Chemical Mathematics.

Chemical Mathematics

Counting by Weighing Small items can be tedious to count.
Items like nuts and bolts are priced by the kilogram after someone initially counts how many are in 1 kg. We will pretend that there are 400 in a kg. This only is done once and keeps the customer from having to do it. There is an example on the next slide.

Counting by Weighing Let’s say that a bolt is priced at $ 5.00 per kg.
The customer grabs a handful of the amount that is needed and it gets weighed. Let’s say the customer needs 1.6 kg of bolts. We would set up the following proportion. 1 kg = kg x = $ 8.00 $ x

Counting by Weighing Or if the customer needs 300 bolts, then they don’t have to count out 300, which would take a long time, they would weigh out .75 kg.

Atomic Mass Atoms are so small, it is difficult to discuss how much they weigh in grams. Use atomic mass units. An atomic mass unit (amu) is one twelfth the mass of a carbon atom. This gives us a basis for comparison. The decimal numbers on the table are atomic masses in amu.

Atomic Mass While the atomic mass unit is important, it is not usually used in day to day problems. They are too small to weigh out So the gram is used to measure out mass. 1 molecule of water weighs 18 amu 1 mole of water weighs 18 grams. For comparision: 1 amu = 1.67 x 10 -24 gram

They are not whole numbers
Because they are based on averages of atoms and of isotopes. We can figure out the average atomic mass from the mass of the isotopes and their relative abundance. Add up the percent as decimals times the masses of the isotopes. It will work for any number of isotopes.

Examples There are two isotopes of carbon 12C with a mass of amu (98.892%), and 13C with a mass of amu (1.108%). What is the atomic wt. of Carbon ? There are two isotopes of nitrogen , one with an atomic mass of amu and one with a mass of amu. What is the percent abundance of each ? All of these are solved with algebra !

Solved Example # 1 There are two isotopes of carbon 12C with a mass of amu (98.892%), and 13C with a mass of amu (1.108%). What is the atomic wt. of Carbon ? Here is how we would set up the equation. ( x ) + ( x ) = Info on 12C  Info on 13C  12.01 grams

Solved Example # 2 There are two isotopes of nitrogen , one with an atomic mass of amu and one with a mass of amu. What is the percent abundance of each ? This one uses the same ideas but we have to look at it differently. We know from the periodic chart that the atomic wt. of Nitrogen is grams.

Solved Example # 2 Here is the set-up
14.01 = ( ● (1-x) ) + ( x ) Look at how the two different percentages were handled. Since they have to total up to 100 % (or 1 as a fraction). We know that if one of the values is = to x, then the other must be 1 – x.

Solved Example # 2 Now we can solve.
14.01 = ( ● (1-x) ) + ( x ) Simplify to 14.01 = – x x .0069 = x x = or .692 % 15N Therefore, The % of 14N is = to % Does this make sense ? Why ?

Formula Weight Alternate names: Molar Mass, Atomic Weight, Molecular Weight, Formula Weight. Use units of grams or grams / mole Using the Periodic Chart, determine the molecular weights of various examples, going from the simplest to the more complex, on the next slide.

Formula Weight – General Info
You can use the terms weight and mass interchangeably here. The most general term is FORMULA WT. Molar mass is another common term. Atomic weight is the weight of an atom. Molecular weight is the weight of a molecule. Ionic weight is the weight of an ion. Etc., etc., etc.

Formula Weight Mass of 1 unit of a substance, called the mole. More on that in a bit Often called molecular weight. (if it is a compound ) To determine the atomic weight of an element, look on the periodic table. To determine the molecular weight of a compound, add up the molar masses of the elements that make it up.

Determining Formula Weight
Sn g NaBr = g Cl x 2 = g K2O (39.1 x 2) + 16 = 94.2 g Al2O3 (27 x 2) + (16 x 3) = 102 g 4 Al2O3 Same as above – coefficients don’t change the formula weight Ca(NO3) (14 + (3 x 16)) = g (NH4)2S 2 (14 + ( 4 x 1)) = g

Rounding Since different periodic charts can have slightly different atomic weights, a range of values are allowed on answers. So rounding is permissible. If the value is within .03 of the next whole number, we can round it.

