s toichiometry general chemistry spring 2010. i ntroduction billions of pounds of chemicals are...
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INTRODUCTION
Billions of pounds of chemicals are produced each year across the world.
These chemicals help manufacture: Medicines Computer chips and electronic instruments Fertilizers and pesticides Glass Paper, plastics, synthetic fibers…….
INTRODUCTION
In order to determine the cost of producing such essential items we use each and every day, chemists and chemical engineers perform calculations based on balanced chemical equations
Which chemical do you think is used the most industrially? Sulfuric acid 165 million mega tons (1000 kg)Value of $8,000,000,000 annually…
yes 8 billion dollars…
Uses of H2SO4
Production of fertilizers,. Used in the manufacture of chemicals such as
hydrochloric acid, nitric acid, sulfate salts, synthetic detergents, dyes and pigments, explosives, and drugs.
It is used in petroleum refining to wash impurities out of gasoline and other refinery products.
Sulfuric acid is used in processing metals. In cleaning iron and steel before plating them with tin or zinc.
Rayon is made with sulfuric acid. It serves as the electrolyte in the lead-acid storage
battery commonly used in motor vehicles (acid for this use, containing about 33% H2SO4).
Stoichiometry? Big Deal…. What are some items you use everyday?
Clothes? Soap and shampoo? CD’s and iPods? Medicine?
Company’s that make these and every other substance do not want the cost to make these items to be higher than the cost they’re sold at. Profit ring a bell? Companies carry out chemical rxns
economically to keep prices down (we hope at least)
Why did I have to learn how to balance an equation?
Chemists use balanced equations as a basis to calculate how much reactant is needed or product is formed in a reaction. If you know the quantity of 1 piece, you can
calculate the quantity of anything else in the reaction
We use grams and molsSometimes L, tons, molecules are used
http://www.theyshoulddothat.com/images/newIpods.jpg
What IS stoichiometry? The calculation of quantities in chemical
reactions We will use a version of the mole road map for
this Equations must be balanced Allows the chemist (you) to keep track of the
amounts of R and P using ratios of moles or particles
http://www.hccbrandon.net/chem1211/mod5/stoichiometryTitle.gif
Still not convinced? Air bags must inflate in less than a second. The effectiveness of air bags is based on the
rapid conversion of a small amount of sodium azide into a large volume of gas.
The gas fills up an air bag, preventing the driver and passengers from incurring any life threatening injuries.
2NaN3(s) 2 Na(s) + 3 N2(g)
Sensors trigger an igniter
Fills up the
airbag
INTERPRETING A CHEMICAL EQUATION
What information can you get from a balanced chemical equation?
Nitrogen monoxide is present in car exhaust. UV light catalyzes the reaction of nitrogen monoxide and O2 to produce nitrogen dioxide smog.
2 NO(g) + O2 (g) 2 NO2(g)
UV
INTERPRETING A CHEMICAL EQUATION
• 2 NO(g) + O2 (g) 2 NO2(g) We see that 2 molecules of NO react w/ 1 molecule
of O2 to produce 2 molecules of NO2. These coefficients indicate relative numbers of
reactant and product molecules… it’s a ratio! The coefficients indicate a ratio of moles, or the
MOLE RATIO, of reactants and products in every balanced chemical equation.
UV
VERIFYING THE CONSERVATION OF MASS LAW
“Mass is neither created nor destroyed during a chemical reaction.”
The combined masses of the reactants must equal the combined masses of the products.
