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- Stoichiometry: Calculations with
Chemical Formulas and Equations
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- A chemical reaction is written down in the form of a chemical equation.
- A chemical equation shows how substances react to form new substances.
- The substances that react are called reactants.
- The substances formed are called products.
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- A chemical equation also shows the relative amounts of reactants
consumed and products formed.
- Stoichiometry is concerned with the relative amounts of reactants
consumed and products formed.
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- Reactants on left side of arrow.
- Products on right side of arrow.
- Phases
- (s) solid
- (l) liquid
- (g) gas
- (aq) aqueous (water solution)
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- Because atoms cannot be created nor destroyed in a chemical reaction,
the total number of each type of atom on both sides of the arrow must be
equal.
- When this is true, the equation is said to be balanced.
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- Chemical equations are balanced by changing coefficients on each species
until the numbers of each type of atom is the same on both sides.
- Never change the subscripted numbers on the formulas when balancing
equations!!
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- Start with the most complicated formula in the equation and assume there
is a “1” in front of it.
- Change all other coefficients to equalize the number of atoms on both
sides of the arrows.
- If you get a fraction, multiply the whole equation by a number to
eliminate the fraction.
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- Balance the following equations:
- H2(g) + Br2(g) → HBr(g)
- K(s) +
H2O(l) →
KOH(aq) + H2(g)
- Al(s) +
H2SO4(aq) → Al2(SO4)3(aq)
+ H2(g)
- S(s)+ HNO3(aq) → H2SO4(aq)+ NO2(g)
+ H2O(l)
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- Combination reactions: 2 or more
reactants combine to form one product.
- 2 Na(s) + Cl2(g) ¾® 2 NaCl(s)
- Decomposition reactions: One
substance breaks up into 2 or more products.
- 2 HgO(s) ¾® 2 Hg(l) +
O2(g)
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- Combustion reactions: reaction of
a substance with oxygen (O2(g)) in a flame.
- Usually involves a hydrocarbon, which reacts with oxygen to form carbon
dioxide and water.
- 2 C8H18(g)
+ 25 O2(g) ¾®
16 CO2(g)
+ 9 H2O(l)
- 2 Mg(s) + O2(g) ¾® 2 MgO(s)
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- The formula weight of a compound is the sum of the atomic weights in its
chemical formula
- Al2(SO4)3
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- The formula weight for a molecule is also called its molecular weight.
- Important note!!!
- Ionic compounds are not considered to be molecules. Ionic compounds consist of ions and
generally contain metal and nonmetal atoms.
- Molecules consist of nonmetal atoms only.
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- What is the formula weight for iron(II)gluconate, Fe(C6H11O7)2
?
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- The composition of a compound may also be expressed in terms of percent
composition by mass of each element in the compound.
- It is defined as shown below:
- Given the formula of a compound, you should be able to calculate percent
by mass of each element.
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- What is the mass percent of cobalt in cobalt(III)nitrate, Co(NO3)3?
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- Amounts of substances in chemical reactions are expressed in units
called “moles”.
- One mole of a substance contains exactly the same number of particles
(atoms, molecules, formula units, etc.) as the number of atoms in
exactly 12 grams of 12C.
- The number of particles in a mole is called Avogadro’s Number:
- NA= 6.022 × 1023 mol-1
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- Mass of one mole of a substance is equal to its molar mass.
- The molar mass of a substance is equal to its formula weight (or atomic
weight for an atom) with units of g/mol (grams per mole).
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- Calculate the molar mass of:
- (a) N2O3;
(b) Ca(C2H3O2)2
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- Amounts of substances in lab usually determined by their mass in grams
(g).
- Molar mass is used to convert between number of moles and the mass in
grams.
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- How many moles are there in 25 g of barium fluoride, BaF2?
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- How many grams are contained in 0.275 mol of dinitrogen tetroxide, N2O4?
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- Avogadro’s number is used to convert
between number of particles (atoms, formula units, or molecules)
and number of moles.
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- Capsaicin is the compound that makes peppers hot. Suppose you have 0.00345 g of
capsaicin, C18H27NO3. How many capsaicin molecules are there
in this sample? How many carbon
atoms are in this sample?
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- Gives the elemental composition of a compound usually in percent by
mass.
- The elemental analysis can be used to determine the empirical formula of
a compound.
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- Cumene is a hydrocarbon, a compound that contains only C and H. It is 89.94%C and 10.06%H by
mass. What is the empirical
formula for cumene? The molar
mass of cumene is 120.2 g/mol, what is the molecular formula for cumene?
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- Sum of all percents is equal to 100% (one mass percent may be omitted).
- Assuming that there is exactly 100 grams of the compound, convert the
mass percents to the number of grams of each element in 100 grams of the
compound.
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- Convert the number of grams of each element into number of moles of each
element in 100 grams of the compound by dividing by the atomic weight of
that element.
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- Find the lowest whole number ratio between the number of moles of each
element.
- Divide by smallest number in the list.
- If you get a fraction, multiply all number of moles by a number that
removes the fraction
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- The smallest whole number ratio gives the empirical formula. Write the empirical formula for the
compound. For cumene, the
empirical formula is C3H4.
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- Can only obtain empirical formula from just the percents by mass.
- In order to get the molecular formula, the molar mass of the compound
must be given. For cumene, the
molar mass of cumene is 120.2 g/mol.