Rounding Here are some that we can round. Hydrogen 1.00794  1
Carbon  Oxygen  Phosphorus  31

Rounding Here are some that we can’t round to a whole number.
Boron or 10.80 Chlorine Calcium or Nickel or

Formula Weight of a Hydrate
MgSO4 ● 7 H2O (hydrates). These are easy if you understand what the formula represents. There are 7 water molecules attached to the single MgSO4 molecule. The raised dot DOES NOT mean multiplication. ( (4 x 16)) + 7 ( 18) = g

Find the formula weight of
CH4 Mg3P2 Ca(NO3)2 Al2(Cr2O7)3 CaSO4 · 2H2O

CH4 16 g Mg3P2 134.9 g Ca(NO3)2 164.08 g Al2(Cr2O7)3 702 g
Answers CH g Mg3P g Ca(NO3) g Al2(Cr2O7) g CaSO4 · 2H2O g

The Mole The mole is a number.
A very large number, but still, just a number. 6.02 x 1023 of anything is a mole. Similar in idea to a dozen. A dozen is always 12 items A mole is always 6.02 x 1023 items A mole is in exactly 12 grams of carbon-12.

Moles  Particles We need to know how to convert the number of moles to the number of particles. We do this by setting up a proportion. Like all proportions, we should put the GIVEN ratio on the right hand side. Like this: 1 mole / 6.02 x 1023 particles = X moles / y particles

Moles  Particles How many particles would be in 2.1 mole of Argon.
Here is the set-up: 1 mole / 6.02 x 1023 particles = X moles / y particles 1 mole / 6.02 x 1023 particles = 2.1 moles / y particles Solving gives us 1.26 x 1024 particles.

The Mole and the Magic Formula
From a practical standpoint, we want to be able to convert grams to moles and moles to grams. THIS IS OF UTMOST IMPORTANCE IN CHEMISTRY !! We need to introduce the most used formula in Chemistry. n = g / M Moles = grams / formula wt.

Solving Mole problems Using simple algebra and n = g / M solve the following: How many moles are in 154 g of CO2 ? How many grams are in 12.6 moles of NaCl ? How many moles are in .255 grams of H2O ? How many grams are in .6 moles of H2SO4 ? What is the formula wt. of a substance if .85 moles of has a weight of 42 grams ?

Solving Mole problems - answers
154 g of CO2 ? n = 154 / n = 3.5 moles 12.6 moles of NaCl ? = g / g = g .255 grams of H2O ? n = .255 / 18 n = n .6 moles of H2SO4 ? .6 = g / g = g What is the formula wt. of a substance if .85 moles of it has a weight of 42 grams ? .85 = 42 / M M = g

Percent Composition Percent of each element in a compound.
If there are two or more atoms of the same type in a compound, you need to include all the atoms. Find the mass of each element, divide by the total mass (formula wt.), multiply by 100. Find the percent composition of CH4 Al2(Cr2O7)3 CaSO4

Percent Composition - answers
CH4 – Formula Wt. of 16 g C 12 / 16 = x 100 % = % Carbon H 4 / 16 = x 100 % = % Hydrogen Al2(Cr2O7)3 – Formula Wt. of 702 grams Al 54 / 702 = x 100 % = % Al Cr 312 / 702 = x 100 % = % Cr O 336 / 702 = x 100 % = % O

Percent Composition - answers
CaSO4 – Formula Wt. of Ca / = x 100 % = % S / = x 100 % = % O 64 / = x 100 % = % Notice that the sums of the percents always will equal 100 % So if a compound such as NaCl is known to be % Na, we can simply subtract that from 100 % to determine that it is % Cl.

Empirical Formula From percent composition, you can determine the empirical formula. Empirical Formula is based on mole ratios and is the lowest ratio of atoms in a molecule. It is important to know that Empirical Formulas are not necessarily real chemical compounds. Why then do we bother ? Because it is a great analytical tool to analyze an unknown compound.

How to determine Empirical Formulas
No matter what unit is given, you need it to be in moles. If moles, you are ready. If grams, use n = g/M If a %, assume 100 grams and use n = g/M After moles are obtained, you have a proper ratio. Simply make sure that it is in whole numbers !

