SCH3U - Unit 3 - Stoichiometry

• Accuracy - The agreement of a particular value with the true value

• Precision - The reproducibility of a value, regardless of if it is accurate

• Significant figures

  • All non-zeros are significant

  • All zeros between non-zero digits are significant

  • Leading zeros are never significant

  • Trailing zeros are only significant if there is a decimal in the number

  • Multiplication and division

    • Solve, and round the final number to the amount of significant figures in the least significant figure

  • Addition and subtraction

    • Solve, and round the final number to the last place value of the least precise value

• Scientific notation

  • Write a value as a factor with 10 to the relevant power

    • Ex. 345 = 3.45 × 10²

Lesson 1 - Percent Composition

• % of element = (Molar mass of element x # of atoms in chemical formula)/molar mass of compound

• For percent composition

  • Calculate the molar mass of the compound

    • Multiply the amu of each value by the number of atoms in the compound

    • This should be in g/mol or g x mol^-1

  • Add all of them together to find the total molar mass of the compound

  • Divide each individual mass by the total mass to find the percent that each element makes up

• Empirical Formula - The relative number of each element in a compound as the most reduced ratio

  • Convert each percent to a percent of 100 grams, if not already given in grams

  • Multiply the grams by the reciprocal of its amu in mol/gram to cancel out the grams

  • Divide all values by the solved value with least number of moles

  • If it is close to a whole number, round to the whole number

  • If it is close to the decimal of a fraction, multiply all values by the denominator of that fraction to only get whole numbers

    • Ex. 1.33 (about 4/3) can be multiplied by 3 to get approximately 4

• Molecular Formula - The actual ratio of elements in a given formula

  • Need the empirical formula and molar mass

    • It is always a whole-number multiple of the ratio in the empirical formula

  • Find the total molecular mass by multiplying each amu, in g/mol, by the number of atoms of the element present in the empirical formula, and add all values together

  • Divide the given molecular mass by the empirical mass to find the multiple of the ratio

  • Multiply each element in the empirical formula by the ratio to get the molecular formula

Lesson 2 - The Mole, “n”

• A balanced chemical equation represents the relationships between reactants and products, as well as atoms and molecules

  • Ex. In CO₂, there is one carbon atom per every molecule, and three molecules total per every molecule

• Mole - The amount of a substance that contains the same number of entities as there are atoms in exactly 12 grams of carbon 12, represented by Avogadro’s number

  • Avogadro’s Number - 1 mol = 6.022 × 10²³ entities

• Molar Mass - The mass of one mole of an element, represented by the mass on the periodic table in grams

• Equations:

  • n = m/M → mole = mass / molar mass

  • n = N/N_A → mole = number of entities / Avogadro’s number

  • n is the moles (mol)

  • m is the mass, in grams

  • M is the molar mass, in g/mol

  • N is the number of entities (amount of molecules or ions)

  • N_A is Avogadro’s number, 6.022 × 10²³

• Solving:

  • When finding the number of atoms in a molecule given the mass, in grams

    • Calculate the molar mass

    • Divide the mass by the molar mass to find moles (n = m/M)

    • Multiply the moles by Avogadro’s number to find the total number of entities (n = N/N_A)

    • Multiply the number of entities by the number of atoms per molecule

  • When finding the mass of an element needed to balance an equation, given the mass of the other reactants

    • Find the molar mass of the other reactants

    • Divide the mass by the molar mass to find the moles of the other reactants (n = m/M)

    • Multiply the ratio of the other reactants to unknown element in the initial equation