Topics in composition stoichiometry include the calculation of: Molar mass Percent Composition Molecular formula Empirical formulas

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1 Composition Stoichiometry Composition Stoichiometry NOTES 1 So far, we ve studied the products of a chemical reaction in terms of their identity. Stoichiometry is a branch of chemistry dealing with quantities. Composition stoichiometry deals with the calculation of how much of each element or ion are present in a compound. We will later study reaction stoichiometry, which uses correctly written chemical formulas in correctly balanced chemical equations to predict the amount of a product that may be obtained. Topics in composition stoichiometry include the calculation of: Molar mass Percent Composition Molecular formula Empirical formulas Molar Mass a review, so skip this if you can work the last problem on the homework, in the format shown. Molar mass is defined as the mass of 1 mole, or X particles of that substance. A particle will be an atom if the substance is an element A particle will be a molecule if the substance is a covalently bonded compound. A particle will be a formula unit if the substance is an ionically bonded compound. A particle will be an ion if the compound is dissolved in water.

2 Composition Stoichiometry NOTES 2 An atom has a mass of 1 atomic mass unit, or AMU, which is a weighted average of the number of protons and neutrons in the nuclei of all the isotopes. The mass of 1 mole of an element is equal to the atomic mass, changing the units from amu to grams. Thus, 1 mole of a substance has a mass equal to it s atomic mass stated in grams: The mass of 1 atom of carbon is amu. The mass of 1 mole of carbon atoms is grams. Thus, the molar mass, or mass of 1 mole, is the mass of 1 mole of the substance. The molar mass of carbon is grams/mole. Check your understanding: 1. What is the mass of 1 mole aluminum? Include units. 2. What is the mass of 1 mole sodium? Include units. 3. How many particles are present in 1 mole calcium atoms? What is the mass of that number of calcium atoms? 4. You are given grams of sulfur. How many moles sulfur do you have? How many atoms? 5. Explain why you used these numbers, and how you found them.

3 Composition Stoichiometry NOTES 3 To find the molar mass of a compound, add together the molar masses of every element in the compound. Consider how many atoms (moles) are present. For example, the setup to find the molar mass of strontium nitrate Sr(NO 3 ) 2 is: Sr: 1 mole Sr X grams Sr = grams Sr 1 mole Sr N: 2 moles N X grams N = grams N 1 mole N O: 6 moles O X grams O = grams O 1 mole O Molar mass of Sr(NO3)2 = grams mole Now, complete Composition Stoichiometry Assignment #1. Use the format shown above for your answers. Use your judgement as to how much practice you need to master the calculation and the format; you ll need it next. Percent Composition Cent means 100. Percent means per 100. Then, 50% means 50 out of 100, or as a ratio: which reduces to 2, or, calculated as a decimal, 0.50 Percents are always calculated as part total

4 Composition Stoichiometry NOTES 4 In composition stoichiometry, the percents of different components in a compound are determined by analyzing the chemical formula by atomic mass. NaCl is 50% Na and 50% if the particles are counted. Because Na and Cl have different masses, the percent by mass is not 50% Na and 50% Cl. Percent composition is sometimes called percent by mass, and is calculated like this: To calculate the percent composition of each element in sodium chloride: First, use the molar mass calculation to find the molar mass. percent: Then, use the masses of each element as part and the molar mass as total to calculate Na: 1 mole Na X g Na = g Na g Na x 100 = 39.33% Na 1 mole g NaCl Cl: 1 mole Cl X g Cl = g Cl g Cl x 100 = % Cl 1 mole g NaCl grams NaCl mole Let s try this for aluminum oxide. Calculate the percent by mass of aluminum and oxygen in aluminum oxide. (Hint: you must write the correct formula first.)

5 Composition Stoichiometry NOTES 5 Now, calculate the percent by mass of each aluminum and water in aluminum sulfate dihydrate. Water is part of the total mass, but is calculated separately, as shown below: The formula is Al2(SO 4 ) 3 2 H 2 O. Al: O: S: H2O In review, percent composition lets you find the composition by mass of a compound if you know its chemical formula and have a periodic table. Calculate the molar mass for each problem in Composition Stoichiometry Assignment #1. Leave space for the percent calculations.

