1 / 36

Chapter 7 Chemical Formula Relationships

Chapter 7 Chemical Formula Relationships. Number of Atoms in a Formula. In writing the formula of a substance, subscript numbers are used to indicate the number of atoms or groups of atoms of each element in the formula unit. Number of Atoms in a Formula.

mervyn
Télécharger la présentation

Chapter 7 Chemical Formula Relationships

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Chapter 7Chemical Formula Relationships

  2. Number of Atoms in a Formula In writing the formula of a substance, subscript numbers are used to indicate the number of atoms or groups of atoms of each element in the formula unit.

  3. Number of Atoms in a Formula How many atoms of each element are in a formula unit of ammonium carbonate? The formula of the ammonium ion is NH4+ The formula of carbonate is CO32– The formula of the ammonium carbonate is (NH4)2CO3. Each element in the parentheses is multiplied by two. Total number of atoms of each element: 2 nitrogen atoms, 8 hydrogen atoms, 1 carbon atom, 3 oxygen atoms

  4. Atomic Mass The average mass of atoms of an element By definition the mass of a carbon-12 atom is 12 u. Atomic Mass

  5. Molecular Mass The sum of the atomic masses of each atom in the molecule. Formula mass The sum of atomic masses in the formula unit Molecular & Formula Mass

  6. Molecular & Formula Mass What is the formula mass of calcium phosphate? Calcium ion: Ca2+ Phosphate ion: PO43– Calcium phosphate: Ca3(PO4)2 Ca 3 × 40.08 u = 120.24 u P 2 × 30.97 u = 61.94 u O 8 × 16.00 u = 128.00 u Ca3(PO4)2 310.18 u

  7. Defenition of Mole The mole is the amount of substance that contains as many elementary entities as there are atoms in exactly 12 grams of carbon-12. In 12 g of carbon-12 there are 6.022 x 10 23 atoms 1 mole of any substance = 6.022 x 10 23 units of that substance. When the mole is used the elementary entities must be specified: atoms, molecules, ions…

  8. Avogadro’s Number NA The number of elementary units in one mole 6.02214179  1023 units/mol The Avogadro constant is a conversion factor between units and mole.

  9. Conversion of Mole to Molecules How many carbon dioxide molecules are in 2.0 moles of carbon dioxide? 2.0 mol CO2 × = 1.2 × 1024 molecules CO2

  10. Molar Mass Molar mass of a substance is the mass in grams of one mole of the substance. Units: g/mol Molar mass of an element is the mass of the element per mole of its atoms.

  11. Atomic mass unit and gram The mass of one atom of carbon-12 is exactly 12 atomic mass units. The mass of one mole of carbon-12 atoms (6.022 x 1023 atoms of carbon-12) is exactly 12 grams (6.022 x 1023 atoms) x (12 u/atom) = 12 grams 6.022 x 1023 u = 1 gram

  12. Molar Mass The mass of one atom of carbon-12 is exactly 12 atomic mass units. The mass of one mole of carbon-12 atoms is exactly 12 grams. This leads to the conclusion: The molar mass of any substance in grams per mole is numerically equal to the atomic, molecular or formula mass of that substance in atomic mass units.

  13. Molar Mass Calculate the mass of one NH3 molecule and the mass of one mole of NH3 molecules. 14.01 u + 3x1.008 u = 17.03 u The mass of one ammonia molecule is 17.03 u. To change from molecular mass to molar mass, change the units from u to g/mol: 17.03 g/mol. One mole of ammonia molecules has a mass of 17.03 g.

  14. Molar Mass

  15. Molar mass, MM, links mass in grams with the number of moles. Avogadro’s number, NA, links the number of moles with the number of particles. Mass, # of Moles, # of Units

  16. Mass, # of Moles, # of Units How many molecules are in 454 g of water? g  moles  molecules 454g x x = 1.52 x 1025 molecules H2O

  17. Mass↔Moles↔ Units How many hydrogen atoms are in 1.0 kg of ammonia? 1.0 kg NH3 × × × × = 1.1 × 1026 atoms H

  18. Percentage Composition Percentage % A = × 100% The percentage composition of a compound is the percentage by mass of each element in the compound.

