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SKILLS Project

SKILLS Project. Mole/Mass Conversions. Mole/Mass Conversions. This unit is devoted specifically to making conversions between moles and masses.

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SKILLS Project

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  1. SKILLS Project Mole/Mass Conversions

  2. Mole/Mass Conversions • This unit is devoted specifically to making conversions between moles and masses. • So, why should we be focusing on this so much? Simple- chemical equations use moles to show the relationships between different substances. Unfortunately, scales don’t measure in terms of moles. • Instead, DA allows us to convert between moles and mass, allowing us to work with chemical equations in the real-world. • Typically, the goal of these conversions predict how much of a substance will be made in a chemical equation, etc.

  3. Note! • If you are having trouble remembering how to format DAs, you should work through the “Basic Dimensional Analysis” and “Molar Mass” presentations prior to reading this one.

  4. Setting Up Molar Conversions • To convert between moles and grams, you follow the same rules as you would any normal DA. • Two types of conversion factors are used here: • g/mol conversions use the molar mass of a substance on the periodic table. • 1 mol = X g • Mol/mol conversions use the balanced chemical equation to show the relationships between different substances. • X mol = Y mol

  5. Example 1: g  mol • Convert 26.7 g of salt, NaCl, to moles. 26.7 mol NaCl ____ 26.7 g NaCl 1 mol NaCl x ___________________________________ 58.44 g NaCl 58.44 Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “mol NaCl” at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: 58.44 g NaCl = 1 mol NaCl (Molar Mass) 0.457 mol NaCl To start this problem, place “26.7 g NaCl” at the beginning of the DA. We will use the molar mass to find moles.

  6. Example 2: mol  g • What is the mass of 5.40 mol of potassium nitrate? 546 g KNO3 ____ 5.40 mol KNO3 101.11 g KNO3 x ___________________________________ 1 mol KNO3 1 Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “g KNO3” at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: 1 mol KNO3 = 101.11 g KNO3 (Molar Mass) 546 g KNO3 To start this problem, place “5.40 mol KNO3” at the beginning of the DA. We will use the molar mass to find grams.

  7. Example 3: g  mol • Convert 73.9 g of hydrogen sulfide, H2S , to moles. 73.9 mol H2S ____ 73.9 g H2S 1 mol H2S x ___________________________________ 34.09 g H2S 34.09 Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “mol H2S” at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: 34.09 g H2S = 1 mol H2S (Molar Mass) 2.17 mol H2S To start this problem, place “73.9 g H2S” at the beginning of the DA. We will use the molar mass to find moles.

  8. Example 4: mol  g • How many grams could be made from 1.43e4 mol LiOH? 3.42e5 g LiOH ____ 1.43e4 mol LiOH 23.95 g LiOH x ___________________________________ 1 mol LiOH 1 Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “g LiOH” at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you know that: 1 mol LiOH = 23.95 g LiOH (Molar Mass) 3.42e5 g LiOH To start this problem, place “1.43e4 mol LiOH” at the beginning of the DA. We will use the molar mass to find grams.

  9. Practice on Your Own • 29.2 g K2O  mol • 1.39 g AgNO3  mol • 4.10e4 g PbSO4  mol • 74.6 g NH4C2H3O2  mol • 990.3 g Ca3(PO4)2  mol 0.310 mol K2O g  mol 8.18e-3 mol AgNO3 135 mol PbSO4 0.968 mol NH4C2H3O3 2.444 mol Ca3(PO4)2

  10. Practice on Your Own • 1.21 mol YSO3 g • 24.9 mol U3O8  g • 4.24e45 mol NF5  g • 8.32 mol Te2At7  g • 4.10 mol Cr2O72-  g 204 g YSO3 mol  g 2.10e4 g U3O8 4.62e47 g NF5 1.44e4 g Te2At7 886 g Cr2O72-

  11. So Far: • We’ve covered the basics behind converting between moles and grams, however, you can do much more. • Moles act as a gateway to many other measurements, including molecules, atoms, liters, and even the moles of other substances. • DAs are only limited to the conversion factors they contain. If you have the conversions, a DA can be as long or short as you like and end in any answer you like.

