stoichiometry: calculations with chemical formulas and equations chapter 3
TRANSCRIPT
An equation• Describes a reaction
• Must be balanced to follow Law of Conservation of Energy
• Can only be balanced by changing the coefficients.
• Has special symbols to indicate state, and if catalyst or energy is required.
Symbols used in equations
• the arrow separates the reactants from the products.
• (s) after the formula –solid
• (g) after the formula –gas
• (l) after the formula –liquid
• (aq) after the formula-aqueous (Dissolved in water, and aqueous solution)
Symbols used in equations
• indicates a reversible reaction
• shows that heat is supplied to the reaction
• is used to indicate a catalyst used supplied, in this case, platinum.
heat ,
Pt
What is a catalyst?
• A substance that speeds up a reaction without being changed by the reaction.
– ex.. enzymes-biological catalysis
protein catalysts
Convert these to equations
• Solid iron (II) sulfide reacts with gaseous hydrogen chloride to form iron (II) chloride and hydrogen sulfide gas.
• Nitric acid dissolved in water reacts with solid sodium carbonate to form liquid water and carbon dioxide gas and sodium nitrate dissolved in water.
Balanced Equation
• Atoms can’t be created or destroyed
• A balanced equation has the same number of each element on both sides of the equation.
Rules for balancing Equations
Write the correct formulas for all the reactants and products
Count the number of atoms of each type appearing on both sides
Balance the elements one at a time by adding coefficients (the numbers in front)
Check to make sure it is balanced.
Never
• Change a subscript to balance an equation
• If you change the formula you are describing a different reaction
• H2O is a different compound than H2O2
• Never put a coefficient in the middle of a formula
• 2 NaCl is okay, Na2Cl is not.
Balance the following equation
• AgNO3 + Cu → Cu(NO3)2 + Ag
• Mg + N2 → Mg3N2
• P + O2 → P4O10
• Na + H2O → H2 + NaOH
• CH4 + O2 → CO2 + H2O
Types of Reactions• There are millions of reactions.
• Fall into several categories.
• Combination• Decomposition• Single Replacement• Double Displacement• Combustion
Combination Reactions
2 elements, or compounds combine to make one compound.
• Ca +O2 CaO
• SO3 + H2O H2SO4
• We can predict the products if they are two elements.
• Mg + N2
Write and balance• Ca + Cl2 →
• Fe + O2 →iron (II) oxide
• Al + O2 →
• Remember that the first step is to write the formula
• Then balance
Decomposition Reactions• decompose = fall apart
• one reactant falls apart into two or more elements or compounds.
• NaCl Na + Cl2
• CaCO3 CaO + CO2
electricity
Double Replacement
• Two things replace each other.
• Reactants must be two ionic compounds or acids.
• Usually in aqueous solution
• NaOH + FeCl3
• The positive ions change place.
• NaOH + FeCl3 Fe+3 OH- + Na+1Cl-1
• NaOH + FeCl3 Fe(OH)3 + NaCl
Double Replacement
• Will only happen if one of the products– doesn’t dissolve in water and forms a solid – or is a gas that bubbles out.– or is a covalent compound usually
water.
Complete and balance
• assume all of the reactions take place.
• CaCl2 + NaOH
• CuCl2 + K2S
• KOH + Fe(NO3)3
• (NH4)2SO4 + BaF2
How to recognize which type
• Look at the reactants
• E + E Combination
• C Decomposition
• C + CDouble replacement
Combustion
• A compound composed of only C H and maybe O is reacted with oxygen
• If the combustion is complete, the
products will be CO2 and H2O.
• If the combustion is incomplete, the
products will be CO and H2O.
Reactions
• Come in 5 types. (We learned 4)
• Can tell what type they are by the reactants.
• Double Replacement happens if the product is a solid, water, or a gas.
The Process
• Determine the type by looking at the reactants.
• Put the pieces next to each other
• Use charges to write the formulas
• Use coefficients to balance the equation.
Moles
• Defined as the number of carbon atoms in exactly 12 grams of carbon-12.
• 1 mole is 6.02 x 1023 particles/molecules.
• Treat it like a very large dozen
• 6.02 x 1023 is called Avogadro's number.
Representative particles
• The smallest pieces of a substance.
• For a molecular compound it is a molecule.
• For an ionic compound it is a formula unit.
• For an element it is an atom.
Types of questions
• How many oxygen atoms in the following?
– CaCO3
– Al2(SO4)3
• How many ions in the following?
– CaCl2– NaOH
– Al2(SO4)3
Types of questions
• How many molecules of CO2 are the in
4.56 moles of CO2 ?
• How many moles of water is 5.87 x 1022 molecules?
• How many atoms of carbon are there in
1.23 moles of C6H12O6 ?
• How many moles is 7.78 x 1024 formula
units of MgCl2?
Measuring Moles
• The amu was one twelfth the mass of a carbon 12 atom.
