hi everyone in previous videos we learned how to think about a balanced chemical equation and these stoichiometric factors that we can write from that equation we've also learned how to do simple calculations to convert from one species to another in a reaction and this video is going to do a little bit more practice performing stoichiometric calculations involving Mass moles and solution molarity so let's look at some examples first we have a question asking what mass of sodium hydroxide NaOH would be required to produce 16 grams of the antacid milk of magnesia so this is magnesium hydroxide by the following reaction and so we're given the balanced chemical equation and we want to go from the mass of the antacid right the milk of magnesia to a mass of sodium hydroxide right so how much sodium hydroxide would we need to produce these 16 grams of the magnesium hydroxide now since both of these species are involved in the chemical equation we know that we can relate the moles of magnesium hydroxide to the number of moles of sodium hydroxide right using the Stoichiometry of the equation but this question is asking us to relate the mass of these two species so if we start with a certain mass of magnesium hydroxide we would first need to figure out how much in terms of the number of moles we have of magnesium hydroxide before we then can relate the moles of magnesium hydroxide to moles of sodium hydroxide and then finally since we're looking for the mass of sodium hydroxide we would need to use the molar mass of that species to figure out the mass in that certain number of moles so let's do this problem all right so we said that we have a desire to produce 6 15 grams of the antacid milk of magnesia so this is magnesium hydroxide all right and according to our flow diagram what we first want to do is convert from grams of magnesium hydroxide to moles of magnesium hydroxide and to do this I need to use my molar mass I can see from where I've started right I have grams of magnesium hydroxide I'm going to want to have that cancel so I'll want grams in my denominator so it is 58.3 grams of magnesium hydroxide in one mole of magnesium hydroxide all right so grams of magnesium hydroxide are going to cancel and now we're in moles so now we can use our stoichiometric Factor now there's different ways for us to write our stoichiometric Factor right depending on which species we put in the numerator in which in the denominator here I can see that all want moles of magnesium hydroxide to cancel so I'll want that species to be in the denominator and I'm trying to relate this to the number of moles of sodium hydroxide so that would be in the numerator and then I need to look at the coefficients for my balanced chemical equation I can see that the coefficient for the magnesium hydroxide is one whereas the coefficient for the sodium hydroxide is 2. so that gives me the numbers to put into my stoichiometric Factor all right so now we've canceled moles of magnesium hydroxide and we are now in moles of sodium hydroxide so it looks like we have one more step the question is asking what mass of sodium hydroxide would be required so we would need to convert from moles of sodium hydroxide to a mass and here I'm going to need to use my molar mass for sodium hydroxide in this case I want moles to cancel right since I have moles here in my numerator I'm going to put moles of NaOH in my denominator so one mole of NaOH is 40.0 grams of NaOH all right so now moles of sodium hydroxide will cancel and the number that I calculate will have the units of grams of sodium hydroxide all right I'm putting these numbers into a calculator I calculated that 22 grams of sodium hydroxide would be required to produce the 16 grams of magnesium hydroxide there's a nice flow diagram in your textbook which helps us think about these more complex calculations involving stoichiometric factors so as we can see in the middle of the flow diagram we're able to relate A and B right if these are substances involved in a chemical reaction and we have a balanced chemical equation for the reaction we should be able to relate the number of moles of a to the number of moles of B using this stoichiometric Factor right based on the coefficients of these substances in the balanced chemical equation but often in the laboratory we don't know how many moles we have instead we measure other parameters right so we might measure the mass of substance a and so just as we did in the last problem we would need to use the molar mass to convert from the mass of a to moles another parameter we might measure instead of mass would be the volume of a pure substance so say we have a liquid substance that's pure we can measure its volume with a graduated cylinder then we would need to figure out how we could get from the volume of the substance to moles and what we would need to use here would be the density so density would help us to go from a volume to a mass and then once we have the mass we can go to moles using the molar mass another parameter we might have instead of moles would be the volume of a solution that contains substance a so for example maybe we have an aqueous solution containing the solute a we know that we have a certain volume of that solution how would we calculate the number of moles of a well to do that we would use the volume and the molarity in particular the volume times the molarity would help us to calculate the moles of that substance as a solute in that solution and then another parameter that you may see in your calculations would be instead of having moles of a we may have the number of particles of a so we know that we can convert from numbers of particles to moles using Avogadro's number and so once we use this we can then relate the moles of a using the stoichiometric factor to moles of B so let's try one more example to make use of what we learned from the flow diagram so in this case we have the information that Automotive airbags inflate when a sample of sodium azide nan3 is very rapidly decomposed and we have our balanced chemical equation right so we have two moles of sodium azide decomposed to give two moles of sodium which is solid and three moles of nitrogen gas so our question is what mass of sodium azide is required to produce this is 2.6 cubic feet or 73.6 liters of nitrogen gas with a density of 1.25 grams per liter all right so we can see that we're trying to relate the nitrogen gas which we see here as a product to the mass of sodium azide the reactant that we start with all right so instead of moles here we have other parameters right we want to find the mass of sodium azide required and here in terms of the nitrogen gas we want to produce 73.6 liters so a volume so where to start in this question we need to start with a specific piece of information that we're given how much we want to produce in this case 73.6 liters of nitrogen gas and let's reference the flow diagram that we saw on the previous slide so if we know a volume right that puts us up here on the flow diagram we can relate the volume to the mass of that substance using the density and then once we have the mass we can use the molar mass to get moles of that substance all right so our flow diagram told us to use the density so we're given the density here it is 1.25 grams per liter now I'm going to want to have liters cancel so I'll put that in the denominator and grams in my numerator and I have 1.25 grams for every liter so now liters of nitrogen cancel this is grams of nitrogen all right and then our flow diagram said once we have grams right we can use the molar mass to convert to moles so the molar mass of this substance right this is nitrogen is going to be 28 grams per mole now I want grams to cancel so I'll put grams in the denominator so we have 28 grams of nitrogen for every one mole of nitrogen and two so now grams of nitrogen cancel and I am in moles of nitrogen so now that I'm in moles I can use my stoichiometric factor to convert from moles of nitrogen to moles of sodium azide from my calculation I can see that I want moles of nitrogen to be in the denominator and so then I want moles of the sodium azide to be in my numerator looking at my balanced chemical equation I can see that the coefficient for nitrogen is three so I'll put that down here and the coefficient for the sodium azide is two so that'll go up in the numerator now moles of nitrogen can cancel and I am in moles of sodium azide let's look at my question again it says what mass of sodium azide is required right so now that we're in moles of sodium azide it looks like we need to finish the problem by converting to Mass so for this we can use our molar mass the molar mass of sodium azide is 65 grams per mole it looks like I need moles to cancel so I'll put that piece of information in the denominator so moles of sodium azide goes in the denominator and the mass of that mole would go in the numerator so 65 grams and moles of sodium azide cancel and we can plug this into our calculator to get the number 142 my unit should be grams and this would be of the substance sodium azide my reactant so when 142 grams of sodium azide react they generate 73.6 liters of nitrogen gas that's quite a large volume so that's going to inflate your airbag and protect you during an automotive accident