Chapter 3 Chemical Calculations

# Chapter 3 Chemical Calculations

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## Chapter 3 Chemical Calculations

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1. Chapter 3 Chemical Calculations “I renamed my dog “Stoichiometry” “Ooooh, I just love mole ratios” • Part A: Compound Calcs • molecular weight • % composition • empirical formula • Part B: Reaction Calcs • mole ratios • mass ↔ moles • limiting reactant

2. Life is full of common counting units.. • A dozen bagels = 12 • A gross of rubber bands = 144 • A ream of paper = 500 • Hogshead = 14,553 in3 The mole is just another one, but unique to chemistry.

3. The MOLE concept 1 CH4 + 2 O2→ CO2 + 2 H2O 1dozen molecules of CH4 reacts with 2dozen molecules of O2 to form products 1thousand molecules of CH4 reacts with 2thousand molecules of O2 to form products 1 109 molecules of CH4 reacts with 2 109 molecules of O2 to form products 1 6.022  1023 molecules of CH4 reacts with 2 6.022  1023 molecules of O2 to form products

4. How many objects are in a mole? 602,213,670,000,000,000,000,000 objects 6.0221367x1023 = “Avogadro’s number” 1 mole of an element has 6.022 x 1023 atoms 1 mole of compound has 6.022 x 1023 molecules Why this number? Number of atoms of 12C in exactly 12g

5. Bicycle as a Molecule 1 frame 2 wheels 1 handlebar 1 seat FW2HS

6. A Mole of Bicycles 1 mole frames 2 mole wheels 1 mole handlebars 1 mole seats 1 mole of FW2HS

7. Figure 3.2: One mole each of various substances.Photo courtesy of American Color. Octanol (C8H17OH) Mercury (II) iodide sulfur Methanol (CH3OH)

8. Calculation of Formula Weight KClO3 K 39.098 amu * 1 = 39.098 amu Cl 35.453 amu * 1 = 35.453 amu O 15.999 amu * 3 = 47.997 amu Formula Weight of KClO3 = 122.548 amu

9. How to use the atomic weight information in the periodic table The average mass of one atom is given as the atomic weight in units of “amu” The mass of one mole of atoms is given as the “molar mass” by replacing the unit “amu” with the unit “gram”

10. What is the molar mass of Fe2O3? 1 molecule Fe: 55.85 amu x 2 = 111.7 amu O: 16.00 amu x 3 = 48.0 amu Total: 159.7 amu 1 mole of molecules Fe: 55.85 g x 2 = 111.7 g O: 16.00 g x 3 = 48.0 g Total: 159.7 g Molecular Weight (1amu=1.661x10-27kg) often used interchangeably Molar Mass

11. Information Contained in the Chemical Formula of Glucose C6H12O6 ( M = 180.16 g/mol) Carbon (C) Hydrogen (H) Oxygen (O) Atoms/molecule of compound Moles of atoms/ mole of compound Atoms/mole of compound Mass/moleculeof compound Mass/mole of compound 6 atoms 12 atoms 6 atoms 6 moles of 12 moles of 6 moles of atoms atoms atoms 6(6.022 x 1023) 12(6.022 x 1023) 6(6.022 x 1023) atoms atoms atoms 6(12.01 amu) 12(1.008 amu) 6(16.00 amu) =72.06 amu =12.10 amu =96.00 amu 72.06 g 12.10 g 96.00 g

12. Calculation of Molar Mass For sucrose (table sugar) the chemical formula is C12H22O11 What is the molar mass (4 sig figs)? C 12.011 amu x 12 H 1.008 amu x 22 O 15.999 amu x 11 342.3 g/mole What is the mass of 1 molecule in amu? In grams?

13. moles Avogadro’s # molar mass mass molecules (or atoms)

14. How many moles... in 10.8 g of boron (B)? in 25.0 g of boron? in 342.3 g of sucrose? in 10.0 g of sucrose? of O in 1.0 mol of sucrose? of O in 2.30 mol of sucrose? of 454 g of NaCl? How many grams... • in 0.50 mol of magnesium (Mg)? • in 2.5 mol of helium (He)? • in 0.75 mol of sucrose? • in 1.0 mol of NaCl? • in 10.0 mol of water? • in 2.30 mole of gold (Au)? • in 0.10 mol NaF?

15. How many moles... in 10.8 g of boron (B)? 1.00 mol in 25.0 g of boron? 2.31 mol in 342.3 g of sucrose? 1.000 mol in 10.0 g of sucrose? 0.0292 mol of O in 1.0 mol of sucrose? 11. mol of O in 2.30 mol of sucrose? 25.3 mol of 454 g of NaCl? 7.77 mol How many grams... • in 0.50 mol of magnesium (Mg)? 12.15 g • in 2.5 mol of helium (He)? 10. g • in 0.750 mol of sucrose? 257 g • in 1.00 mol of NaCl? 58.44 g • in 10.0 mol of water? 180. g • in 2.30 mole of gold (Au)? 453 g • in 0.10 mol NaF?4.2 g

16. moles Avogadro’s # molar mass mass molecules (or atoms)

17. A sample of nitric acid, HNO3, contains 0.253 mol HNO3. How many grams is this? First, find the molar mass of HNO3: 1 H 1(1.008) = 1.008 1 N 1(14.01) = 14.01 3 O 3(16.00) = 48.00 (2 decimal places) 63.02 g/mol 63.018

18. Mass - Mole Relationships of a Compound For an Element For a Compound Mass (g) of Element Mass (g) of compound Moles of Element Amount (mol) of compound Amount (mol) of compound Molecules (or formula units of compound) Atoms of Element

