CHE 1: Math Prep - Scientific Notation and Significant Figures

Course Overview and Logistics

• Course: CHE 1 (Prep for Gen Chem), Section B, UC Davis, Fall Quarter 2026.

• Instructor: Yu Wang.

• Lecture Schedule: Lecture 1 presented on Wednesday, 9/23.

• Required Readings: Math Toolbox Sections 1.1 and 1.2 in the course textbook.

• Syllabus Requirement: Review the official course syllabus thoroughly.

• ALEKS Online Homework: Complete "Math Prep" and the assigned portion of "Chapter 1" (the remainder of Chapter 1 is covered the following Monday). Both assignments are due next Wednesday before lecture at $1:00\,\text{PM}$.

Working with Extreme Quantities in Chemistry

• Chemistry frequently deals with quantities that are exceptionally large or infinitesimally small, making standard decimal notation unwieldy and prone to counting errors.

• Macroscopic Quantities of Atoms:

  • A single mole ($1\,\text{mol}$) of carbon contains approximately $602,200,000,000,000,000,000,000$ atoms ($6.022 \times 10^{23}$ atoms).

• Microscopic Mass of Individual Atoms:

  • A single carbon atom has an approximate mass of $0.0000000000000000000000199\,\text{g}$.

• Calculation of Molar Mass:

  • Finding the mass of $1\,\text{mol}$ of carbon requires multiplying these two extreme numbers:     $(602,200,000,000,000,000,000,000\,\text{atoms}) \times (0.0000000000000000000000199\,\text{g/atom}) \approx 12.0\,\text{g}$

  • Writing and calculating with strings of dozens of zeros in conventional notation is inefficient and introduces counting mistakes.

Fundamentals of Scientific Notation

• Purpose of Scientific Notation:

  • Replaces long strings of leading or trailing zeros with a numeric exponential power of $10$ to prevent counting errors and simplify calculations.

• Standard Mathematical Form:

  • Any number is written as:     $C \times 10^n$

  • Coefficient ($C$): A decimal number restricted to the interval $1 \le C < 10$. It is obtained by moving the decimal point until exactly one non-zero digit remains to the left of the decimal point.

  • Exponent ($n$): A whole integer representing the exact number of places the decimal point was shifted.

• Directional Rules for Exponent Sign:

  • When the decimal point is moved to the left, the exponent is positive ($n > 0$).

  • When the decimal point is moved to the right, the exponent is negative ($n < 0$).

Converting Normal Notation to Scientific Notation

• Step 1: Categorize the Original Magnitude:

  • (a) If the number is $10$ or larger: The decimal point must move to the left, resulting in a positive exponent ($n > 0$).

  • (b) If the number is between $1$ and $10$ ($1 \le \text{number} < 10$): No conversion is strictly necessary because the exponent is zero ($n = 0$).

  • (c) If the number is smaller than $1$: The decimal point must move to the right, resulting in a negative exponent ($n < 0$).

• Step 2: Determine the Value of $n$:

  • Count the exact number of digit positions the decimal point must cross to produce a coefficient $C$ satisfying $1 \le C < 10$.

  • Assign the magnitude of shifts to $n$, combining it with the sign identified in Step 1.

• Practice Conversions:

  • Converting $90065.00$:

  • Magnitude category: Greater than $10$, so $n > 0$.

  • Movement: Decimal point shifts left across $4$ places ($0, 6, 5, 0$) to sit after $9$.

  • Result: $9.006500 \times 10^4$.

  • Converting $0.0000834$:

  • Magnitude category: Smaller than $1$, so $n < 0$.

  • Movement: Decimal point shifts right across $5$ places ($0, 0, 0, 0, 8$) to sit after $8$.

  • Result: $8.34 \times 10^{-5}$.

  • Converting $6.3$:

  • Magnitude category: Between $1$ and $10$, so no shift is required ($n = 0$).

  • Result: $6.3$ (or $6.3 \times 10^0$).

Addition and Subtraction in Scientific Notation

• Step 0: Ensure Both Numbers Are in Scientific Notation:

  • Convert all components if presented in normal decimal notation.

• Step 1: Equalize Exponents to the Larger Exponential Power:

  • Compare exponents and convert the term with the smaller exponent to match the larger exponent.

  • Adjust the smaller number's coefficient by moving its decimal point to the left by $\Delta n$ places, where $\Delta n = n_{\text{larger}} - n_{\text{smaller}}$.

• Step 2: Perform the Operation on Coefficients:

  • Add or subtract the coefficients ($C$) directly while keeping the shared exponent ($10^n$) unchanged.

• Step 3: Re-normalize the Exponential Power (if necessary):

  • If the resulting coefficient $C \ge 10$ or $C < 1$, adjust the decimal point and exponent to return to proper scientific notation.

• Worked Example ($0.0000325 + 0.0123$):

  • Convert to scientific notation:     $0.0000325 = 3.25 \times 10^{-5}$     $0.0123 = 1.23 \times 10^{-2}$

  • Align exponents to the larger value ($10^{-2}$ vs. $10^{-5}$, where $-2 > -5$):

  • Difference: $-2 - (-5) = 3$ places.

