Math for Chemistry

Scientific Notation

Chemistry deals with very large and very small numbers. We use scientific notation to handle them.

Calculation rules:

Operation	Rule	Example
Multiply	Multiply coefficients, add exponents	$(2 \times 10^3)(3 \times 10^4) = 6 \times 10^7$
Divide	Divide coefficients, subtract exponents	$(6 \times 10^8) \div (2 \times 10^3) = 3 \times 10^5$

Significant Figures

Value	Sig figs	Count
$12.3$	1, 2, 3	3
$0.0045$	4, 5	2 (leading zeros don’t count)
$1.20 \times 10^3$	1, 2, 0	3 (trailing zero is significant)

Unit Conversions

Common conversions in chemistry:

Conversion	Relationship
L ↔ mL	$1\ \mathrm{L} = 1000\ \mathrm{mL}$
g ↔ kg	$1\ \mathrm{kg} = 1000\ \mathrm{g}$
Pa ↔ atm	$1\ \mathrm{atm} = 1.013 \times 10^5\ \mathrm{Pa}$
°C ↔ K	$T\ \mathrm{(K)} = t\ \mathrm{(°C)} + 273$

Dimensional Analysis

Example: How many mol is $36\ \mathrm{g}$ of water? (Molar mass $18\ \mathrm{g/mol}$)

$$\frac{36\ \mathrm{g}}{18\ \mathrm{g/mol}} = 2.0\ \mathrm{mol}$$

The $\mathrm{g}$ cancels, leaving $\mathrm{mol}$ — correct!

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Check Your Understanding

Q1 How many significant figures does 0.00350 have?

2345

Q2 $(3.0 \times 10^4) \times (2.0 \times 10^{-2})$ equals?

$6.0 \times 10^2$$6.0 \times 10^6$$5.0 \times 10^2$$6.0 \times 10^{-8}$

Q3 Convert 25°C to Kelvin:

248 K273 K298 K325 K

答え合わせ

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Exercises

Q1. Calculate, paying attention to significant figures: $1.5 \times 10^3 + 2.50 \times 10^2$

Solution

$1.5 \times 10^3 + 0.250 \times 10^3 = 1.750 \times 10^3$

For addition, round to the least precise place. $1.5 \times 10^3$ is certain only to the hundreds place:

$\approx 1.8 \times 10^3$

Q2. A 500 mL solution contains 4.0 g of NaOH. Find the molar concentration. ($M_{\mathrm{NaOH}} = 40.0$)

Solution

$n = \frac{4.0\ \mathrm{g}}{40.0\ \mathrm{g/mol}} = 0.10\ \mathrm{mol}$

$c = \frac{n}{V} = \frac{0.10\ \mathrm{mol}}{0.500\ \mathrm{L}} = 0.20\ \mathrm{mol/L}$