Unit 6: The Mathematics of Chemical Formulas x (6.02 x ) 6.02 x mass: 2 g

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1 Unit 6: The Mathematics of Chemical Formulas # of molecules # of H atoms # of O atoms x (6.02 x ) 6.02 x mass: 18 g mass: 2 g mass: 16 g molar mass: the mass of one mole of a substance PbO 2 Pb: 1 (207.2 g) = g O: 2 (16.0 g) = 32.0 g g HNO 3 H: 1 (1.0 g) = 1.0 g N: 1 (14.0 g) = 14.0 g 63.0 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 NH 4 1+ PO 4 3 P: 1 (31.0 g) = 31.0 g g (NH 4 ) 3 PO 4 O: 4 (16.0 g) = 64.0 g

2 percentage composition: the mass % of each element in a compound g element % of element x 100 molar mass of compound Find % composition: PbO 2 Pb: 1(207.2 g) = g Pb g 86.6% Pb O: 2(16.0 g) = 32 g O g 13.4% O (NH 4 ) 3 PO 4 N: 3(14.0 g) = 42.0 g P g 28.2% N H: 12(1.0 g) = 12.0 g H g 8.1% H P: 1(31.0 g) = 31.0 g P g 20.8% P O: 4(16.0 g) = 64.0 g O g 43.0% O zinc acetate Zn 2+ CH 3 COO 1 Zn(CH 3 COO) 2 Zn: 1(65.4 g) = 65.4 g Zn g 35.7% Zn C: 4(12.0 g) = 48.0 g C g 26.2% C H: 6(1.0 g) = 6.0 g H g 3.3% H O: 4(16.0 g) = 64.0 g O g 34.9% O g

3 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. 1mol Y 45.5 g Y mol Y g Y 1mol Cl 54.5 g Cl mol Cl g Cl YCl 3 A ruthenium/sulfur compound is 67.7% Ru. Find its empirical formula. 1mol Ru 67.7 gru mol Ru g Ru 1mol S 32.3 g S mol S g S RuS 1.5 Ru 2 S 3 1 3

4 A g sample of a technetium/oxygen compound contains g of Tc. Find the empirical formula. 1mol Tc g Tc mol Tc g Tc 1mol O 6.33 g O mol O g O TcO 3.5 Tc 2 O 7 A compound contains 4.63 g lead, 1.25 g nitrogen, and 2.87 g oxygen. Name the compound. 1mol Pb 4.63 gpb mol Pb gpb 1mol N 1.25 gn mol N gn 1mol O 2.87 g O mol O g O 1 PbN 4 O 8 Pb(NO 2 ) 4 Pb? 1 4 NO 2 lead (IV) nitrite plumbic nitrite

5 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. 1mol C 92.3 g C 7.69 mol C g C 1mol H 7.7 gh 7.7 mol H gh 1 1 CH mm emp = 13 g n = 78 g 13 g = 6 C 6 H 6 A compound has g nitrogen, g oxygen, and molar mass 92 g. Find molecular formula. 1mol N gn 1.881mol N gn NO 2 1mol O g O mol O g O mm emp = 46 g n = 92 g 46 g = 2 N 2 O 4

6 Mole Calculations 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 STP = 22.4 L = 22.4 dm 3 What mass is 1.29 mol ferrous nitrate? X g Fe(NO Fe 2+ 1 NO 3 Fe(NO 3 ) g 1mol 3 ) mol 232 gfe(no3 ) 2 How many molecules is 415 L sulfur dioxide at STP? X m'cules SO 23 1mol 6.02 x 10 m'cules 415 L 22.4 L 1mol x 10 m cules SO 2

7 What mass is 6.29 x m cules aluminum sulfate? Al 3+ SO 2 4 Al 2 (SO 4 ) mol g X g 6.29 x 10 m'c 3580 g Al2 (SO4 ) x 10 m'c 1mol At STP, how many g is 87.3 dm 3 of nitrogen gas? 3 1mol 28.0 g X g N dm 109 g N dm 1mol How many m cules is 315 g of iron (III) hydroxide? Fe 3+ OH 1 Fe(OH) 3 X m'cules 315 g 1mol g x 10 m'cules 1mol 1.78 x 10 m cules Fe(OH) 3 24 How many atoms are in 145 L of CH 3 CH at STP? X m'cules 145 L X atoms 3.90 x x 10 m'cules 1mol 9 atoms m'c 3.51 x 10 1molecule 1mol 22.4 L x 10 atoms 24 m'c

8 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 Uses: dessicants in leather goods, electronics, vitamins hydrate: an anhydrous salt with the water attached -- symbolized by MN.? Examples: CuSO. 4 5 BaCl. 2 2 Na 2 CO FeCl. 3 6 MN HEAT MN hydrate anhydrous salt water Finding the Formula of a Hydrate 1. Find the # of g of MN and # of g of. 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. +

9 Find formula of hydrate for each problem. sample s mass before heating = 4.38 g (MN.? ) sample s mass after heating = 1.93 g (MN) molar mass of anhydrous salt = 85 g 1mol MN 1.93 gmn mol MN gmn 1mol H2O 2.45 gh2o mol H2O gh2o MN. 6 A. beaker = g B. beaker + sample before heating = g C. beaker + sample after heating = g molar mass of anhydrous salt = g A B C 1mol MN 3.57 gmn mol MN gmn 1mol H2O 3.96 gh2o 0.22 mol H2O gh2o MN. 8

10 A. beaker = g B. beaker + sample before heating = g C. beaker + sample after heating = g molar mass of anhydrous salt = 128 g A B C 1mol MN 4.20 gmn mol MN gmn 1mol H2O 2.36 gh2o mol H2O gh2o MN. 4 Find % water and % anhydrous salt (by mass). 4 (18 gh2o) % H2O 36.0% H2O 4 (18 g ) 128 g OR (and 64.0% MN) 2.36 gh2o % H2O 36.0% H2O (and 64.0% MN) 2.36 g 4.20 g

11 Find % comp. of iron (III) chloride. Fe 3+ Cl 1 FeCl 3 Fe: 1(55.8 g) = 55.8 g Fe g 34.4% Fe Cl: 3(35.5 g) = g Cl g 65.6% Cl g A compound contains g C and g H. Its molar mass is 58 g. Find its molecular formula. 1mol C g C mol C g C 1mol H gh mol H gh empirical CH 2.5 C 2 H 5 mm emp = 29 g n = 58 g 29 g = 2 C 4 H 10 At STP, how many g is 548 L of chlorine gas? 1mol 71.0 g X g Cl2 548 L 1740 g Cl L 1mol

12 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 = g beaker + sample after heating = g formula of salt Sr 2+ Cl 1 SrCl 2 molar mass = g 1mol 73.0 g SrCl mol SrCl g 1mol H2O 49.7 gh2o 2.761mol H2O gh2o SrCl 2. 6