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Moles and Formula Mass The Mole 1 dozen = 1 gross = 1 ream = 1 mole = 12 144 500 6.022 x 10 23 There are exactly 12 grams of carbon-12 in one mole of.

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Presentazione sul tema: "Moles and Formula Mass The Mole 1 dozen = 1 gross = 1 ream = 1 mole = 12 144 500 6.022 x 10 23 There are exactly 12 grams of carbon-12 in one mole of."— Transcript della presentazione:

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2 Moles and Formula Mass

3 The Mole 1 dozen = 1 gross = 1 ream = 1 mole = x There are exactly 12 grams of carbon-12 in one mole of carbon-12.

4 Avogadro’s Number x is called “Avogadro’s Number” in honor of the Italian chemist Amadeo Avogadro ( ). Amadeo Avogadro I didn’t discover it. Its just named after me!

5 Calculations with Moles: Converting moles to grams How many grams of lithium are in 3.50 moles of lithium? 3.50 mol Li = g Li 1 mol Li 6.94 g Li 45.1

6 Calculations with Moles: Converting grams to moles How many moles of lithium are in 18.2 grams of lithium? 18.2 g Li = mol Li 6.94 g Li 1 mol Li 2.62

7 Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 3.50 moles of lithium? 3.50 mol = atoms 1 mol 6.02 x atoms 2.07 x 10 24

8 Calculations with Moles: Using Avogadro’s Number How many atoms of lithium are in 18.2 g of lithium? 18.2 g Li = atoms Li 1 mol Li x atoms Li 1.58 x g Li1 mol Li (18.2)(6.022 x )/6.94

9 Calculating Formula Mass Calculate the formula mass of magnesium carbonate, MgCO g g + 3(16.00 g) = g

10 Calculating Percentage Composition Calculate the percentage composition of magnesium carbonate, MgCO 3. From previous slide: g g + 3(16.00 g) = g

11 Formulas  molecular formula = (empirical formula) n [n = integer]  molecular formula = C 6 H 6 = (CH) 6  empirical formula = CH Empirical formula: the lowest whole number ratio of atoms in a compound. Molecular formula: the true number of atoms of each element in the formula of a compound.

12 Formulas (continued) Formulas for ionic compounds are ALWAYS empirical (lowest whole number ratio). Examples: NaClMgCl 2 Al 2 (SO 4 ) 3 K 2 CO 3

13 Formulas (continued) Formulas for molecular compounds MIGHT be empirical (lowest whole number ratio). Molecular: H2OH2O C 6 H 12 O 6 C 12 H 22 O 11 Empirical: H2OH2O CH 2 O C 12 H 22 O 11

14 Empirical Formula Determination 1.Base calculation on 100 grams of compound. 2.Determine moles of each element in 100 grams of compound. 3.Divide each value of moles by the smallest of the values. 4.Multiply each number by an integer to obtain all whole numbers.

15 Empirical Formula Determination Adipic acid contains 49.32% C, 43.84% O, and 6.85% H by mass. What is the empirical formula of adipic acid?

16 Empirical Formula Determination (part 2) Divide each value of moles by the smallest of the values. Carbon: Hydrogen: Oxygen:

17 Empirical Formula Determination (part 3) Multiply each number by an integer to obtain all whole numbers. Carbon: 1.50 Hydrogen: 2.50 Oxygen: 1.00 x Empirical formula: C3H5O2C3H5O2

18 Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 1. Find the formula mass of C 3 H 5 O 2 3(12.01 g) + 5(1.01) + 2(16.00) = g

19 Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = g 2. Divide the molecular mass by the mass given by the emipirical formula.

20 Finding the Molecular Formula The empirical formula for adipic acid is C 3 H 5 O 2. The molecular mass of adipic acid is 146 g/mol. What is the molecular formula of adipic acid? 3(12.01 g) + 5(1.01) + 2(16.00) = g 3. Multiply the empirical formula by this number to get the molecular formula. (C 3 H 5 O 2 ) x 2 = C 6 H 10 O 4

21 Convertire in moli(M): a)- 0,18 Kg di clorato di bario:

22 Scrivo la reazione per ottenere il clorato di bario: 2HClO 3 + BaSO Ba(ClO 3 ) 2 + H 2 SO 4 Determino il peso molecolare del clorato di bario: Ba = g Cl = g O = g. 96 p.m.= g. 304,4 Kg. 0,18 = g. 180 Determino le moli corrispondenti a Kg 0,18 di clorato di bario: n° Moli = g : p.m. = g. 180 / 304,4 = 0, M 0,59 M Risposta a) 0,18 Kg. di clorato di bario corrispondono a 0,59 M del medesimo sale.

23 Convertire: a)- 3 moli di acqua in Kg ed in molecole.

24 Soluzione a) Sapendo che: p.m. (H 2 O) = 2H = 2 O = 16 p.m. H 2 O = 18 Determino il n° di moli, grammi e di molecole di H 2 O : n° moli = 3 g. = 3 x 18 = g. 54 (grammi di acqua) = Kg 0,054 n° molecole = n° moli x N = 3x6, = 18, = 1, (molecole) Risposta a) 3 moli = Kg. 0,054 = molecole 18,

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