  19. Percentage Composition Determine the percentage composition of calcium fluoride Ca F2. Solution: In one mole of Ca F2 1x(40.08 g Ca) + 2x(19.00 g F) = 78.08 g CaF2 (Solution continued on the next slide)

  20. Percentage Composition  100 = 51.33 % Ca  100 = 48.67% F Check: 51.33% + 48.67% = 100.00%

  21. Empirical Formula Empirical Formula The simplest ratio of atoms of the elements in a compound. The empirical formula of C2H4 is CH2. Likewise, the empirical formula of C3H6 is CH2. All compounds with the general formula CnH2n have the same empirical formula and therefore the same percentage composition.

  22. Empirical Formula Write the empirical formulas of benzene, C6H6, and octane, C8H18. Look for the simplest whole-number ratio of elements: For C6H6, the 6/6 ratio can be reduced to 1/1: CH. For C8H18, the 8/18 ratio can be divided by 2 on top and bottom to be reduced to 4/9: C4H9.

  23. Find Empirical Formula To find empirical formula, you need to find the ratio of atoms of the elements ratio of atoms = ratio of moles of atoms

  24. How to Find an Empirical Formula • Find the masses of different elements in a sample of the compound. • Convert the masses into moles of atoms. • Determine the ratio of moles of atoms. • Express the moles ratio as the smallest possible ratio of integers. • Write the empirical formula, using the number in the integer ratio as the subscript in the formula.

  25. Find Empirical Formula What is the empirical formula of a compound that has 85.6% carbon, 14.4 % hydrogen? Solution: It is usually helpful to organize the calculations in a table with the following headings: Mole Formula Empirical Element Grams Moles Ratio Ratio Formula

  26. Find Empirical Formula Mole Formula Empirical Element Grams Moles Ratio Ratio Formula C 85.6 1 7.13 1 H 14.4 2 CH2 14.3 2.01

  27. Find Empirical Formula What is the empirical formula of a compound that analyzes as 20.0% carbon, 2.2% hydrogen, and 77.80% chlorine? Mole Formula Empirical Element Grams Moles Ratio Ratio Formula C 20.0 1.67 1 3 H 2.2 2.2 1.3 4 Cl 77.8 2.19 1.31 4 C3H4Cl4

  28. Molecular Formula The molecular formula of a compound can be found by determination of the number of empirical formula units in the molecule.

  29. How to Find the Molecular Formula • Determine the empirical formula of the compound. • Calculate the molar mass of the empirical formula unit. • Divide the molar mass of the compound by the molar mass of the empirical formula unit to get n, the number of empirical formula units per molecule.

  30. Molecular Formula What is the molecular formula of a compound with the empirical formula C2H5and a molar mass of 58.12 g/mol? The molar mass of the empirical formula unit is 2(12.01 g/mol C) + 5(1.008 g/mol H) =29.06 g/mol The number of empirical formula units per molecule is (C2H5)2 = C4H10

  31. Find Molecular Formula • An unknown compound is found to be 40.0% of carbon, 6.71% of hydrogen and the remainder is oxygen. The molar mass of the compound is 180.16 g/mol. Find the empirical and molecular formulas of the compound.

  32. First Find Empirical Formula Mole Formula Empirical Element Grams Moles Ratio Ratio Formula C 40.0 3.33 1 1 H 6.71 6.66 2 2 O 53.3 3.33 1 1 CH2O

  33. Calculate Molar Mass/Empirical Formula Mass The molar mass of the empirical formula unit is (12.01 g/mol C) + 1(1.008 g/mol H) + (16.00 g/mol O) = 30.03 g/mol The number of empirical formula units per molecule is (CH2O) 6 = C6H12O6

  34. Molecular Formula Another way to find molecular formula is to consider one mole of the compound (180.16 g of compound). The numbers of moles are also numbers of atoms in the molecule. The masses of carbon, hydrogen and oxygen are : 40.0% x 180.16 g = 72.06 g of C 6.71% x 180.16 g = 12.09 g of H 53.3% x 180.16 g = 96.03 g of O

  35. Molecular Formula Element Grams Moles Molecular Formula C 72.06 6 H 12.09 12 O 96.03 6 C6H12O6

  36. Homework 7, 15, 21, 23, 43, 57, 61, 63, 65, 68, 74.

More Related