CHEM 101 Chapter 01 Notes: The Study of Change — 1.7 Measurement, 1.8 Handling Numbers, 1.9 Dimensional Analysis

Measurement and Quantities (1.7)

• Different instruments measure different properties in chemistry labs:

  • Meterstick measures length.

  • Buret, pipet, graduated cylinder, and volumetric flask measure volume.

  • Balance measures mass.

  • Thermometer measures temperature.

• Two types of properties measured by tools:

  • Macroscopic: can be determined directly (visible to naked eye).

  • Microscopic: on atomic/molecular scale; require indirect methods.

• Measured quantities are expressed as a number followed by a unit.

  • Example: $16.43\;\mathrm{g}$

• Concepts of measurement foundations and units are built on the SI system (Systeme International).

SI Base Units (Table 1.2) and Prefixes (Table 1.3)

• SI Base Units (base quantities):

  • Length: $\mathrm{meter}\quad (m)$

  • Mass: $\mathrm{kilogram}\quad (kg)$

  • Time: $\mathrm{second}\quad (s)$

  • Electrical current: $\mathrm{ampere}\quad (A)$

  • Temperature: $\mathrm{kelvin}\quad (K)$

  • Amount of substance: $\mathrm{mole}\quad (mol)$

  • Luminous intensity: $\mathrm{candela}\quad (cd)$

• Common prefixes (examples):

  • tera- (T): $10^{12}$

  • giga- (G): $10^{9}$

  • mega- (M): $10^{6}$

  • kilo- (k): $10^{3}$

  • deci- (d): $10^{-1}$

  • centi- (c): $10^{-2}$

  • milli- (m): $10^{-3}$

  • micro- (\mu): $10^{-6}$

  • nano- (n): $10^{-9}$

  • pico- (p): $10^{-12}$

• Examples:

  • $1\;\mathrm{terameter}\ (\mathrm{Tm}) = 1\times 10^{12}\;\mathrm{m}$

  • $1\;\mathrm{gigameter}\ (\mathrm{Gm}) = 1\times 10^{9}\;\mathrm{m}$

  • $1\;\mathrm{kilometer}\ (\mathrm{km}) = 1\times 10^{3}\;\mathrm{m}$

Macroscopic vs Microscopic Measurements

• Instruments provide measurements for:

  • Macroscopic properties: direct measurements (e.g., length, volume visible at a glance).

  • Microscopic properties: require indirect methods (e.g., molecular-scale properties).

Volume and Mass Conversions (Volume focus)

• Common volume units and relationships:

  • $1\ \mathrm{cm^3} = 1\ \mathrm{mL}$

  • $1\ \mathrm{L} = 1\ \mathrm{dm^3}$

  • $1\ \mathrm{m^3} = 10^{3}\ \mathrm{L}$

  • $1\ \mathrm{m^3} = 10^{6}\ \mathrm{cm^3}$

• Practical conversions:

  • $1\ \mathrm{m^3} = (10^2\ \mathrm{cm})^3 = 10^6\ \mathrm{cm^3}$

  • $1\ \mathrm{m^3} = (10^{1}\ \mathrm{dm})^3 = 10^{3}\ \mathrm{dm^3}$

Density and Mass–Volume Relations

• Density is a derived quantity:$d = \frac{m}{V}$

• SI derived unit for density: $\mathrm{kg\;m^{-3}}$

• Common conversions for density:

  • $1\ \mathrm{g\;cm^{-3}} = 1\ \mathrm{g\;mL^{-1}} = 10^{3}\ \mathrm{kg\;m^{-3}}$

  • Therefore, $1\ \mathrm{g\;cm^{-3}} = 10^{3}\ \mathrm{kg\;m^{-3}}$

• Example: Given density and volume,

  • If density is $d = 21.5\;\mathrm{g\;cm^{-3}}$ and volume is $V = 4.49\;\mathrm{cm^{3}}$, then mass is

  • $m = d\times V = 21.5\;\mathrm{g\;cm^{-3}} \times 4.49\;\mathrm{cm^{3}} \approx 96.5\;\mathrm{g}$

• Example: Gold density problem (Example 1.1):

  • Given: mass $m = 301\;\mathrm{g}$, volume $V = 15.6\;\mathrm{cm^{3}}$

  • Density: $d = \frac{m}{V} = \frac{301}{15.6}\approx 19.3\;\mathrm{g\;cm^{-3}}$

Example Calculations: Density

• Example 1.1 (Gold):

  • Compute density: $d = \frac{m}{V} = \frac{301}{15.6} \approx 19.3\;\mathrm{g\;cm^{-3}}$

• Example 1.2 (Mercury):

  • Density: $d = 13.6\;\mathrm{g\;mL^{-1}}$

  • Mass for $V = 5.50\;\mathrm{mL}$: $m = d\times V = 13.6\times 5.50 = 74.8\;\mathrm{g}$

Temperature Scales and Conversions

• Temperature scales:

  • Kelvin (K) is the absolute scale; no degree symbol; 0 K is absolute zero.

  • Celsius (°C) and Fahrenheit (°F) are related via conversion formulas.

