Hydrates, Percent Composition, and Empirical and Molecular Formulas

Transcription

1 Hydrates, Percent Composition, and Empirical and Molecular Formulas

2 Hydrates Hydrates are ionic cmpds (salts) that have water molecules bound to their ions. Examples: CuSO 4 5H 2 O Fe(NO 3 ) 3 9H 2 O CoCl 2 6H 2 O SnCl 2 2H 2 O Water can sometimes be removed by heating. A hydrate that loses all of its water becomes anhydrous ( without water )

3 Hydrates Anhydrous cobalt(ii) chloride, CoCl 2 Cobalt(II) chloride hexahydrate, CoCl 2 6H 2 O

4 Naming Hydrates Name ionic compound. Add prefix + hydrate. CoCl 2 6H 2 O = cobalt(ii) chloride hexahydrate CuSO 4 5H 2 O = copper(ii) sulfate pentahydrate Fe(NO 3 ) 3 9H 2 O = iron(iii) nitrate nonahydrate SnCl 2 2H 2 O = tin(ii) chloride dihydrate

5 Molar Mass of Hydrates Add appropriate number of moles of water: Molar mass of CuSO 4 5H 2 O: 1 x Cu = 1 x g/mol = g/mol 1 x S = 1 x g/mol = g/mol 4 x O = 4 x g/mol = g/mol 5 x H 2 O = 5 x g/mol = g/mol Total = g/mol Molar mass of anhydrous CuSO 4 : 1 x Cu = 1 x g/mol = g/mol 1 x S = 1 x g/mol = g/mol 4 x O = 4 x g/mol = g/mol Total = g/mol

6 Percentage Composition Percentage Composition percent of a cmpd s mass that is made of each element.

7 Percentage Composition What is the % Comp. of H 2 O? First, find the molar mass. 2 x H = 2 x 1.01 g/mol = 2.02 g/mol 1 x O = 1 x g/mol = g/mol Total = g/mol % H = (2.02 / 18.02) x 100% = 11.2% % O = (16.00 / 18.02) x 100% = 88.79% What does that mean? In 100 grams of water, there are 11.2 grams of hydrogen and grams of oxygen.

8 Empirical Formula Empirical Formula A formula that gives the simplest whole-number ratio of the atoms of each element in a compound. Molecular Formula H 2 O 2 C 6 H 12 O 6 CH 3 O CH 3 OOCH = C 2 H 4 O 2 Empirical Formula HO CH 2 O CH 3 O CH 2 O

9 Determine the empirical formula for a compound containing g Cl and g Ca. Steps 1. Find mole amounts. 2. Divide each mole by the smallest mole.

10 1. Find mole amounts g Cl x 1 mol Cl = mol Cl g Cl g Ca x 1 mol Ca = mol Ca g Ca

11 2. Divide each mole by the smallest mole. Cl = mol Cl = 2.00 mol Cl Ca = mol Ca = 1.00 mol Ca Ratio 1 Ca: 2 Cl Empirical Formula = CaCl 2

12 A compound weighing g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Hint Percent to mass Mass to mole Divide by small Multiply til whole

13 A compound weighing g consists of 72.2% magnesium and 27.8% nitrogen by mass. What is the empirical formula? Percent to mass: Mg (72.2%/100)* g = g N (27.8%/100)* g = g Mass to mole: Mg g * ( 1 mole ) = 8.86 mole 24.3 g N g * ( 1 mole ) = 5.92 mole g Divide by small: Mg mole/5.92 mole = 1.50 N mole/5.92 mole = 1.00 mole Multiply til whole: Mg 1.50 x 2 = 3.00 N 1.00 x 2 = 2.00 Mg 3 N 2

14 Molecular Formula The molecular formula gives the actual number of atoms of each element in a molecular compound. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Mass. 3. Divide the molar mass by the EFM. 4. Multiply empirical formula by factor. Find the molecular formula for a compound whose molar mass is ~ and empirical formula is CH 2 O EFM = g /62.03 = (CH 2 O 3 ) = C 2 H 4 O 6

15 Find the molecular formula for a compound that contains 4.90 g N and 11.2 g O. The molar mass of the compound is 92.0 g/mol. Steps 1. Find the empirical formula. 2. Calculate the Empirical Formula Mass. 3. Divide the molar mass by the EFM. 4. Multiply empirical formula by factor.

16 Empirical formula. A. Find mole amounts g N x 1 mol N = mol N g N 11.2 g O x 1 mol O = mol O g O

17 B. Divide each mole by the smallest mole. N = = 1.00 mol N O = = 2.00 mol O Empirical Formula = NO 2 Empirical Formula Mass = g/mol

18 Molecular formula Molar Mass = 92.0 g/mol = 2.00 Emp. Formula Mass g/mol Molecular Formula = 2 x Emp. Formula = N 2 O 4

19 A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula?

20 A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? g C (48.38/100)* g = g g H (8.12/100)* g = g g O (43.5/100)* g = g mole C g * ( 1 mole ) = mol g mole H g * ( 1 mole ) = mol 1.01 g mole O g * ( 1 mole ) = mol g

21 A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? From last slide: mol C, mol H, mol O C 21.29/14.27 = 1.49 H 42.49/14.27 = 2.98 (esentially 3) O 14.27/14.27 = 1.00 C 1.49 x 2 = 3 H 3 x 2 = 6 C 3 H 6 O 2 O 1 x 2 = 2

22 A g compound containing only carbon, hydrogen, and oxygen is found to be 48.38% carbon and 8.12% hydrogen by mass. The molar mass of this compound is known to be ~ g/mol. What is its molecular formula? From last slide: Empirical formula = C 3 H 6 O 2 EFM = Molar mass = = ~3 EFM (C 3 H 6 O 2 ) = C 9 H 18 O 6