19. Calculating the Number of Moles and Atoms in a Given Mass of Element Problem: Tungsten (W) is the element used as the filament in light bulbs, and has the highest melting point of any element 3680oC. How many moles of tungsten, and atoms of the element are contained in a 35.0 mg sample of the metal? Plan: Convert mass into moles by dividing the mass by the atomic weight of the metal, then calculate the number of atoms by multiplying by Avogadro’s number! Solution: Converting from mass of W to moles: Moles of W = 35.0 mg W x = 0.00019032 mol 1.90 x 10 - 4 mol NO. of W atoms = 1.90 x 10 - 4 mol W x = = 1.15 x 1020 atoms of Tungsten 1 mol W 183.9 g W 6.022 x 1023 atoms 1 mole of W

20. Calculating the Moles and Number of Formula Units in a Given Mass of Cpd. Problem: Sodium Phosphate is a component of some detergents. How many moles and formula units are in a 38.6 g sample? Plan: We need to determine the formula, and the molecular mass from the atomic masses of each element multiplied by the coefficients. Solution: The formula is Na3PO4. Calculating the molar mass: M = 3x Sodium + 1 x Phosphorous = 4 x Oxygen = = 3 x 22.99 g/mol + 1 x 30.97 g/mol + 4 x 16.00 g/mol = 68.97 g/mol + 30.97 g/mol + 64.00 g/mol = 163.94 g/mol Converting mass to moles: Moles Na3PO4 = 38.6 g Na3PO4 x (1 mol Na3PO4) 163.94 g Na3PO4 = 0.23545 mol Na3PO4 Formula units = 0.23545 mol Na3PO4 x 6.022 x 1023 formula units 1 mol Na3PO4 = 1.46 x 1023 formula units

21. Mass Moles \$ Gold (Au) is now valued at \$___ per gram. (197.0 g Au/mol Au) How rich would you be if you had a gold brick composed of 5.0 moles of gold atoms? Suppose you had 2.5 x 1022 gold atoms. How much would this be worth?

22. Analyzing % composition by mass Which has more gold, 100 g of gold (III) sulfide or 100 g of gold (I) oxide ? Au2S3: 490.2 g/mol Au2O: 410.0 g/mol

23. Calculating % composition “% composition by mass” of O in H2O: total mass of O in H2O total mass of H2O  100 % O in H2O = (by mass)

24. Flow Chart of Mass Percentage Calculation Moles of X in one mole of Compound M (g / mol) of X Mass (g) of X in one mole of compound Divide by mass (g) of one mole of compound Mass fraction of X Multiply by 100 Mass % of X

25. Using % composition by mass 1) a sample of clear liquid (3.0000 g) is found to contain 0.1778 g H and 2.8222 g O 2) a sample of clear liquid (5.0000 g) is found to contain 0.5595 g H and 4.4405 g O 3) a sample of clear liquid (7.0000 g) is found to contain 0.2946 g H and 4.6769 g O Are these the same or different compounds?

26. Atomic mass H: 1.008 C: 12.01 O: 16.00

27. Calculating Mass Percentage and Masses of Elements in a Sample of a Compound - I Problem: Sucrose (C12H22O11) is common table sugar. (a) What is the mass percent of each element in sucrose? ( b) How many grams of carbon are in 24.35 g of sucrose? (a) Determining the mass percent of each element: mass of C = 12 x 12.01 g C/mol = 144.12 g C/mol mass of H = 22 x 1.008 g H/mol = 22.176 g H/mol mass of O = 11 x 16.00 g O/mol = 176.00 g O/mol 342.296 g/mol Finding the mass fraction of C in Sucrose& % C : Total mass of C 144.12 g C mass of 1 mole of sucrose 342.30 g Cpd Mass Fraction of C = = = 0.421046 To find mass % of C = 0.421046 x 100% = 42.105%

28. Calculating Mass Percents and Masses of Elements in a Sample of Compound - II (a) continued Mass % of H = x 100% = x 100% = 6.479% H Mass % of O = x 100% = x 100% = 51.417% O (b) Determining the mass of carbon: Mass (g) of C = mass of sucrose X( mass fraction of C in sucrose) Mass (g) of C = 24.35 g sucrose X = 10.25 g C mol H x M of H 22 x 1.008 g H mass of 1 mol sucrose 342.30 g mol O x M of O 11 x 16.00 g O mass of 1 mol sucrose 342.30 g 0.421046 g C 1 g sucrose

29. Chemical Formulas Empirical Formula- Shows the relative number of atoms of each element in the compound. It is the simplest formula, and is derived from masses of the elements. Molecular Formula - Shows the actual number of atoms of each element in the molecule of the compound. Structural Formula - Shows the actual number of atoms, and the bonds between them ; that is, the arrangement of atoms in the molecule.

30. Empirical and Molecular Formulas Empirical Formula - The simplest formula for a compound that agrees with the elemental analysis! The smallest set of whole numbers of atoms. Molecular Formula - The formula of the compound as it exists, it may be a multiple of the Empirical formula.

31. Climate Change Climate change is a significant shift in temperature and weather patterns around the world. While some changes are normal, the vast majority of scientists agree that our activities are causing dramatic changes to the Earth’s climate.

32. Over the last three hundred years we have radically increased our use of energy sources like oil, coal and natural gas. Burning these fuels releases carbon dioxide into the atmosphere. A little bit of carbon dioxide is a good thing – it keeps the planet warm and habitable. Now, however, we are putting so much carbon dioxide into the atmosphere that the planet is getting too warm. This problem is also called the greenhouse effect .

33. Greenhouse effect module