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The Mole and Stoichiometry

The Mole and Stoichiometry. p. 167 – 201 and p. 223 – 259. Review of concepts. What is the Law of Definite Proportions? How do we calculate relative atomic mass? Do question p. 167 #5 What do the co-efficients in a balanced chemical equation refer to?. The Mole.

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The Mole and Stoichiometry

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  1. The Mole and Stoichiometry p. 167 – 201 and p. 223 – 259

  2. Review of concepts • What is the Law of Definite Proportions? • How do we calculate relative atomic mass? • Do question p. 167 #5 • What do the co-efficients in a balanced chemical equation refer to?

  3. The Mole • What is a mole? (Abbreviated to mol) • SI unit, which is a lab-sized unit of a substance that is a quantity of matter in chemistry • It is like terms we use like “dozen”, which means 12, “pair”, which means 2, etc. • ***1 mol of any substance is equal to the formula mass or molecular mass of that substance.

  4. The Mole • The formula mass is the mass of a formula unit. • For example: 1 mol of carbon (C) atoms = 12.01 g/mol 1 mol of sodium hydroxide (NaOH) = 40.00 g/mol 1 mol of aluminum carbonate (Al2(CO3)3) = 233.99 g/mol

  5. The Mole • Percent Composition • The percentage by weight of an element in a compound is the same as the number of grams of that element present in 100 grams of the compound

  6. The Mole • Determining a chemical formula begins with finding what elements are present. This work is called qualitative analysis. • Then the ratios by mass of these elements are determined by measuring how much of each element can be obtained from a given mass of the compound. Such experimental work is called quantitative analysis.

  7. The Mole • The clearest way to understand percentage by weight is to view it as representing the number of grams of a particular element in 100 g of the compound. • Problem: A sample of a liquid with a mass of 8.657 g was decomposed into its elements to give 5.217 g of carbon, 0.9620 g of hydrogen, and 2.478 g of oxygen. What is the percentage by weight of each element in the liquid?

  8. The Mole • If we have the experimental percentage composition of an unknown substance, we can possibly identify the unknown by comparing its percentage composition to those of known substance. • To make this comparison, you have to be able to calculate theoretical percentages, the percentage composition of a compound based on its formula. • Problem: What is the percentage by weight of each element in CaO?

  9. The Mole • Mol Calculations • We have defined a mol as, “SI unit, which is a lab-sized unit of a substance that is a quantity of matter in chemistry”. • Also, one mol is equal to the formula mass of a substance • Since all mols are not equal, we need to calculate the corresponding masses for each mol quantity we have, especially in chemical reactions.

  10. The Mole • Avagadro’s Number • The number of formula units in one mole is so large that we really can’t fully appreciate it, but one mole of any pure substance contains 6.02 x 1023 formula units. • This number is called Avogadro’s number in honour of Amadeo Avogadro (1776-1856) and is given the symbol N.

  11. The Mole • Anologies • Shopping and the price you pay • Your mark on a test • Comparing Apples to Oranges • The Mol Ladder

  12. Using the Mole • Empirical Formulas • A compound’s empirical formula tells what elements are present and in what ratio by atoms (or moles) • The next step to finding the formula of a compound is to use the percentages of each element to calculate the ratios of the different atoms present.

  13. Using the Mole • The chemical formula that gives these ratios by atoms in their smallest whole numbers is called the empirical formula. (the simplest formula) HO – a 1 to 1 ration of hydrogen atoms to oxygen atoms. Hydrogen peroxide, however, consists of individual molecules that are made of two hydrogen atoms and two oxygen atoms. The formula of the molecule – the molecular formula – is therefore H2O2. • Water, on the other hand, has an empirical formula of H2O. The molecular formula is the same, since the ratio of hydrogen to oxygen is always 2:1.

  14. Using the Mole • Molecular Formulas • A molecular formula goes beyond the information in an empirical formula to give the actual composition of a molecule • In an earlier exercise, you found an empirical formula of HgCl, a 1 to 1 ratio of mercury atoms to chlorine atoms. • But that ratio could also be expressed in the formula Hg2Cl2, or Hg3Cl3, etc. • The formula of calomel is, in fact, not HgCl but Hg2Cl2, and that kind of formula is called a molecular formula.

  15. Using the Mole • A molecular formula is a chemical formula that gives the actual composition of a molecule. • To find out that calomel is Hg2Cl2 and not HgCl all that is needed is one more piece of data, the formula weight of the compound. • The formula weight of calomel, as determined by experiment, is 472.09. The formula HgCl corresponds to a formula weight of 236.04. That’s half of 472.09. • The only way to double 236.04 and have the same basic ratio of atoms of mercury to atoms of chlorine is to double all the atoms in HgCl and write the formula as Hg2Cl2.

  16. Using the Mole • The formula of a hydrate • A hydrate is a compound formed from a solution of crystals in water. When the compound crystallizes, water molecules adhere to the crystal. • The formula is a ratio of the compound to the water. The formula mass has to take into consideration the water.

  17. Using the Mole • How can we determine the formula of a hydrate? • To determine the formula experimentally, we need: • to heat the substance • find the mass of water lost • find the mols of each • compare the mol ratios of the compound to the water

  18. Stoichiometry • Stoichiometry is the study of the relationships or ratios between two or more substances undergoing a physical or chemical change (chemical reaction). • A balanced chemical equation provides the same kind of quantitative information that a recipe does. In a cookie recipe, the ingredients are the reactants and the cookies are the products.

  19. Stoichiometry • A balanced chemical equation tells us what amounts of reactants to mix and what amounts of product to expect. • Chemists use balanced chemical equations as a basis to calculate how much reactant is needed or product is formed in a reaction.

  20. Stoichiometry • The numbers in a chemical equation are the number of moles reacting. We can convert the number of moles to grams to find the mass of the substances we are using. The ratios are the same as what you did above. • In a calculation, this is called using a Mol Ratio.

  21. Stoichiometry • Example Consider the following equation: H2(g) + N2(g) NH3(g) If you react 6 mols of nitrogen:

  22. Stoichiometry • Stoichiometric Calculations • We do not find the mass of substances in mols. We use grams. • We start off with grams and end up measuring our products in grams. We use the mols and chemical equation to tell us how much is going to react and the ratio between reactants and products.

  23. Stoichiometry • The steps for stoichiometry are: • Balance the equation • Find the mols of the substance you are given • Use a mol ratio between what you are given and what you are finding. • Find the mols of what you are finding in the question. • Find the mass of what you are finding in thequestion.

  24. Stoichiometry • Analytical chemistry is the study of the separation, identification, and quantification of the chemical components of natural and artificial materials.[1] • Qualitative analysis gives an indication of the identity of the chemical species in the sample and quantitative analysis determines the amount of one or more of these components. The separation of components is often performed prior to analysis.

  25. Stoichiometry • When we experimentally determine the mass of the product formed, and make a comparison between the theoretical yield and actual yield, we are performing analytical methods • Theoretical yield • – the amount of substance used or produced, based upon the calculation • Actual yield • – the amount of substance used or produced experimentally (REALLY MADE!!)

  26. Stoichiometry • Percent yield • – the ratio of the actual yield to the theoretical yield • Percent error • – the value given in a percentage, as to how close you were to the theoretical value. You can be above or below the value, depending on the experimental errors that occurred.

  27. Stoichiometry • Limiting Reactants • In a chemical reaction, one substance will be used up totally leaving some of the other substance behind. • The substance (reactant) used up first will determine the amount of product produced. This is called the limiting reactant. • The left over reactant is in excess.

  28. Stoichiometry

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