Overview
Does your empirical formula for magnesium oxide match the true formula for magnesium oxide, MgO? If your calculated empirical formula matches the expected empirical formula, what steps during the experiment do you think contributed to the success of your experiment? If your calculated empirical formula does not match the expected empirical formula, what parts or steps of the experiment do you think can be improved to ensure better results? Explain your reasoning in at least 500 words. Include your calculated empirical formula from your Data Submission
The molecular formula for Epsom salt has seven waters of hydration. Does your calculated waters of hydration ratio support this formula? If your calculated waters of hydration match the expected waters of hydration, what steps during the experiment do you think contributed to the success of your experiment? If your calculated waters of hydration do not match the expected waters of hydration, what parts or steps of the experiment do you think can be improved to ensure better results?
IntroductionFinding the Empirical Formula for Magnesium OxideThe basis for all chemical reactions is founded upon scientific laws. The first is the law of conservation of mass and energy, which states that matter and energy cannot be created or destroyed, but may be converted from one to the other. The second major law is the law of definite proportions, which states that a chemical compound contains exactly the same proportion of elements by mass. We can see this by looking at the molar mass of a compound. The molar massof a compound is the mass in grams of one mole of atoms or molecules of that compound. The mole (mol) is simply a grouped counting term used to group things together according to equation 1 below. This is similar to how the term “dozen” is used to group 12 things together to simplify counting larger numbers. The average molar mass of each element is listed on the periodic table, generally below the element’s symbol. For example, one mole of water molecules is comprised of two moles of hydrogen atoms and one mole of oxygen atoms. From the periodic table, we see that the molar mass of hydrogen atoms is 1.00794 g/mol, and the molar mass of oxygen is 15.9994 g/mol. Thus, the molar mass of water is 18.0153 g/mol. The law of definite proportions states that water is always made of approximately 8/9 oxygen and 1/9 hydrogen by mass. (Equation 1)Another fundamental law to chemistry is the law of multiple proportions. The law of multiple proportions states that if two elements form more than one compounds between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will be ratios of small whole numbers. Hydrogen and oxygen form two common compounds, water and hydrogen peroxide. In order to form water, 2 grams of hydrogen must react with 16 grams of oxygen. In order to form hydrogen peroxide, 2 grams of water must react with 32 grams of oxygen. The ratio of the mass of oxygen that react with hydrogen are 32:16, or 2:1. When applied to Dalton’s atomic theory, it becomes clear that water must have one oxygen atom while hydrogen peroxide has two atoms of oxygen.Stoichiometry is based upon all three of these fundamental laws in order to mathematically work back and forth within a balanced chemical equation. Equation 2 shows the unbalanced reaction of sucrose with elemental oxygen to form carbon dioxide and water. Equation 2 as written violates the law of conservation of mass and energy because the two sides of the reaction do not contain the same number of atoms of each element. In order to correct this, the equation must be balanced. The balanced chemical equation is given as equation 3. Upon inspection of equation 3, you can see that the number of atoms of each element are equal on both sides of the reaction. Equation 3 can be read as 2 molecules of sucrose react with 35 molecules of elemental oxygen to form 24 molecules of carbon dioxide and 22 molecules of water. Because the mole is a counting term, we can also read this as 2 moles of sucrose reacts with 35 moles of oxygen to form 24 moles of carbon dioxide and 22 moles of water.C12H22O11+O2? CO2+ H2OC12H22O11+O2? CO2+ H2O(Equation 2)2C12H22O11+35O2?24CO2+22H2O2C12H22O11+35O2?24CO2+22H2O(Equation 3)Since equation 3 is a balanced equation, we cannot change the molar coefficients for any compound involved in the reaction as this would violate the law of conservation of mass and energy. The molar ratio of each compound is constant. The relation between sucrose and carbon dioxide in this reaction will always have a 2:24 relationship. Because the chemical equation is balanced, and the molar ratio between all species involved in the reaction must be constant, if the reaction vessel contains a disproportion of reactants, then only those that are is proportion react. For example, if a reaction vessel contains 1 mole of sucrose and 35 moles of oxygen, then only 17 moles of oxygen will react with the sucrose to form carbon dioxide and water. The remaining 17 moles of oxygen will remain unreacted. This was determined because there