Chem Study Notes 1

Science and Measurement

What is Chemistry?

• Chemistry is defined as the study of matter and energy.

• It involves the transformation of matter and energy, often resulting in new matter.

Importance of Measurement and Units

• Necessity of Units: Just as a cookie recipe requires specific units (e.g., cups, teaspoons) for ingredients to achieve a good result, chemistry experiments require accurate units of measurement.

• Crucial Role: Units are the most important aspect of chemistry. Always pay attention to your units to ensure good chemistry and accurate results.

Units of Measurement

Metric/SI Base Units

• The metric system uses prefixes to scale units, making them larger or smaller than the base unit.

  • Examples: $1 \text{ kg} = 1000 \text{ g}$, $1 \text{ cm} = 0.01 \text{ m}$

• SI Base Unit for Mass: The SI base unit for mass is kilograms ($\text{kg}$), not grams ($\text{g}$).

• Common Prefixes: Be familiar with common metric prefixes used in chemistry.

Derived Units

• Derived units are obtained by combining base units through mathematical operations.

• In chemistry, calculations involve both numbers and units. Following unit math first is the easiest way to get the correct answer.

• Area: Calculated by multiplying two length units.

  • Example: $3 \text{ cm} \times 3 \text{ cm} = 9 \text{ cm}^2$

• Volume: Calculated by multiplying three length units.

  • Example: $2 \text{ m} \times 2 \text{ m} \times 2 \text{ m} = 8 \text{ m}^3$

  • Common Conversions: $1 \text{ m}^3 = 1000 \text{ L}$; $1 \text{ cm}^3 = 1 \text{ mL}$. This means $1 \text{ mL} = 1 \text{ cm}^3$.

• Density ($d$):

  • Relates the mass of an object to its volume.

  • Formula: $\text{Density} = \frac{\text{Mass}}{\text{Volume}}$

  • Extremely important concept in chemistry.

  • Example: Density of water ($d_{\text{water}}$) is approximately $1.0 \text{ g/mL}$.

Temperature Scales

• Three primary temperature scales are used in the United States:

  1. Fahrenheit ($^{\circ}\text{F}$):

     • Water freezes at $32^{\circ}\text{F}$.

     • Water boils at $212^{\circ}\text{F}$.

  2. Celsius ($^{\circ}\text{C}$):

     • Water freezes at $0^{\circ}\text{C}$.

     • Water boils at $100^{\circ}\text{C}$.

  3. Kelvin (K) (SI scale):

     • Starts at $0 \text{ K}$, defined as absolute zero.

     • A change of $1 \text{ K}$ is equivalent to a change of $1^{\circ}\text{C}$.

• Conversion Formulas:

  • Fahrenheit to Celsius: $^{\circ}\text{C} = (^{\circ}\text{F} - 32) \times \frac{5}{9}$

  • Celsius to Fahrenheit: $^{\circ}\text{F} = (^{\circ}\text{C} \times \frac{9}{5}) + 32$

  • Celsius to Kelvin: $\text{K} = ^{\circ}\text{C} + 273.15$

Measurement and Uncertainty

Making Measurements

• When making measurements with a measuring tool, always estimate one digit beyond the smallest scale division.

  • Example: If a ruler has divisions of $0.1 \text{ cm}$, and a shell ends slightly past $4.0 \text{ cm}$, an appropriate measurement might be $4.09 \text{ cm}$ (estimating the $0.01 \text{ cm}$ digit).

Accuracy and Precision

• Repeated Measurements: Scientific measurements are usually repeated three or more times to obtain an average value, which is likely closer to the true value than any single measurement.

• Accuracy: The closeness of the average of a set of measurements to the true (accepted) value.

• Precision: The closeness of all values within a set of measured values to one another (reproducibility).

Significant Figures (Sig Figs)

• Definition: Significant figures are those digits in a number that include all certain (known) digits plus a final digit having some uncertainty (the estimated digit).

• Exact vs. Inexact Numbers:

  • Exact Numbers: Have NO uncertainty and an infinite number of significant figures.

    • Conversion Factors: Those derived from definitions ($60 \text{ s} = 1 \text{ min}$; $1000 \text{ mL} = 1 \text{ L}$; $\text{K} = 273 + ^{\circ}\text{C}$).

    • Counted Quantities: The number of discrete items (e.g., $392$ students).

  • Inexact Numbers: All experimentally measured data have uncertainty and are considered inexact numbers.

• Rules for Determining Significant Figures:

  1. Non-zero digits: All non-zero digits are significant.

     • Examples: $24$ (2 sig figs), $563$ (3 sig figs), $1.5 \times 10^5$ (2 sig figs), $14.45$ (4 sig figs).

