Ch. 5 Stoichiometry

Conversion Factors

mole (units: mol): chemistry unit measurement representing $6.05\cdot10^{23}$ atoms or molecules of a substance
Avogadro's number: $6.02\cdot10^{23}$
molar mass: gram amount of each mole of a substance; equal to atomic mass on periodic table
- molar mass of compounds is the sum of molar masses of each atom in the compound
the number of atoms in a substance is in the right ratio
the number of atoms/molecules in a balanced chemical equation is in the right ratio

Example

In the equation $\ce{2C6H6 + 15O2 -> 12CO2 + 6H2O}$ :
$\pu{2 mol } \ce{C6H6} = \pu{15 mol } \ce{O2}$
$\pu{5 mol } \ce{O2} = \pu{4 mol } \ce{CO2}$
Other ratios similarly apply

Molarity (units: $M=\pu{mol/L}$ ): moles of a solute dissolved per liter of solution
Density (units: $\pu{g/cm^3}$ or $\pu{g/mL}$ ): grams of a substance per cubic centimeter (or milliliter)

Possible conversions

Example 1

How many $\ce{N}$ atoms are in $\pu{1000 g}$ of $\ce{(NH4)3PO4}$ ?

Convert grams of the molecule to moles of the molecule using the molar mass ( $\pu{149.1 g/mol}$ ),
then to moles of nitrogen ( $\pu{3 mol \ce{N}}/\pu{mol \ce{(NH4)3PO4}}$ ), then to atoms.
$\pu{1000 g \ce{(NH4)3PO4}}\cdot\frac{\pu{1 mol \ce{(NH4)3PO4}}}{\pu{149.1 g \ce{(NH4)3PO4}}} \cdot \frac{\pu{3 mol \ce{N}}}{\pu{1 mol \ce{(NH4)3PO4}}}\cdot \frac{6.02\cdot10^{23} \pu{\ce{N} atoms}}{\pu{mol \ce{N}}}=1.21\cdot10^{25} \pu{\ce{N} atoms}$

Example 2

How many moles of water will be formed from the complete combustion of $\pu{2.50 mol}$ of $\ce{CH4}$ ?

The balanced equation will be
$\ce{CH4 + 2O2 -> CO2 + 2H2O}$

So, we have
$\pu{2.50 mol \ce{CH4}}\cdot \frac{\pu{2 mol \ce{H2O}}}{\pu{1 mol \ce{CH4}}} = \pu{5 mol \ce{H2O}}$

Limiting Reactants

some problems ask to determine the limiting reactants: the substance that is totally consumed in a reaction
the other substance is the excess reactant(s)

Example

If $\pu{45.5 mL}$ of $0.200 M \ce{AgNO3}$ (molar mass $= \pu{169.9 g/mol}$ ) is mixed with $\pu{35.8 mL}$ of $0.436 M \ce{Na2CrO4}$ (molar mass $= \pu{161.9 g/mol}$ ), how many grams of the precipitate $\ce{Ag2CrO4}$ (molar mass $= \pu{331.7 g/mol}$ )?

The balanced chemical equation is
$\ce{2AgNO3(aq) + Na2CrO4(aq) -> Ag2CrO4(s) + 2NaNO3(aq)}$
We must determine which reactant would produce a lower theoretical hield of silver chromate.
The silver nitrate is completely consumed with
$\pu{45.5 mL \ce{AgNO3}}\cdot \frac{\pu{0.200 mol \ce{AgNO3}}}{\pu{1000 mL \ce{AgNO3}}}\cdot \frac{\pu{1 mol \ce{Na2CrO4}}}{\pu{2 mol \ce{AgNO3}}}\cdot \frac{\pu{1000 mL \ce{Na2CrO4}}}{\pu{0.436 mol \ce{Na2CrO4}}}=\pu{10.4 mL \ce{Na2CrO4}}$
Since we have more sodium chromate than that, the limiting reactant must be silver nitrate.
Thus, we convert silver nitrate to silver chromate
$\pu{45.5 mL \ce{AgNO3}}\cdot \frac{\pu{0.200 mol \ce{AgNO3}}}{\pu{1000 mL \ce{AgNO3}}}\cdot \frac{\pu{1 mol \ce{Ag2CrO4}}}{\pu{2 mol \ce{AgNO3}}}\cdot \frac{\pu{331.7 g \ce{Ag2CrO4}}}{\pu{1 mol \ce{Ag2CrO4}}}=\pu{1.51 g \ce{Ag2CrO4}}$

Titration

a technique to determine the concentration of a substance of an unknown concentration
done by adding a controlled amount of an indicator until the color changes, indicating a completed reaction

Example

What is the molarity of a $\pu{30.00 mL}$ solution of $\ce{Fe^2+}$ if it takes $\pu{4.53 mL}$ of $\pu{0.687 } M$ $\ce{MnO4-}$ to titrate it?
$\ce{5Fe^2+ + MnO4- + 8H+ -> Mn^2+ + 5Fe^3+ + 4H2O}$

We convert $\pu{0.687 } M$ $\ce{MnO4-}$ to molarity of $\ce{Fe^2+}$
$\frac{\pu{0.687 mol \ce{MnO4-}}}{\pu{1000 mL \ce{MnO4-}}}\cdot \frac{\pu{5 mol \ce{Fe^2+}}}{\pu{1 mol \ce{MnO4-}}}\cdot \frac{\pu{4.53 mL \ce{MnO4-}}}{\pu{30.00 mL \ce{Fe^2+}}}\cdot \frac{\pu{1000 mL \ce{Fe^2+}}}{\pu{1 L \ce{Fe^2+}}}=\pu{0.519 mol/L \ce{Fe^2+}}=\pu{0.519 } M \ce{Fe^2+}$

Ch. 5 Stoichiometry

Conversion Factors

Limiting Reactants

Titration

Other Types of Problems

Percent Composition

Empirical Formulas

Molecular Formulas