  • Shift the decimal point of $3.25$ to the left by $3$ digits:       $3.25 \times 10^{-5} = 0.00325 \times 10^{-2}$

  • Add coefficients together:     $(0.00325 \times 10^{-2}) + (1.23 \times 10^{-2}) = (0.00325 + 1.23) \times 10^{-2} = 1.23325 \times 10^{-2}$

  • Check normalization: Since $1 \le 1.23325 < 10$, the final result is $1.23325 \times 10^{-2}$.

Multiplication and Division in Scientific Notation

• Step 0: Convert Terms to Scientific Notation:

  • Put all terms into standard $C \times 10^n$ format.

• Step 1: Compute Coefficients:

  • Multiply or divide the coefficients ($C$) directly.

• Step 2: Combine Exponents Using Index Laws:

  • For multiplication, add the exponents:     $10^{n_1} \times 10^{n_2} = 10^{n_1 + n_2}$

  • For division, subtract the denominator exponent from the numerator exponent:     $\frac{10^{n_1}}{10^{n_2}} = 10^{n_1 - n_2}$

• Step 3: Combine and Re-normalize:

  • Combine the new coefficient and power of $10$.

  • If $C \ge 10$ or $C < 1$, re-normalize following the exponential adjustment rules.

• Worked Example ($0.0000325 \times 0.0123$):

  • Scientific notation conversion:     $0.0000325 = 3.25 \times 10^{-5}$     $0.0123 = 1.23 \times 10^{-2}$

  • Multiply coefficients:     $3.25 \times 1.23 = 3.9975$

  • Add exponents:     $10^{-5} \times 10^{-2} = 10^{(-5) + (-2)} = 10^{-7}$

  • Combine terms:     $3.9975 \times 10^{-7}$

  • Normalization check: Since $1 \le 3.9975 < 10$, no adjustment is required.

Exponential Adjustment Rules for Non-Normalized Coefficients

• Case 1: When Coefficient $C \ge 10$:

  • The decimal point must move to the left by the number of places needed to satisfy $1 \le C < 10$.

  • Each shift to the left increases the exponent ($+1$ per shift).

  • Example:     $983.5 \times 10^4 = 9.835 \times 10^{4 + 2} = 9.835 \times 10^6$

• Case 2: When Coefficient $C < 1$:

  • The decimal point must move to the right by the number of places needed to satisfy $1 \le C < 10$.

  • Each shift to the right decreases the exponent ($-1$ per shift).

  • Example:     $0.09835 \times 10^4 = 9.835 \times 10^{4 - 2} = 9.835 \times 10^2$

Measurement Uncertainty and Significant Figures

• Scientific Meaning vs. Mathematical Exactness:

  • In mathematics, numbers are abstract and exact. In physical science, numbers represent measurements tied to measuring instruments with finite precision.

  • Any measured number contains a limited quantity of reliably known digits plus one final estimated/uncertain digit.

• Semi-Truck Scale Case Study:

  • A semi-truck weighs $6.8\,\text{tons}$ on a highway truck scale. Two drivers weighing a combined $130.8\,\text{kg}$ board the truck.

  • Conversion factor: $1\,\text{ton} = 1000\,\text{kg}$, so the drivers weigh $0.1308\,\text{tons}$.

  • Mathematical addition: $6.8\,\text{tons} + 0.1308\,\text{tons} = 6.9308\,\text{tons}$.

  • Scientific reality: The scale only reads to the tenths place ($\pm 0.1\,\text{ton}$). The truck's true weight could be $6.84\,\text{tons}$ or $6.76\,\text{tons}$ ($6.8???\,\text{tons}$).

  • Adding $0.1308\,\text{tons}$ to an unknown hundredths and thousandths place yields an unknown value: $6.8???\,\text{tons} + 0.1308\,\text{tons} = 6.9???\,\text{tons}$.

  • Claiming $6.9308\,\text{tons}$ falsely implies measurement precision that does not exist. Scientifically, $6.8\,\text{tons} \ne 6.8000\,\text{tons}$.

• Laboratory Balance Sensitivity and Percent Yield:

  • In a chemical synthesis with a theoretical yield of $38.5412\,\text{g}$, an experimenter collects $17\,\text{g}$ of product on a coarse balance.

  • Calculator output:     $\frac{17\,\text{g}}{38.5412\,\text{g}} \times 100\% = 44.108642\ldots\%$

  • If a more precise balance measures the same product as $17.1\,\text{g}$, the yield becomes $44.368105\ldots\%$.

  • If the product weighs $16.9\,\text{g}$, the yield becomes $43.849180\ldots\%$.

  • Retaining digits past the first two has zero scientific validity because experimental inaccuracy causes those digits to fluctuate wildly.

Rules for Identifying Significant Figures

• Rule 1: Non-Zero Digits:

  • All non-zero digits are always significant.