• Since the mole is the number of atoms in 12 grams of carbon-12,
• the decimal number on the periodic table is also the mass of 1 mole of those atoms in grams.
Gram Atomic Mass
• The mass of 1 mole of an element in grams.
• 12.01 grams of carbon has the same number of pieces as 1.008 grams of hydrogen and 55.85 grams of iron.
• We can write this as 12.01 g C = 1 mole
Examples
• How much would 2.34 moles of carbon weigh?
• How many moles of magnesium in 24.31 g of Mg?
• How many atoms of lithium in 1.00 g of Li?
• How much would 3.45 x 1022 atoms of U weigh?
What about compounds?
• in 1 mole of H2O molecules there are two
moles of H atoms and 1 mole of O atoms
• To find the mass of one mole of a compound – determine the moles of the elements they have– Find out how much they would weigh– add them up
What about compounds?
• What is the mass of one mole of CH4?
• 1 mole of C = 12.01 g• 4 mole of H x 1.01 g = 4.04g
• 1 mole CH4 = 12.01 + 4.04 = 16.05g
• The Gram Molecular mass of CH4 is 16.05g
• The mass of one mole of a molecular compound.
Atomic Mass
• The mass of one mole of an ionic compound.
• Calculated the same way.
• What is the atomic mass of Fe2O3?
• 2 moles of Fe x 55.85 g = 111.70 g
• 3 moles of O x 16.00 g = 48.00 g
• The atomic mass = 111.70 g + 48.00 g = 159.70g
Molar Mass
• The generic term for the mass of one mole.
• The same as gram molecular mass, gram formula mass, and gram atomic mass.
Examples
• Calculate the molar mass of the following.
• Na2S
• N2O4
• C
• Ca(NO3)2
• C6H12O6
• (NH4)3PO4
Molar Mass
• The number of grams of 1 mole of atoms, ions, or molecules.
• We can make conversion factors from these.
• To change grams of a compound to moles of a compound.
For example
• How many moles is 5.69 g of NaOH?
5 69. g 1 mole
40.00 = 0.142 mol NaOH
g
need to change grams to moles for NaOH 1mole Na = 22.99g 1 mol O = 16.00 g
1 mole of H = 1.01 g 1 mole NaOH = 40.00 g
Examples
• How many moles is 4.56 g of CO2 ?
• How many grams is 9.87 moles of H2O?
• How many molecules in 6.8 g of CH4?
• 49 molecules of C6H12O6 weighs how much?
Mole to Mole conversions• How many moles of O2 are produced
when 3.34 moles of Al2O3 decompose?
• 2 Al2O3 Al + 3O2
3.34 moles Al2O3 2 moles Al2O3
3 mole O2 = 5.01 moles O2
Your Turn• 2C2H2 + 5 O2 4CO2 + 2 H2O
• If 3.84 moles of C2H2 are burned, how
many moles of O2 are needed?
• How many moles of C2H2 are needed
to produce 8.95 mole of H2O?
• If 2.47 moles of C2H2 are burned, how
many moles of CO2 are formed?
For example...• If 10.1 g of Fe are added to a solution of
Copper (II) Sulfate, how much solid copper would form?
• Fe + CuSO4 Fe2(SO4)3 + Cu
• 2Fe + 3CuSO4 Fe2(SO4)3 + Cu
10.1 g Fe
55.85 g Fe
1 mol Fe= 0.181 mol Fe
2Fe + 3CuSO4 Fe2(SO4)3 + 3Cu
0.181 mol Fe2 mol Fe
3 mol Cu= 0.272 mol Cu
0.272 mol Cu1 mol Cu63.55 g Cu
= 17.3 g Cu
More Examples• To make silicon for computer chips they
use this reaction
• SiCl4 + 2Mg 2MgCl2 + Si
• How many grams of Mg are needed to make 9.3 g of Si?
• How many grams of SiCl4 are needed
to make 9.3 g of Si?
• How many grams of MgCl2 are
produced along with 9.3 g of silicon?
For Example• The U. S. Space Shuttle boosters use
this reaction
• 3 Al(s) + 3 NH4ClO4
Al2O3 + AlCl3 + 3 NO +
6H2O
• How much Al must be used to react
with 652 g of NH4ClO4 ?
• How much water is produced?
• How much AlCl3?
Gases• Many of the chemicals we deal with are
gases.
• They are difficult to weigh.
• Need to know how many moles of gas we have.
• Two things effect the volume of a gas– Temperature and pressure
Standard Temperature and Pressure (STP)
• 0ºC and 1 atm pressure
• At STP 1 mole of gas occupies 22.4 L
• Called the molar volume
• Avogadro's Hypothesis - at the same temperature and pressure equal volumes of gas have the same number of particles.
For Example• If 6.45 grams of water are decomposed,
how many liters of oxygen will be produced at STP?
• 2H2O 2H2 + O2
6.45 g H2O 18.02 g H2O1 mol H2O
2 mol H2O1 mol O2
1 mol O2
22.4 L O2
Example• How many liters of CO2 at STP will be
produced from the complete combustion of 23.2 g C4H10 ?
