Unit 6: The Mathematics of Chemical Formulas # of H2O molecules # of H atoms # of O atoms 1 2 1 2 4 2 3 6 3 100 200 100 6.02 x 1023 2 (6.02 x 1023) 6.02 x 1023 mass: 18 g mass: 2g mass: 16 g molar mass: the mass of one mole of a substance PbO2 Pb: 1 (207.2 g) = 207.2 g O: 2 (16.0 g) = 32.0 g H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0 g O: 3 (16.0 g) = 48.0 g ammonium N: 3 (14.0 g) = 42.0 g phosphate H: 12 (1.0 g) = 12.0 g HNO3 NH41+ PO43– (NH4)3PO4 P: 1 (31.0 g) = 31.0 g O: 4 (16.0 g) = 64.0 g 239.2 g 63.0 g 149.0 g percentage composition: the mass % of each element in a compound % of element g element x 100 molar mass of compound Find % composition: PbO2 Pb: 1(207.2 g) = 207.2 g Pb 239.2 g 86.6% Pb O: 2(16.0 g) = 32 g O 239.2 g 13.4% O (NH4)3PO4 N: 3(14.0 g) = 42.0 g P 149.0 g 28.2% N H: 12(1.0 g) = 12.0 g H 149.0 g 8.1% H P: 1(31.0 g) = 31.0 g P 149.0 g 20.8% P O: 4(16.0 g) = 64.0 g O 149.0 g 43.0% O zinc acetate Zn2+ CH3COO1– Zn(CH3COO)2 Zn: 1(65.4 g) = 65.4 g Zn 183.4 g 35.7% Zn C: 4(12.0 g) = 48.0 g C 183.4 g 26.2% C H: 6(1.0 g) = 6.0 g H 183.4 g 3.3% H O: 4(16.0 g) = 64.0 g O 183.4 g 34.9% O 183.4 g Finding an Empirical Formula from Experimental Data 1. Find # of g of each element. 2. Convert each g to mol. 3. Divide each “# of mol” by the smallest “# of mol.” 4. Use ratio to find formula. A compound is 45.5% yttrium and 54.5% chlorine. Find its empirical formula. 1 mol Y 45.5 g Y 0.512 mol Y 0.512 1 88.9 g Y 1mol Cl 54.5 g Cl 1.535 mol Cl 0.512 3 35.5 g Cl YCl3 A ruthenium/sulfur compound is 67.7% Ru. Find its empirical formula. 1mol Ru 67.7 g Ru 0.670 mol Ru 0.670 1 101.1 g Ru 1 mol S 32.3 g S 1.006 mol S 0.670 1.5 32.1 g S RuS1.5 Ru2S3 A 17.40 g sample of a technetium/oxygen compound contains 11.07 g of Tc. Find the empirical formula. 1mol Tc 11.07 g Tc 0.113 mol Tc 0.113 1 97.9 g Tc 1mol O 6.33 g O 0.396 mol O 0.113 3.5 16.0 g O TcO3.5 Tc2O7 A compound contains 4.63 g lead, 1.25 g nitrogen, and 2.87 g oxygen. Name the compound. 1mol Pb 4.63 g Pb 0.0223 mol Pb 0.0223 1 207.2 g Pb 1mol N 1.25 g N 0.0893 mol N 0.0223 4 14.0 g N 1 mol O 2.87 g O 0.1794 mol O 0.0223 8 16.0 g O PbN4O8 Pb(NO2)4 lead (IV) nitrite plumbic nitrite Pb? 4 NO21– To find molecular formula… 1). Find empirical formula. 2). Find molar mass of empirical formula. 3). Find n = mm molecular mm empirical 4). Multiply all parts of empirical formula by n. A carbon/hydrogen compound is 7.7% H and has a molar mass of 78 g. Find its molecular formula. 1 mol C 92.3 g C 7.69 mol C 7.69 1 12.0 g C 1 mol H 7.7 g H 7.7 mol H 7.69 1 1.0 g H mmemp = 13 g n = 78 g 13 g = 6 CH C6H6 A compound has 26.33 g nitrogen, 60.20 g oxygen, and molar mass 92 g. Find molecular formula. 1mol N 26.33 g N 1.881 mol N 1.881 1 14.0 g N 1mol O 60.20 g O 3.763 mol O 1.881 2 16.0 g O mmemp = 46 g n = 92 g 46 g = 2 Mole Calculations N2O4 NO2 New Points about Island Diagram: a. Diagram now has four islands. b. “Mass Island” now for elements or compounds c. “Particle Island” now for atoms or molecules d. “Volume Island”: for gases only 1 mol @ STP = 22.4 L = 22.4 dm3 What mass is 1.29 mol ferrous nitrate? Fe2+ 179.8 g X g Fe(NO3 ) 2 1.29 mol 1 mol NO31– Fe(NO3)2 232 g Fe(NO3 ) 2 How many molecules is 415 L sulfur dioxide at STP? 