NO + O2 NO2
2 NO + O2 2 NO2
Which one obeys the Law of Conservation of Mass? THIS IS WHY YOU BALANCE AN EQUATION
MOLE-MOLE RELATIONSHIPS Remember, the coefficients in a chemical equation
indicate the mole ratio of the reactants and products Consider the synthesis reaction of nitrogen
and oxygen to give nitrogen monoxide:N2(g) + O2(g) 2 NO(g)
We see that 1 mol of nitrogen reacts with 1 mol of oxygen to produce 2 mol of nitrogen monoxide
Mole-Mole
N2(g) + O2(g) 2 NO(g)We can write several mole ratios
for this equation1 mol N2 : 1 mol O2
1 mol N2 : 2 mol NO1 mol O2 : 2 mol NO
Each can become a unit factor to convert between units
MOLE-MOLE RELATIONSHIPS
N2(g) + O2(g) 2 NO(g) How many moles of oxygen react with 2.25 mol of
nitrogen? To cancel units, we will use a mole ratio as our
unit factor: 1 mol O2 / 1 mol N2
2.25 mol N2 1 mol O2 = 2.25 mol O2 1 1 mol N2
Mole-Mole Relationships
N2(g) + O2(g) 2 NO(g) Calculate moles of nitrogen monoxide
produced by the reaction if 2.25 mol of N2 are used.
Plan: 1. Use the mole ratio 2 mol NO/1 mol N2 as the
unit factor.
2.25 mol N2 2 mol NO
= 4.50 mol NO
1 1 mol N2
MOLE- MOLE RELATIONSHIPS
Whenever we have a balanced chemical equation, we can always convert from moles of one substance to moles of another substance using a mole ratio as a unit factor.
MOLE-MOLE RELATIONSHIPS
Carbon monoxide is produced in a furnace by passing oxygen gas over hot coal. The balanced equation is: 2 C(s) + O2(g) 2 CO(g) How many moles of oxygen react with 2.50 mol
of carbon? Mole ratio is 2 mol C / 1 mol O2
2.50 mol C 1 mol O2 = 1.25 mol
O2 1 2 mol C
Mole-Mole relationships
2 C(s) + O2(g) 2 CO(g)How many moles of CO are
produced from 2.50 mol of carbon?Mole ratio is 2 mol C / 2 mol CO
2.50 mol C 2 mol CO= 2.50 mol CO
1 2 mol C
STOICHIOMETRY
We will apply mole ratios in order to relate quantities of reactants and products.
Stoichiometry is a term used to refer to the relationship between quantities in a chemical reaction according to a balanced chemical equation.
We can determine mass-mass relationships, mass-volume relationships (w/ a gas), and volume-volume relationships (w/ 2 gases)
MASS A MASS B PROBLEMS
An unknown mass of substance is calculated from a given mass of reactant or product.
After balancing the equation, proceed as follows:1. Convert the given mass to moles using the
molar mass of the substance as a unit factor2. Convert the moles of the given to moles of the
unknown using the mole ratio (coefficients)3. Convert the moles of unknown to grams using
the molar mass of the unknown as a unit factor.
1) GIVEN MASS MOLES 2) GIVEN MOLES UNKNOWN MOLES 3) UNKNOWN MOLES UNKNOWN MASS
Calculate the mass of tin(IV)chloride produced by the reaction of 1.25 g of metallic tin with yellow chlorine gas.
Sn(s) + 2 Cl2(g) SnCl4(s) First verify the equation is balanced, then calculate the moles
of tin. The molar mass of Sn is 118.71 g/mol.
Next, find the moles of SnCl4 using the mole ratio (1 mol Sn = 1 mol SnCl4)
Last, calculate the mass of product using the molar mass of SnCl4 (260.51 g/mol)
1.25 g Sn 1 mol Sn= 0.0105 mol Sn 1 118.71 g Sn
0.0105 mol Sn 1 mol SnCl4= 0.0105 mol SnCl4
1 1 mol Sn
0.0105 mol SnCl4 260.51 g = 2.74 g SnCl4 1 1 mol
DO YOU HAVE TO COMPLETE THESE IN 3 SEPARATE STEPS?
Once you get comfortable with the calculations, this is what it looks like as one step:
1.25 g Sn x x x = 2.74 g SnCl4
Until then, complete them in three separate steps:
1. Given mass moles (using molar mass)
2. Given moles unknown moles (using mole ratio)
3. Unknown moles unknown mass (using molar mass)
1 mol Sn118.71 g Sn
1 mol SnCl4 1 mol Sn
260.51 g SnCl4 1 mol SnCl4
Mass-Mass
Calculate the mass of potassium iodide, KI required to yield 1.78 g mercury(II)iodide, HgI2.
