Density Calculations and Factor-Label Method Study Notes

Liquid solutions dissolved in water exhibit a density very close to $1\,\text{g/mL}$ .
Standard Equivalence: $1\,\text{g}$ of water is equivalent to $1\,\text{mL}$ of water ( $1\,\text{g} = 1\,\text{mL}$ ).
Experimental Measurement: Density is a measured physical property rather than an exact defined quantity.
Significant Figures Rules for Water Density:
- When performing calculations, the density of water is treated as having three significant figures: $1.00\,\text{g/mL}$ .
- The volume denominator of $1\,\text{mL}$ in density expressions is defined exactly ( $1\,\text{mL}$ exact).
- In a working expression such as $\frac{1.00\,\text{g H}_2\text{O}}{1\,\text{mL H}_2\text{O}}$ , three significant figures are in effect for the density relationship.

Purpose: The factor-label method (dimensional analysis) uses units to establish mathematical operations for problem-solving.
Rules for Setup:
1. Identify the requested target unit.
2. Identify all provided given information.
3. Single-Unit Rule: When presented with more than one piece of information, always start the calculation with the quantity that possesses a single unit (e.g., mass in $\text{g}$ or volume in $\text{L}$ ), rather than a compound unit.
Properties of Compound Units:
- Any quantity expressed with two units (e.g., $\text{g/mL}$ ) functions as a conversion factor.
- Conversion factors can be written in two reciprocal forms depending on unit cancellation requirements:
- Form 1: $\frac{2.698\,\text{g}}{1\,\text{mL}}$
- Form 2: $\frac{1\,\text{mL}}{2.698\,\text{g}}$
- Starting a problem directly with a conversion factor leads to selecting the incorrect orientation $50\%$ of the time. Starting with the single-unit quantity allows unit cancellation to dictate the correct orientation.

Problem Context: Determine the volume in milliliters ( $\text{mL}$ ) contained in one can of Coke made of aluminum.
Given Data:
- Density of aluminum: $2.698\,\text{g/mL}$
- Mass of an empty Coke can: $16.38\,\text{g}$
Resolution Procedure:
- Target Unit: Milliliters ( $\text{mL}$ ).
- Starting Quantity: $16.38\,\text{g}$ (the given value with a single unit).
- Apply Conversion Factor: Orient the density of aluminum so that grams appear on the bottom to cancel given grams: $\text{Volume} = 16.38\,\text{g} \times \frac{1\,\text{mL}}{2.698\,\text{g}}$
- Calculation: $\text{Volume} = \frac{16.38}{2.698}\,\text{mL} = 6.071\,\text{mL}$

Evaluation of Incorrect Method:
- Starting directly with density or dividing mass incorrectly yields: $\frac{2.698\,\text{g/mL}}{16.38\,\text{g}} = 0.1647\,\frac{1}{\text{mL}}$
Unit Errors vs. Target Units:
- The resulting unit $\frac{1}{\text{mL}}$ (inverse milliliters) is mathematically distinct from $\text{mL}$ .
- Just as $\frac{1}{3} \neq 3$ , $\frac{1}{\text{mL}} \neq \text{mL}$ .
Physical Plausibility Check:
- A calculated volume of $0.1647\,\text{mL}$ corresponds to approximately $15$ drops of liquid.
- Fabricating an aluminum beverage can out of only $15$ drops of aluminum would result in an impossibly thin structure, demonstrating that $6.071\,\text{mL}$ is the physically reasonable answer.
Mathematical Relationship via Reciprocals:
- Taking the reciprocal (inverse) of the inverted unit result converts $\frac{1}{\text{mL}}$ back to $\text{mL}$ : $\frac{1}{0.1647\,\text{mL}^{-1}} = 6.071\,\text{mL}$

Problem Context: Calculate the mass in grams ( $\text{g}$ ) of ethanol needed to obtain a target volume of $2.0\,\text{L}$ of ethanol.
Given Data:
- Target volume: $2.0\,\text{L}$
- Density of ethanol: $0.789\,\text{g/mL}$
Resolution Procedure:
1. Target Unit: Grams ( $\text{g}$ ).
2. Starting Value: $2.0\,\text{L}$ (single-unit quantity).
3. Unit Conversion Step: Convert volume from liters ( $\text{L}$ ) to milliliters ( $\text{mL}$ ) to align with the density units: $\text{Conversion Factor: } 1\,\text{mL} = 0.001\,\text{L}$ $\text{Volume in mL} = 2.0\,\text{L} \times \frac{1\,\text{mL}}{0.001\,\text{L}} = 2000\,\text{mL}$
4. Apply Density Conversion Factor: $\text{Mass} = 2000\,\text{mL} \times \frac{0.789\,\text{g}}{1\,\text{mL}} = 1578\,\text{g}$
Estimation and Plausibility Verification:
- $2.0\,\text{L}$ equals $2000\,\text{mL}$ .
- If density were $1.0\,\text{g/mL}$ , $2000\,\text{mL}$ would yield $2000\,\text{g}$ .
- Because the density of ethanol ( $0.789\,\text{g/mL}$ ) is slightly less than $1.0\,\text{g/mL}$ , the calculated mass should be slightly less than $2000\,\text{g}$ . The result of $1578\,\text{g}$ fits this ballpark.
Application of Significant Figures:
- The starting measurement $2.0\,\text{L}$ contains two significant figures.
- The calculated value $1578\,\text{g}$ must be rounded to two significant figures.
- Final Correct Answer: $1600\,\text{g}$ .

Question: Is the density of water treated as having three significant figures based on the numbers used?
Answer: Yes. When using the density of water in working problems, it is treated as having three significant figures ( $1.00\,\text{g/mL}$ ). This value is measured rather than defined. However, the $1\,\text{mL}$ in the density expression is defined exactly.