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Chemical Quantities

Chemical Quantities. Unit 6 Chapter 10, Section 10.2 Mole-Mass and Mole-Volume Relationships. Objectives. When you complete this presentation, you will be able to … describe how to convert the mass of a substance to the number of moles of a substance, and moles to mass.

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Chemical Quantities

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  1. Chemical Quantities Unit 6 Chapter 10, Section 10.2 Mole-Mass and Mole-Volume Relationships

  2. Objectives • When you complete this presentation, you will be able to … • describe how to convert the mass of a substance to the number of moles of a substance, and moles to mass. • identify the volume of a quantity of gas at STP (Standard Temperature and Pressure).

  3. Introduction • How many jelly beans are in the jar? • How might you approach the problem? • Count each jelly bean • Estimate the volume of each bean • Estimate the mass of each bean

  4. Introduction • As chemists, we can’t count the number of individual atoms in a reaction (why?). • We rely on other relationships, such as … • moles → mass and mass → moles • moles → volume and volume → moles • These relationships allow us to accurately estimate the number of atoms or molecules involved in a chemical reaction.

  5. The Mole-Mass Relationship • The relationship between the mass of a material and the number of moles of matter is based on the molar mass of the matter. • where Mis molar mass, mis mass, and nis number of mols. m m = = or or m = n × M M n n M

  6. The Mole-Mass Relationship • For example: • The molar mass of NaCl is 58.5 g/mol • The mass of 3.00 mol of NaCl is given by • Let’s try Sample Problem 10.5 (page 298) m = n × M = (3.00 mol)(58.3 g/mol) = 176 g

  7. Sample Problem 10.5 The aluminum satellite dishes shown below are resistant because the aluminum reacts with oxygen in the air to form a coating of aluminum oxide (Al2O3). This tough, resistant coating prevents any further corrosion. What is the mass of 9.45 mol of aluminum oxide? Known: n = 9.45 mol M = [(2 × 27.0) + (3 × 16.0)] g/mol M= 102.0 g/mol Unknown: m = ? g Al2O3 m = n × M = (9.45)(102.0) g m = 964 g Al2O3

  8. The Mole-Mass Relationship • For another example: • The molar mass of Na2SO4 is 142.1 g/mol • The mols of 10.0 g of Na2SO4 is given by • Let’s try Sample Problem 10.6 (page 299) 10.0 g m = = n M 142.1 g/mol = 7.04 × 10-2mol

  9. Sample Problem 10.6 When iron is exposed to air, it corrodes to form red-brown rust. Rust is iron(II) oxide(Fe2O3). How many mols of iron(II) oxide are contained in 92.2 g of pure Fe2O3? Known: m = 92.9 g M = [(2 × 55.8) + (3 × 16.0)] g/mol M= 159.6 g/mol Unknown: n = ? mol Fe2O3 m n = = 0.578 mol = 92.9 g M 159.6 g/mol

  10. The Mole-Mass Relationship • Practice Problems: • What is the mass of 2.50 mols of hydrogen gas, H2? • What is the mass of 0.100 mols of glucose, C6H12O6? • What is the mass of 12.0 mols of water, H2O? • What is the mass of 0.250 mols of methane gas, CH4? m = n × M mH2 = nH2 × MH2 = (2.50 mol)(2.02 g/mol) = 5.05 g mglucose = nglucose × Mglucose = (0.100 mol)(180 g/mol) = 18.0 g mH2O = nH2O × MH2O = (12.0 mol)(18.0 g/mol) = 216 g mCH4 = nCH4 × MCH4 = (2.50 mol)(16.0 g/mol) = 4.00 g

  11. The Mole-Mass Relationship m n = • Practice Problems: • How many mols are in 3.25 g of H2? • How many mols are in 9.00 g of C6H12O6? • How many mols are in 100. g of H2O? • How many mols are in 9.60 g of CH4? M mH2 3.25 g mH2O 100.g mCH4 9.60 g mC6H12O6 9.00 g nCH4 = = = 0.600 mol nH2O = = = 5.56 mol nC6H12O6 = = = 0.0500 mol nH2 = = = 1.61 mol MH2 2.02 g/mol MH2O 18.0 g/mol MCH4 16.0 g/mol MC6H12O6 180 g/mol

