# SKILLS Project Mole/Mass Conversions. This unit is devoted specifically to making conversions between moles and masses. So, why should we be focusing.

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SKILLS Project 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.

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.

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

To complete this problem, you know that: 58.44 g NaCl = 1 mol NaCl (Molar Mass) Everything checks out, so multiply the top and divide by the bottom to find your answer. ___________________________________ 1 mol NaCl 26.7 mol NaCl 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. To start this problem, place “26.7 g NaCl” at the beginning of the DA. We will use the molar mass to find moles. 58.44 g NaCl Example 1: g  mol Convert 26.7 g of salt, NaCl, to moles. _ _ _ _ 26.7 g NaCl x 58.44 0.457 mol NaCl

Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “g KNO 3 ” at the end. Everything checks out, so multiply the top and divide by the bottom to find your answer. 546 g KNO 3 To start this problem, place “5.40 mol KNO 3 ” at the beginning of the DA. We will use the molar mass to find grams. 1 To complete this problem, you know that: 1 mol KNO 3 = 101.11 g KNO 3 (Molar Mass) ___________________________________ 101.11 g KNO 3 546 g KNO 3 1 mol KNO 3 Example 2: mol  g What is the mass of 5.40 mol of potassium nitrate? _ _ _ _ 5.40 mol KNO 3 x

To complete this problem, you know that: 34.09 g H 2 S = 1 mol H 2 S (Molar Mass) Everything checks out, so multiply the top and divide by the bottom to find your answer. ___________________________________ 1 mol H 2 S 73.9 mol H 2 S Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “mol H 2 S” at the end. To start this problem, place “73.9 g H 2 S” at the beginning of the DA. We will use the molar mass to find moles. 34.09 g H 2 S Example 3: g  mol Convert 73.9 g of hydrogen sulfide, H 2 S, to moles. _ _ _ _ 73.9 g H 2 S x 34.09 2.17 mol H 2 S

To complete this problem, you know that: 1 mol LiOH = 23.95 g LiOH (Molar Mass) Everything checks out, so multiply the top and divide by the bottom to find your answer. 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. 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. 1 ___________________________________ 23.95 g LiOH 3.42e5 g LiOH 1 mol LiOH Example 4: mol  g How many grams could be made from 1.43e4 mol LiOH? _ _ _ _ 1.43e4 mol LiOH x

Practice on Your Own 1.29.2 g K 2 O  mol 2.1.39 g AgNO 3  mol 3.4.10e4 g PbSO 4  mol 4.74.6 g NH 4 C 2 H 3 O 2  mol 5.990.3 g Ca 3 (PO 4 ) 2  mol g  m o l 0.310 mol K 2 O 8.18e-3 mol AgNO 3 135 mol PbSO 4 0.968 mol NH 4 C 2 H 3 O 3 2.444 mol Ca 3 (PO 4 ) 2

Practice on Your Own 1.1.21 mol YSO 3  g 2.24.9 mol U 3 O 8  g 3.4.24e45 mol NF 5  g 4.8.32 mol Te 2 At 7  g 5.4.10 mol Cr 2 O 7 2-  g m o l  g 204 g YSO 3 2.10e4 g U 3 O 8 4.62e47 g NF 5 1.44e4 g Te 2 At 7 886 g Cr 2 O 7 2-

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.

63.55 g Cu 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) ___________________________________ 63.55 7.91e24 atoms 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. 1 mol Cu 5.03e26 atoms Cu Example 5: g  atoms How many copper atoms would be found in 835 g of Cu? _ _ _ _ _ _ _ _ To start this problem, place the “835 g Cu” at the beginning. We’ll turn grams into moles and moles into atoms. 835 g Cu 1 mol Cu 6.022e23 atoms Cu x x x

1.84e24 atoms PbAsO 4 ___________________________________ To complete this problem, you need to know that: 6.022e23 atoms = 1 mol (Avogadro’s number) 1 mole PbAsO 4 = 346.12 g PbAsO 4 (Molar mass) Everything checks out, so multiply the top and divide by the bottom to find your answer. To start this problem, place the “5.31e20 atoms PbAsO 4 ” at the beginning. We’ll turn atoms into moles and moles into grams. Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “g PbAsO 4 at the end. 1 mol PbAsO 4 Example 6: atoms  g What is the mass of 5.31e20 atoms of Lead (III) arsenate? _ _ _ _ _ _ _ _ 5.31e20 atoms PbAsO 4 6.022e23 atoms PbAsO 4 1 mol PbAsO 4 346.12 g PbAsO 4 x x x 6.022e23 0.305 g PbAsO 4

2.37e23 molecules Li 3 P To complete this problem, you need to know that: 51.79 g Li 3 P = 1 mol Li 3 P (Molar mass) 1 mol = 6.022e23 atoms (Avogadro’s number) To start this problem, place the “20.4 g Li 3 P ” 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 Li 3 P” at the end. 51.79 g Li 3 P Everything checks out, so multiply the top and divide by the bottom to find your answer. ___________________________________ 51.79 1 mol Li 3 P 1.23e25 molecules Li 3 P Example 7: grams  molecules How many molecules of lithium phosphide would be found in 20.4 g of Li 3 P? _ _ _ _ _ _ _ _ 20.4 g Li 3 P 1 mol Li 3 P 6.022e23 molecules Li 3 P x x x

___________________________________ 1.40e37 g H 2 O To complete this problem, you need to know that: 6.022e23 molecules = 1 mol (Avogadro’s number) 1 mole H 2 O = 18.02 g H 2 O (Molar mass) Everything checks out, so multiply the top and divide by the bottom to find your answer. To start this problem, place the “7.78e35 molecules H 2 O” at the beginning. We’ll turn molecules into moles and moles into grams. Check your work and cancel out units. If you’ve set up this DA correctly, we should be left with “g H 2 O at the end. 1 mol H 2 O Example 8: molecules  g What is the mass of 7.78e35 molecules of water? _ _ _ _ _ _ _ _ 7.78e35 molecules H 2 O 6.022e23 molecules H 2 O 1 mol H 2 O 18.02 g H 2 O x x x 6.022e23 2.33e13 g H 2 O

Practice on Your Own 1.2.57e23 molecules Th 2 O  g 2.8.310e12 molecules NiO 3  g 3.1.72e83 atoms Ir  g 4.5.45e24 molecules Al 4 C 3  g 5.7.72e30 atoms Hg  g p a r t i c l e s  g 205 g Th 2 O 1.472e-9 g NiO 3 5.49e61 g Ir 1.30e3 g Al 4 C 3 2.57e9 g Hg

Practice on Your Own 1.45.6 g BH 4  molecules 2.20.2 g Ti  atoms 3.7.04e-19 g H 2 CO 3  molecules 4.33.52 g NaOH  molecules 5.4.78 g Ba 2+  atoms g  p a r t i c l e s 1.85e24 molecules BH 4 2.54e23 atoms Ti 6830 molecules H 2 CO 3 5.046e23 molecules NaOH 2.10e22 atoms Ba 2+

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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