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Stoichiometry. Lets Review- Moles and Particles. 1 mole = 6.022 x 10 23 particles (formula units/ molecules/atoms). Multiply by 6.022 x 10 23. Moles. Moles. Particles. Divide by 6.022 x 10 23. Particles. Another Way to Look at It:.
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Lets Review- Moles and Particles • 1 mole = 6.022 x 1023 particles (formula units/ molecules/atoms) Multiply by 6.022 x 1023 Moles Moles Particles Divide by 6.022 x 1023 Particles
Another Way to Look at It: • N = Number of particles (molecules, formula units, atoms) • n = amount (in mol) • NA = Avagadro constant (mol-1) N = n x NA
For Example • A sample contains 1.25 mol of nitrogen dioxide. • How many molecules are in the sample? Molecules of NO2 = 1.25 mol x (6.022 x 1023 molecules/mol) = 7.52 x 1023 molecules • How many atoms are in the sample? 7.52 x 1023 molecules x (3 atoms/ molecule) = 2.26 x 1024 atoms in 1.25 mol of NO2 N = n x NA There is 1 atom of nitrogen and 2 atoms of oxygen in every molecule of NO2
Lets Review- Molar Mass • A mole of an element has a mass in grams that is numerically equivalent to the elements average atomic mass. • Molar Mass (M)= mass of one mole of a substance (g/mol) • Molar mass of an element can be found by looking at the periodic table • Molar mass of a compound can be found by totalling the mass of all elements in the compound • What is the molar mass of beryllium oxide? MBeO = MBe + MO = 9.01 g/mol + 16.00 g/mol =25.01 g/mol
Particles to Mass • What is the mass of 0.750 mol of CO2 gas? • 1. MCO2 = 2 x (16.00g/mol) + 12.01 g/mol = 44.01 g/mol • 2. m = (0.750 mol)(44.01 g/mol) = 33.0g Divide by 6.022 x 1023 Multiply by M Mass Moles Divide by M Multiply by 6.022 x 1023 Particles
Particles to Mass • How many molecules of iodine chloride, ICl, are in a 2.74 x 10-1 g sample? • 1. MICl =126.90 g/mol + 35.45 g/mol = 162.36 g/mol • 2. n = (2.74 x 10-1 g)/(162.36 g/mol) = 1.69 x 10-3 mol • 3. (1.69 x 10-3 mol)(6.022 x 1023 molecules/mol) = 1.01 x 1021 molecules Multiply by M Divide by 6.022 x 1023 Mass Moles Divide by M Multiply by 6.022 x 1023 Particles
Pressure • What is pressure? • Collisions • Depend on: • Temperature • Volume • PV=nRT
Standard Temperature and Pressure • Since V depends on P and T, you must state both when describing a gases V. • STP: • Average atmospheric pressure at sea level (101.3 kPa) • Freezing point of water (OoC, 273K)
Molar Volume • Joseph Louis Gay-Lussac (1778-1853) • Law of combining volumes: when gases react the volume of reactants and products (measured at equal T and P) are always whole number ratios + 2L H2 1L O2 2L H2O
Molar Volume • Amedeo Avogadro (1776-1856) • Realized he could relate the volume of a gas to the amount that was present (from mass) • Avogadro’s Hypothesis: equal volumes of all ideal gases at the same temperature and pressure contain the same number of molecules • In other words: 1 mole of any gas has the same volume as 1 mole of any other gas at STP.
Thought Lab- Molar Volume of Gases • Two students in a lab decided to calulate the molar volume of carbon dioxide, oxygen and methane gas. They measured the mass of an empty,150 mL syringe and then the mass of the syringe + gas. They repeated this procedure for each gas. • The experiment was carried out in a room maintained at STP (273 K and 101.3 kPa). Their results are in the table on the next slide:
Molar Volume • It turns out that the molar volume of any gas at STP is 22.4 L/mol! • Well, actually, that’s not entirely true…. • The molar volume of 22.4 L/mol is assumed for ideal gases- which are hypothetical gases that don’t take up space and do not attract one another • Real gases do take up space and attract each other somewhat- so the molar volume would vary slightly from 22.4 L/mol….but it’s close • We will always assume the molar volume of a gas is 22.4L/mol at STP!
Example: • What is the volume of 3.0 mol of nitrogen dioxide gas at STP?
Try This: • Suppose you have 44.8 L of methane gas at STP. • 1. How many moles are present? • 2. What is the mass of the gas? • 3. How many molecules of the gas is present?