Empirical Problem If a sample is found to be 30.43 % N, and 69.56 % O
What is its empirical formula ? 1. Since we need to get moles, we need to remember that the percent composition is an intensive property and doesn’t change no matter what the size of the sample is.

Empirical Formula So let’s choose a 100 gram sample to simplify the math. So  % N of a 100 gram sample = grams of Nitrogen Doing the same thing to the other element gives us grams of Oxygen

Empirical Formula So we have the following: 30.43 g N 69.56 g O
Convert to moles using n = g / M 2.17 n N n O This is the ratio of atoms in the compound, but there is clearly a problem if we write it as the following : N2.17O4.34 We can’t have a fraction of an atom !

Empirical Formula So we need to keep the same ratio of atoms but have them expressed as a whole number. We can do this by dividing each by the lowest, so it will look like this. This gives us a ratio of 1:2 So the empirical formula is NO2

Empirical Formula A gram sample of a compound (vitamin C) composed of only C, H, and O is burned completely with excess O g of CO2 and g of H2O are produced. What is the empirical formula? Follow the steps and analyze the problem. A general rule in Chemistry is that if a reaction is taking place it is a good idea to write it out. It might not help – but it will never hurt !!

Empirical Formula – Solved Problem
The equation is: CxHyOz + O2  CO2 + H2O When you analyze the problem you see that the Carbon in the CO2 had to come from the Carbon in the Hydrocarbon. The Hydrogen in the water had to come from the Hydrogen in the Hydrocarbon. That is our clue to proceed. This is a very representative type of AP Chemistry Problem for this topic.

So if we have grams of CO2 we can use our knowledge of percent composition to determine how much Carbon is in this reaction. CO2 is % C so that is grams of Carbon in the sample. The reactant must contain this amount of Carbon. Doing the same for the H in H2O gives us grams of H in the reactant.

.0818 g. of C g. of H = grams Since the total mass of the reactant is .200 g. the remaining mass must be the mass of the oxygen. = g. of Oxygen are in the reactant. Since it is in grams, the 100 g. sample is not needed. We can solve directly for moles.

So this is what we have. The reagent is made up of the following: .0818 g C g H g O Let’s convert to moles: .0068 n C n H n O We can’t have a formula of C.0068H.0091O.0068 because we can’t have a fraction of an atom.

Here is our current formula C.0068H.0091O.0068 We need to express the same relationship in whole numbers. We do that by dividing each value by the lowest. (that is in this problem) .0068/ / /.0068 so we get the following 1 : 1.33 : 1

so we get the following 1 : 1.33 : 1 What we have to do is to finish the work to get all whole numbers in the formula. It helps to know your decimal fractional equivalents. We can rewrite the ratio above as 1 : 1 & 1/3 : 1

1 : 1 & 1/3 : 1 If we multiply each factor by 3 we would get 3 : 4 : 3 which can be written as C3H4O3 and we have our empirical formula

Empirical Formula We need to remember that what we solved for is the empirical formula, or the ratio of atoms in the compound. C3H4O3 may or may not be the actual formula for Vitamin C. To determine this we will need more information, which brings us to the next topic.

Empirical To Molecular Formulas
Empirical is lowest ratio of atoms in a formula. Molecular is the actual molecule. What we need is the Formula Weight. This will be given to you. Ratio of empirical to molar mass will tell you the molecular formula. Must be a whole number because molecules are made up of whole number of atoms.

Empirical to Molecular Formulas
Here is a generic example. Let’s say we determined an empirical formula to be CH2 and we are given that the formula weight of the actual compound is 154 g. First we should look at how many possible molecular formulas that have an empirical formula of CH2 there are.

Let’s start by listing all the possible molecular formulas. CH2 C6H12 C11H22 C2H4 C7H14 C12H24 C3H6 C8H16 C13H26 C4H8 C9H18 C14H28 C5H10 C10H20 C15H30 We can now compare the given molecular weight with all the possible weights of the formulas above. This can get tedious very quickly !!

