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The Mole

The Mole. The Chemist’s Package. How do we group things?. Pairs, Dozen, ??? Chemists group chemicals into moles . A mole of any chemical is the same amount. Just like a dozen is always 12…. From one to another… Moles to Moles. We can change from one substance to another using a ratio.

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The Mole

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  1. The Mole The Chemist’s Package

  2. How do we group things? • Pairs, • Dozen, • ??? • Chemists group chemicals into moles. • A mole of any chemical is the same amount. • Just like a dozen is always 12…

  3. From one to another… Moles to Moles • We can change from one substance to another using a ratio. • Usually a part to a whole • This ratio is a mole ratio. • It converts from moles of one thing to moles of another. • ____moles X = ____moles Y • Ex. H2O : Part to Whole • 2 moles H = 1 mole H2O • 1 mole O = 1 mole H2O

  4. How many are there? Moles to Particles • Moles can also be converted to tell us how many particles there are. • Particles can be atoms (atom), ions (ion), molecules(molec), and formula units (fmu). • There are always 6.022 x 1023 particles in a mole. • This number is known as Avagadro’s Number. • Ex. H2O: 1 mole H2O = 6.022 x 1023molec H2O • Ex. Na: 1 mole Na = 6.022 x 1023 atoms Na

  5. How much is it? Moles to Grams • The mass of one mole is a chemical substance’s molar mass. • Each substance has its own molar mass that can be determined using the periodic table. • The molar mass of an element is equal to its atomic mass. • Ex. Na: 1 mol Na = 22.99 g Na • The molar mass of a compound is equal to the sum of the atomic masses of the elements that are in the compound. • Ex. H2O: 1 mole H2O = 18.02 g H2O (2H + 1O)

  6. How big is it? Moles to Volume • The volume of a mole of gas under standard conditions is always the same. • This is not true under nonstandard conditions or for liquids and solids. • Standard Conditions of Temperature and Pressure (STP) are: • 273 K and 101.3 kPa • At STP, 1 mole of gas has a volume of 22.4 L. • This is called molar volume. • Ex. O2 at STP: 1 mole O2 = 22.4 L O2 • Ex. CO2 at STP: 1 mole CO2 = 22.4 L CO2

  7. Conversion Factors • These are all conversion factors. • Mole to Mole – Mole Ratio • Mole to Particle – Avagadro’s Number • Mole to Mass – Molar Mass • Mole to Volume – Molar Volume

  8. Mole Map

  9. Practice Problems Mole to Mole • How many moles of carbon are in 2.0 moles of sucrose, C12H22O11? • Known: 2.0 moles sucrose Unknown: ? moles carbon • Conversion Factor: Mole Ratio 1 molC12H22O11 = 12 mol C 2.0 mol C12H22O11 12 mol C = 24 mol C 1 mol C12H22O11 • How many moles of sucrose will contain 12 moles of hydrogen? • K: 12 mol H UK: ? Mol C12H22O11 • CF: Mole Ratio - 22 mol H = 1 mol C12H22O11 12 mol H 1 mol C12H22O11 = 0.55 mol C12H22O11 22 mol H

  10. Practice Problems Mole to Particle • How many moles are equal to 2.41 x 1024 formula units of sodium chloride? • K: 2.41 x 1024fmuNaCl UK: ? mol NaCl • CF: Avagadro’s Number - 1 mol NaCl = 6.022 x 1023fmuNaCl • How many moles are equal to 9.03 x 1024 atoms of mercury? • K: 9.03 x 1024 atoms Hg UK: ? mol Hg • CF: 1 mol Hg = 6.022 x 1023 atoms Hg • How many atoms are equal to 4.50 moles of copper? • K: 4.50 mol Cu UK: ? atoms Cu • CF: 1 mol Cu = 6.022 x 1023 atoms Cu • How many molecules are equal to 100.0 moles of carbon dioxide? • K: 100.0 moles CO2 UK: ? molec CO2 • CF: 1 mol CO2 = 6.022 x 1023molec CO2

  11. Practice Problems Moles to Particles (2 steps) • How many atoms of oxygen are in 3.65 moles of sucrose? • K: 3.65 mol C12H22O11 UK: ? atoms O • CF: Mole Ratio 1 mol C12H22O11 = 11 mol O • CF: Avagadro’s Number 1 mol O= 6.022 x 1023 atoms O  3.65 mol C12H22O11 11 mol O 6.022 x 1023 atoms O = 1 mol C12H22O11 1 mol O = 2.42 x 1025 atoms O • How many atoms are in 1.00 mole of sucrose?

