Mathematics of Chemistry Topic 3 Mathematics of Chemistry aka Stoichiometry

Topic 3-Outline Formula mass/Gram-formula mass Molecular Mass/Gram-molecular mass Percent Composition -hydrates Empirical and Molecular Formulas Mole grams moles liters Math of Equations a. mole-mole problems b. gram-gram problems c. volume-volume problems d. mixed problems

A term used to refer to the mass of a mole of any substance Definitions to Know Molar mass- A term used to refer to the mass of a mole of any substance aka: formula mass/gram-formula mass molecular mass/gram-molecular mass Mole- The amount of a substance that contains 6.02 x 1023 (Avogadro’s #) representative particles of that substance Just like 1 dozen = 12 representative particles 1 baker’s dozen = 13 representative particles 1 gross = 144 representative particles 1 ream = 500 representative particles 1 mole = 6.02 x 1023 representative particles

How do we calculate a molar mass? C6H12O6 1. Make sure the formula is written correctly 2. Determine the number of atoms of each element. Carbon (C)-6 atoms total Hydrogen (H)-12 atoms total Oxygen (O)-6 atoms total 3. Multiply the # of atoms by the mass # C = 6 atoms x 12 = 72 H = 12 atoms x 1 = 12 O = 6 atoms x 16 = 96 4. Add these numbers and this is the molar mass 72 + 12 + 96 = 180 g/mole

Let’s try: (NH4)2SO4 CH3COOH N = 2 atom x 14 = 28 C = 2 atoms x 12 = 24 H = 4 atoms x 1 = 4 H = 8 atoms x 1 = 8 O = 2 atoms x 16 = 32 S = 1 atom x 32 =32 O = 4 atoms x 16 = 64 Total = 60 g/mol Total = 132 g/mol CuSO4.5H2O Ca(NO3)2 Cu = 1 atom x 64 = 64 Ca = 1 atom x 40 = 40 S = 1 atom x 32 = 32 N = 2 atoms x 14 = 28 O = 9 atoms x 16 = 144 O = 6 atoms x 16 = 96 H = 10 atoms x 1 = 10 Total = 250 g/mol Total = 164 g/mol

Percent Composition-see Table T % element = mass of element in compound x 100 total mass of compound Ex: H2SO4 % H = 2 x 100 = 2.04 % H 98 % S = 32 x 100 = 32.65 % S 98 % O = 64 x 100 = 65.31 % O 98

Let’s try: (NH4)2SO4 % N = 28 x 100 = 21.21 % N 132 % H = 8 x 100 = 6.06 % H 132 % S = 32 x 100 = 24.24 % S 132 % O = 64 x 100 = 48.48 % O 132

Let’s try it with information presented a different way! How much gold can be obtained from 500 g of AuCl? 1. Find the percent of Au in the compound AuCl. % Au = 197 x 100 = 84.91 % Au 232 Nothing has changed so far but now here is the additional step: Multiply the mass of the sample by the % of element and divide by 100 (500 g) (84.91)/100 = 424.55 g Au

You must be able to do this with the information presented Percent of Water in a Hydrate You must be able to do this with the information presented one of two ways: Formula 2. Lab Data Formula: CoCl2 . 6H2O %H2O = 108 x100 = 45.57 % H2O 237

Lab Data: Mass of Hydrate = 5.00 g Mass of Anhydrous = 3.61 g % H2O = ? %H2O = 1.39 x 100 = 27.8 % H2O 5 Mass of water = 5.00 g – 3.61 g = 1.39 g H2O Anhydrous-with all water removed Na2SO4.10H2O-hydrated salt Na2SO4 - anhydrous

A formula with the lowest whole number ratio of elements in a compound Empirical Formula A formula with the lowest whole number ratio of elements in a compound You must be able to calculate an empirical formula A substance is made up of 94.1 % oxygen and 5.9 % hydrogen. What is its empirical formula? 1. Divide by the mass number of each element H = 5.9 = 5.9 1 O = 94.1 = 5.88 16

