Empirical Formula, Molecular Formula, Percent Composition, & Stoichiometry Class Notes Mr. Brown-2012 The Mole (Molar Mass) 1. Each atom (element) on the Periodic Table has a certain mass to it. Since all elements are different from one another, they each have different atomic masses. The average atomic mass for each element on the Periodic Table is the amount of grams in 1 mole of that element called the molar mass of that element or atom. For example: 1 mole of H atoms = 1.01 grams 1 mole of He atoms = 4.00 grams 1 mole of Li atoms = 6.94 grams 1 mole of Be atoms = 9.01 grams 2. Since there are different versions of each atom called isotopes (each having a slightly different mass) the average atomic mass of an element is given on the table as a decimal number (as shown above for H, He, Li, and Be atoms).

Molecular Formula 1. The molecular formula for a compound tells the types and numbers of atoms that chemically make up a compound. Remember a compound is a substance where 2 or more different types of elements are combined in a mole to mole ratio. 2. For example here are some molecular (chemical) formulas for some compounds: H2O 2 Hydrogen atoms and 1 Oxygen atom Mg (NO3)2 1 Magnesium atom, 2 Nitrogen atoms, 6 Oxygen atoms CaCl2 1 Calcium atom, 2 Chlorine atoms Ca3(PO4)2 3 Calcium atoms, 2 Phosphorus atoms, 8 Oxygen atoms MgSO4 1 Magnesium atom, 1 Sulfur atom, 4 Oxygen atoms AlCl3 1 Aluminum atom, 3 Chlorine atoms 3. Now to calculate the molar mass (what one mole of each of the previous chemical formulas has in grams per one mole) Count the number and types of each atom that make up the molecule and multiply by the atom’s average atomic mass from the periodic table.

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Examples: A) Find the molar mass for Mg(NO3)2 1 Magnesium atom x 24.31 grams= 24.31 grams 2 Nitrogen atoms x 14.01 g = 28.02 g 6 Oxygen atoms x 16.00=96.00 g Take the sum of all the masses to get = 148.33 g (this is the molar mass of Mg(NO3)2) B) Find the molar mass for CaCl2 1 Calcium atom x 40.08 g = 40.08 g 2 Chlorine atoms x 35.43g = 70.86 Take the sum of all the masses to get = 110.94 g (this is the molar mass of CaCl2)

Empirical Formula 1. The molecular formula indicates the types and number of atoms that make up a chemical compound. The chemical (molecular) formula is a multiple of a much simpler formula called the empirical formula. The empirical formula is simply the lowest reduced subscripts that make up a molecular formula. For example: The following molecular formulas: C2H4O2, C4H8O4, C6H12O6, C200H400O200, all are based on the empirical formula CH2O, which is the lowest reduced subscript formula. 2. When solving for the empirical formula of a compound follow the steps below: a) When given the grams of each element in a compound, convert the grams amount into moles of each element by dividing by the molar mass of that element. b) If you are given percentages of each element in a compound, assume you have a 100 gram sample of the compound and change the % sign into the grams of the element and follow step a) above. c) Once you know the # of moles of each element get all the mole ratios to their lowest ratio by dividing all the mole amounts by the lowest answer. d) You have now calculated the empirical formula for the compound. e) If your molar ratios are now whole numbers such as 1.5 moles, multiply by the decimal amount to all mole numbers to get them to whole numbers. For example is the molar ratio for one element is 2.5, multiply all half to get rid of the 0.5 and all numbers should now be to whole numbers. 3. Empirical Formula Sample Problem: A compound was analyzed in a lab to determine its empirical formula. Decomposition of the compound at standard temperature and pressure produced 3.60 g of carbon, 16.8 L hydrogen gas, and 2.80 L of oxygen gas. a) What is the empirical formula for this compound? b) The molar mass of the compound is 174.24 g/mol. What is its molecular formula?

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Solution: a) You are given the mass of carbon and Liters of hydrogen and oxygen in the molecule. The first step is to convert all of these values to moles of each element because we can only compare them in a mole to mole ratio. 3.60 g carbon x 1 mole carbon/12.01 g = 0.3 moles carbon 16.8 L hydrogen x 1 mole hydrogen/22.4 L = 0.750 moles hydrogen 2.80 L oxygen x 1 mole oxygen/22.4L = 0.125 moles of oxygen Now you have the molar ratios of each element in the compound but they are all in decimal form. Let’s get them to whole number ratios (which are easier to work with than decimals) by dividing all the moles by the lowest number in this case 0.125 moles for the oxygen. By doing this we get whole number ratios! 0.3 moles carbon 0.750 moles hydrogen 0.125 moles of oxygen