  12. The Mole-Volume Relationship • We saw earlier that 1 mol of solidsor liquidsare not necessarily the same volume. • 1 mol of C6H12O6 is a much largervolume than 1 mol of H2O. • The story is different with gases. • 1 mol of O2 is nearly the same volume as 1 mol of H2under the same conditions. • 1 mol of O2is nearly the same volume as 1 mol of CO2under the same conditions.

  13. The Mole-Volume Relationship • Why is this? • In 1811, Amadeo Avogadro proposed that equal volumesof gases, under similar conditions, contain the equal numbers of molecules. • This is called Avogadro’s hypothesis. • The individual moleculesof each gas are different sizes and masses. • But, in gases, the distances between the molecules is much largerthan the molecule size.

  14. The Mole-Volume Relationship • When talking about gases we use the phrase “under similar conditions” a lot. • That is because temperatureand pressurehave a big effect on the volumeof gases. • We will find out more about this in the next unit. • To compare different gases, we put them at a Standard Temperature and Pressure. • This is called “STP.”

  15. The Mole-Volume Relationship • Standard temperature and pressure: • Standard temperature is 0°C (273.16 K). • Standard pressure is 1atm (101.3kPa). • Don’t worry about the units right now. • The important thing is that if we say that two gases are at STP, then they are “under similar conditions.”

  16. The Mole-Volume Relationship • At STP, 1 mol of a gas has a volume of 22.4 L. • The relationship between the volumeof a gas and the number of mols of a gas is given by • where Vis the volume of the gas in liters (L) and nis the amount of gas in mols 22.4 L 1 mol V = n × or n = V × 1 mol 22.4 L

  17. The Mole-Volume Relationship • For example • the volume of 1.25 mols of O2 gas at STP is • the number of mols of H2 in 67.2 L at STP is • Let’s try Sample Problem 10.7 (page 301). 1 mol 22.4 L = 67.2 L × 1.25 mol × = 28.0 L 22.4 L 1 mol 1 mol 22.4 L n = V × V = n × 22.4 L 1 mol = = 3.00 mol

  18. Sample Problem 10.7 Sulfur dioxide (SO2) is a gas produced by burning coal. It is an air pollutant and one of the causes of acid rain. Determine the volume, in liters, of 0.60 mol of SO2 gas at STP. Known: n = 0.60 mol Unknown: V = ? L SO2 22.4 L 22.4 L V = n × = 0.60 mol × 1 mol 1 mol V = 13 L SO2

  19. The Mole-Volume Relationship 22.4 L V = n × • Practice Problems: • What is the volume of 3.25 mols of H2 at STP? • What is the volume of 0.250 mols of O2 at STP? • What is the volume of 12.5 molsof Cl2 at STP? • What is the volume of 0.0100 molsof CH4? 1 mol 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L VCH4 = nCH4× = 0.0100 mol × = 0.224 L VO2 = nO2× = 0.250 mol × = 5.60 L VCl2 = nCl2× = 12.5 mol × = 280. L VH2 = nH2× = 3.25 mol × = 72.8 L 1 mol1 mol 1 mol1 mol 1 mol1 mol 1 mol1 mol

  20. The Mole-Volume Relationship 1 mol n = V × • Practice Problems: • How many molsof H2are in 1.00 L at STP? • How many mols of O2are in 50.0 L at STP? • How many mols of C2H2are in 0.250 L at STP? • How many mols of F2are in 11.2 L at STP? 22.4 L 1 mol 1 mol 1 mol 1 mol 1 mol 1 mol 1 mol 1 mol nC2H2 = VC2H2× = 0.250 L × = 5.60 mol nF2 = VF2× = 11.2 L × = 0.500 mol nO2 = VO2× = 50.0 L × = 2.23 mol nH2 = VH2× = 1.00 L × = 0.0446 mol 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L