Here is a much quicker way. Determine the empirical formula by the method shown earlier. Set up the following: Molecular Wt. Empirical Wt. 3. This will always give us a whole number. (or very close due to rounding)

In our sample problem here, the math would work like this: 154 / 14 = 11 Where the molecular formula was given as The empirical formula of CH2 is solved for an empirical wt. of 14 Notice the answer is a whole number.

So what we want is the 11th version of the empirical formula. The key is to know what it will look like, sometimes students format it the wrong way. There are three possibilities but only one of them are correct.

Here are the three options: 11 CH2 (CH2)11 C11H22 The 1st one is incorrect. This is saying that there are 11 units of a small CH2 molecule. The second one is also incorrect. This is simply bad form all around. Some organic groups can be written this way sometimes, but not an entire formula.

So that leaves us with the third option as our choice. C11H22 has a formula weight of 154 g (12 x 11) + (22 x 1) It can be ‘factored down’ to CH2 which is the empirical formula in question.

Molecular Formula – Solved Example
A compound is made of only nitrogen and oxygen. It is 30.4 % N by mass. Its formula weight is 92 g/mol. What is its formula ? We should notice that we don’t have both percentages, but we can easily determine the % of oxygen. 100 – 30.4 = 69.6 % Oxygen

We have % N and 69.6 % O Find the empirical formula – assume 100 g of sample 30.4 g N & g O Moles 2.17 n n Determine the ratio: We divide each by the lowest and get an empirical formula of NO2

We determine the empirical weight of NO2 to be 46 g/n. 92 / 46 = 2 So the molecular formula is N2O4

Chemical Equations Are chemical sentences.
Describe what happens in a chemical reaction. Reactants ® Products Equations should be balanced. Balanced means having the same number of each kind of atoms on both sides because ... To obey the Law of Conservation of Mass.

Chemical Equations Here are some guides to balancing equations.
Make sure that all species are correctly written. Remember that you can only change the coefficients, which is the number of the species. Changing the subscript changes the identity of the species and therefore changes the chemical equation. Start with the largest compound and assume that it has a coefficient of one. Only balance one atom at a time. Save atoms that are found in multiple species to the end.

Balancing equations CH4 + O2  CO2 + H2O 1 C 1 4 H 2 2 O 3 Reactants
Products 1 C 1 4 H 2 2 O 3

Balancing equations CH4 + O2  CO2 + 2 H2O 1 C 1 4 H 2 4 4 2 O 3
Reactants Products 1 C 1 4 H 2 4 4 2 O 3

Balancing equations CH4 + 2 O2  CO2 + 2 H2O 1 C 1 4 H 2 4 4 4 2 O 3
Reactants Products 1 C 1 4 H 2 4 4 4 2 O 3

Abbreviations & Symbols
(s) or Solid (g) or Gas ( l ) Liquid (aq) Aqueous heat Heat added as a catalyst catalyst Any other catalyst

Practice Ca(OH)2 + H3PO4  H2O + Ca3(PO4)2 Cr + S8  Cr2S3
KClO3(s)  KCl (g) + O2(g) Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas.

Bal. Equation Answers Ca(OH)2 + H3PO4 ® H2O + Ca3(PO4)2
Start with Ca3(PO4)2 because it is the largest. 3 Ca(OH)2 + H3PO4 ® H2O + Ca3(PO4)2 3 Ca(OH)2 + 2 H3PO4 ® H2O + Ca3(PO4)2 Oxygen is not a good choice to balance next because it is found in every species. 3 Ca(OH)2 + 2 H3PO4 ® 6 H2O + Ca3(PO4)2

Cr + S8 ® Cr2S3 Bal. Equations Answers 16 Cr + 3 S8 ® 8 Cr2S3
_____________________________________________________________________________________________________________________________________________________________________________ We have 8 S  3 S so we can use lowest common multiple Cr + 3 S8 ® 8 Cr2S3 This leaves Cr to balance. 16 Cr + 3 S8 ® 8 Cr2S3

Bal. Equations Answers 2 KClO3(s)  2 KCl (g) + 3 O2(g)
Note: The phases don’t affect the balancing ____________________________________________________________________________________________________________________________________________ Start with KClO3 It is the largest. 2 KClO3(s)  KCl (g) + 3 O2(g) We notice that we have 2 K and 2 Cl on the left. It fits nicely with the KCl product. 2 KClO3(s)  2 KCl (g) + 3 O2(g)

Bal. Equations Answers Solid iron(III) sulfide reacts with gaseous hydrogen chloride to form solid iron(III) chloride and hydrogen sulfide gas. The only difference here is that we have to use our formula writing skills to correctly write out a skeleton (unbalanced) equation. Fe2S3 + HCl  FeCl3 + H2S

Balancing Equations If we write a formula incorrectly one of two things can happen (none of them good). 1. We will find it impossible to balance. 2. We will be able to balance it, but we won’t be balancing the desired equation.

