0:02 hi everyone in previous videos we
0:04 learned how to think about a balanced
0:05 chemical equation and these
0:07 stoichiometric factors that we can write
0:09 from that equation we've also learned
0:12 how to do simple calculations to convert
0:14 from one species to another in a
0:17 reaction and this video is going to do a
0:19 little bit more practice performing
0:21 stoichiometric calculations involving
0:25 Mass moles and solution molarity
0:27 so let's look at some examples first we
0:30 have a question asking what mass of
0:33 sodium hydroxide NaOH would be required
0:37 to produce 16 grams of the antacid milk
0:39 of magnesia so this is magnesium
0:42 hydroxide by the following reaction and
0:45 so we're given the balanced chemical
0:50 equation and we want to go from the mass
0:52 of the antacid right the milk of
0:56 magnesia to a mass of sodium hydroxide
1:00 right so how much sodium hydroxide would
1:03 we need to produce these 16 grams of the
1:05 magnesium hydroxide
1:08 now since both of these species are
1:10 involved in the chemical equation we
1:13 know that we can relate the moles of
1:16 magnesium hydroxide to the number of
1:19 moles of sodium hydroxide right using
1:23 the Stoichiometry of the equation but
1:26 this question is asking us to relate the
1:28 mass of these two species so if we start
1:31 with a certain mass of magnesium
1:34 hydroxide we would first need to figure
1:37 out how much in terms of the number of
1:39 moles we have of magnesium hydroxide
1:42 before we then can relate the moles of
1:45 magnesium hydroxide to moles of sodium
1:48 hydroxide and then finally since we're
1:51 looking for the mass of sodium hydroxide
1:54 we would need to use the molar mass of
1:58 that species to figure out the mass in
2:00 that certain number of moles so let's do
2:02 this problem all right so we said that
2:07 we have a desire to produce 6 15 grams
2:11 of the antacid milk of magnesia so this
2:19 all right and according to our flow
2:22 diagram what we first want to do is
2:24 convert from grams of magnesium
2:27 hydroxide to moles of magnesium
2:30 hydroxide and to do this I need to use
2:35 I can see from where I've started right
2:38 I have grams of magnesium hydroxide I'm
2:41 going to want to have that cancel so
2:44 I'll want grams in my denominator so it
2:53 grams of magnesium hydroxide in one mole
2:59 all right so grams of magnesium
3:02 hydroxide are going to cancel and now
3:05 we're in moles so now we can use our
3:10 now there's different ways for us to
3:11 write our stoichiometric Factor right
3:14 depending on which species we put in the
3:16 numerator in which in the denominator
3:19 here I can see that all want moles of
3:22 magnesium hydroxide to cancel so I'll
3:25 want that species to be in the denominator
3:32 and I'm trying to relate this to the
3:35 number of moles of sodium hydroxide so
3:41 and then I need to look at the
3:43 coefficients for my balanced chemical
3:46 equation I can see that the coefficient
3:48 for the magnesium hydroxide is one
3:50 whereas the coefficient for the sodium
3:53 hydroxide is 2. so that gives me the
3:55 numbers to put into my stoichiometric Factor
3:56 Factor
3:59 all right so now we've canceled moles of
4:04 magnesium hydroxide and we are now in
4:06 moles of sodium hydroxide so it looks
4:08 like we have one more step the question
4:11 is asking what mass of sodium hydroxide
4:13 would be required
4:16 so we would need to convert from moles
4:19 of sodium hydroxide to a mass and here
4:22 I'm going to need to use my molar mass
4:25 for sodium hydroxide in this case I want
4:27 moles to cancel right since I have moles
4:29 here in my numerator I'm going to put
4:32 moles of NaOH
4:38 in my denominator so one mole of NaOH is
4:47 all right so now moles of sodium
4:50 hydroxide will cancel and the number
4:53 that I calculate will have the units of
4:57 grams of sodium hydroxide
4:58 all right I'm putting these numbers into
5:02 a calculator I calculated that 22 grams
5:05 of sodium hydroxide would be required to
5:07 produce the 16 grams of magnesium hydroxide
5:14 there's a nice flow diagram in your
5:16 textbook which helps us think about
5:18 these more complex calculations
5:20 involving stoichiometric factors
5:22 so as we can see in the middle of the
5:26 flow diagram we're able to relate A and
5:28 B right if these are substances involved
5:31 in a chemical reaction and we have a
5:33 balanced chemical equation for the
5:36 reaction we should be able to relate the
5:38 number of moles of a to the number of
5:41 moles of B using this stoichiometric
5:44 Factor right based on the coefficients
5:46 of these substances in the balanced
5:47 chemical equation
5:49 but often in the laboratory we don't
5:53 know how many moles we have instead we
5:55 measure other parameters right so we
5:59 might measure the mass of substance a
6:01 and so just as we did in the last
6:04 problem we would need to use the molar
6:07 mass to convert from the mass of a to moles
6:09 moles
6:12 another parameter we might measure
6:16 instead of mass would be the volume of a
6:18 pure substance so say we have a liquid
