1. After completing this lesson you should be able to : Balanced equations show the mole ratio(s) of reactants and products. The molar volume is the same for all gases at the same temperature and pressure. Calculations involving molar volume. Determination of quantities of reactants and/or products using balanced equations and moles. The molar volume (in units of litres mol-1 ) is the same for all gases at the same temperature and pressure. The volume of a gas can be calculated from the number of moles and vice versa. Determination of quantities of reactants and/or products using balanced equation and:  gram formula masses (GFM) and mass  moles  concentrations and volumes of solutions  masses of solutes  molar volumes and volumes for gaseous reactions The concentration of a solution can be expressed in mol l-1. Balanced equations can be used in conjunction with concentrations and volumes of solutions and/or masses of solutes to determine quantities of reactants and/or products. The volumes of reactant and product gases can be calculated from the number of moles of each reactant and product.

2. b) Calculation of the mass or volume of products Learning intention Learn how the theoretical mass or volume of product can be calculated from the balanced reaction equation.

3. Avogadro's number Avogadro’s constant, symbol N A or L, is the number of ‘elementary particles’ in one mole of any substance. It has the value of 6.02 x 10 23 COPY

4. It is a number, just as is "dozen"; you can think of Avogadro's number as the "chemist's dozen". It is a huge number, far greater in magnitude than we can visualize An Avogadro's number of standard soft drink cans would cover the surface of the earth to a depth of over 200 miles. If we were able to count atoms at the rate of 10 million per second, it would take about 2 billion years to count the atoms in one mole. Things to understand about Avogadro's number 6.02 x 10 23 602 billion trillion 602,000,000,000,000,000,00 0,000 COPY

5. Ideal Gases and The mole

6. Amadeo AvogadroAmadeo Avogadro (1766-1856) Avogadro’s hypothesis states that “equal volumes of different gases, under STP, contain equal numbers of molecules.” Avogadro's number is one of the fundamental constants of chemistry. 6.02 x 10 23 602 billion trillion 602,000,000,000,000,000,00 0,000 Avogadro never knew his own number; it was named in his honor by a French scientist in 1909. Its value was first estimated by Josef Loschmidt, an Austrian chemistry teacher, in 1895. COPY

7. Molar Gas Volume One mole of all gases at the same temperature and pressure will have the same volume. 1 mole of oxygen 1 mole of nitrogen 1 mole of carbon dioxide Comparing gas volumes 5 litres 2.5 litres Since the larger ‘balloon’ has twice the volume it must have twice the number of moles of gas within it. It must also have twice as many particles. COPY

8. Mole and gas volume The molar volume of a gas is its volume per mole, litre mol -1. It is the same for all gases at the same temperature and pressure. The value, though, is temperature and pressure dependent. The molar volume of all gases is approximately 24 litre mol -1 at 20 o C and 22.4 litre mol -1 at 0 o C. Volume (l) Molar Volume (l) nx COPY

9. Determining molar volume experimentally A round bottom flask can be evacuated and then weighed. It is then reweighed after being filled with a gas. The volume of the flask can be found by filling it with water and emptying it into a measuring cylinder. This can be repeated using other gases. The molar volume can then be calculated from the data obtained. Calculate the molar volume of carbon dioxide from the following data obtained at 20ºC and 1 atmosphere of pressure. Mass of empty flask = 107.49g Mass of flask + carbon dioxide = 108.37g Volume of flask = 480cm 3 Calculation Mass of carbon dioxide = 108.37 – 107.49 = 0.88g GFM of CO 2 = 44g 0.88g of CO 2 occupies a volume of 480 cm 3. 44g  X 24000 cm 3

10. Measuring Molar gas volume Finding out the molar volume by experiment. Method 1 Density = mass/volume Or mass = density x volume 1. Weight the flask with air in it. Use the density of air and the volume of the flask to calculate the mass of air in the flask and hence the weight of the empty flask. Mass of flask + air = g Mass of air = density x volume = 1.29 x 10 -3 x........................... = g Mass of empty flask = g 2. The flask is filled with the gas and reweighed. Calculate the mass of gas. Higher Chemistry Eric Alan and John Harris

11. Measuring Molar gas volume O 2 OxygenN 2 NitrogenCO 2 Carbon dioxide Mass of empty flask (g) Mass of flask + gas (g) Mass of gas (g) Volume of flask i.e. volume of gas (l) Mass of one mole (g) Volume of 1 mole 3. The volume of the flask can be measured by filling with water and emptying into a measuring cylinder. Example calculation You have found that...................... g of O 2 occupies.................... Litres.........................g ↔..................... Litres 32g ↔. Litres Therefore 1 mole, 32g of O 2 will occupy........................ Litres Higher Chemistry Eric Alan and John Harris