  12. Practice Problems Molar Mass • Determine the molar mass of the following compounds. • Carbon dioxide CO2 : 1 Carbon + 2 Oxygen = 12.01g + 2(16.00 g) = 44.01 g CO2 • Sulfur trioxide SO3 : 1 Sulfur + 3 Oxygen = • Bromine Br2 : 2 Bromine = • Sodium hydroxide NaOH : 1 Sodium + 1 Oxygen + 1 Hydrogen = • Barium nitrate Ba(NO3)2 : 1 Barium + 2 Nitrogen + 6 Oxygen =

  13. Practice Problems Moles to Mass • Calculate the mass in grams of 0.250 moles of the following compounds. K: 0.250 moles UK: ? grams • Sucrose • Sodium chloride • Potassium permanganate • Calcium sulfide • Lithium chlorate

  14. Practice Problems Moles to Mass • Calculate the number of moles in 100.0 grams of the following compounds. K: 100.0 grams UK: ? moles • Sucrose CF: • Sodium chloride CF: • Potassium permanganate CF: • Calcium sulfide CF: • Lithium chlorate CF:

  15. Practice Problems Moles to Volume • What is the volume in liters of 8.35 moles of sulfur trioxide gas at STP? K: 8.35 mol SO3 UK: ? L SO3 CF: 1 mol SO3 = 22.4 L SO3 • What is the volume in liters of 0.750 moles of carbon dioxide gas at STP? K: 0.750 mol CO2 UK: ? L CO2 CF: 1 mol CO2 = 22.4 L CO2 • How many moles are there in 52.5 liters of oxygen gas at STP? K: 52.5 L O2 UK: ? mol O2 CF: 1 mol O2 = 22.4 L O2 • How many moles are there in 15.0 liters of nitrogen gas at STP? K: 15.0 L N2 UK: ? mol N2 CF: 1 mol N2 = 22.4 L N2

  16. Multi-Step Problems • Same conversions… just more than one at a time… • Don’t solve them any differently… • Identify your known and unknown. • Identify the conversion factors you need. • Let the Mole Map do the work for you… • Solve.

  17. Multi-Step Practice • How many grams of carbon dioxide are there in 4.20 L of carbon dioxide at STP? • What is the volume of 245 grams of water vapor at STP? • How many formula units are there in 63.4 grams of barium hydroxide?

  18. Multi-Step Practice • What is the volume of 6.21 x 1031 molecules of sulfur trioxide gas at STP? • What is the mass of 5.23 x 1018 atoms of gold? • How many molecules of silicon dioxide gas does it take to fill a 6.0 L container at STP? 

  19. Multi-Step Practice • How many atoms of carbon are there in 4.24 grams of carbon tetrabromide? • How many liters of carbon dioxide gas are there if you have 4.7 x 1020 atoms of oxygen? • If you have 3.62 x 1024 atoms of nitrogen, then what is the maximum number of grams of dinitrogenpentoxide that can be formed?

  20. Empirical Formulas • Empirical Formulas are the LOWEST whole number ratio of elements in a compound. • Sometimes this is the same as the chemical formula • Examples: • C2H6 → CH3 • C6H12O6 → CH2O • Empirical Formulas can be determined experimentally from the percent composition of a compound.

  21. Empirical Formula Determination STEPS • Assume you have 100 g of the substance (makes the math easier because everything is a straight percent). • Consider the amounts you are given as being in units of grams. • Convert the grams to moles for each element. • Find the smallest whole number ratio of moles for each element by dividing all values by the smallest value.

  22. Example: Find the empirical formula for a compound consisting of 63% Mn and 37% O Solution for Finding the Empirical Formula • Assuming 100 g of the compound, there would be 63 g Mn and 37 g O • Convert grams to moles. (Use molar mass from the periodic table.) • 63 g Mn × (1 mol Mn)/(54.94 g Mn) = 1.1 mol Mn • 37 g O × (1 mol O)/(16.00 g O) = 2.3 mol O • Find the smallest whole number ratio by dividing the number of moles of each element by the number of moles for the element present in the smallest molar amount. (In this case there is less Mn than O, so divide by the number of moles of Mn) • 1.1 mol Mn/1.1 = 1 mol Mn • 2.3 mol O/1.1 = 2.1 mol O • The best ratio is Mn:O of 1:2 and the formula is MnO2The empirical formula is MnO2

  23. Empirical Formula Calculation Worksheet 1. 88.8% Cu and 11.2% O • Assume you have 100 grams. • 88.8 g Cu and 11.2 g O • Convert grams to moles • 88.8g Cu 1 mol Cu = 1.40 mol Cu 63.55g Cu • 11.2g O 1 mol O = 0.700 mol O 16.00g O • Find the smallest whole number ratio. • 1.40 mol Cu / 0.700 = 2 • 0.70 mol O /0.700 = 1 • So the empirical formula is Cu2O.

  24. Molecular Formula from Empirical Formula • The empirical formula is the lowest ratio of elements and not always the actual ratio of elements in a compound. • The actual ratio (molecular formula) can be determined using the molar mass of the actual compound and the molar mass of the empirical formula. • The molar mass of the molecular formula is experimentally determined (Given). • The molar mass of the empirical formula can be determined from the periodic table. • The molecular mass is divided by the empirical mass and equals a whole number. • The whole number is multiplied by the subscripts in the molecular formula to give the empirical formula.

  25. Example: Find the molecular formula for a compound that has an empirical formula of CH2 and a molecular mass of 41.5g. • The molecular mass is 41.5 grams. • Given • The empirical mass is 14.03 grams. • C + 2H = 12.01 + 2(1.008) = (14.026)14.03 • 41.5 / 14.03 = (2.96) 3 • 3 (CH2) = C3H6 • The molecular formula is C3H6 .

  26. Molecular Formula Practice Use Empirical Formulas from Worksheet with the following Molecular Masses. • 290 g/mol • 178 g/mol • 26 g/mol • 160 g/mol • 34 g/mol • 160 g/mol • 138 g/mol • 240 g/mol • 245 g/mol • 230 g/mol

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