All divided by 0.125 moles =

3 moles carbon 6 moles hydrogen 1 mole oxygen

Our empirical formula is C3H6O The molar mass for our empirical formula will be: 3 carbon atoms x 12.01 g = 36.03 grams carbon 6 hydrogen atoms x 1.01 g= 6.06 grams hydrogen 1 oxygen atoms x 16.00 g = 16.00 grams oxygen 36.03 g carbon + 6.06 g hydrogen + 16.00 g oxygen = 58.09 grams / 1 mole of C3H6O b) Now the molecular formula will be a multiple of the empirical formula C 3H6O. If the molar mass of the empirical formula is 58.09 g, then our molecular formula’s molar mass must be a multiple of that because it is all the same elements and ratios. Divide the molecular formula’s molar mass (174.24 g) by the empirical formula’s molar mass (58.09) and that quotient answer tells you how many times the empirical formula divides into the molecular formula and multiply the empirical formula subscripts by that value to get the molecular formula which the problem is essentially asking you to figure out. 174.25 g / 58.09 g = 3 The molecular formula is a ratio 3 times that of the empirical formula. C3H6O x 3 = C9H18O3

Percent Composition 1. Percent composition is the percentage that a certain atom makes up by mass in the chemical compound that it is a part of. 2. To solve for percent composition:

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a) Find the mass that the atom makes up within the compound it is a part of and divide its mass by the molar mass of the compound that it is in and multiply by 100% to make the decimal a percentage. Examples: A) Find the percent composition for each atom that makes up the compound: Mg(NO 3)2 1 Magnesium atom x 24.31 grams= 24.31 grams 2 Nitrogen atoms x 14.01 g = 28.02 g 6 Oxygen atoms x 16.00=96.00 g Take the sum of all the masses to get = 148.33 g (this is the molar mass of Mg(NO3)2) % Mg = 24.31 g Mg/ 148.33 g Mg(NO3)2 =0.1639 x 100% = 16.39% Mg by mass % N = 28.02 g N/ 148.33 g Mg(NO3)2 = 0.1889 x 100% = 18.89% N by mass % O = 96.00 g O/ 148.33 g Mg(NO3)2 = 0.6472 x 100% = 64.72% O by mass

Mole Conversions 1. When converting between moles, grams, molecules and Liters of a gas do the following rules: a) Whatever the unit given, always convert to moles. If given grams of a substance, divide by the substance’s molar mass to get to moles. b) If given Liters of a gas, the magic numbers is that 1 mole of any gas occupies a volume of 22.4 Liters, so if given Liters divide it by 22.4 L to get to moles of that gas. c) If given molecules, 1 mole of any substance contains 6.02 x 10 23 molecules. So, if you’re given molecules of a substance divide by 6.02 x 10 23 to get moles of that substance. d) If given moles of a substance and asked to convert to another unit (grams, Liters, or molecules) simply do the opposite math operation as mentioned above. 2. For a reference use the table flow chart below for more help.

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2. Mole Conversion Sample Problem: Convert 80 grams of NH3 gas to: a) Liters b)moles c) molecules Solution: a) First you must calculate the molar mass of NH3 by adding up the grams of hydrogen and nitrogen that make it up. The molar mass of NH3 = 17.04 grams/ 1 mole of . Convert 80 grams NH3 by dividing by its molar mass 17.04 g.

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80 g NH3 x 1 mole NH3/17.04 g = 4.69 moles of NH3 Next to convert to L, multiple 4.69 moles of NH3 by the conversion factor 22.4 L/ 1 mole of gas = 105. 06 L of NH3 gas. b) 4.69 moles NH3 from step a c) To calculate molecules NH3 gas, multiply 4.69 moles of NH3 by Avogadro’s number (6.02 x 1023 molecules/1 mole). 4.69 moles NH3 x 6.02 x 10 23 molecules /1 mole NH3 = 2.82 x 10 24 molecules of NH3.

Stoichiometry 1. Stoichiometry is a branch of chemistry where chemists can calculate how much reactant or product in a chemical reaction will be needed to react and/or formed in the reaction. It is simply related to the mole concept and mole conversions of relating different reactants and products in a reaction to each other in a mole to mole ratio. I’ll say that again it is vital that you only compare substances in a mole to mole ratio, not gram to gram or Liter to Liter but in a mole to mole ratio!!! 2. The molar ratios in the equation that compare substances in the equation come from the coefficients that are in the final balanced equation. 3. Steps in Solving Stoich Problems:

a) Balance the given equation FIRST!! Use the appropriate coefficients in front of each substance’s chemical formula to balance the numbers of atoms in the equations that reside on the left and right side of the reaction arrow. b) Once you balance the equation take the given amount of the substance and convert it to moles using the mole conversion rules mentioned above. c) Compare the moles you calculated to the moles of the other substance that you are trying to calculating in the problem by comparing the 2 substances in a mole to mole ratio using the coefficients from the balanced chemical equation. d) Once you now know the moles of the substance, you may be asked to convert it to grams, Liters, and/or molecules and here again use your mole conversion rules mentioned previously. Stoich Sample Problem: Using the following equation answer the questions below:

Al + H2SO4  Al2(SO4)3 + H2 a) If 8 moles of Al react, how many moles of H2 gas will be produced? b) 4.5 moles of H2SO4 makes how many moles of Al2(SO4)3? c) 4 grams Al would produce how many moles of H2 gas? d) If 6.02 x 10 23 molecules of H2 were made how many molecules of H2SO4 were needed to react? Solution:

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a) First, like all stoich problems the chemical equation given must be balanced by placing the correct number of coefficients in front of the chemical formulas in the equation. By doing so, we get a balanced equation of:

2Al + 3H2SO4  Al2(SO4)3 + 3H2 Next, we are given 8 moles of Al and we need to compare it to moles of H 2 by looking at the balanced equation. Always start off with what you have in moles and multiple it by the molar ratio of the two compounds you are comparing in the problem. So we have: 8 moles Al x 3 moles H2/2 moles Al = 12 moles of H2 gas produced!

b) We are given 4.5 moles of H2SO4 and asked how many moles of Al2(SO4)3 will be produced so we must multiply our given by the molar ratios from the balanced equation of H2SO4 and Al2(SO4)3. 4.5 moles H2SO4 x1 mole Al2(SO4)3 / 3 moles H2SO4 = 1.5 moles of Al2(SO4)3 produced. c)We are given 4 g of Al, so first to compare it to H2, we must convert 4 g Al to moles of Al by dividing by the molar mass of Al: 4 g Al x 1 mole Al/26.98 g = 0.15 moles of Al Next we compare moles Al to moles H2: 0.15 moles Al x 3 moles H2/2 moles Al=0.23 moles H2 d) Finally we are given 6.02 x 10 23 molecules of H2 gas and asked how many moles of H2SO4 will be needed to react, we always convert the given information to moles first, then compare it mole to mole with the other chemical in the problem. 6.02 x 1023 molecules H2 x 1 mole H2/6.02 x 10 23 molecules = 1 mole H2 1 mole H2 x 3 moles H2SO4/3 moles H2=1 mole H2SO4 1 mole H2SO4 x 6.02 x 10 23 molecules/ 1 mole H2SO4 = 6.02 x 10 23 molecules H2SO4

Limiting Reactant 1. Many times substances in a chemical reaction (equation) will not react in a perfect mole to mole ratio based on the balanced chemical equation. The amount of the substance that fully reacts and gets totally used up is called the limiting reactant because it is the substance that you are limited by and can only make so much product as its amount will allow you. For example, if you had all the bread in the world but only one piece of turkey you are limited by the turkey and can only make one sandwich. The bread on the other

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hand is the reactant that is not totally consumed in the chemical reaction and is called the excess reactant, because it is leftover in excess. Limiting Reactant/ Excess Reactant Sample Problem Using the balanced equation below answer the problems below:

2Al + 3H2SO4  Al2(SO4)3 + 3H2 4 moles of Al react with 7 moles of H2SO4: a) How many moles of Al2(SO4)3 can be made? b) What is your limiting reactant? c) What is your excess reactant? d) How many grams of Excess Reactant are leftover (unreacted) from the reaction? Solution: a) & b) As always start off with a balanced chemical equation which is given to us, so that work is done. Compare both of your reactant amounts to the same product in this case Al2(SO4)3. Then find out how much products will be produced from each individual reactant. Whichever reactant yields the least amount of product that is your limiting reactant. 4 moles Al x 1 mole Al2(SO4)3 / 2 moles Al= 2 moles Al2(SO4)3 7 moles H2SO4 x 1 mole Al2(SO4)3 / 3 moles H2SO4 = 2.33 moles Al2(SO4)3 The fact that Al gave less moles means you are limited by the amount of Al that you have so Aluminum is you limiting reactant. And only 2 moles of Al2(SO4)3 can be produced. c) The only other reactant you have is H2SO4, so it is your excess reactant meaning not all of it reacted during the chemical equation. d) The final problems here is to find out how much of your excess reactant was needed to fully react with your limiting reactant and then subtract that amount from the given amount of excess reactant to how much excess reactant was leftover. We have: 4 moles Al (limiting reactant) x 3 moles H2SO4/ 2 moles Al = 6 moles of H2SO4 (excess reactant) needed to fully consume the 4 moles of Al. We needed 6 moles H2SO4 of to fully consume Al, but we were given 7 moles of H2SO4 in the beginning of the problem. So of the 7 moles H2SO4given, 6 moles H2SO4 consumed the 4 moles of Al. The remainder is 7 moles H2SO4– 6 moles H2SO4 = 1 mole H2SO4.

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1 mole H2SO4 x 98.09 g / 1 mole H2SO4= 98.09 grams of H2SO4 leftover and unreacted (still floating around in the reaction).

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