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NCSU CH 101 - Introduction to Stoichiometry and Converstion factors in Chemical Equations

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CH 101 1st Edition Lecture 1 Outline of Current Lecture I Basic Chemistry Definitions A Converting from Grams to Moles and Grams to Molecules II Balancing Chemical Equations III Finding the Molar Mass of a Molecule IV Introduction to Stoichiometry A Converting from Grams to Moles Current Lecture Appendix A I Basic Chemistry Definitions Atom The smallest component of an element An atom is not the smallest form of matter but it is the smallest form of matter that still has elemental properties Oxygen is an atom but O 2 is not an atom In a molecule like HNO2 there is 1 Hydrogen atom 1 Nitrogen atom and 2 Oxygen atoms Element A group of the same type of atoms H Hydrogen and Ne Neon are elements and can be found on the periodic table There are certain elements that are diatomic and are only considered an element when bonded in pairs N2 O2 F2 Cl2 Br2 I2 and At2 Molecule Two or more atoms stuck together A molecule can be composed of the same element or of different elements O2 and H2O are both molecules Compound Two or more different atoms stuck together A compound is always a molecule but a molecule is not always a compound H2O and CO2 are both compounds but O2 is not Avogadro s number the mole Avogadro s number is used to determine the amount of atoms or molecules in a certain amount of grams or moles of a substance Similar to the word dozen the word mole is a representation of an amount In the way that a dozen 12 a mole 6 022x10 23 A mole of any substance equals 6 022x1023 atoms In 1 mole of HNO2 molecules there is 1 mole of Hydrogen 1 mole of Nitrogen and 2 moles of Oxygen A Converting from Grams to Moles and Grams to Molecules Ex How many Na atoms are in 65 0 grams of Na 65 0 grams of Na Start with what you know 65 0 g Na 1 mole Na 22 99 g Na Multiply by the ratio of moles to grams of Na so that the grams cancel 1 mole Na has 22 99 g this is the atomic mass of Na and 1 mole of any element has 6 022x10 23 atoms These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute 65 0 g Na 1 mole Na 22 99 g Na 6 022x1023 atoms of Na 1 mole Na Now multiply by the ratio of atoms to moles and 1 mole of any element has 6 022x10 23 atoms 65 0 g Na 6 002x1023 atoms Na 22 99 g Na 1 703x1024 atoms Na When you are finding atoms expect to get a large number as the answer Also this equation can be simplified into 65 0g Na 6 002x10 23 atoms Na 22 99 g Na 1 703x1024 The same equation is used to convert from grams to molecules except you use the molar mass of the molecule instead of the atomic mass Ex Find the number of molecules in a 2 56 gram sample of SO 2 2 56 g SO2 1 mole SO2 64 063 g SO2 Multiply by the ratio of moles to grams of SO 2 grams of SO2 is found by calculating the molar mass 32 065 2 15 99 64 063 2 56 g SO2 1 mole SO2 64 063 g SO2 6 022x1023 molecules SO2 1 mole SO2 2 41x1022 molecules SO2 II Balancing Chemical Equations Balancing equations are important because you have to know the amount of each substance needed to produce the desired product The easiest way to balance an equation is to 1 Start with the most complicated molecule or the molecule that contains the most elements 2 Try to find a coefficient that will allow the amount of atoms in the reactants equal to the amount of atoms in the product 3 Leave the diatomic molecules such as O 2 or elements such as H for last because the coefficients of these won t affect anything else in the equation Ex Balance the equation O2 Al Al2O3 In this equation there are two oxygen atoms and one aluminum atom on the reactant side and on the product side there are two Aluminum atoms and three oxygen atoms Start plugging in coefficients to try and balance the equation O2 2 Al Al2O3 Now there are two Aluminum atoms on each side but the Oxygen atoms aren t balanced try a different coefficient and this time start with the Al 2O3 2 O2 4 Al 2 Al2O3 This is a balanced equation III Finding the Molar Mass of a Molecule The molar mass is the mass of a certain element or substance The molar mass of an element is found on the periodic table under the element s symbol Oxygen s molar mass is 15 999 as found on the periodic table To calculate the molar mass of a compound you simply add the molar mass of the elements that make up the compound after multiplying them by their subscripts Ex Find the molar mass of Al2 SO4 3 Molar mass of Al 26 982 First find the molar mass of each element Molar mass of S 32 065 Molar mass of O 15 999 2 26 982 3 32 065 12 15 999 Plug in the molar mass of each element and multiply each molar mass by the subscript notice that SO4 is multiplied by three so 3 S and 3 O 4 342 147 Answer the molar mass of Al2 SO4 3 IV Introduction to Stoichiometry Using ratios between quantities to determine the amount of a substance after a reaction or the forming of a compound These ratios are used to determine the amount of a substance when it undergoes some kind of change like in a chemical reaction Ex 1 dollar 100 pennies the ratio is 1 meaning that they are equal amounts So how many pennies are in 56 dollars Since the ratio is 1 there would be 56 pennies to every 56 dollars What happens when the ratio is not 1 1 the process becomes more challenging In this case you would have to take the given amount of substance 1 and multiply it by the ratio of substance 2 to substance 1 so that the units of substance 1 cancel each other and you find the correct amount of substance 2 Ex 1 dollar 20 Nickels the ratio is 1 because if someone has 1 dollar they have just as much as someone else who has 20 nickels So how many nickels are in 55 dollars Take the amount of dollars given to you 55 and set it equal to the ratio of nickels to dollars so that the dollars cancel each other X amount of substance 1 ratio of substance 2 ratio of substance 1 Y amount of substance 2 55 dollars 20 nickels 1 dollar nickels Set up your equation 55 dollars 20 nickels 1 dollar nickels The dollars …


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NCSU CH 101 - Introduction to Stoichiometry and Converstion factors in Chemical Equations

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