Balancing Equations A balanced equation can be used to describe a reaction in moles, molecules and atoms. NEVER GRAMS ! In other words: 2 H2 + O2  2 H2O means 2 molecules (or moles) of Hydrogen react with 1 molecule (or mole) of Oxygen to form 2 molecules (or moles) of Water It doesn’t mean 2 grams of H + 1 gram of O to form 2 grams of water !!

Stoichiometry Given an amount of either starting material or product, determining the other quantities. We must have a balanced equation. Use conversion factors from Formula Weights (g - mole) Balanced Equation (mole - mole) Keep track by setting up proportions in moles.

Stoichiometry Example # 1
One way of producing O2(g) involves the decomposition of potassium chlorate into potassium chloride and oxygen gas. A 25.5 g sample of Potassium chlorate is decomposed. How many moles of O2(g) are produced ? How many grams of potassium chloride ? How many grams of oxygen ?

First we need a balanced equation: KClO3  KCl + O2 is the unbalanced equation. 2 KClO3  2 KCl + 3 O2 is balanced. We need to organize the problem. Here is one suggestion how to do it. 25.5 grams ? ? 2 KClO3  2 KCl + 3 O2 This clarifies what we have and what we need.

25.5 grams ? ? 2 KClO3  2 KCl + 3 O2 Convert to moles  n of KClO3 We need a proportion 2 KClO3 / 3 O2 = n KClO3 / x x = moles of O2 Using n = g / m grams of O2

To get the KCl, we follow a similar procedure. 2 KClO3 / 2 KCl = n KClO3 / x x = moles of KCl Using n = g / m gives us g of KCl Let’s verify by the Law of Cons. Of Mass Total mass of reactants = 25.5 g Total mass of products = = g Our answers look good !

A piece of aluminum 5.11 cm. by 3.23 cm. x cm. is dissolved in excess HCl(aq). How many grams of H2(g) are produced ? We have some preliminary work to do. The volume of the Aluminum piece is mL OK - That’s nice, I suppose ! What is now the next step ?

We can use the Density of Al to get the mass. D = m / V = g / .629 mL mass = 1.70 g. Here is our balanced equation: 2 Al + 6 HCl  2 AlCl3 + 3 H2 1.7 grams of Al = moles Our proportion is going to contain Al because we have info on it and H because that is what we are looking for.

Here is the proportion: 2 Al / 3 H2 = n / x x = n of H2 Convert to grams using n = g / m We obtain grams of H2 as our answer.

How many grams of each reactant are needed to produce 15 grams of iron from the following reaction ? Fe2O3(s) + Al(s) ® Fe(s) + Al2O3(s) First we balance: Fe2O3(s) + 2 Al(s) ® 2 Fe(s) + Al2O3(s) We also need to know moles of Fe n = g / m  n = 15 / n = .267 n Fe

Now that we have moles we can move forward with the problem. We can do the Fe2O3 first Our proportion is: 1 Fe2O3 / 2 Fe = x / .267 n Fe Solving gives us moles of Fe2O3 Converting to grams gives us g of Fe2O3 Doing similar work for the Al gives us 2 n Al / 2 n Fe = x / .267 n Fe x = .267 n of Al .267 n Al  grams of Al

K2PtCl4 + NH3 ® Pt(NH3)2Cl2 + KCl What mass of Pt(NH3)2Cl2 can be produced from 65 g of K2PtCl4 ? How much KCl will be produced ? First we must balance K2PtCl NH3 ® Pt(NH3)2Cl2 + 2 KCl Following the steps gives as in the previous problem: 65 g of K2PtCl4  moles