was only 1 mole of sucrose in the reaction vessel. We therefore call sucrose in this instance the limiting reactant. The limiting reactant is simply the reactant in the least amount that can be used in the reaction. The law of definite proportions can be applied within the individual molecules to the molar ratio of each atom within that molecule. Sucrose will always have a 12:22 molar relationship between carbon and hydrogen.Experimentally, we can only find one of two things about a molecule: the simple molar ratio of atoms within the molecule, or the molar mass of the molecule. Let’s look at benzene as an example. The molar ratio of carbon to hydrogen is 6:6 in benzene. Mathematically, the ratio of 6:6 is identical to a 1:1 ratio. This means that experimentally, we can only determine a molar ratio of 1:1 for carbon and hydrogen in benzene. Therefore, our experimental (or empirical) formula is CH, which is simply the most reduced mole ratio between the two elements in this compound. It is possible for multiple compounds to have the same empirical formula. For example, cyclobuta-1,3-diene and cycloocta-1,3,5,7-tetraene both have empirical formulas of CH, but have different molecular formulas. Without knowing the experimental molar mass of benzene, we cannot determine its molecular formula.In this experiment, you will find the empirical formula of magnesium oxide by reacting elemental magnesium with the oxygen in the air. Since the amount of oxygen in the air is so vast compared to the amount of magnesium you will be reacting, magnesium is your limiting reactant. Because the air is only 20 % elemental oxygen and approximately 79 % elemental nitrogen, a side reaction will occur within your crucible. The magnesium will also react with the nitrogen to form magnesium nitride. In order to remove the magnesium nitride from the same so that you only have magnesium oxide, water can be added to the reaction vessel. The magnesium nitride will react with the water to form ammonia and magnesium hydroxide. Magnesium hydroxide is unstable at high temperatures and will decompose to magnesium oxide and water. Because of the law of conservation of mass and energy, you can determine the molar ratio of magnesium to oxygen in magnesium oxide by using the mass of the magnesium you start with and the mass of the product, since the only mass added to the product is atomic oxygen.To determine the molar ratio of magnesium to oxygen, you need to convert the mass of magnesium and the mass of oxygen to the number of moles of each element. This can be done using Equation 4 below. From there, you simply need to find the ratio of the moles of magnesium to the moles of oxygen. Your TA will help you find the smallest whole number ratio based on your data when you are performing this experiment. (Equation 4)Finding the Waters of Hydration of an Ionic SaltIonic compounds are classified as being made up of a cation and an anion. Many ionic compounds that are soluble in water can be separated from the water by boiling off the water through a process called distillation. Sometimes, the binding energy between the water molecules and the ions is so strong that through simple distillation, not all of the water will boil off, or evolve. This leaves behind a hydrated version of that ionic compound. Common hydrated ionic compounds include sodium acetate trihydrate, sodium thiosulfate pentahydrate, and barium chloride dihydrate. While most hydrated salts have integer values for the waters of hydration, it is possible to have fractional values.In this experiment, you will find the waters of hydration for the hydrated magnesium sulfate salt, also known as Epsom salt.The waters of hydration are found by determining the molar ratio between the anhydrous salt and the water evolved from heating. To find the molar ratio of these two entities, you will need to first find the number of moles of the anhydrous salt and the moles of water that evolved during the dehydration process. The formula for finding these values is given as Equation 5 below. Once you know the moles of salt and moles of water, you can find the simple ratio between these two values in order to find the waters of hydration. If you need help, your TA will help you find the smallest whole number ratio based on your data when you are performing this experiment. (Equation 5)Launch External ToolGlossary of Chemical Formulas and NamesChemical NameChemical FormulaWaterH2OH2OSucroseC12H22O11C12H22O11Carbon dioxideCO2CO2Elemental oxygenO2O2BenzeneC6H6C6H6Elemental nitrogenN2N2AmmoniaNH3NH3Magnesium hydroxideMg(OH)2Mg(OH)2Sodium acetate trihydrateNaC2H3O2?3H2ONaC2H3O2?3H2OSodium thiosulfate pentahydrateNa2S2O3?5H2ONa2S2O3?5H2OBarium chloride dihydrateBaCl2?2H2OBaCl2?2H2OMagnesium sulfateMgSO4MgSO4Experiment ObjectivesDemonstrate the safe use of an open flame to heat a reaction.Demonstrate the safe handling of hot glassware and equipment through the use of various tongs and protective equipment.Additional ResourcesFormula Mass from OpenStax ChemistryDetermining Empirical Formulas from OpenStax Chemistry