According to a set of standards, a certain type of substance can contain a maximum of 0.001% phosphorus by mass....

1. TRANSLATE the problem information

• Given information:
  • Maximum allowable phosphorus: \(\mathrm{0.001\%}\) by mass
  • Sample mass: \(\mathrm{140}\) grams
  • Need to find: Maximum mass of phosphorus in this sample

• What this tells us: We need to calculate \(\mathrm{0.001\%}\) of \(\mathrm{140}\) grams

2. TRANSLATE the percentage to decimal form

• Convert \(\mathrm{0.001\%}\) to decimal: \(\mathrm{0.001\% = 0.001/100 = 0.00001}\)
• This decimal represents the fraction of the total mass that can be phosphorus

3. SIMPLIFY by calculating the maximum phosphorus mass

• Maximum phosphorus mass = \(\mathrm{0.00001 \times 140}\)
• Using calculator: \(\mathrm{0.00001 \times 140 = 0.0014}\) grams

Answer: \(\mathrm{0.001}\) grams (or .0014)

Why Students Usually Falter on This Problem

Most Common Error Path:

Weak TRANSLATE skill: Students confuse percentage conversion, treating \(\mathrm{0.001\%}\) as \(\mathrm{0.001}\) instead of \(\mathrm{0.00001}\)

They might calculate: \(\mathrm{0.001 \times 140 = 0.14}\), which is 100 times too large. This leads to confusion since this answer seems unreasonably high for such a small allowable percentage, causing them to abandon systematic solution and guess.

Second Most Common Error:

Poor SIMPLIFY execution: Students correctly identify the need to find \(\mathrm{0.001\%}\) of \(\mathrm{140}\) but make decimal multiplication errors

They might misplace decimal points during calculation, getting answers like \(\mathrm{0.14}\) or \(\mathrm{0.014}\) instead of \(\mathrm{0.0014}\). This computational error leads to selecting an incorrect numerical answer.

The Bottom Line:

The key challenge is managing very small percentages and their decimal conversions. Students must be extremely careful with decimal place values when converting tiny percentages (\(\mathrm{0.001\%}\) requires moving the decimal point 5 places to the left total).