  2. Zeros between non-zero digits (middle zeros): Are significant.

     • Examples: $107$ (3 sig figs), $4.502$ (4 sig figs), $1001$ (4 sig figs).

  3. Zeros to the left of the first non-zero digit (leading zeros): Are not significant.

     • Examples: $0.11$ (2 sig figs), $0.011$ (2 sig figs), $0.0101$ (3 sig figs).

  4. Zeros at the end of the number (trailing zeros): Are significant only if there is a decimal anywhere in the number.

     • Examples: $1.0$ (2 sig figs), $1.100$ (4 sig figs), $0.100$ (3 sig figs).

     • Numbers like $100$ (with an implied decimal) are ambiguous without a decimal point explicitly shown or scientific notation.

• Why Significant Figures Matter: They reflect the precision of a measurement. If a measurement is less precise (fewer sig figs), any calculations using it cannot be more precise.

Significant Figures in Calculations

• Rounding Rule: Only round at the very end of a multi-step calculation to avoid premature rounding errors.

• Addition and Subtraction:

  • Report the answer to the fewest decimal places of the original numbers.

  • Example: $6.52$ (2 decimal places) $+ 8.1$ (1 decimal place) $= 14.62 \rightarrow 14.6$

• Multiplication and Division:

  • Report the answer to the fewest significant figures of the original numbers.

  • Example: $4.6$ (2 sig figs) $* 7.64$ (3 sig figs) $= 35.144 \rightarrow 35$

• Exact Numbers: Are not subject to error and should not be used to determine significant figures in calculations. Treat them as having infinite significant figures.

• Practice Examples:

  • $22.34 + 3.03 = 25.37$ (retains 2 decimal places).

  • $3.821 \times 62 = 236.902 \rightarrow 240 \text{ or } 2.4 \times 10^2$

    • Original numbers: $3.821$ (4 sig figs), $62$ (2 sig figs). Answer must have 2 sig figs.

  • $12.8 / 32.08 = 0.3990$

    • Original numbers: $12.8$ (3 sig figs), $32.08$ (4 sig figs). Answer must have 3 sig figs.

  • $450 - 68.3 = 381.7 \rightarrow 380$

    • Original numbers: $450$ (0 decimal places), $68.3$ (1 decimal place). Answer must have 0 decimal places.

  • $(3.817 + 4.58) \times 1.890 = 8.397 \times 1.890 = 15.8703 \rightarrow 15.9$

    • $3.817 + 4.58 = 8.397$ (intermediate has 2 decimal places, but keep extra digits for calculation).

    • $8.397$ (4 sig figs) $\times 1.890$ (4 sig figs). Answer must have 4 sig figs.

  • $(80.21 - 79.93) / 65.22 = 0.28 / 65.22 = 0.000429316 \rightarrow 0.00043$

    • $80.21 - 79.93 = 0.28$ (intermediate has 2 decimal places, which means 2 sig figs for the subtraction result).

    • $0.28$ (2 sig figs) $/ 65.22$ (4 sig figs). Answer must have 2 sig figs.

  • $92.12 \times 0.912 + 233.02 = 84.01104 + 233.02 = 317.03104 \rightarrow 317.0$

    • $92.12$ (4 s.f.) $\times 0.912$ (3 s.f.) $= 84.01104$ (intermediate calculation, keep extra digits for now, but its significant figures are limited by $0.912$ to 3 s.f., meaning the $84.0$ part is significant).

    • $84.01 + 233.02$ (The result of multiplication has its last significant digit in the hundredths place. The $233.02$ has its last significant digit in the hundredths place.) The sum should be rounded to the hundredths place result, or in this case, the first decimal place for $84.0$ becomes the limit for the sum, leading to $317.0$ or $317.03$. Here the slide states $317.0$, implying the $84.01$ being rounded to $84.0$ with the last significant digit at the tenth's place (for addition/subtraction rule after multiplication).

Dimensional Analysis (Factor-Label Method)

• Conversion Factors: Fractions where the numerator and denominator are equal but expressed in different units.

  • Examples: $1 \text{ minute} = 60 \text{ seconds}$; $1 \text{ inch} = 2.54 \text{ cm}$; $1 \text{ mL} = 1 \text{ cm}^3$.

  • Can be written in two forms, both equal to $1$:

    • $\frac{60 \text{ seconds}}{1 \text{ minute}} = 1$

    • $\frac{1 \text{ minute}}{60 \text{ seconds}} = 1$

• Key Points:

  • Multiplying anything by $1$ does not change its value.

  • Multiplying by a conversion factor will change the units.

  • Anything divided by itself equals $1$. (Units will cancel out).