23 1 mol 6.02 x 10 m' cules 25 X m' cules SO 2 415 L 1.12 x 10 1 mol 22.4 L m’cules SO2 What mass is 6.29 x 1024 m’cules aluminum sulfate? Al3+ SO42– Al2(SO4)3 1 mol 342.3 g X g 6.29 x 1024 m' c 3580 g Al 2 (SO 4 ) 3 23 6.02 x 10 m' c 1 mol At STP, how many g is 87.3 dm3 of nitrogen gas? 1 mol 28.0 g X g N2 87.3 dm3 109 g N2 3 22.4 dm 1 mol How many m’cules is 315 g of iron (III) hydroxide? Fe3+ OH1– Fe(OH)3 1 mol 6.02 x 1023 m' cules 1.78 x 1024 X m' cules 315 g 1 mol m’cules 106.8 g Fe(OH)3 How many atoms are in 145 L of CH3CH2OH at STP? 23 1 mol 6.02 x 10 m' cules 24 X m' cules 145 L 3.90 x 10 m' c 1 mol 22.4 L 9 atoms 25 X atoms 3.90 x 1024 m' c 3.51 x 10 atoms 1 molecule Hydrates and Anhydrous Salts anhydrous salt: an ionic compound (i.e., a salt) that attracts water molecules and forms loose chemical bonds with them; symbolized by MN “anhydrous” = “without water” “dessicants” in leather goods, Uses: electronics, vitamins hydrate: an anhydrous salt with the water attached -- symbolized by MN . ? H2O Examples: H2O H2O H2O MN H2O H2O H2O CuSO4 . 5 H2O BaCl2 . 2 H2O Na2CO3 . 10 H2O FeCl3 . 6 H2O HEAT hydrate MN + H2O H2O H2O H2O H O 2 H2O anhydrous salt water Finding the Formula of a Hydrate 1. Find the # of g of MN and # of g of H2O. 2. Convert g to mol. 3. Divide each “# of mol” by the smallest “# of mol.” 4. Use the ratio to find the hydrate’s formula. Find formula of hydrate for each problem. sample’s mass before heating = 4.38 g (MN . ? H2O) sample’s mass after heating = 1.93 g (MN) molar mass of anhydrous salt = 85 g 1 mol MN 1.93 g MN 0.0227 mol MN 0.0227 1 85 g MN 1mol H2O 0.1361 mol H2O 0.0227 6 2.45 g H2O 18 g H2O MN . 6 H2O A. beaker = 46.82 g A B. beaker + sample before heating = 54.35 g B C. beaker + sample after heating = 50.39 g C molar mass of anhydrous salt = 129.9 g 1 mol MN 3.57 g MN 0.0275 mol MN 0.0275 1 129.9 g MN 1mol H2O 0.22 mol H2O 0.0275 8 3.96 g H2O 18 g H O 2 MN . 8 H2O A. beaker = 47.28 g A B. beaker + sample before heating = 53.84 g B C. beaker + sample after heating = 51.48 g C molar mass of anhydrous salt = 128 g 1 mol MN 4.20 g MN 0.0328 mol MN 0.0328 1 128 g MN 1mol H2O 0.1311 mol H2O 0.0328 4 2.36 g H2O 18 g H2O MN . 4 H2O Find % water and % anhydrous salt (by mass). % H2O 4 (18 g H2O) 36.0% H2O (and 64.0% MN) 4 (18 g ) 128 g OR… % H2O 2.36 g H2O 36.0% H2O (and 64.0% MN) 2.36 g 4.20 g Find % comp. of iron (III) chloride. Fe3+ Cl1– FeCl3 Fe: 1(55.8 g) = 55.8 g Fe 162.3 g 34.4% Fe Cl: 3(35.5 g) = 106.5 g Cl 162.3 g 65.6% Cl 162.3 g A compound contains 70.35 g C and 14.65 g H. Its molar mass is 58 g. Find its molecular formula. 1 mol C 70.35 g C 5.863 mol C 5.863 1 12.0 g C 1 mol H 14.65 g H 14.65 mol H 5.863 2.5 1.0 g H empirical CH2.5 C2H5 mmemp = 29 g n = 58 g 29 g = 2 At STP, how many g is 548 L of chlorine gas? C4H10 1 mol 71.0 g X g Cl 2 548 L 1740 g Cl 2 22.4 L 1 mol Strontium chloride is an anhydrous salt on which the following data were collected. Find formula of hydrate. beaker = 65.2 g beaker + sample before heating = 187.9 g beaker + sample after heating = 138.2 g formula of salt Sr2+ Cl1– SrCl2 molar mass = 158.6 g 1mol 73.0 g SrCl2 0.460 mol SrCl 2 0.460 1 158.6 g 1mol H2O 2.761 mol H2O 0.460 6 49.7 g H2O 18 g H O 2 SrCl2 . 6 H2O