• What volume of oxygen will be required?
Example• How many liters of CH4 at STP are
required to completely react with 17.5 L
of O2 ?
• CH4 + 2O2 CO2 + 2H2O
17.5 L O2 22.4 L O2 1 mol O2
2 mol O2 1 mol CH4
1 mol CH4 22.4 L CH4
= 8.75 L CH4
22.4 L O2 1 mol O2
1 mol CH4 22.4 L CH4
Density of a gas• D = m /V
• for a gas the units will be g / L
• We can determine the density of any gas at STP if we know its formula.
• To find the density we need the mass and the volume.
• If you assume you have 1 mole than the mass is the molar mass (PT)
• At STP the volume is 22.4 L.
The other way• Given the density, we can find the molar
mass of the gas.
• Again, pretend you have a mole at STP, so V = 22.4 L.
• m = D x V
• m is the mass of 1 mole, since you have 22.4 L of the stuff.
• What is the molar mass of a gas with a density of 1.964 g/L?
• 2.86 g/L?
Limiting Reagent• If you are given one dozen loaves of bread, a
gallon of mustard and three pieces of salami, how many salami sandwiches can you make (rhetorical question).
• The limiting reagent is the reactant you run out of first.
• The excess reagent is the one you have left over.
• The limiting reagent determines how much product you can make
How do you find out?• Do two stoichiometry problems.
• The one that makes the least product is the limiting reagent.
• For example
• Copper reacts with sulfur to form copper ( I ) sulfide. If 10.6 g of copper reacts with 3.83 g S how much product will be formed?
• If 10.6 g of copper reacts with 3.83 g S. How many grams of product will be formed?
• 2Cu + S Cu2S
10.6 g Cu 63.55g Cu 1 mol Cu
2 mol Cu 1 mol Cu2S
1 mol Cu2S
159.16 g Cu2S
= 13.3 g Cu2S
3.83 g S 32.06g S 1 mol S
1 mol S 1 mol Cu2S
1 mol Cu2S
159.16 g Cu2S
= 19.0 g Cu2S
= 13.3 g Cu2S
Cu is Limiting Reagent
Your turn
• If 10.1 g of magnesium and 2.87 L of HCl gas are reacted, how many liters of gas will be produced?
• How many grams of solid?
• How much excess reagent remains?
Your Turn II
• If 10.3 g of aluminum are reacted with 51.7 g of CuSO4 how much copper will be produced?
• How much excess reagent will remain?
Yield
• The amount of product made in a chemical reaction.
• There are three types
• Actual yield- what you get in the lab when the chemicals are mixed
• Theoretical yield- what the balanced equation tells you you should make.
• Percent yield = Actual x 100 % Theoretical
Example
• 6.78 g of copper is produced when 3.92 g of Al are reacted with excess copper (II) sulfate.
• 2Al + 3 CuSO4 Al2(SO4)3 + 3Cu
• What is the actual yield?
• What is the theoretical yield?
• What is the percent yield?
Details
• Percent yield tells us how “efficient” a reaction is.
• Percent yield can not be bigger than 100 %.
The Empirical Formula• The lowest whole number ratio of elements
in a compound.
• The molecular formula the actual ratio of elements in a compound.
• The two can be the same. –CH2 empirical formula–C2H4 molecular formula–C3H6 molecular formula–H2O both
Calculating Empirical• Just find the lowest whole number ratio
• C6H12O6
• CH4N
• It is not just the ratio of atoms, it is also the ratio of moles of atoms.
• In 1 mole of CO2 there is 1 mole of carbon and 2 moles of oxygen.
• In one molecule of CO2 there is 1 atom of C and 2 atoms of O.
Calculating Empirical• We can get ratio from percent composition.
• Assume you have a 100 g.
• The percentages become grams.
• Can turn grams to moles.
• Find lowest whole number ratio by dividing by the smallest.
Example
• Calculate the empirical formula of a compound composed of 38.67 % C, 16.22 % H, and 45.11 %N.
• Assume 100 g so
• 38.67 g C x 1mol C = 3.220 mole C 12.01 gC
• 16.22 g H x 1mol H = 16.09 mole H 1.01 gH
• 45.11 g N x 1mol N = 3.219 mole N 14.01 gN
Example
• The ratio is 3.220 mol C = 1 mol C 3.219 mol N 1 mol N
• The ratio is 16.09 mol H = 5 mol H 3.219 mol N 1 mol N
• C1H5N1
• A compound is 43.64 % P and 56.36 % O. What is the empirical formula?
• Caffeine is 49.48% C, 5.15% H, 28.87% N and 16.49% O. What is its empirical formula?
Empirical to molecular• Since the empirical formula is the lowest ratio
the actual molecule would weigh more.
• By a whole number multiple.
• Divide the actual molar mass by the the mass of one mole of the empirical formula.
• Caffeine has a molar mass of 194 g. what is its molecular mass?