2. Divide by the smaller number obtained in step #1. 5.88 H = 5.9 = 1 5.88 These whole numbers are the subscripts: therefore: empirical formula = OH 1,6 diaminohexane is used to make nylon. What is the empirical formula of this compound if it is 62.1% carbon, 13.8% hydrogen & 24.1% nitrogen? C= 62.1 = 5.18 12 H = 13.8 = 13.8 1 N = 24.1 = 1.72 14 5.18 = 3.01 1.72 13.8 = 8.02 1.72 1.72 = 1 1.72 C3H8N ANSWER:

What if the numbers obtained in Step 2 are not whole numbers? A compound is analyzed and found to contain 25.9 % nitrogen and 74.1 % oxygen. What is the empirical formula of the compound? O = 74.1 = 4.63 16 1. N = 25.9 = 1.85 14 N = 1.85 = 1 1.85 O = 4.63 = 2.50 1.85 NO2.5 but…definition of empirical formula states WHOLE NUMBERS…this cannot be the answer So…what do we do?

Multiply the subscripts by the LOWEST possible number until we get a whole number…. These are our new subscripts 1 N x 2 = 2 2.5 O x 2 = 5 ANSWER: N2O5

Remember: a molecular formula is the true ratio of atoms in a compound How to determine a molecular formula Remember: a molecular formula is the true ratio of atoms in a compound Find the compound’s empirical formula Divide the molar mass by the mass of the EF Multiply the EF by the number obtained in Step 2. This is your molecular formula

A compound is made up of 94.1 % O and 5.9 % H. Its molar mass is 34 g. What is its empirical and molecular formula? 1. Find the empirical formula O = 94.1/16 = 5.88 H = 5.9/1 = 5.9 Divide by the smaller number from step 1 O = 5.88/5.88 = 1 H = 5.9/5.88 = 1 EF = OH

Find mass of EF OH = 17 g Divide the molar mass (from problem) from the empirical formula mass (calculated) 34/17 = 2 This must be a whole number Multiply the EF by the whole number obtained OH x 2 = O2H2

The compound methyl butanoate smells like apples. Its percent composition is 58.8 % C, 9.8 % H and 31.4 % O and its molar mass is 102 g/mol. What is its empirical formula? What is its molecular formula? Has to be a whole number so lets multiply C = 2.5 x 2 = 5 H = 5 x 2 = 10 O = 1 x 2 = 2 EF = C5H10O2 C = 58.8/12 =4.9 H = 9.8/1 = 9.8 O = 31.4/16 = 1.96 Divide by smallest C = 4.9/1.96 = 2.5 H = 9.8/1.96 = 5 O = 1.96/1.96 = 1 Find mass of EF C = 5 atoms x 12 = 60 H = 10 atoms x 1 = 10 O = 2 atoms x 16 = 32 3. EF = C2.5H5O 6. Mass of EF = 60 + 10 + 32 = 102 g

Here’s the new step: Divide the molar mass (given in the problem) by The mass of the EF (which you just calculated) 102/102 = 1 Multiply the EF by the number just obtained C5H10O2 x 1 = C5H1OO2 C5H10O2 is both the empirical and molecular formula

Another way to do it comes from the fact that they give you the empirical formula and the molar mass and you have to find the molecular formula. These are real easy! What is the molecular formula of the compound if its empirical formula is CH2O and its molar mass is 90 g/mol? You have the empirical formula…you don’t have to figure it out so just determine its mass. C = 1 atom x 12 = 12 H = 2 atoms x 1 = 2 O = 1 atom x 16 = 16 Total = 30 g 2. Divide molar mass/EF mass 90/30 = 3 3. Multiply EF formula x number from Step 2 4. (CH2O) x 3 = C3H6O3 …this is molecular formula

Mole Island /GMM /22.4 Liters Moles Grams X 22.4 X GMM Multiply to get off Mole Island Divide to go onto Mole Island

How many moles are in 100 g of C6H12O6? START WITH WHAT YOU KNOW!!!!!!!!!!!!! 100 g C6H12O6 x 1mole C6H12O6 = .56 mole 180 g How many liters are in 45 g of H2O 45 g H2O x 1 mol H2O x 22.4 L H2O = 56 L H2O 18 g 1 mol

Stoichiometry-Math of Chemistry There are 4 types of problems: 1. mole-mole 2. mass-mass 3. volume-volume 4. mixed The key to any math problem in chemistry is to always get to the mole. This is the only means of comparison!