12. Measuring Molar gas volume Finding out the molar volume by experiment. Method 2

13. Measuring Molar gas volume Finding out the molar volume by experiment. Method 2 Calculate the molar gas volume ____ g magnesium has produced ____cm 3 hydrogen ____ g ↔ ________ cm 3 24 g ↔ _________ The volume of one mole of H 2 gas is ________ Litres

14. Why? It may seem surprising that the molar volume is the same for all gases, even at the same temperature and pressure. In a gas the molecules have much more kinetic energy and are relatively far apart so the volume of the gas does not depend on the sizes of the particles. In a gas at room temperature and pressure the molecules only occupy about 0.1% of the volume of a gas. The rest is empty space! COPY

15. Calculations involving molar gas volume At room temperature and pressure the molar gas volume is 24 litres mol -1. Worked example 1. Calculate the volume of 0.025 moles of oxygen. 1 mole  24 litres 0.025  0.025 / 1 x 24 = 0.6 litres Worked example 2. Calculate the number of moles in 72 litres of hydrogen. 1 mole  24 litres 72 / 24 x 1  72 litres = 3 moles COPY

16. Calculations involving molar gas volume What is the mass of steam in 180 cm 3 of the gas, when the molar volume is 24 litres mol -1 ? 0.135 g COPY

17. Worked example 1.The equation below shows the reaction between calcium carbonate and hydrochloric acid. CaCO 3 (s) + 2HCl(aq)  CaCl 2 (aq) + CO 2 (g) + H 2 O(l) 20g of calcium carbonate reacts with excess hydrochloric acid. Calculate the volume of carbon dioxide gas formed. (Take the molar volume to be 23.0 litre mol -1 ) CaCO 3 (s) + 2HCl(aq)  CaCl 2 (aq) + CO 2 (g) + H 2 O(l) Write the balanced equation Show mole ratio 1 mol  1mol Change moles into required units 100 g  23.0 litres Use proportion 20 g  20 / 100 x 23.0 litre 4.6 litres Calculations involving mass and volume COPY

18. The molar volume is the volume occupied by one mole of a gas. Worked example 1. In an experiment the density of carbon dioxide was measured and found to be 1.85 g l -1. Calculate the molar volume of carbon dioxide. 1.85 g  1 litre So 1 mole, 44 g  44 / 1.85 x 1 = 23.78 litres Worked example 2. A gas has a molar volume of 24 litres and a density of 1.25 g l -1. Calculate the mass of 1 mole of the gas. 1 litre  1.25g So 1 mole, 24 litre  24 / 1 x 1.25 = 30 g Calculations involving density and molar volume COPY

19. Calculations for you to try. 1.Calculate the number of moles in 0.36 litres of argon (molar gas volume = 24 litres mol -1 ). 2.Calculate the volume of 0.04 moles of CO 2. (molar gas volume = 24 litres mol -1 ). 4. A gas has a density of 2.74 g l -1 and a molar volume of 23.4 litre mol -1. Calculate the molecular mass of the gas. 3. Under certain conditions oxygen has a density of 1.44 g l -1. Calculate the molar volume of oxygen under these conditions. 22.22 litres 64.1 g 0.96L 0.015moles COPY

20. Gas Volume Calculations N 2 (g) + 2O 2 (g)  2NO 2 (g) 1mole 2moles Calculate the volume of nitrogen dioxide gas produced when 100cm 3 of nitrogen is sparked in excess oxygen 200cm 3 of NO 2 100cm 3 1vol 2vol 200cm 3 COPY

21. Gas Volume Calculations CH 4 (g) + 2O 2 (g)  CO 2 (g) + 2 H 2 0 (l) Balanced equation What volume of C0 2, is produced if 100 cm 3 of O 2 is used to completely to burn some CH 4 gas? Combustion of methane 50cm 3 of CO 2 COPY

22. Gas Volume Calculations CO (g) + ½ O 2 (g)  CO 2 (g) Balanced equation 50 cm 3 of C0, is burned with 20cm 3 of oxygen. a)Which gas is in excess b)What is the volume and composition of the resulting gas mixture CO in excess 10cm 3 CO + 40cm 3 CO 2 = 50cm 3 of gas COPY

23. Gas Volume Calculations C 3 H 8 (g) + 5O 2 (g)  3CO 2 (g) + 4H 2 0 (l) Balanced equation 10 cm 3 of propane gas is mixed with 75cm 3 of O 2 and the mixture exploded. a)Calculate the volume and the composition of the resulting gas mixture. b)What will be the change in volume when the resulting gas mixture is shaken with dilute sodium hydroxide solution. Combustion of propane 25cm 3 of O 2 + 30cm 3 of CO 2 = 55cm 3 of gas Volume will decrease by 30cm 3 (CO 2 ) removed COPY