• Rule 2: Leading Zeros Before Decimal Point:

  • A zero alone in front of a decimal point is not significant (e.g., in $0.205$, the leading zero is merely a placeholder).

• Rule 3: Leading Zeros After Decimal Point:

  • Zeros to the right of a decimal point but preceding the first non-zero digit are not significant placeholders (e.g., in $0.00205$, the first three zeros are not significant; the value has $3$ significant figures: $2, 0, 5$).

• Rule 4: Trailing Zeros After Decimal Point:

  • All zeros to the right of the decimal point that appear after non-zero digits are significant (e.g., $0.2050000$ has $7$ significant figures).

• Rule 5: Trailing Zeros in Whole Numbers:

  • Zeros at the end of a number to the left of an implied decimal point are not significant unless an explicit decimal point is written:

  • $25600$ contains $3$ significant figures.

  • $25600.$ contains $5$ significant figures.

• Significant Figures in Scientific Notation:

  • Exponential multiplier terms ($10^n$) do not contribute to significant figure counts; only digits in the coefficient $C$ are counted.

• Significant Figure Counting Practice:

  • $26$: Contains $2$ significant figures (both non-zero digits).

  • $19628.00$: Contains $7$ significant figures ($5$ non-zero digits plus $2$ trailing zeros after the decimal point).

  • $0.003416$: Contains $4$ significant figures (leading zeros are placeholders; $3, 4, 1, 6$ are significant).

  • $9 \times 10^{19}$: Contains $1$ significant figure (only the digit $9$ counts; $10^{19}$ is ignored).

  • $1.2407661 \times 10^{-2}$: Contains $8$ significant figures (all $8$ digits in the coefficient are significant; $10^{-2}$ is ignored).

Calculation Rules for Significant Figures

• Addition and Subtraction Rule:

  • Round off the final result to the same decimal place as the least precise measurement in the problem (the first uncertain digit position, or fewest decimal places).

  • Worked Practice Problem:     $14.668 + 12.2 + 0.1355 = 27.0035$

  • Precision breakdown:

    • $14.668$ has $3$ decimal places (thousandths place).

    • $12.2$ has $1$ decimal place (tenths place) — this is the least precise value.

    • $0.1355$ has $4$ decimal places (ten-thousandths place).

  • Rounding: Round $27.0035$ to the tenths place ($1$ decimal place).

  • Final Answer: $27.0$.

• Multiplication and Division Rule:

  • The final result must have the same number of significant figures as the measurement with the fewest significant figures in the calculation.

  • Worked Practice Problem:     $\frac{(5.5 \times 10^{-8})(4 \times 10^{10})}{6.65 \times 10^{45}} = 3.30827\ldots \times 10^{-43}$

  • Significant figure count breakdown:

    • $5.5 \times 10^{-8}$ has $2$ significant figures.

    • $4 \times 10^{10}$ has $1$ significant figure — this is the limiting precision.

    • $6.65 \times 10^{45}$ has $3$ significant figures.

  • Rounding: Round the unrounded output to $1$ significant figure.

  • Final Answer: $3 \times 10^{-43}$.

Comprehensive Practice: Circumference Flight Duration

• Problem Statement:

  • Earth's equatorial circumference is $40,075\,\text{km}$. An aircraft travels at an average ground speed of $925\,\text{km/h}$. Calculate how many hours it takes for the aircraft to complete one full flight around the equator.

• Step 0: Conversion to Scientific Notation:

  • Distance ($d$): $40,075\,\text{km} = 4.0075 \times 10^4\,\text{km}$ ($5$ significant figures).

  • Speed ($v$): $925\,\text{km/h} = 9.25 \times 10^2\,\text{km/h}$ ($3$ significant figures).

• Step 1: Division Formulation ($t = \frac{d}{v}$):

  • Divide coefficients:     $\frac{4.0075}{9.25} \approx 0.43324324\ldots$

  • Divide exponential powers of $10$:     $\frac{10^4}{10^2} = 10^{4 - 2} = 10^2$

  • Unadjusted intermediate expression:     $0.43324324\ldots \times 10^2\,\text{h}$

• Step 2: Exponential Normalization:

  • Since $C = 0.43324324\ldots < 1$, shift the decimal point $1$ digit to the right to obtain $4.3324324\ldots$.

  • Decrease the exponent by $1$:     $10^{2 - 1} = 10^1$

  • Normalized intermediate form:     $4.3324324\ldots \times 10^1\,\text{h}$

• Step 3: Application of Significant Figure Rules:

  • The division involves a $5$-sig-fig numerator ($40,075$) and a $3$-sig-fig denominator ($925$).

  • The final answer must be restricted to $3$ significant figures (limited by $925\,\text{km/h}$).

  • Round $4.3324324\ldots \times 10^1\,\text{h}$ to $3$ significant figures:

  • In scientific notation: $4.33 \times 10^1\,\text{h}$.

  • In standard decimal notation: $43.3\,\text{h}$.