  21. The Mole-Volume Relationship • Density and Molar Mass • The density of matter is mass per unit volume. • where ρ = density (g/L), m = mass, and V = volume. • We can use our mole-mass and mole-volume relationships to determine density. m ρ = V m = n × M 22.4 L 22.4 L V = n × 1 mol 1 mol m n × M M 1 mol = = = M × ρ = 22.4 L V 22.4 L n × 1 mol

  22. The Mole-Volume Relationship • Density and Molar Mass • If we know the molar mass of a gas, then we can calculate its density. • If we know the density of a gas, then we can calculate its molar mass. 1mol 22.4 L ρ = M × M = ρ × 22.4 L 1 mol

  23. The Mole-Volume Relationship • Density and Molar Mass • For example, find the density of H2 at STP. • For example, find the molar mass of a gas with a density of 3.17 g/L at STP. = 0.0902 g/L = 71.0 g/mol 22.4 L 1mol 22.4 L 1mol ρ = M × = 2.02 g/mol× M = ρ × = 3.17 g/L × 1 mol 22.4 L 1 mol 22.4 L

  24. Sample Problem 10.8 The density of a gaseous compound containing carbon and oxygen is found to be 1.964 g/L at STP. What is the molar mass of this compound Known: ρ = 1.964 g/L Unknown: M = ? g/mol 22.4 L 22.4 L M = ρ × = 1.964 g/L × 1 mol 1 mol M = 44.0 g/mol What do you think the gas is? Why?

  25. The Mole-Volume Relationship 1 mol ρ = M × • Practice Problems: • What is the density of F2 at STP? • What is the density of CH4at STP? • What is the density of Ne at STP? • What is the density of Rnat STP? 22.4 L 1 mol 1 mol 1 mol 1 mol 1 mol 1 mol 1 mol 1 mol ρRn = MRn× = 222 g/mol × = 9.91 g/L ρNe = MNe× = 20.2 g/mol × = 0.902 g/L ρCH4 = MCH4× = 16.0 g/mol × = 0.714 g/L ρF2 = MF2× = 38.0 g/mol × = 1.70 g/L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L

  26. The Mole-Volume Relationship 22.4 L M = ρ × • Practice Problems: • Find the molar mass of a gas with a density of 1.34 g/L at STP. • Find the molar mass of a gas with a density of 2.59 g/L at STP. • Find the molar mass of a gas with a density of 1.78 g/L at STP. • Find the molar mass of a gas with a density of 6.52 g/L at STP. 1 mol 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L 22.4 L M = ρ × = 6.52 g/L × = 146 g/mol M = ρ × = 1.78 g/L × = 39.9 g/mol M = ρ × = 2.59 g/L × = 58.0 g/mol M = ρ × = 1.34 g/L × = 30.0 g/mol 1 mol1 mol 1 mol1 mol 1 mol1 mol 1 mol1 mol

  27. Summary • The relationship between the mass of a material and the number of moles of matter is based on the molar mass of the matter. • where M is molar mass, m is mass, and n is number of mols. or or m = n × M m m = = M n n M

  28. Summary • To compare different gases, we put them at a Standard Temperature and Pressure. • This is called “STP.” • Standard temperature is 0°C (273.16 K). • Standard pressure is 1 atm (101.3 kPa). • The relationship between the volume of a gas and the number of mols of a gas is given by • where V is the volume of the gas in liters (L) and n is the amount of gas in mols 22.4 L 1 mol V = n × n = V × 1 mol 22.4 L or

  29. Summary • We can use our mole-mass and mole-volume relationships to determine density. • If we know the molar mass of a gas, then we can calculate its density. • If we know the density of a gas, then we can calculate its molar mass. 1mol 22.4 L ρ = M × M = ρ × 22.4 L 1 mol

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