DAT General Chemistry: Scientific Notation, SI Units, and Temperature Scales

DAT High-Yield Concept

Scientific Notation: Format as N × 10ⁿ where 1 ≤ |N| < 10. Moving the decimal left yields a positive exponent; moving right yields a negative exponent.
7 SI Base Units: Meter ( $\pu{m}$ ) for length, kilogram ( $\pu{kg}$ ) for mass, second ( $\pu{s}$ ) for time, kelvin ( $\pu{K}$ ) for thermodynamic temperature, mole ( $\pu{mol}$ ) for amount of substance, ampere ( $\pu{A}$ ) for electric current, and candela ( $\pu{cd}$ ) for luminous intensity.
Temperature Conversions: $T_{\text{K}} = T_{\text{degC}} + 273.15$ and $T_{\text{degF}} = 1.8 \times T_{\text{degC}} + 32$ . Kelvin is an absolute scale with no negative values and never uses a degree symbol.
Volume & Cubed Unit Conversions: Remember that $\pu{1 L} = \pu{1 dm^3} = \pu{1000 mL} = \pu{1000 cm^3}$ . When converting cubed length units, cube the conversion factor!

1. Scientific Notation

Scientific notation handles extremely large or small quantities in chemistry, such as Avogadro's number ( $\pu{6.022 \times 10^{23} mol^{-1}}$ ) or proton radius. A number is written as $N \times 10^n$ , where $1 \le |N| < 10$ and $n$ is an integer.

When multiplying terms, multiply the coefficients and add exponents: $(a \times 10^x) \times (b \times 10^y) = (a \cdot b) \times 10^{x+y}$ . When dividing, divide coefficients and subtract exponents: $(a \times 10^x) / (b \times 10^y) = (a / b) \times 10^{x-y}$ .

Worked Example 1: Scientific Notation Calculation
Problem: A chemical reaction produces

\pu{0.0000456 mol}

\ce{H2SO4}

in a total volume of

\pu{0.00200 L}

. Express the mole count in scientific notation and calculate the molar concentration.
Solution:

Move the decimal 5 places right to get $N = 4.56$ : $\pu{0.0000456 mol} = \pu{4.56 \times 10^{-5} mol}$ .
Convert volume to scientific notation: $\pu{0.00200 L} = \pu{2.00 \times 10^{-3} L}$ .
Calculate molarity: $\text{Molarity} = \frac{\pu{4.56 \times 10^{-5} mol}}{\pu{2.00 \times 10^{-3} L}} = \left(\frac{4.56}{2.00}\right) \times 10^{-5 - (-3)} = \pu{2.28 \times 10^{-2} M}$

Exam Trap: Forgetting to adjust the coefficient so that

1 \le |N| < 10

after operations (e.g., writing

\pu{0.228 \times 10^{-1} M}

instead of

\pu{2.28 \times 10^{-2} M}

2. SI Base Units and Metric Prefixes

The International System of Units (SI) defines seven fundamental base units from which all other chemical units (like density, $\pu{g/cm^3}$ , or pressure, $\pu{Pa}$ ) are derived.

Metric Prefix Mnemonic: Great Many kingdoms have dark dungeons containing many μnseen nasty parasites.
Giga (

10^9

), Mega (

10^6

), kilo (

10^3

), hecto (

10^2

), deka (

10^1

), deci (

10^{-1}

), centi (

10^{-2}

), milli (

10^{-3}

), micro (

10^{-6}

), nano (

10^{-9}

), pico (

10^{-12}

Prefix	Symbol	Factor	Example Equivalent
Mega-	M	$10^6$	$\pu{1 MJ} = \pu{10^6 J}$
Kilo-	k	$10^3$	$\pu{1 kg} = \pu{1000 g}$
Centi-	c	$10^{-2}$	$\pu{1 cm} = \pu{0.01 m}$
Milli-	m	$10^{-3}$	$\pu{1 mg} = \pu{0.001 g}$
Micro-	μ	$10^{-6}$	$\pu{1 \mu g} = \pu{10^{-6} g}$
Nano-	n	$10^{-9}$	$\pu{1 nm} = \pu{10^{-9} m}$
Pico-	p	$10^{-12}$	$\pu{1 pm} = \pu{10^{-12} m}$

Worked Example 2: Metric Conversion
Problem: Convert a sample mass of

\pu{350. mg}

\ce{NaCl}

into kilograms (

\pu{kg}

).
Solution:

\pu{350. mg} \times \left(\frac{\pu{1 g}}{\pu{1000 mg}}\right) \times \left(\frac{\pu{1 kg}}{\pu{1000 g}}\right) = \pu{3.50 \times 10^{-4} kg}

Exam Trap: Misplacing exponents by using

\pu{10^3 mg} = \pu{1 g}

incorrectly inverted in the denominator, or failing to cube unit factors when converting volumes (e.g.,

\pu{1 m^3} = (10^2 \text{ cm})^3 = \pu{10^6 cm^3}

3. U.S. Customary Units in Chemistry Problems

Although scientific measurements rely on the metric/SI system, DAT questions occasionally test conversion between U.S. Customary units and SI units.

Length: $\pu{1 in} = \pu{2.54 cm}$ (exact)
Mass: $\pu{1 lb} = \pu{453.59 g}$
Volume: $\pu{1 gal} = \pu{3.785 L}$ ; $\pu{1 fl oz} = \pu{29.57 mL}$

Worked Example 3: Customary to Metric Conversion
Problem: An aqueous solution has a volume of

\pu{1.50 gal}

. Calculate its volume in liters (

\pu{L}

).
Solution:

\pu{1.50 gal} \times \left(\frac{\pu{3.785 L}}{\pu{1 gal}}\right) = \pu{5.68 L}

Exam Trap: Confusing fluid ounces (volume,

\pu{1 fl oz} \approx \pu{29.57 mL}

) with avoirdupois mass ounces (

\pu{1 oz} \approx \pu{28.35 g}

4. Temperature Scales and Conversions

Temperature measures the average kinetic energy of particles. Three scales are commonly referenced: Fahrenheit, Celsius, and Kelvin.

Three major temperature scales: Celsius, kelvin and Fahrenheit with their respective boiling points, freezing points and absolute zero points. (Source: Emeka Udenze · CC0 · Wikimedia Commons)

Key conversion formulas:

T_{\text{K}} = T_{\text{degC}} + 273.15

T_{\text{degF}} = 1.8 \times T_{\text{degC}} + 32

T_{\text{degC}} = \frac{T_{\text{degF}} - 32}{1.8}

Reference Point	Fahrenheit ( $\pu{degF}$ )	Celsius ( $\pu{degC}$ )	Kelvin ( $\pu{K}$ )
Absolute Zero	$\pu{-459.67 degF}$	$\pu{-273.15 degC}$	$\pu{0 K}$
Water Freezing Point	$\pu{32 degF}$	$\pu{0 degC}$	$\pu{273.15 K}$
Water Boiling Point	$\pu{212 degF}$	$\pu{100 degC}$	$\pu{373.15 K}$
Degree Magnitude Ratio	1.8 $\pu{degF}$ per 1 $\pu{degC}$	1 $\pu{degC}$ per 1 $\pu{K}$	1 $\pu{K}$ per 1 $\pu{degC}$

Worked Example 4: Temperature Conversion
Problem: A clinical patient presents with a body temperature of

\pu{98.6 degF}

. Express this temperature in

\pu{degC}

and

\pu{K}

.
Solution:

Convert Fahrenheit to Celsius: $T_{\text{degC}} = \frac{98.6 - 32}{1.8} = \frac{66.6}{1.8} = \pu{37.0 degC}$
Convert Celsius to Kelvin: $T_{\text{K}} = 37.0 + 273.15 = \pu{310.15 K}$

Exam Trap: Subtracting 32 after dividing by 1.8 (violating order of operations) or writing Kelvin with a degree symbol (

\pu{^\circ K}

is incorrect; write

\pu{K}

Critical Thinking Questions

Why must gas law calculations (such as $PV = nRT$ ) always use temperature expressed in Kelvin rather than Celsius or Fahrenheit?
If a student converts a surface area of $\pu{10.0 cm^2}$ to square meters by dividing by 100, what logical error did they make, and what is the correct numerical value?
At what numeric temperature are the values on the Celsius and Fahrenheit scales identical? Show the calculation.

References

Sources returned by Google Search grounding during lesson verification.

1.3: Measurement and Scientific Notation - Chemistry LibreTexts — chem.libretexts.org
1.7: Temperature Units and Conversions - Chemistry LibreTexts — chem.libretexts.org

Scientific Notation, SI Units, and Temperature Scales

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