• Key conversion formulas:

  • $K = C + 273.15$

  • $F = \frac{9}{5}C + 32$

  • $C = \frac{5}{9}(F - 32)$

• Quick notes:

  • Kelvin temperatures cannot be negative.

  • 0 K is the lowest possible temperature theoretically attainable.

• Examples:

  • Convert $172.9^{\circ}\mathrm{F}$ to °C:

  • $C = \frac{5}{9}(F - 32) = \frac{5}{9}(172.9 - 32) \approx 78.3^{\circ}\mathrm{C}$

  • Convert (a) -141°C to °F:

  • $F = \frac{9}{5}(-141) + 32 \approx -95.8^{\circ}\mathrm{F}$

  • Convert (b) -452°F to °C:

  • $C = \frac{5}{9}(-452 - 32) = \frac{5}{9}(-484) \approx -268.9^{\circ}\mathrm{C}$

  • Convert (c) Mercury melting point -38.9°C to K:

  • $K = C + 273.15 = -38.9 + 273.15 = 234.25\ \,\mathrm{K}$

Scientific Notation

• Purpose: express very large or very small numbers compactly.

• Form: $m \times 10^{n}$ where 1 ≤ m < 10 and n is an integer.

• Examples:

  • Carbon: atoms in 12 g of carbon ~$6.022\times 10^{23}$ atoms.

  • Mass of a single carbon atom: $1.99\times 10^{-23}\;\mathrm{g}$

• Useful rules:

  • Positive exponent means number > 1; negative exponent means number < 1.

  • Moving decimal point corresponds to adjusting the exponent.

• Common conversions:

  • $1.5\times 10^{4} = 15000$

  • $5.8\times 10^{-4} = 0.00058$

• Calculator usage:

  • express $3.45\times 10^{6}$ as a decimal: on calculator use 3.45\times 10^6 or 3.45E6 → $3.45\times 10^{6}$

• Scientific notation to numbers examples:

  • $5.68762\times 10^{2}\;=\;568.762$

  • $7.72\times 10^{-6}\;=\;0.00000772$

Handling Numbers (1.8)

• In addition/subtraction:

  1. Write each quantity with the same exponent (align decimal places).

  2. Add/subtract the decimal numbers.

  3. Keep the least number of decimal places from the inputs.

  • Example: $4.31\times 10^{4} + 0.39\times 10^{4} = 4.70\times 10^{4}$

  • Practical steps:

  • Identify the smaller exponent and adjust the decimal point accordingly.

• In multiplication:

  1. Multiply the coefficients and add exponents.

  2. Maintain the same power-of-10 scale in the result.

  • Example: $(4.0\times 10^{-5})\times(7.0\times 10^{3}) = (4.0\cdot 7.0)\times 10^{-5+3} = 28\times 10^{-2} = 2.8\times 10^{-1}$

• In division:

  1. Divide the coefficients and subtract exponents.

  • Example: $(8.5\times 10^{4})\div (5.0\times 10^{9}) = (8.5/5.0)\times 10^{4-9} = 1.7\times 10^{-5}$

• Significance and uncertainty in numbers are governed by significant figures rules (see below).

Significant Figures

• Definition: significant figures are the digits that carry meaning about precision; the last digit is uncertain.

• Rule: certain digits plus the first uncertain digit determine the number of significant figures.

• Basic rules:

  • Any nonzero digit is significant: 1.234 kg has 4 sig figs.

  • Zeros between nonzero digits are significant: 606 m has 3 sig figs.

  • Leading zeros are not significant: 0.08 L has 1 sig fig.

  • Trailing zeros to the right of the decimal point are significant: 2.0 mg has 2 sig figs.

  • Zeros to the right of the first nonzero digit in numbers less than 1 are not always significant unless specified by a decimal point: 0.004020 g has 4 sig figs.

  • In numbers without a decimal point, trailing zeros may be ambiguous: 3400 (2 sig figs) vs 3400 (4 sig figs) is not determined from the number alone; a decimal point (3400.) or a bar/notation would clarify.

• Exact numbers have infinite significant figures (e.g., numbers defined by definitions or counting objects).

  • Example: averaging three measured lengths with one or more exact counts can affect the final sf count.

  • Example given: average of 6.64, 6.68, and 6.70 was computed as 6.67333… and reported with appropriate significant figures; exact counts (e.g., 3) contribute infinite SF in principle, but the final reported SF is dictated by the measured values.

• Examples (from the notes):

  • Determine the number of significant figures:

  • (a) 394 cm → 3 SF

  • (b) 5.03 g → 3 SF

  • (c) 0.714 m → 3 SF

  • (d) 0.052 kg → 2 SF

  • (e) 2.720×10^{22} atoms → 4 SF

  • (f) 3000 mL → often 1 SF unless otherwise indicated

Addition and Subtraction with Significant Figures

• Rule: the answer cannot have more decimal places than the least precise measurement.

• Examples:

  • $89.332\;\text{ + }\;1.1 = 90.4$ (one decimal place)

  • $3.70\; - \;2.9133 = 0.7867 \rightarrow 0.79$ (two decimal places)

Multiplication and Division with Significant Figures

• Rule: the number of significant figures in the result equals the least number of significant figures among the factors.