1 n K2PtCl4 / 1 n Pt(NH3)2Cl2 = / x x = moles of Pt(NH3)2Cl2 Convert to grams (FW = g)  g FOR KCl 1 n K2PtCl4 / 2 n KCl = / x x = moles of KCl .3132 n of KCl  g of KCl

Limiting Reagent The concept of limiting reagent is that if you have two or more reactants in a chemical equation. When they start reacting, one of them are often going to run out first. This occurs in non – chemical situations as well. If we are making bicycles and have 40 seats and 50 handle bars, we can only make 40 bicycles because once the seats run out, we can’t make any more, no matter how many handle bars you have.

Limiting Reagent The term limiting reagent is used for the material that runs out first. Since the reaction stops when a reactant is missing, the limiting reagent governs the amount of product produced. The other material that has some left over is said to be in excess. Sometimes abbreviated as inxs.

Limiting Reagent - Summary
Reactant that determines the amount of product formed. It governs the reaction. The one you run out of first. Makes the least amount of product. If it is not easily apparent which reactant is the limiting reagent, you can always set up 2 proportions and see which one fits. One will fit and the other one won’t – ALWAYS.

Limiting reagent There are two possible ways to work a Limiting Reagent problem 1. To determine the limiting reagent requires that you do two stoichiometry problems. Figure out how much product each reactant makes. The one that makes the least is the limiting reagent.

Limiting Reagent 2. The second way is to set up two proportions and analyze them to determine which is the limiting reagent. This is the preferred way and the method that we will explore in class. However, either way is acceptable.

Limiting Reagent – Example
Ammonia is produced by the following reaction N2 + H2 ® NH3 What mass of ammonia can be produced from a mixture of 500 g N2 and 100 g H2 ? How much unreacted material remains?

Limiting Reagent – Example
The first thing we need is a balanced equation. N2 + 3 H2 ® 2 NH3 500 g of N2 = moles 100 g of H2 = 50 moles The key is setting up the proportions, and equally important is how to analyze them to obtain the desired information.

Limiting Reagent N2 + 3 H2 ® 2 NH3 500 g of N2 = 17.857 moles
100 g of H2 = 50 moles At first glance it is tempting to say that the N2 must be the limiting reagent simply because there is much less of it. However, that is not often the case.

Limiting Reagent What we need to do is to set up proportions.
The balanced equation is: N2 + 3 H2 ® 2 NH3 1 n N2 / 3 n H2 = n of N2 / x n of H2 You create the proportion by using the coefficients from the balanced equation on the left hand side, and the information in the problem on the right hand side.

Limiting Reagent 1 n N2 / 3 n H2 = 17.857 n of N2 / x n of H2
It is very important to realize what this proportion is asking for. It is solving for how much H2 would be needed IF all the N2 reacts. We need to solve for x and then analyze the answer. Doing the math gives us moles of H2

Limiting Reagent If we need moles of H2 but we only have 50 moles of H2 what does that mean ? We need more H2 than we have, so that is the material in short supply. That makes the H2 the limiting reagent and the material that controls the reaction.

Limiting Reagent Just to verify that it wasn’t just luck in choosing the correct material, let’s set up the proportion using the other reagent. 1 n N2 / 3 n H2 = x n of N2 / 50 n of H2 Solving for x gives us moles of N2 needed. However since we have n of N2 given to us initially, we have more N2 than we need and so N2 is in excess (INXS). So that makes the H2 the Limiting Reagent.

Limiting Reagent So now we can see that the key is not choosing the ‘magic’ material to put in the proportion – it doesn’t matter which one we choose. The key is analyzing what we get when we solve the proportion and use it accordingly. We need to use the limiting reagent in the proportion because it governs the reaction. When it runs out – the reaction stops.

