Collecting a gas over water…

1. Collecting a gas over water… Sometimes a gas is produced in a chemical reaction…. That gas is bubbled through water and collected in a separate tube or container The total pressure inside the container, though, is not just the gas Some water has evaporated and is adding pressure to the gas The total pressure inside the container, then, is the pressure of the gas and the pressure of the water vapor that has evaporated The top of the container contains the gas produced from the reaction and water vapor

2. So how do we account for this in a problem…? • We do the same thing we always do – except one small change in the pressure! • Write a chemical reaction: • 2 KClO3 (s) -----------> 2 KCl (s) + 3 O2 (g) • Convert to moles: • 1.300 g KClO3 x 1 mole KClO3/122.55 g = .0106 moles KClO3 • Set up a ratio from the reaction: • .0106 moles KClO3 x 3 moles O2/ 2 moles KClO3 = .0159 moles O2 • PV/nT = R • However, we need to adjust the pressure – it tells us in the problem that the total pressure inside the container is 745 mm Hg – this pressure is both the gas created and the water that has evaporated! Oxygen can be prepared from heating potassium chlorate. How many moles of oxygen would be obtained from 1.300 grams of potassium chlorate? If the oxygen is collected over water at 230C and a total pressure of 745 mm Hg, what volume would the oxygen occupy?

3. KClO3 (s) -----------> KCl (s) + 3 O2 (g) • .0159 moles O2 has to be converted into liters to solve the problem • However, we are not at standard temperature and pressure • We need to use the ideal gas law • The pressure given in the problem - 745 mm Hg – is the pressure of the O2and the water that has evaporated and is adding to the pressure! • We have to subtract out the water vapor pressure from the total pressure to get the pressure of just the gas!

4. We need to subtract out the water vapor pressure…. • We need a table that tells us what the evaporated pressure of water is at 230C! • The table to the right tells us that at 230C, the vapor pressure due to water evaporating is 21.0 mm Hg. • We need to subtract this from the total pressure – this will tell us how much of the pressure is due to just the gas! • Pgas = 745 mm Hg – 21.0 mm Hg • Pgas = 724 mm Hg • We use this pressure to finish the problem!

5. KClO3 (s) -----------> KCl (s) + 3 O2 (g) • PV/nT = R = .0821 L atm/ K mole • Make sure you convert all units appropriately! • Pgas = 724 mm Hg x 1 atm/760 mm Hg = .953 atm • T = 230C + 273 = 296 K • n = .0159 moles O2 • V = ? • (.953 atm)(x)/(.0159 moles O2)(296 K) = .0821 L atm/ K mole • V = .405L O2

6. Density of gases and Dumas’s Law • Years ago, it was easier to calculate the density of a gas in order to help identify it, rather than figure out moles! Moles were not easy to calculate. • Density = mass/volume • If you look at the ideal gas law: • PV/nT = R • And rearrange: • PV = nRT • n, or moles, is equal to mass/molar mass (Mm) • We know this from the mole map! Old lab – figuring out moles, hard. Figuring out density, easy.

7. Density of gases and Dumas’s Law • PV = nRT • Substituting in mass/Mm for n gives you: • PV = mass x R x T • Mm • Multiplying both sides by Mm gives you: • Mm x P x V = mass x R x T • Dividing both sides by V, you get: • Mm x P = mass x R x T • V • mass/V is density, so this can be simplified into: • Mm x P = D x R x T Jean Baptist Andre’ Dumas

8. Mm x P = D x R x T • This is called Dumas’ Law! • It is more useful sometimes than the traditional gas law because it has density in it, which is easy for a scientist to measure! • Let’s look at a homework problem! • What is the density of elemental oxygen in grams per liter at 250C and .850 atm? • Using Dumas’ Law, we are solving for D • Elemental oxygen, or O2, has a molar mass of 32 g/mole • T needs to be converted into Kelvin – T = 250C + 273 = 298 K • (32 g/mole)(.850 atm) = (D)(.0821 L atm/K mole)(298 K) • Solving, we get D = 1.11 g/L • The only units left are g/L, units for density!