Chem chapter 6 - stan pauls chem notes PDF

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Avogardo’s number: 6.002 * 10^23 Converting between moles and number of atoms ● 1 mol = 6.022*10^23 atoms ● 6.022*10^23 atoms = 1 mol Converting between grams and number of atoms A silver ring contains 1.1 * 10^22 silver atoms. How many moles of silver are in the ring? 1. Given: 1.1 * 10^22 Ag atoms Find: mol AG 2. Conversion factor is avogardo’s number - 1 mol Ag/6.022 * 10^23 Ag atoms 3. Solve 1.1*10^22 Ag atoms * 1 mol Ag/6.022 *10^23 Ag atoms = 1.8 * 10^-2 mol of Ag Converting between grams and moles of an element ● The mass of 1 mol of atoms of an element is its molar mass. The molar mass is numerically equal to atomic mass in amu. Ex: copper has an atomic mass of 63.55 amu, therefore 1 mol of copper atoms has a mass of 63.55. How many moles of carbon in a 0.58 gram diamond (diamonds are pure carbon) 1. Given: 0.58 g of C find: mol C 2. Find conversion factor: carbon has atomic number of 12, so 1 mol/12.01 g 3. Solve: 0.58 g C * 1 mol C/12.01 g C = 4.8*10^-2 mol C Converting between grams and moles for compounds 1. Find the molar mass (which is equal to the formula mass) a. Formula mass of CO2 = 1(atomic mass of C = atomic number) + 2(atomic mass of O) i. 1(12.01 amu) + 2(16.00 amu) = 44.01 amu b. Molar mass = 44.01 g/mol 2. Find the number of moles in a 22.5 g sample of dry ice (solid CO2) a. Given: 22.5 CO2 find: mol CO2 b. Use the molar mass as a conversion factor 1 mol of CO2/44.01 g of CO2 3. Solve a. 22.5 g * 1 mol CO2/44.01 g = 0.511 mol CO2 Converting between Mass of a compound and number of molecules Find the number of CO2 molecules in a sample of dry ice with a mass of 22.5 grams Another type of question with the same conversion factor: what is the mass of 4.78 mol of 10^24 NO2 molecules/?

1. Calculate moles (see above) and then multiply by avogardo’s number CO2 molecules/1 mol of CO2 a. 22.5 g of CO2 * 1 mol CO2/44.01 g CO2 * 6.022 * 10^23 CO2 molecules / 1 mol of CO2 = 3.08 * 10^23 g of CO2 molecules Converting between grams of a compound and grams of a constituent element Determine the number of moles of O in 1.7 mol of CaCO3 1. Given: 1.7 mol CaCO3 Find: mol O 2. Determine conversion factor from the chemical formula which indicates three O atoms for every CaCO3 unit so 3 mol O/1 mol CaCO3 3. Solve: 1.7 mol CaCO3 * 3 mol O/1 mol CaCO3 = 5.1 mol O For longer problems, the way to set it up is mass compound -> moles compound -> moles element -> mass element. Example of where to use this below Find the mass of sodium in 15g of NaCl 1 mole of sodium in 1 mol of NaCl G NaCl -> mol NaCl -> mol Na -> g Na 15 g * 1 mol NaCl/58.44 g NaCl * 1 mol Na/1 mol NaCl * 22.9 g Na/1 mol Na = 5.9 g Na Using mass percent composition as a conversion factor Mass percent: element’s percentage of the total mass of the compound Mass percent of element X = mass of X in a sample of the compound/mass of the sample o the compound * 100% Suppose a 0.358 g sample of chromium reacts with oxygen to form 0.523 g of the metal oxide. Calculate the mass percent of chromium Mass percent Cr = mass Cr/mass metal oxide * 100% 0.358 g/0.523 g * 100% = 68.5% The FDA recommends that adults consume less than 2.4 g of sodium per day. How many grams of sodium chloride can you consume and still be within the FDA guidelines? Sodium chloride is 39% sodium by mass? 1. Given: 2.4 g of Na Find: g NaCl 2. 2.4 g of Na * 100 g NaCl/29 g Na = 6.2 g of NaCl

Mass percent composition from a chemical formula Mass percent of element X = mass of X in a sample of the compound/mass of the sample o the compound * 100% Calculate the mass percent composition of Cl in the chlorofluorocarbon CCl2F2 Mass percent Cl = 2 * molar mass Cl / molar mass CCL2F2 * 100% We multiply the molar mass of Cl by 2 because the chemical formula has a subscript of 2 Molar mass of CCL2F2 = 1(12.01)+2(35.45)+2(19.00) = 120.91 g/mol Since Mass percent Cl = 2 * molar mass Cl / molar mass CCL2F2 * 100% 2*25.45g/120.91 * 100% =58.64% Obtaining empirical formula from experimental data Empirical formula: gives the smallest whole number ratio of each type of atom in a molecule Example: you decompose a compound containing nitrogen and oxygen in the lab and produce 24.5g of nitrogen and 70.0 g of oxygen. Calculate the empirical formula of the compound 1. Write down or calculate the masses present in a sample of the compound a. Given: 24.5 N and 70.0 g O find: empirical formula 2. Convert each of the masses in step 1 to moles by using the appropriate molar mass for each element as a conversion factor. a. 24.5 g N * 1 mol N/14.01 g N = 1.75 mol N b. 70.0 g * 1 mol O/16.00 g O = 4.38 mol O 3. Write down a psudoformula for the compound using the moles of each element from step 2 as the subscripts a. N1.75O4.38 b. N1.75/1.75O4.38/1.75 c. N1O2.5 4. If the subscripts are not whole numbers multiply by the smallest whole number that will give you a whole number (check book if you need a reference) a. N1O2.5*2 = N2O5 Obtaining empirical formula from reaction data A 3.24 g sample of titanium reacts with oxygen to form 5.40g of the metal oxide. What is the empirical formula of the metal oxide? 1. Given: 3.24 g of Ti 5.40 g metal oxide Find: empirical formula 2. Write down the mases of each element present in a sample of the compound

Calculating Molecular Formulas for compounds Calculating Molecular Formula From Emprical Formula and Molar Mass Molecular formula = empirical formula * n N = molar mass/empirical formula molar mass Example: naphthalene is a compound containing carbon and hydrogen that is used in mothballs. Its empirical formula is C5H4 and its molar mass is 128.16 g/mol. What is its molecular formula? 1. Given: empirical formula = C5H4 and molar mass 126.16 g/mol Find: molecular formula 2. Determine n by molar mass/empricial formula molecular mass a. Empirical formula molar mass 5(12.01) + 4(1.01) = 64.09 g/mol b. 128.16/64.09 g mol = 2. c. N = 2 3. Multiply molecular empirical formula by N a. C5H4 * 2 = C10H8...