3/22/2017. Chapter 8. Chemical Composition. Counting by Weighing. Section 8.1

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1 Chapter 8 Chemical Composition Section 8.1 Counting by Weighing 2 1

2 Section 8.1 Counting by Weighing A pile of marbles weigh g. 10 marbles weigh g. How many marbles are in the pile? g Avg. Mass of 1 Marble = 10 marbles = 3.76 g / marble g = 105 marbles 3.76 g 3 Section 8.1 Counting by Weighing Averaging the Mass of Similar Objects Example: What is the mass of 1000 jelly beans? 1. Not all jelly beans have the same mass. 2. Suppose we weigh 10 jelly beans and find: Bean Mass 5.1 g 5.2 g 5.0 g 4.8 g 4.9 g 5.0 g 5.0 g 5.1 g 4.9 g 5.0 g 3. Now we can find the average mass of a bean. 4. Finally we can multiply to find the mass of 1000 beans! 1000 x 5.0g = g 4 2

3 Section 8.2 Atomic Masses: Counting Atoms by Weighing Atoms have very tiny masses so scientists created a unit to avoid using very small numbers. 1 atomic mass unit (amu) = g The average atomic mass for an element is the weighted average of the masses of all the isotopes of an element. 5 Section 8.2 Atomic Masses: Counting Atoms by Weighing Average Atomic Mass for Carbon Even though natural carbon does not contain a single atom with mass 12.01amu, for our purposes, we can treat carbon as though it is composed of only one type of atom with a mass of 12.01amu. This enables us to count atoms of natural carbon by weighing a sample of carbon. 6 3

4 Section 8.3 The Mole The number equal to the number of carbon atoms in grams of carbon. 1 mole of anything = x units of that thing (Avogadro s number). 1 mole C = x C atoms = g C 23 1 mol C C atoms or C atoms 1 mol C C atoms g C or g C C atoms 1 mol C g C or g C 1 mol C 7 Section 8.3 The Mole A sample of an element with a mass equal to that element s average atomic mass (expressed in g) contains one mole of atoms ( atoms). Comparison of 1-Mol Samples of Various Elements 8 4

5 Section 8.3 The Mole Concept Check Determine the number of copper atoms in a g sample of copper Cu atoms 9 Section 8.3 The Mole Concept Check Which of the following is closest to the average mass of one atom of copper? a) g b) g c) g d) g e) x g 1 mol Cu g Cu Cu atom = g Cu Cu atoms 1 mol Cu 10 5

6 Section 8.3 The Mole Concept Check Calculate the number of iron atoms in a 4.48 mole sample of iron mol Fe atoms Fe 1 mol Fe Fe atoms 11 Section 8.3 The Mole Concept Check Which of the following g samples contains the greatest number of atoms? a) Magnesium b) Zinc c) Silver 12 6

7 Section 8.3 The Mole Rank the following according to number of atoms (greatest to least): a) g of silver b) 70.0 g of zinc c) 21.0 g of magnesium b) a) c) 13 Section 8.4 Learning to Solve Problems Conceptual Problem Solving Where are we going? Read the problem and decide on the final goal. How do we get there? Work backwards from the final goal to decide where to start. Reality check. Does my answer make sense? Is it reasonable? 14 7

8 Section 8.4 Learning to Solve Problems Mole Mass Calculation What is the mass of 1.33 moles of titanium, Ti? 1. We want grams. 2. We have 1.33 moles of titanium. 3. Use the molar mass of Ti: g/mol moles mass 1.33 mole Ti x g Ti 1 mole Ti = 63.7 g Ti Section 8.5 Molar Mass Mass in grams of one mole of the substance: Molar Mass of N = g/mol Molar Mass of H 2 O = g/mol ( g) g Molar Mass of Ba(NO 3 ) 2 = g/mol g + ( g) + ( g) 16 8

9 Section 8.5 Molar Mass What is the molar mass of nickel(ii) carbonate? a) g/mol b) g/mol c) g/mol d) g/mol The formula for nickel(ii) carbonate is NiCO 3. Therefore the molar mass is: (16.00) = g/mol. 17 Section 8.5 Molar Mass Molar (formula) Mass Substance Formula Mass (g/mol) Carbon, C 12 Oxygen, O 2 32 Carbon dioxide, CO 2 44 Sucrose, C 12 H 22 O