• The Process:

  1. Write down what you know (your initial measurement with its unit).

  2. Determine the necessary conversion factor(s) and their orientation. Arrange the conversion factor so that the unit you want to cancel is in the denominator, and the unit you want to convert to is in the numerator.

  3. Multiply the numerators and divide by the denominators, ensuring units cancel out to leave the desired unit.

  • General form: $\text{Measurement (old unit)} \times \frac{\text{Conversion Factor}}{\text{ (to new unit)}} = \text{Measurement (new unit)}$

• Examples:

  • Work/Pay:

    • If you work $20$ hours at $15\text{/hour}$: 20 \text{ hours} \times \frac{15\$}{1 \text{ hour}} = 300\

    • If you spend 300\ on Nintendo Switch games at $50\text{/game}$: 300\ \times \frac{1 \text{ game}}{50\} = 6 \text{ games}

  • Seconds in a week:

    • $1 \text{ week} \times \frac{7 \text{ days}}{1 \text{ week}} \times \frac{24 \text{ hours}}{1 \text{ day}} \times \frac{60 \text{ minutes}}{1 \text{ hour}} \times \frac{60 \text{ seconds}}{1 \text{ minute}} = 604800 \text{ seconds}$

  • Usain Bolt's speed (ft/s to mile/hr):

    • Given: $65.6 \text{ ft}$ in $1.61 \text{ s}$, so speed is $\frac{65.6 \text{ ft}}{1.61 \text{ s}}$.

    • Conversions needed: $1 \text{ mile} = 5280 \text{ ft}$, $1 \text{ min} = 60 \text{ s}$, $1 \text{ hr} = 60 \text{ min}$.

    • $\frac{65.6 \text{ ft}}{1.61 \text{ s}} \times \frac{1 \text{ mi}}{5280 \text{ ft}} \times \frac{60 \text{ s}}{1 \text{ min}} \times \frac{60 \text{ min}}{1 \text{ hr}} = 27.799… \text{ mi/hr}$

    • With significant figures (limit 3 from $65.6$ and $1.61$): $27.8 \text{ mi/hr}$.

  • Football field length (yards to meters to nanometers):

    • Given: $120 \text{ yards}$.

    • Conversion factor: $1 \text{ yd} = 0.9144 \text{ m}$ (inexact).

    • Length in meters: $120 \text{ yd} \times \frac{0.9144 \text{ m}}{1 \text{ yd}} = 109.728 \text{ m}$.

    • With significant figures (limit 2 from $120$ if trailing zero is not significant, or 3 if it is; $120.$ would be 3. The slide implies 2 sig figs for the answer '2.4 x 10^2' so let's use 2): $1.1 \times 10^2 \text{ m}$.

    • Length in nanometers: Using the non-rounded value: $109.728 \text{ m} \times \frac{10^9 \text{ nm}}{1 \text{ m}} = 1.09728 \times 10^{11} \text{ nm}$.

    • With significant figures (from the 1.1 x 10^2 m, or original 120 yd -- assuming 2 sig figs): $1.1 \times 10^{11} \text{ nm}$.

  • Mass of Molybdenum block (volume to mass):

    • Given: Block dimensions $2.0 \text{ yd} \times 3.0 \text{ yd} \times 4.0 \text{ yd}$, Density ($d$) $= 10.2 \text{ g/cm}^3$.

    • Step 1: Calculate Volume:

      • $V = 2.0 \text{ yd} \times 3.0 \text{ yd} \times 4.0 \text{ yd} = 24 \text{ yd}^3$

    • Step 2: Convert Volume and Calculate Mass:

      • Use conversion factors: $1 \text{ yd} = 91.44 \text{ cm}$; $\frac{10.2 \text{ g}}{1 \text{ cm}^3}$ (density); $1 \text{ kg} = 1000 \text{ g}$.

      • $\text{Mass} = 24 \text{ yd}^3 \times (\frac{91.44 \text{ cm}}{1 \text{ yd}})^3 \times \frac{10.2 \text{ g}}{1 \text{ cm}^3} \times \frac{1 \text{ kg}}{1000 \text{ g}}$

      • $\text{Mass} = 24 \text{ yd}^3 \times \frac{762464.21 \text{ cm}^3}{1 \text{ yd}^3} \times \frac{10.2 \text{ g}}{1 \text{ cm}^3} \times \frac{1 \text{ kg}}{1000 \text{ g}}$

      • $\text{Mass} = 187163.0 \text{ kg}$

      • With significant figures (limit 2 from dimensions, or 3 from density and $91.44\text{ cm}$ if it's considered inexact from the context of previous examples): $1.9 \times 10^5 \text{ kg}$ (using 2 sig figs from $24 \text{ yd}^3$).

Problem-Solving Strategy for General Chemistry

1. Identify the Concept/Problem Type: Digest the