Percent by Mass & Empirical Formulas

Mass percent basics and conversion factors

• In percent-by-mass problems, the denominator is 100. If you change the denominator, you complicate things; best to start from a base of $100\%$ or $100\ \text{g}$ to extract conversion factors easily.
• A conversion factor comes directly from the data (percentages). For example, a data point like 25% corresponds to a factor $\frac{25}{100}$ or, when using masses, $\frac{25\ \text{units}}{100\ \text{units}}$.
• You can pull conversion factors for any pair of units provided they are both mass units (lbs, kg, g, mg, µg, etc.).
• The pie chart or data chart is just a representation of these conversion factors; treat it as a source of ratios, not something scary.

Dimensional analysis with mass-to-mass conversions

• Start with a known quantity and multiply by a conversion factor to reach the goal quantity, ensuring units cancel.
• Example structure (conceptual): starting amount × (desired unit / starting unit) = target amount, with units canceling along the way.
• Important practice: attach numbers to their units (e.g., $85\ \text{mg}$ must go with $\text{mg}$, not separated). This avoids unit-flip errors.
• When converting via percentages from a pie chart, you can form factors like $\frac{\text{volume of A}}{\text{volume of total}} = \frac{X}{100}$, then use the appropriate starting volume to get the target volume.

Practical examples (soil and mineral matter)

• Problem setup: given mineral matter per soil, convert to soil mass using the ratio from data.
• Key steps:
  • Pick a starting point (e.g., 100 mg of mineral matter or 100 mg soil).
  • Pull the conversion factor from the data (e.g., from the pie chart).
  • Set up dimensional analysis so the unwanted quantity cancels and the desired quantity remains.
  • Compute the target mass (e.g., soil mass) from the starting amount.
• Takeaway: the pie chart is just a streamlined way to read the conversion factors; the math is standard dimensional analysis with consistent mass units.

Equal masses vs equal moles (big caution)

• Equal masses do not imply equal moles in general.
• Equal moles do not imply equal masses either.
• Always base calculations on moles when using atomic/molar masses, and on masses when using percent-by-mass data, then connect them via molar masses.

Empirical formulas: concept and workflow

• Empirical formula = simplest whole-number mole ratio of elements in a sample.
• When you see empirical formula problems, think in terms of moles.
• Two common pathways:
  • Path A: start with a logical amount (often 100 g) to convert to moles and find the simplest ratio.
  • Path B: use direct mole ratios from the compound’s formula (e.g., CuSO$4$ has 1 S per 1 CuSO$4$; 4 O per CuSO$_4$).
• Steps to obtain empirical formula from percent composition:
  1) Assume a sample size (commonly $100\ \text{g}$). Then the mass of each element equals its percent.
  2) Convert each element mass to moles: $n<em>i = \frac{m</em>i}{M<em>i}$ where $M</em>i$ is the atomic/molar mass.
  3) Divide all $n<em>i$ by the smallest $n</em>i$ to get a simple whole-number ratio.
  4) Write the empirical formula with those integers.
• Important note: if you get fractions, divide all by the smallest fraction to simplify, then round to the nearest whole numbers as appropriate.

Example: empirical formula from a copper sulfate problem

• Given a pure CuSO$_4$ sample, to find sulfur mass percent:
  • Start with a convenient amount, say $100\ \text{g CuSO}_4$.
  • Use the formula to relate moles: 1 mole CuSO$4$ contains 1 mole S, so the moles of S equal moles of CuSO$4$.
  • Molar masses: $M<em>{CuSO</em>4} \approx 159.6\ \text{g/mol},\ M_{S} \approx 32.07\ \text{g/mol}.$
  • Moles: $n<em>{CuSO</em>4} = \frac{100}{159.6},\ n<em>{S} = n</em>{CuSO_4}.$
  • Mass of S: $m<em>S = n</em>S \times M_S.$
  • Percent S: $\%S = \frac{m_S}{100}\times 100\%.$
• Alternative path (mole ratio): use the known ratio from the formula, e.g., 4 O per CuSO$_4$ to find oxygen mass fraction, then convert to percent by mass.
• Resulting percentage for S in CuSO$_4$ isapproximately 20% (example value from the session). The exact number follows from the same steps above.

Empirical vs molecular formula

• Empirical formula shows the simplest whole-number ratio (e.g., C:H:O = 1:1:2 for some compounds).
• Molecular formula shows the actual numbers of atoms in a molecule and may be a multiple of the empirical formula.
• Example:
  • If empirical formula is CH$2$O, a molecule could be CH$2$O, C$2$H$4$O, C$3$H$6$O, etc.
  • To determine the actual molecular formula, compare the compound’s true molar mass to the empirical formula mass and multiply the empirical formula by the appropriate integer factor.
• Common pitfall: an empirical formula like HO is not necessarily the actual formula; it’s the simplest ratio. The actual molecule could be H$2$O, HO$2$, etc., depending on the molar mass.

Quick guidelines and recap

• When you see a percentage problem, think in terms of 100 units as your starting point to extract conversion factors.
• Always use the proper units; attach numbers to their units so cancellations occur correctly in dimensional analysis.
• For mass percent problems, you can start with 100 g or 100% and work with moles via molar masses to get empirical formulas.
• Remember: equal masses ≠ equal moles; equal moles ≠ equal masses.
• Use empirical formulas to identify the simplest ratio, then determine the molecular formula by matching molar masses.

Quick reference formulas

• Moles to mass: $n<em>i = \frac{m</em>i}{M_i}$
• Mass to moles (element): $n<em>i = \frac{m</em>i}{M_i}$
• Empirical formula from moles: divide all $n_i$ by the smallest to get integers.
• Empirical to molecular: $M<em>{Molecule} = n \times M</em>{Empirical}$; solve for integer $n$ by comparing with the actual molar mass.
• For a compound with known formula ratio, e.g., CuSO$4$: there are 4 O per CuSO$4$; 1 S per CuSO$_4$.