6:21 substance that's pure we can measure its
6:24 volume with a graduated cylinder then we
6:27 would need to figure out how we could
6:29 get from the volume of the substance to
6:32 moles and what we would need to use here
6:34 would be the density so density would
6:37 help us to go from a volume to a mass
6:39 and then once we have the mass we can go
6:43 to moles using the molar mass
6:46 another parameter we might have instead
6:50 of moles would be the volume of a
6:53 solution that contains substance a so
6:56 for example maybe we have an aqueous
7:00 solution containing the solute a we know
7:02 that we have a certain volume of that
7:04 solution how would we calculate the
7:07 number of moles of a well to do that we
7:10 would use the volume and the molarity in
7:12 particular the volume times the molarity
7:15 would help us to calculate the moles of
7:17 that substance as a solute in that solution
7:18 solution
7:20 and then another parameter that you may
7:23 see in your calculations would be
7:26 instead of having moles of a we may have
7:30 the number of particles of a so we know
7:31 that we can convert from numbers of
7:35 particles to moles using Avogadro's
7:38 number and so once we use this we can
7:41 then relate the moles of a using the
7:45 stoichiometric factor to moles of B
7:48 so let's try one more example to make
7:50 use of what we learned from the flow
7:52 diagram so in this case we have the
7:54 information that Automotive airbags
7:58 inflate when a sample of sodium azide
8:01 nan3 is very rapidly decomposed and we
8:03 have our balanced chemical equation
8:06 right so we have two moles of sodium
8:09 azide decomposed to give two moles of
8:12 sodium which is solid and three moles of
8:14 nitrogen gas
8:17 so our question is what mass of sodium
8:21 azide is required to produce this is 2.6
8:26 cubic feet or 73.6 liters of nitrogen
8:30 gas with a density of 1.25 grams per liter
8:32 liter
8:33 all right so we can see that we're
8:36 trying to relate the nitrogen gas which
8:40 we see here as a product to the mass of
8:44 sodium azide the reactant that we start with
8:45 with
8:49 all right so instead of moles here we
8:51 have other parameters right we want to
8:53 find the mass of sodium azide required
8:57 and here in terms of the nitrogen gas we
9:01 want to produce 73.6 liters so a volume
9:03 so where to start in this question we
9:04 need to start with a specific piece of
9:07 information that we're given how much we
9:10 want to produce in this case
9:18 and let's reference the flow diagram
9:21 that we saw on the previous slide
9:24 so if we know a volume right that puts
9:27 us up here on the flow diagram we can
9:30 relate the volume to the mass of that
9:32 substance using the density and then
9:33 once we have the mass we can use the
9:35 molar mass to get moles of that substance
9:41 all right so our flow diagram told us to
9:43 use the density
9:45 so we're given the density here it is
9:48 1.25 grams per liter
9:50 now I'm going to want to have liters
9:52 cancel so I'll put that in the
9:55 denominator and grams in my numerator
9:59 and I have 1.25 grams for every liter so
10:03 now liters of nitrogen cancel this is
10:06 grams of nitrogen
10:08 all right and then our flow diagram said
10:11 once we have grams right we can use the
10:14 molar mass to convert to moles
10:16 so the molar mass of this substance
10:20 right this is nitrogen is going to be 28
10:23 grams per mole now I want grams to
10:26 cancel so I'll put grams in the denominator
10:32 so we have 28 grams of nitrogen for
10:41 so now grams of nitrogen cancel and I am
10:43 in moles of nitrogen so now that I'm in
10:46 moles I can use my stoichiometric factor
10:49 to convert from moles of nitrogen to
10:52 moles of sodium azide
10:54 from my calculation I can see that I
10:56 want moles of nitrogen to be in the denominator
11:02 and so then I want moles
11:05 of the sodium azide
11:08 to be in my numerator looking at my
11:10 balanced chemical equation I can see
11:12 that the coefficient for nitrogen is
11:15 three so I'll put that down here and the
11:17 coefficient for the sodium azide is two
11:20 so that'll go up in the numerator
11:23 now moles of nitrogen can cancel and I
11:26 am in moles of sodium azide let's look
11:28 at my question again it says what mass
11:31 of sodium azide is required right so now
11:33 that we're in moles of sodium azide it
11:35 looks like we need to finish the problem
11:39 by converting to Mass
11:41 so for this we can use our molar mass
11:45 the molar mass of sodium azide is 65
11:48 grams per mole it looks like I need
11:50 moles to cancel so I'll put that piece
11:58 so moles of sodium azide goes in the
12:00 denominator and the mass of that mole
12:08 and moles of sodium azide cancel
12:10 and we can plug this into our calculator
12:14 to get the number 142 my unit should be
12:16 grams and this would be of the substance
12:21 sodium azide my reactant so when 142
12:24 grams of sodium azide react they
12:27 generate 73.6 liters of nitrogen gas
12:29 that's quite a large volume so that's
12:32 going to inflate your airbag and protect
