 Download Download Presentation Relationship between mass, moles and molecules in a compound

# Relationship between mass, moles and molecules in a compound

Download Presentation ## Relationship between mass, moles and molecules in a compound

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1. Amount (moles) # molecules or Formula units Relationship between mass, moles and molecules in a compound X molar mass (__g__ mole) Mass (g) X 6.022 x 1023 (units mole) Grams moles = moles gram moles units = units or molecules mole moles grams = grams mole Mullis

2. Molar mass • Molar mass of a substance = mass in grams of one mole of the substance. • A compound’s molar mass is NUMERICALLY equal to its formula mass. 2 mol H x 1.01 g H = 2.02 g H 1 mol H 1 mol O x 16.00 g O = 16.00 g O 1 mol O molar mass H2O = 18.02 g/mol • Formula mass H2O = 18.02 amu • Molar mass H2O = 18.02 g/mol Mullis

3. Molar Mass Example What is the molar mass of K2SO4? 2 mol K x 39.10 g K = 78.20 g K 1 mol K 1 mol S x 32.10 g S = 32.07 g S 1 mol S 4 mol O x 16.00 g O = 64.00 g O 1 mol O molar mass K2SO4 = 174.27 g/mol How many moles of each element are present in this compound? 2 mol K, 1 mol S, 4 mol O Mullis

4. What is the molar mass of C6H12O6? 6 mol C x 12.01 g C = 72.06 g C 1 mol C 12 mol H x 1.01 g H = 12.12 g H 1 mol H 6 mol O x 16.00 g O = 96.00 g O 1 mol O molar mass C6H12O6 = 180.18 g/mol How many moles of each element are present in this compound? 6 mol C, 12 mol H, 6 mol O Mullis

5. Converting to grams from moles How many moles of glucose are in 4.15x10-3 g C6H12O6? 4.15x10-3 g x 1 mol C6H12O6= 2.30 x 10-5 mol C6H12O6 180.18 g How many molecules of glucose are in 4.15x10-3 g C6H12O6? 2.30 x 10-5 mol C6H12O6x 6.022 x 10 23 molecules = 1 mol (2.30 x 6.022)(10(-5+23)) = 13.90 x 10 –18 molecules = 1.39 x 10 –19 molecules Mullis

6. What is the mass in grams of 6.25 moles copper (II) nitrate? Cu 2+ NO3 - : formula isCu(NO3)2 Find molar mass of Cu(NO3)2first. 1 mol Cu x 63.55 g Cu = 63.55 g Cu 1 mol Cu 2 mol N x 14.01 g N = 28.02 g N 1 mol N 6 mol O x 16.00 g O = 96.00 g O 1 mol O molar mass Cu(NO3)2 = 187.57 g/mol Now find mass in grams of 6.25 moles: 6.25 moles x 187.57 g= 1172 g Ans. 1170 g Cu(NO3)2 1 mol Mullis

7. Atoms and Ions Within Compounds • How many carbon atoms are in one mole of C2H6? 1 mole C2H6 | 2 moles C | 6.022 x 1023 atoms = 1.204 x 1024 atoms |1 mole C2H6 | 1 mole C • How many MOLES of carbon atoms are in one mole of C2H6? 1 mole C2H6 | 2 moles C = 2 moles C atoms |1 mole C2H6 • How many moles of hydroxide ions are in one mole of calcium hydroxide? How many moles of Ca2+? 1 mole Ca(OH)2 | 2 moles OH--=2 moles hydroxide ions |1 mole Ca(OH)2 1 mole Ca(OH)2 | 1 mole Ca2+=1 mole calcium ions |1 mole Ca(OH)2 Mullis

8. Percentage Composition % Composition is the % by mass of each element in a compound. Find the percentage composition of sodium chloride. Na+ Cl - : formula isNaCl 1 mol Na x 22.99 g Na = 22.99 g Na 1 mol Na 1 mol Cl x 35.45 g Cl = 35.45 g Cl 1 mol Cl molar mass NaCl = 58.44 g/mol 22.99 g Na x 100 = 39.34 % Na 58.44 g NaCl 35.45 g Cl x 100 = 60.66 % Cl 58.44 g NaCl Mullis

9. Find the percentage composition of sodium nitrate. Na+ NO3 - : formula isNaNO3 1 mol Na x 22.99 g Na = 22.99 g Na 1 mol Na 1 mol N x 14.01 g N = 14.01 g N 1 mol N 3 mol O x 16.00 g O = 48.00 g O 1 mol O molar mass NaNO3 = 85.00 g/mol 22.99 g Na x 100 = 27.05 % Na 85.00 g NaNO3 14.01 g N x 100 = 16.48 % N 85.00 g NaNO3 48.00 g O x 100 = 56.47% O 85.00 g NaNO3 Mullis

10. Percentage Composition • Why mass instead of moles? • Isn’t 2/3 of the water molecule hydrogen? • Moles indicate the amounts of each atom needed to make the molecule stable from an electron standpoint. 2.02 g H x 100 = 11.21 % H 18.02 g H2O O H H Mullis

11. Empirical Formula Use % composition to convert to original formula: • Assume 100 g sample, so % = grams • Convert grams to moles for each element • Divide the number of moles for each element by the smallest number of moles • The result for each type of element is its subscript in the empirical formula. • The order of elements is usually: Organics: C,H,O,N Inorganics: Metal, nonmetal, oxygen • Keep 4 decimal places when dividing numbers. If the result has a decimal between .2 and .8, may need to multiply all numbers by the number needed to get a whole number. Ex: 3.5 should be multiplied by 2 to get 4. Then multiply all other elements by the same number. Mullis

12. Example: Empirical and Molecular formula What is the empirical formula of a compound with 54.82% C, 5.624% H, 32.45% O, 7.104% N? 54.82 g C| 1 mole C = 4.568 mole C / .5074 = 9 C | 12 g C 5.624 g H| 1 mole H =5.624 mole H / .5074 = 11 H | 1 g H C9H11O4N 32.45 g O| 1 mole O = 2.028 mole O / .5074 = 4 O | 16 g O 7.104 g N| 1 mole N =0.5074 mole N / .5074 = 1 N | 14 g N If a compound has this same composition but its molecular weight is 394 g/mol, what is its molecular formula? MW. C9H11O4N = 197 g/mol 394/197 = 2 so molecular formula is C18H22O8N2 Mullis

13. Oxidation Numbers • Used to indicate the general distribution of electrons among the bonded atoms in molecular compounds or polyatomic ions. • Analogous to charges in ionic compounds. • An oxidation number is assigned to each element. • Assign the ones you know 1st. • Find the others based on the numbers it takes to make the charge equal to the charge of the ion or compound. (A compound has a charge of zero.) Mullis

14. Oxidation Numbers: Rules • Pure element = 0 • F = -1 • O = -2 (except in peroxides and bonds with halogens) • H = +1 (except in bonds with metals) • The more electronegative element = same (-) charge as its anion • The less electonegative element = same (+) charge as its cation • The sum of a compound or polyatomic ion’s oxidation numbers is equal to its charge. Mullis