(from the problem)
- Determine the empirical formula weight.
Divide this formula weight into the molar mass, you should get
very close to a whole number.
- Multiply the empirical formula by this whole number to get the molecular
formula.
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- For cumene, the empirical formula is C3H4. It has an empirical formula weight of
40.06 g/mol (=3×12.01 + 4×1.008).
- Dividing into molar mass:
- Molecular formula is C9H12.
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- Butanedioic acid is 40.7%C, 5.1%H and 54.2%O by mass. It has a molar mass of 118 g/mol. What is the molecular formula for
butanedioic acid?
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- Convert the masses of CO2 and H2O into the mass of
C and H contained in the compound.
- Subtract the masses of C and H from the mass of the sample to obtain the
mass of O in the compound.
- Convert all masses to number of moles.
- Find smallest whole number ratio between the elements to get the
empirical formula.
- Use the molar mass to get the molecular formula.
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- An unknown compound is known to contain only carbon, hydrogen, and
oxygen in its formula. When
0.0956 g of the unknown compound is combusted, 0.1356 g CO2
and 0.0833 g H2O are obtained. The molar mass of the compound is 62.1
g/mol. What is the molecular
formula of the compound?
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- The coefficients in a balanced chemical reaction is called the stoichiometry
of the reaction.
- N2(g) + 3 H2(g) ¾® 2 NH3(g)
- Can obtain stoichiometrically equivalent amounts:
- 1 mol N2 = 3 mol H2 = 2 mol NH3
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- Can use stoichimetrically equivalent amounts to produce mole ratios
which can be used to convert moles of one substance to another.
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- Given the reaction of nitrogen and hydrogen to produce ammonia, how many
moles of N2 will react with 4.5 mol of H2? How many moles of NH3 will
be produced from 4.5 mol of H2?
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- Asks for the amount of one substance in a reaction given the amount of
another substance in the reaction.
- Make sure the reaction is balanced before proceeding!!
- Identify the known quantities and the unknown quantities
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- Convert the known quantity to number of moles of the known.
- Convert moles of the known to moles of the unknown using a mole ratio
obtained from the balanced chemical reaction.
- Convert moles of the unknown to the appropriate units of the unknown
quantity.
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- Iron ore is converted to iron metal in a reaction with carbon monoxide:
- 2 Fe2O3(s)
+ 3 CO(g) → 4 Fe(l) +
3 CO2(g)
- How many moles of CO are required to react with 6.2 mol Fe2O3? How many moles of Fe and CO2
will be produced in this reaction?
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- Chromium metal reacts with oxygen to form chromium(III)oxide:
- 4 Cr(s) + 3 O2(g) → 2 Cr2O3(s)
- How many grams of Cr2O3 are produced when 0.175 g
of Cr is reacted with excess oxygen?
How many grams of O2 are consumed in this reaction?
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- In most cases, a reaction is run by mixing the reactants together and
starting the reaction.
- The reaction proceeds until one of the reactants is totally used
up. This reactant is the limiting
reactant or limiting reagent.
- The amount of limiting reactant determines the amount of product made.
- The limiting reactant is the reactant that produces the least amount of
product.
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- Given a chemical reaction and the amounts of all the reactants and asked
for the maximum amount of product made.
- Calculate how much product that each reactant can make assuming there is
enough of the other reactants to react with it. (Stoichiometry problem)
- The reactant that makes the smallest amount of product is the limiting
reactant.
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- The amount of product that can be made from the limiting reactant is the
maximum amount of product made.
- Can also calculate the amounts of the other reactants, called excess
reactants, used up by a stoichiometry problem from the limiting
reactant.
- Subtracting the amount of the excess reactant used up from the initial
amount of excess reactant gives the amount of excess reactant left over
when the reaction is over.
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- Aluminum chloride is produced by treating aluminum metal with chlorine:
- 2 Al(s) + 3 Cl2(g) → 2 AlCl3(s)
- A reaction is begun with 2.70 g of Al and 4.05 g of Cl2. What is the maximum amount, in grams,
of AlCl3 produced?
Which reactant is the excess reactant and how many grams of it is
left over after the reaction is complete?
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- One of the steps in the production of nitric acid, HNO3, is
the reaction of ammonia with oxygen:
- 4 NH3(g) + 5 O2(g) → 4 NO(g) +
6 H2O(g)
- In a certain experiment, 1.50 g of NH3 is reacted with 2.75 g
of O2. What is the
maximum amount, in grams, of NO produced?
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- Yield: the amount of product
produced usually in grams.
- Theoretical Yield: the
calculated amount of product produced (by a stoichiometry problem)
- Actual Yield: the actual amount
of product produced in the laboratory.
- Actual yield is always less than or equal to theoretical yield.
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- Because actual yield is always less than or equal to theoretical yield,
then %Yield is always less than or equal to 100%. Cannot be greater than 100%!!
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- Ammonia gas can be prepared by the following reaction:
- CaO(s) + 2 NH4Cl(s) →
- 2 NH3(g) + H2O(g)
+ CaCl2(g)
- Suppose that 112 g of CaO is reacted with 224 g of NH4Cl. After the reaction is performed, only
16.3 g of NH3 is obtained from the reaction. What is the %yield for this reaction?
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Notes