2 KI + Hg(NO3)2 HgI2 + 2 KNO3
1.78 g HgI2
1 mol HgI2 2 mol KI 166.00g KI= 1.30 g KI 1 454.39 g
HgI21 mol HgI2
1 mol KI
g HgI2 Molar mass Mole ratio Molar massg KI
Mass Mass Concept Map
Mass of Given
Moles of given Moles of Unknown
Mass of unknown
convert to moles using molar mass of given
use molar ratio
convert to mass using molar mass of unknown
MASS-VOLUME PROBLEMS
An unknown volume of gas is calculated from a given mass of reactant or product.
After balancing the equation, proceed as follows:1. Convert the given mass to moles using the molar
mass as a unit factor.2. Convert the moles of the given to moles of the
unknown using the mole ratio (coefficients)3. Convert the moles of unknown to liters using the
molar volume (at STP it’s 22.4 L/mol) as a unit factor.
Of course this can be reversed; we can find the mass of an unknown substance from a given volume of gas.
1) GIVEN MASS MOLE2) GIVEN MOLES UNKNOWN MOLES 3) UNKNOWN MOLES VOLUME
0.165 g of aluminum metal reacts with dilute hydrochloric acid. What is the volume of hydrogen gas produced at STP?
2 Al(s) + 6 HCl(aq) 2 AlCl3 + 3 H2(g) First, calculate the moles of aluminum (26.98 g/mol)
Next, use the mole ratio to find the moles of H2. (2 mol Al = 3 mol H2)
0.165 g Al
1mol Al= 0.00611 mol Al
1 26.98 g Al
0.00611 mol Al
3 mol H2
= 0.00917 mol H2
1 2 mol Al
1) GIVEN MASS MOLE 2) GIVEN MOLES UNKNOWN MOLES 3) UNKNOWN MOLES VOLUME
Last, multiply by the molar volume, 22.4 L/mol, to get the volume of H2 gas produced.
1 step:
0.00917 mol H2 22.4 L H2
= 0.205 L H2
1 1 mol H2
0.165 g Al 1 mol Al 3 mol H2 22.4 L H2
= 0.205 L H2
1 26.98 g Al 2 mol Al 1 mol H2
MASS-VOLUME PRACTICE
Baking soda can be used as a fire extinguisher. When heated, it decomposes to carbon dioxide gas which can smother a fire. If a sample of NaHCO3 (84.01 g/mol) produces 0.500 L of CO2 at STP, what is the mass of the sample?
2 NaHCO3(s) Na2CO3(s) + H2O(g) + CO2(g)
Verify the equation is balanced, find the moles of CO2 by using 22.4 L/mol.
0.500 L CO2 1 mol CO2 2 mol NaHCO3 84.01 g NaHCO3 = 3.75 g
NaHCO3
1 22.4 L CO2 1 mol CO2 1 mol NaHCO3
Mass-Volume Concept Map
Mass of given
Moles of given
Moles of Unknown
Volume of unknown gas
Convert to moles using molar mass of given
Use molar ratio
Convert to volume using molar volume of a gas
VOLUME-VOLUME PROBLEMS
We can convert from a given volume of gas to an unknown volume of gas in a single conversion using the mole ratio.
Sulfuric acid is produced by the conversion of sulfur dioxide to sulfur trioxide using heat and a platinum catalyst. The sulfur trioxide is then passed through water to produce sulfuric acid.
2 SO2(g) + O2(g) 2 SO3(g)
Calculate the liters of sulfur trioxide produced from 37.5 L of SO2. From the balanced equation we see that 2 volumes of SO2 = 2
volumes of SO3.