Limiting Reagent 1 n N2 / 3 n H2 = x n of N2 / 50 n of H2
So let’s go back and finish the problem. Here is our proportion: 1 n N2 / 3 n H2 = x n of N2 / 50 n of H2 Solving for x gives us n of N2 reacting. If we are asked to determine the amount of material in excess, we can simply subtract the amount given minus the amount reacted. 17.86 – = 1.19 moles of N2 We can convert to grams if needed. (33.32 g)

N2 + 3 H2 ® 2 NH3 Limiting Reagent
Here is what we would do to find how much of the product is formed. N2 + 3 H2 ® 2 NH3 Use the limiting reagent in the proportion. For this problem, we established that it was H2 2 n NH3 / 3 n H2 = x n of NH3 / 50 n of H2 Solving for x gives us n of NH3 Convert to grams as needed: g NH3

Limiting Reagent We only need to worry about the Limiting Reagent if the problem contains information about 2 or more reactants. If only information about one reactant is present, it is assumed that it is the Limiting Reagent and everything else is in excess.

More about Stoichiometry
Sometimes they might simplify a problem by saying one of the materials in excess. Then we simply use the other material, which is going to be the one with information included with it. Only one material can be the Limiting Reagent. We can also verify our work by comparing the amounts we obtained and verifying the Law of Conservation of Mass.

More about Stoichiometry
Here is a standard problem we worked earlier. 2 KClO3  2 KCl + 3 O2 We had g of KClO3 We solved for 9.99 grams of O2 and grams of KCl. Total mass of products = = g Mass of Reactants = Mass of Products Our answers look good !

More on Stoichiometry Let’s do the same thing for a limiting reagent problem. N2 + 3 H2 ® 2 NH3 We had a 500 g. of N2 and 100 g. of H2 (600 g. total). We solved the problem and got g of NH3. We appear to be violating the Law of Conservation of matter.

More on Stoichiometry N2 + 3 H2 ® 2 NH3 500 g + 100 g  566.61 g
But we have material in excess (N2). The excess material doesn’t react and essentially just sits there in the vessel. We solved for the amount in excess earlier and got g. 600 – = grams, which is close enough with rounding.

Vocabulary The excess reactant is the one you don’t run out of.
The amount of product you make is the yield. The theoretical yield is the amount you would make if everything went perfect. This is obtained by doing a stoichiometry problem and sometimes called the stoichiometric yield. The observed yield is what you make in the lab.

Vocabulary There are many terms for these two ideas.
The amount that is the amount that we are supposed to make can be called the: Book value Real value Accepted Value Actual Value Theoretical Value

Percent Yield % yield = Observed x 100% Theoretical
Or stated another way % yield = what you got x 100% what you should have got

Percent Yield – Example # 1
Aluminum burns in bromine producing aluminum bromide. During a laboratory 6.0 g of aluminum reacts with excess bromine. The student got 50.3 g of aluminum bromide as a product. What is the percent yield ? We need a balanced chemical equation 2 Al + 3 Br2  2 AlBr3

Here is our equation: 2 Al + 3 Br2  2 AlBr3 6 g of Al = .222 n of Al. Set up a proportion: 2 Al / 2 AlBr3 = .222 / x x = .222 moles of AlBr3 (59.21 g ) 50.3 / = % Be aware that not all reactions go to completion for a variety of reasons.

Years of experience have proven that the percent yield for the following reaction is 74.3% Hg + Br2 ® HgBr2 If 10.0 g of Hg and 9.00 g of Br2 are reacted, how much HgBr2 will be produced ? If the reaction did go to completion, how much excess reagent would be left ? We need to do the Limiting Reagent problem first to determine how much should be made.

10 g of Hg = moles of Hg 9 g of Br2 = moles of Br2 Hg is the Limiting reagent (due to the 1:1 ) 1 n Hg / 1 n HgBr2 = / x x = moles of HgBr2 (17.98 g of HgBr2) The yield is 74.3 % so we would expect to obtain x g = grams of HgBr2

Examples Commercial brass is an alloy of Cu and Zn. It reacts with HCl by the following reaction Zn(s) + 2 HCl(aq) ® ZnCl2 (aq) + H2(g) Cu does not react. When g of brass is reacted with excess HCl, g of ZnCl2 are eventually isolated. What is the composition of the brass ? First think of what you are solving for. Plan the problem out.

Putting it Together ZnCl2 is % Zn so g of ZnCl2 contains grams of Zn The total sample weighs grams so the mass of the Cu in the sample must be g .4592 / = % Note that there are many different types of Brass and the percentage varies accordingly.

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