6 Composition Stoichiometry NOTES 6 Empirical Formulas An empirical formula is a chemical formula showing the lowest ratio of all elements in a compound. The subscripts in the formula are in simplest terms. For example, C 6 H 12 O 6 is not an empirical formula. It is a molecular formula because it represents the actual number of atoms in the molecule. CH 2 O is an empirical formula because it shows the simplest ratio of atoms in the compound. Empirical formulas are calculated in a process that reverses Percent Composition calculations. In Percent Composition, the formula is used to find the amounts of components. The masses of components as determined in a laboratory analysis can be used to predict a chemical formula. Process: Start with an analysis from lab work. This is either given to you, or is the data you collect. If the analysis is given in percents, assume that you have 100 grams of the substance and use percents as grams. Convert the masses to moles, starting with the grams (or percent) given. Write the correct unit and the symbol for the substance with each mass/mole. Write the formula with the moles, written to at least 4 digits, as subscripts. Select the smallest mole value and divide each mole subscript by this smallest value to make 1 the smallest number in the ratio. If the quotients (new mole values) are within 0.1 of a whole number, round them. And you have the new formula. If they are more than 0.1 moles away, multiply each by the same number until all mole subscripts are whole numbers. You will need to refer to this list of steps as you work until you can do the problems without looking, so keep it handy.

7 Composition Stoichiometry NOTES 7 In steps: Using this set of steps, work through this problem. Start with an analysis from lab work. This is either given to you, or is the data you collect. Find the empirical formula of a hydrocarbon with the following composition: 48.64% carbon 8.16% hydrogen 43.20% oxygen If the analysis is given in percents, assume that you have 100 grams of the substance and use percents as grams g C 8.16 g H g O Convert the masses to moles, starting with the grams (or percent) given. Write the correct unit and the symbol for the substance with each mass/mole g C X 1 mole C = moles C g C 8.16 g H X 1 mole H = moles H 1.01 g H g O X 1 mole O = moles O g C

8 Composition Stoichiometry NOTES 8 Write the formula with the moles, written to at least 4 digits, as subscripts. C H O Select the smallest mole value and divide each mole subscript by this smallest value to make 1 the smallest number in the ratio. C H O becomes C H O 1 If the quotients (new mole values) are within 0.1 of a whole number, round them. And you have the new formula. If they are more than 0.1 moles away, multiply each by the same number until all mole subscripts are whole numbers. Hydrogen, at 2.989, can be rounded to 3. Oxygen becomes 1. Carbon, however is so must be multiplied by 2 to become a whole number. Multiply each subscript be 2 so that the ratio does not change, and you have 2 (C H O 1 ) or C 3 H 6 O 2 In review, empirical formulas are found by starting with experimentally determined masses and finding the chemical formulas. Work the problems on Composition Stoichiometry Assignment #2.

9 Composition Stoichiometry NOTES 9 Molecular formula a formula showing the types and numbers of atoms combined in a single molecule of a molecular compound. It is a whole number multiple of the empirical formula. The relationship between a compound s empirical and molecular formula can be written as: x(empirical formula) = molecular formula also x(empirical formula mass) = molecular formula mass Process: To determine the molecular formula of a compound, you must know the compound s formula mass. Divide the molecular mass by the empirical formula mass to determine the whole number multiple (x). You may have to find the empirical formula in order to obtain the empirical formula mass. Example: The empirical formula of a compound of phosphorus and oxygen was found to be P2O5. Experimentation shows that the molar mass of this compound is g. What is the compounds molecular formula. Empirical Formula = P2O5 Empirical Mass = 2 moles P X 31.0 g P = 62 grams P 1 mole P 5 moles O X 16.0 g O = 142g Molecular Mass = g So: 2 (P2O5) => (P2O5) => (P2O5) => P4O10 Work the problems on Composition Stoichiometry Assignment #3.

10 Composition Stoichiometry NOTES 10 Checking your understanding: Explain your answers to the questions below as if you were explaining them to a friend who had missed class be very clear. 1. What calculations, learned earlier this year, are a basis for finding percent composition? 2. How are percent composition calculation and an empirical calculation the reverse of one another? Which specific calculations are done as reverse processes in thee two calculations?