reminder: 1 mol of anything e.g. oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

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Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23. Let’s look at this pile of carbon in the form of charcoal. I bet you there is a bunch of carbon moles in there. Let’s see how much. - PowerPoint PPT Presentation

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Page 1: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23
Page 2: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 1023

Page 3: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

Let’s look at this pile of carbon in the form of charcoal. I bet you there is a bunch of carbon moles in there. Let’s see how much.

Page 4: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

Now, this clump represents one mole. Can you estimate how many do you see in the whole bunch? Well, it’s hard to visualize. Let’s do some math.

Since this is 1 mole, that means it weighs 12.01 g. Why? Because, according to the periodic table, 1 mole of Carbon weighs 12.01 g. Well, how many moles are present in this whole bunch? Let’s weigh it first.

Page 5: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

0.00 g12.01 g6.02 x 1023

Page 6: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

0.00 g

So, we weigh the whole bunch and it turns out to be 15.50 g.

15.50 g

Page 7: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

1 mol

? mol

Page 8: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

So, there are two ways to solve this problem:

a) Use proportions

b) Use the n= m / mMm

a) Use proportions:

Since 1 mol of C = 12.01 g

? mol = 15.50 g

? mol = 1 mol of C x 15.50 g12.01 g

= 1.29 mol

Page 9: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

b) Let’s use the formula n= m / mMm

Step 1: What do we have?

m: 15.50 g

mMm: 12.01 g/mol

N: ?

Step 2: You can use a triangle

m n: m . mMm

n x mMm n: 15.50 g 12.01 g/mol

n: 1.29 mol

Page 10: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

Another example: let’s say there are 4 moles of water in this beaker. How many grams would that be?

Page 11: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

Well, there are a couple of steps. We are trying to find the mass of 4 moles of water, we need the molar mass of water.

Step 1: Find molar mass

H2O: 2(1) + 16 = 18 g/mol

Step 2: Let’s find the mass, I’ll use the proportions method

1 mol of H2O = 18 g

4 mol of H2O = ? g

? g = 4 mol of H2O x 18 g/mol 1 mol of H2O

? g = 72 g

What about the number of molecules?

Page 12: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

Well, for the number of molecules, let’s see what happens.

1 mol of H2O = 6.02 x 1023 molecules of H2O

4 mol of H2O = ? molecules

? molecules = 4 mol of H2O x (6.02 x 1023 molecules of H2O) 1 mol of H2O

= 2.408 x 1024 molecules of H2O

Page 13: Reminder: 1 mol of anything e.g. Oxygen, carbon, dollars, toys, etc. is 6.02 x 10 23

Okay, solve the following problems:

a) How many moles of Ca(OH)2 are present in a 200 g sample of Ca(OH)2?

b) How many molecules of MgF2 are present in a 100 g sample of MgF2?