the remaining steel wool. Determine and record the mass of any leftover steel wool. (If the previous lab was not completed, then take a 1 g sample of steel wool. Record the mass. Soak the steel wool in acetone and then pat dry on a paper towel. Allow to rust over the next few hours or days.) iii) Determine and record the difference in the mass of dried remains minus the mass of any left over steel wool, this is the mass of product. You are trying to determine the mass of the rust and not any un-rusted steel wool. iv) Record the original mass of the steel wool in Table 1. Subtract the mass of any remaining steel wool (the un-rusted steel wool) from the original mass (before rusting) and record the value in Table 1. This is the mass of iron contained in the product. v) Subtract the mass of iron contained in the product from the mass of the product and record it in Table 1. This is the mass of the oxygen contained in the product. vi) Calculate the percentage of iron and oxygen in the sample and record them in Table 2. Remember that the mass of rust = mass of steel wool unrusted + mass of oxygen reacted. vii) The remains of the steel wool and its product may be disposed of in household waste. Table 1 Current Steel Wool Mass, g Leftover Steel Wool, g Product Mass, g Original Steel Wool (before reaction), g Iron, g Oxygen, g

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