  12. Example 5: g  atoms • How many copper atoms would be found in 835 g of Cu? ___________________________________ ____ ____ 835 g Cu 6.022e23 atoms Cu x x 5.03e26 atoms Cu 1 mol Cu 63.55 x 63.55 g Cu 1 mol Cu Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “atoms Cu” at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you need to know that: 63.55 g Cu = 1 mol Cu (Molar mass) 1 mol = 6.022e23 atoms (Avogadro’s number) To start this problem, place the “835 g Cu” at the beginning. We’ll turn grams into moles and moles into atoms. 7.91e24 atoms Cu

  13. Example 6: atoms  g • What is the mass of 5.31e20 atoms of Lead (III) arsenate? ___________________________________ ____ ____ x x 5.31e20 atoms PbAsO4 1.84e24 atoms PbAsO4 1 mol PbAsO4 346.12 g PbAsO4 x 6.022e23 6.022e23 atoms PbAsO4 1 mol PbAsO4 Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “g PbAsO4 at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you need to know that: 6.022e23 atoms = 1 mol (Avogadro’s number) 1 mole PbAsO4 = 346.12 g PbAsO4 (Molar mass) 0.305 g PbAsO4 To start this problem, place the “5.31e20 atoms PbAsO4” at the beginning. We’ll turn atoms into moles and moles into grams.

  14. Example 7: grams  molecules • How many molecules of lithium phosphide would be found in 20.4 g of Li3P? x ___________________________________ ____ x ____ 20.4 g Li3P 6.022e23 molecules Li3P 1 mol Li3P 1.23e25 molecules Li3P x 51.79 51.79 g Li3P 1 mol Li3P To start this problem, place the “20.4 g Li3P ” at the beginning. We’ll turn grams into moles and moles into molecules. Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “molecules Li3P” at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you need to know that: 51.79 g Li3P = 1 mol Li3P (Molar mass) 1 mol = 6.022e23 atoms (Avogadro’s number) 2.37e23 molecules Li3P

  15. Example 8: molecules  g • What is the mass of 7.78e35 molecules of water? ___________________________________ ____ ____ x x 7.78e35 molecules H2O 1.40e37 g H2O 1 mol H2O 18.02 g H2O x 6.022e23 6.022e23 molecules H2O 1 mol H2O Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “g H2O at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. To complete this problem, you need to know that: 6.022e23 molecules = 1 mol (Avogadro’s number) 1 mole H2O = 18.02 g H2O(Molar mass) 2.33e13 g H2O To start this problem, place the “7.78e35 molecules H2O” at the beginning. We’ll turn molecules into moles and moles into grams.

  16. Practice on Your Own • 2.57e23 molecules Th2O  g • 8.310e12 molecules NiO3  g • 1.72e83 atoms Ir  g • 5.45e24 molecules Al4C3  g • 7.72e30 atoms Hg  g 205 g Th2O particles  g 1.472e-9 g NiO3 5.49e61 g Ir 1.30e3 g Al4C3 2.57e9 g Hg

  17. Practice on Your Own • 45.6 g BH4 molecules • 20.2 g Ti  atoms • 7.04e-19 g H2CO3  molecules • 33.52 g NaOH  molecules • 4.78 g Ba2+  atoms 1.85e24 molecules BH4 g  particles 2.54e23 atoms Ti 6830 molecules H2CO3 5.046e23 molecules NaOH 2.10e22 atoms Ba2+

  18. Congratulations! • You have mastered the ability to convert between moles, mass, and particles. • Remember, this skill will allow you to use chemistry in the lab, not just on paper. • Later, you will learn to pair this skill with chemical reactions to perform more advanced (and useful!) calculations. • Good luck and keep on going!

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