College Chemistry - Unit 9: Stoichiometry

trying out this knowt thing, seems kinda good so far though

stoichiometry

a branch of chemistry concerned with the relative amounts of reactants and products in a chemical reaction

balanced equation

coefficients show the mole proportion in each compound; must have same number of each type of atom on both sides of the equation

how to balance an equation

count the number of each atom on both sides

systematically adjust the coefficients to balance each atom

recount after each adjustment

repeat until balanced

special case scenarios when balancing equations

if an atom is in both products, count from both places and save it for last

if a polyatomic ion is unchanged between reactants and products, count it as a unit

if OH is one side and H2O is on the other, count H2O as H-OH

if it is a combustion reaction (products are CO2 and H2O), balance O last and use the fraction method

fraction method

use when there is a combustion reaction

the coefficient for O will be a fraction

multiply by 2 to cancel out the fraction

mole ratio

coefficients in a balanced equation

[coefficient of the unknown] / [coefficient of the given]

mass stoichiometry

grams of given → moles given using molar mass

moles given → moles answer using mole ratio

moles answer→ grams answer using molar mass

volume stoichiometry (gases)

L given → moles given using ideal gas law (PV = nRT)

moles given → moles answer using mole ratio

moles answer → L answer using ideal gas law

R = .08206 for atm, 8.31 for kPa

at standard temperature (273 K) and pressure (1 atm) [STP]: 1 mol = 22.4L

volume stoichiometry (solutions)

L given → moles given using molarity (mol/L)

moles given → moles answer using mole ratio

moles answer → L answer using molarity

limiting reactant

when reactants are not included in the exact proportions of the balanced equation, the reactant that is entirely consumed

limiting reactant calculation

convert given → moles given → moles answer

whichever is least is the limiting reactant

complete the stoichiometry calculation to determine how much product can be made

theoretical yield

found by doing a stoichiometry calculation

actual yield

found by measuring

percent yield

[actual yield] / [theoretical yield] * 100 to make it a percent

theoretical yield should be in same units as actual yield

combustion analysis

used to find unknown compounds contain carbon and hydrogen by combining them with oxygen

steps for combustion reaction

grams CO2 → mol CO2 (molar mass), mol CO2 → mol C (formula ratio: 1 C to 1 CO2)

grams H2O → mol H2O (molar mass), mol H2O → mol H (formula ratio: 2 H to 1 H2O)

divide moles of each element by the smallest mole value → use for the subscripts of the empirical formula

molecular formula

shows the exact number of each element in a compound

ex: molecules

empirical formula

shows the simplest formula of elements in a compound;

ex: ionic compounds

converting empirical formula to molecular formula

calculate the molar mass of the empirical formula

find n ([molar mass from molecular formula] / [molar mass from the empirical formula])

n * [empirical formula] = [molecular formula]

heteroatom

additional atoms other than C and H, like N, O, S, and/or halogens, in combustion analysis

combustion analysis with heteroatoms

grams CO2 → mol CO2 (molar mass) → mol C (formula ratio) → grams C (molar mass)

grams H2O → mol H2O (molar mass) → mol H (formula ratio) → grams H (molar mass)

mass (sample) - [mass (C) + mass (H)] = mass of heteroatom

grams heteroatom → mol heteroatom (molar mass)

divide moles of each element by moles of heteroatom

find molecular formula