10 Section 8.5 Molar Mass Consider equal mole samples of N 2 O, Al(NO 3 ) 3, and KCN. Rank these from least to most number of nitrogen atoms in each sample. a) KCN, Al(NO 3 ) 3, N 2 O b) Al(NO 3 ) 3, N 2 O, KCN c) KCN, N 2 O, Al(NO 3 ) 3 d) Al(NO 3 ) 3, KCN, N 2 O 19 Section 8.5 Molar Mass Consider separate gram samples of each of the following: H 2 O, N 2 O, C 3 H 6 O 2, CO 2 Rank them from greatest to least number of oxygen atoms. H 2 O, CO 2, C 3 H 6 O 2, N 2 O 20 10

11 Section 8.5 Molar Mass How many grams of fluorine are contained in one molecule of boron trifluoride? a) g b) g c) g d) g 1 molecule BF 3 3 atoms F 1 mol F g F 23 1 molecule BF atoms F 1 mol F 3 21 Section 8.6 Percent Composition of Compounds Mass percent of an element: For Carbon in C 2 H 5 OH: 24.02g C x 100 = % C in C 2 H 5 OH 46.07g C 2 H 5 OH 22 11

12 Section 8.6 Percent Composition of Compounds The percent composition of a compound lists the mass percent of each element. For example, the percent composition of water, H 2 O, is 11% hydrogen and 89% oxygen. All water contains: 2.02 g H g H 2 O g O g H 2 O x 100% = 11.2% H x 100% = 88.79% O Section 8.6 Percent Composition of Compounds Morphine, derived from opium plants, has the potential for use and abuse. It s formula is C 17 H 19 NO 3. What percent, by mass, is the carbon in this compound? a) 12.0 % b) 54.8 % c) 67.9 % d) 71.6 % [ ] ( ) + ( ) ( ) ù éë = = 71.6% C û 24 12

13 Section 8.6 Percent Composition of Compounds Consider separate gram samples of each of the following: H 2 O, N 2 O, C 3 H 6 O 2, CO 2 Rank them from highest to lowest percent oxygen by mass. H 2 O, CO 2, C 3 H 6 O 2, N 2 O 25 Section 8.7 Formulas of Compounds Empirical Formulas The empirical formula of a compound is the simplest whole number ratio of the atoms present in the compound. The empirical formula can be found from the percent composition of the compound

14 Section 8.8 Calculation of Empirical Formulas Steps for Determining the Empirical Formula of a Compound A gaseous compound containing carbon and hydrogen was analyzed and found to consist of 83.65% carbon by mass. Determine the empirical formula of the compound. 1. Obtain the mass of each element present (in grams). Assume you have 100 g of the compound % C = g C ( ) 16.35% H = g H 27 Section 8.8 Calculation of Empirical Formulas Steps for Determining the Empirical Formula of a Compound 2. Determine the number of moles of each type of atom present. 1 mol C g C = mol C g C 1 mol H g H = mol H g H 28 14

15 Section 8.8 Calculation of Empirical Formulas Steps for Determining the Empirical Formula of a Compound 3. Divide the number of moles of each element by the smallest number of moles to convert the smallest number to 1. If all of the numbers so obtained are integers (whole numbers), these are the subscripts in the empirical formula. If one or more of these numbers are not integers, go on to step mol C = mol mol H = mol 29 Section 8.8 Calculation of Empirical Formulas Steps for Determining the Empirical Formula of a Compound 4. Multiply the numbers you derived in step 3 by the smallest integer that will convert all of them to whole numbers. This set of whole numbers represents the subscripts in the empirical formula. C: 1 3 = 3 H: = 7 The empirical formula is C H

16 Section 8.8 Calculation of Empirical Formulas The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). What is the empirical formula? C 3 H 5 O 2 31 Section 8.9 Calculation of Molecular Formulas The molecular formula is the exact formula of the molecules present in a substance. The molecular formula is always an integer multiple of the empirical formula. Molecular formula = (empirical formula) n where n is a whole number 32 16

17 Section 8.9 Calculation of Molecular Formulas Continued Example A gaseous compound containing carbon and hydrogen was analyzed and found to consist of 83.65% carbon by mass. The molar mass of the compound is 86.2 g/mol. You determined the empirical formula to be C 3 H 7. What is the molecular formula of the compound? Molar mass of C 3 H 7 = g/mol 86.2 g/mol g/mol = 2 Molecular formula (empirical formula)n = C 3 H 7 2 = C 6 H Section 8.9 Calculation of Molecular Formulas The composition of adipic acid is 49.3% C, 6.9% H, and 43.8% O (by mass). The molar mass of the compound is about 146 g/mol. The empirical formula is C 3 H 5 O 2. What is the molecular formula? C 6 H 10 O