Mole-Mole Problems How many moles of NaCl can be produced when 3 moles of Na reacts with chlorine? 1. Always write a balanced equation: 2Na + Cl2  2NaCl 2. Start with what you know 3 mol Na x 2 mol NaCl = 2 mol Na 3 mol NaCl

How many moles of oxygen can be made from the decomposition of 15 moles of KClO3? 1. Write a balanced equation 2KClO3  2KCl + 3O2 2. Start with what you know….. 15 mol KClO3 x 3 mol O2 2 mol KClO3 = 22.5 mol O2

How many grams of iron (III) chloride can be made from Mass-Mass Problems Now you are given a mass (grams) and asked to find an answer in mass (grams) How many grams of iron (III) chloride can be made from the reaction of 100 g of iron with chlorine? 1. Write a balanced equation 2Fe + 3Cl2 2 FeCl3 Start with what you know….but now you have to get into moles before you can do the math! 100 g Fe x 1 mol Fe x 2 mol FeCl3 x 161 g = 56 g Fe 2 mol Fe 1 mol FeCl3 287.5 g of FeCl3

How many grams of acetylene are produced by adding water to 5 How many grams of acetylene are produced by adding water to 5.0 g of CaC2? 1. Write a balanced equation: CaC2 + 2H2O  C2H2 + Ca(OH)2 2. Start with what you know….. 5 g CaC2 x 1 mol CaC2 x 1 mol C2H2 x 26 g = 64 g 1 mol CaC2 1 mol C2H2 2.03 g Of C2H2

What volume of CO reacting with oxygen, would Volume-Volume Problems Now the information is given in a volume unit and the answer is wanted in a volume unit. Remember…you still must enter the mole to solve any math question!!!!!!!!!! What volume of CO reacting with oxygen, would produce 200 liters of CO2? 1. Start with a balanced equation 2CO + O2  2CO2 2. Start with what you know!!! 200 L CO2 x 1 mol CO2 x 2 mol CO x 22.4 L O2 = 22.4 L CO2 2 mol CO2 1 mol O2 200 L of CO

Lets try another How many liters of SO2 are needed to react with 216 liters of O2 to form SO3? 1. Write a balanced equation: 2SO2 + O2  2SO3 2. Start with what you know……………. 216 L O2 x 1 mol O2 x 2 mol SO2 x 22.4 L SO2 = 22.4 L O2 1 mol O2 1 mol SO2 432 L of SO2

Now…they give you information in moles, mass or Mixed Problems Now…they give you information in moles, mass or volume and want an answer in either moles, mass or volume Calcium carbonate reacts with hydrochloric acid to produce carbon dioxide, calcium chloride and water. What volume of carbon dioxide will be produced when 80 g of calcium carbonate reacts? 1. Write a balanced equation CaCO3 + 2HCl  CO2 + CaCl2 + H2O 2. Start with what you know 80 g CaCO3 x 1 mol CaCO3 x 1 mol CO2 x 22.4 L CO2 = 100 g CaCO3 1 mol CaCO3 1 mol CO2 17.92 L of CO2

Lets try another For the reaction: 2Na + 2H2O  2NaOH + H2, suppose 23 grams of sodium reacts. How many moles of hydrogen will be produced? 1. Already balanced….HURRAH! 2. Start with what you know…. 23 g Na x 1 mol Na x 1 mol H2 = 23 g Na 2 mol Na .5 mole H2