Pt/
37.5 L SO2 2 L SO3
= 37.5 L SO3 1 2 L SO2
SULFURIC ACID CONTINUED
Calculate how much oxygen gas needs to react with 37.5 L SO2 in order to produce that amount of sulfur trioxide.2 SO2(g) + O2(g) 2 SO3(g)
Summary, for volume-volume problems, you only need to use the volume-ratio.
37.5 L SO2 1 L O2
= 18.8 L O2 1 2 L SO2
LIMITING REACTANTS
Recipe: ¼ cup butter 1 10.5 oz bag of mini-marshmallows 5 cup crispy rice cereal
Given that we have 1 box of crispy rice cereal (approx. 10 c. cereal), 1 stick of butter (1/2 c. butter per stick) and 1 bag (10.5 oz) of marshmallows, how many recipes can we make? ONE
LIMITING REACTANTS
What did we run out of first?? Marshmallows limited how many rice-crispy
treats we could make. It’s called the limiting reactant.
If we go and get more butter, can we make more recipes?
If we go and get more cereal, can we make more recipes?
If we go and get more marshmallows, can we make more recipes?
NO
NO
YES!
LIMITING REACTANTS
In problems where we are given the amounts of two reactants, we determine the limiting reactant by:
1. Calculate the mass of product that can be produced from the first reactant.
2. Calculate the mass of product that can be produced from the second reactant. Repeat with each reactant… hope that it’s not a big
reaction…
3. State the limiting reactant (the one that produces the least amount of product) and the corresponding mass of product formed.
LIMITING REACTANTS Lets think of this as a balanced equation. Assume the recipe
looks like this:
3 Sugar + Butter + 8 Cereal Rice Crispy Treats Using the above equation, if I had 3 moles of sugar, 2 moles of
butter, and 14 moles of cereal, how many recipes could I make? 3 mol sugar 1 mol treats
= 1 mol treats 1 3 mol sugar
2 mol butter 1 mol treats= 2 mol treats 1 1 mol butter
16 mol cereal 1 mol treats= 1.75 mol treats 1 8 mol cereal
What is the limiting reactant? In other words, what do we run out of first?
SUGAR!!! We can only make one recipe w/ that amount of
sugar. It limits the number of recipes we can make.
LIMITING REACTANTS
3 C6H12O6 + CH3(CH2)14COOH + 8(C6H12O6)12 Rice Crispy Treats
Using the above equation, if I had 986 g of sugar (C6H12O6), 155 g of butter (CH3(CH2)14COOH), and 12,000 g of cereal ((C6H12O6)12), how many recipes (moles) could I make?
Sugar
Butter
Cereal
986 g sugar 1 mol sugar 1 mol treats= 1.83 mol treats
1 180 g sugar 3 mol sugar
155 g butter 1 mol butter 1 mol treats= 0.61 mol treats
1 256 g butter 1 mol butter
12,000 g cereal
1 mol cereal 1 mol treats= 0.69 mol treats 1 2160 g
cereal8 mol cereal
Last question… You determined the limiting reactant in the previous problem to be
butter. Calculate how many grams of sugar are needed in order to completely react with that amount of butter (155 g butter). Why? Butter is leaving excess cereal and sugar which is wasted money
($$CHA-CHING!) Start with the grams of your limiting reactant (butter) and do a mass-mass
calculation to determine grams of sugar.
We only need 327 g of sugar to “react” with the butter to make these treats.
155 g butter 1 mol butter 3 mol sugar 180 g sugar= 327 g sugar
1 256 g butter 1 mol butter 1 mol sugar
Ok…one more question
How many grams of sugar would be left over?We have 986 g of sugar.We only need 327 g sugar.
986 g – 327 g = 659 g sugar is left over!
Manufacturers need to do these calculations all the time to be sure they’re not wasting anything.
Percent Yield
When the product from a reaction is less than expected, a percentage is calculated
Theoretical yield The maximum amount of product that could be
formed from given amounts of reactions This is calculated by you on paper!
Actual yield The amount of product that actually forms in
laboratory from an experiment Percent Yield- ratio of the two
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