• Examples:

  • $4.51\times 3.6666 = 16.536366 \rightarrow 16.5$ (3 sig figs)

  • $6.8 \div 112.04 = 0.0606926 \rightarrow 0.061$ (2 sig figs)

• Exercise (Example 1.5) includes several operations to apply these rules; practice the same way:

  • (a) 12,343.2 g + 0.1893 g

  • (b) 55.67 L - 2.386 L

  • (c) 7.52 m × 6.9232

  • (d) 0.0239 kg ÷ 46.5 mL

  • (e) 5.21 × 10^{3} cm + 2.92 × 10^{2} cm

Accuracy vs Precision

• Accuracy: how close a measurement is to the true value.

• Precision: how close a set of measurements are to each other.

• Visual examples (from the notes):

  • Accurate and precise (both good).

  • Precise but not accurate (clustered around wrong value).

  • Not accurate and not precise (scattered).

Dimensional Analysis (1.9)

• A method to convert between units by using relationships between units that express the same quantity.

• Steps:

  1. Read the problem; identify given quantities and what needs to be solved.

  2. Find the appropriate equation that relates the given information to the unknown.

  3. Check units, sign, and significant figures; check reasonableness.

  4. Carry units through calculations; verify that all units cancel except the desired unit(s).

• Approach:

  • Use conversion factors to move from the given unit to the desired unit.

  • You may use string conversion factors; you do not need every relationship as long as the start and end units are connected.

  • Algebraic form: given quantity × (conversion factor) = desired quantity; verify the units cancel properly.

• Example dimensional analysis problems (conceptual):

  • Convert 2.4 km to millimeters.

  • Convert speed: 343 m/s to mph (and provided steps with mile and hour conversions).

• Practical use: ensures unit consistency and helps catch mistakes in complex chemistry calculations.

Worked Examples and Practice Problems (selected from the notes)

• 2.4 km to mm:

  • 1 km = 10^{3} m and 1 m = 10^{3} mm, thus 1 km = 10^{6} mm.

  • $2.4\ \text{km} = 2.4 \times 10^{6}\ \mathrm{mm}$

• Speed of sound conversion:

  • $343\ \mathrm{m\,s^{-1}} \times \frac{1\ \mathrm{mi}}{1609\ \mathrm{m}} \times \frac{60\ \mathrm{s}}{1\ \mathrm{min}} \times \frac{60\ \mathrm{min}}{1\ \mathrm{hour}} = 343 \times \frac{3600}{1609} \ \mathrm{mi\,hour^{-1}} \approx 767\ \mathrm{mi\,hour^{-1}}$

• Example 1.6 (glucose intake):

  • Given: daily intake $0.0833\ \mathrm{lb}$. Convert to milligrams.

  • 1 lb = 453.6 g; thus mass in grams: $0.0833\times 453.6 = 37.79\ \mathrm{g}$

  • Convert to mg: $37.79\ \mathrm{g} \times 10^{3} \frac{\mathrm{mg}}{\mathrm{g}} = 3.779\times 10^{4}\ \mathrm{mg}$

• Example 1.7 (volume conversion):

  • A storage tank volume: $275\ \mathrm{L} = 0.275\ \mathrm{m^3}$

• Example 1.8 (density conversion):

  • Given density = $0.808\ \mathrm{g\,cm^{-3}}$. Convert to $\mathrm{kg\,m^{-3}}$.

  • Since $1\ \mathrm{g\,cm^{-3}} = 1000\ \mathrm{kg\,m^{-3}}$, then

  • $0.808\ \mathrm{g\,cm^{-3}} = 808\ \mathrm{kg\,m^{-3}}$

Quick Reference: Common Relations (Summary)

• Mass–volume density relation: $d = \frac{m}{V}$

• Volume units: $1\ \mathrm{cm^3} = 1\ \mathrm{mL},\quad 1\ \mathrm{dm^3} = 1\ \mathrm{L},\quad 1\ \mathrm{m^3} = 10^{3}\ \mathrm{L}$

• Density unit conversion: $1\ \mathrm{g\,cm^{-3}} = 10^{3}\ \mathrm{kg\,m^{-3}}$

• Temperature conversions: $K = C + 273.15,\quad F = \frac{9}{5}C + 32,\quad C = \frac{5}{9}(F-32)$

• Scientific notation: $m\times 10^{n}$ with coefficient m ∈ [1,10) and integer n

• Significant figures rules (recap): leading zeros not significant; zeros between nonzero digits significant; trailing zeros after decimal significant; addition/subtraction keeps least decimal places; multiplication/division keeps least SF; exact numbers have infinite SF

• Dimensional analysis workflow: set up conversion factors to cancel original units and leave the desired unit; check for consistency and reasonableness

Note: The material above follows the provided transcript content and aims to consolidate the key ideas, formulas, rules, and worked examples into a comprehensive study note set suitable for exam preparation. It includes all major topics (Measurement, SI units, prefixes, density, volume, temperature scales, scientific notation, handling numbers, significant figures, dimensional analysis) and representative examples and rules as presented in the transcript.