Significant figures and rounding rules: Exercises — Chemistry deep dive

Level 1 — Recognition

Goal: just count and classify digits.

Recall Solution 1.1

Apply the counting rules digit by digit.

(a) $0.00506$ — the leading $0.00$ are placeholders that only locate the decimal point, so they are not significant. Then $5$ , the sandwiched $0$ , and $6$ are all significant. → 3 sig figs.
(b) $4500.$ — the decimal point is written, so all trailing zeros are deliberate. $4, 5, 0, 0$ → 4 sig figs. (Without the point, $4500$ would be ambiguous.)
(c) $2.0080$ — $2$ significant; the two middle zeros are sandwiched → significant; trailing $0$ after decimal → significant. $2, 0, 0, 8, 0$ → 5 sig figs.
(d) $0.100$ — leading $0$ not significant; $1$ significant; two trailing zeros after a decimal → significant. → 3 sig figs.

Recall Solution 1.2

In $N \times 1 0^{n}$ every digit of $N$ is significant, which is exactly why we use it (see Scientific notation and orders of magnitude).

$0.00506 = 3 digits \Rightarrow 5.06 \times 1 0^{- 3}$ .
$2.0080 = 5 digits \Rightarrow 2.0080 \times 1 0^{0}$ (keep all five zeros/digits).

Level 2 — Application

Goal: apply one rounding or one operation rule.

Recall Solution 2.1

Find the first dropped digit (the "decider").

(a) $0.0034567$ — significant digits are $3, 4, 5, 6, 7$ . Keep $3, 4, 5$ ; decider is the next digit $6 > 5$ → round up → $3.46 \times 1 0^{- 3}$ . Answer: $0.00346$ .
(b) $12.750$ — keep $1, 2, 7$ ; decider is $5$ with a trailing $0$ after it (nothing non-zero follows) → round half to even. The kept digit is $7$ (odd) → bump to $8$ . Answer: $12.8$ .
(c) $12.650$ — keep $1, 2, 6$ ; decider $5$ , nothing non-zero after → round to even. Kept digit $6$ is already even → leave it. Answer: $12.6$ .

Recall Solution 2.2

(a) Multiplication → fewest sig figs. Raw $= 6.28$ . Factors: $3.14$ (3 sf), $2.0$ (2 sf). $min = 2$ , so we keep 2 sig figs ( $6.3$ ). The digit being dropped is the last one, $8$ ; since $8 > 5$ → round the kept $2$ up to $3$ . Answer: $6.3$ .
(b) Addition → fewest decimal places. Raw $= 91.469$ . Decimal places: $88.4$ (1), $3.06$ (2), $0.009$ (3). Fewest $= 1$ . Round to 1 decimal place: the first dropped digit is $6 > 5$ → round up. Answer: $91.5$ .

Level 3 — Analysis

Goal: decide WHICH rule governs, and spot exact numbers.

Recall Solution 3.1

Total $= 6 \times 47.5 = 285.0$ mL raw.

The " $6$ " is a counted (exact) number → infinite sig figs, so it does not cap the answer. (Why: counting 6 beakers is not an estimate.)
The only measured factor is $47.5$ (3 sig figs). So the answer keeps 3 sig figs.
$285.0 \to$ 3 sig figs is $285$ . Answer: $285$ mL (or $2.85 \times 1 0^{2}$ mL).

Recall Solution 3.2

This is addition, so we need decimal places — but only after writing both numbers on the same scale.

$4.0 \times 1 0^{2} = 400.$ — known to the tens place (the last honest digit is the $0$ in the tens column; it is good only to $\pm 10$ ).
$1.25 \times 1 0^{1} = 12.5$ — known to the tenths place.
Raw sum $= 412.5$ . The coarser term is $400.$ , honest only to the tens place. So we round to the tens place: $412.5 \to 410$ .
Answer: $4.1 \times 1 0^{2}$ . (Why: adding a sharp small number to a fuzzy big number cannot sharpen the fuzzy one — mud plus clean water is still muddy.)

Level 4 — Synthesis

Goal: chain multiple operations, carry guard digits, round once.

Recall Solution 4.1

Follow Density and derived quantities: $ρ = m / V$ , $V = L^{3}$ .

$V = 2.1 5^{3} = 9.938375 cm^{3}$ — carry all digits (guard digits), don't round yet.
$ρ = 63.42/9.938375 = 6.38129 \dots g/cm^{3}$ .
Sig-fig audit: $m$ has 4 sf; $L = 2.15$ has 3 sf, and cubing it keeps 3 sf, so $V$ is 3 sf. Division → $min (4, 3) = 3$ .
Round $6.38129$ to 3 sig figs: the first dropped digit is the 4th sig digit $1 < 5$ → keep. Answer: $6.38$ g/cm³.

Recall Solution 4.2

From The mole and molar mass calculations: $n = m / M$ .

$n = 18.02/44.0 = 0.40954 \dots mol$ (keep the extra guard digits).
Sig figs: $18.02$ (4 sf) $\div$ $44.0$ (3 sf) → $min = 3$ .
Round $0.40954$ to 3 sig figs: significant digits are $4, 0, 9$ ; the first dropped digit is $5$ with $4$ non-zero after it → round up. $409 \to 410$ .
Answer: $0.410$ mol (the trailing $0$ is significant — 3 sig figs).

Level 5 — Mastery

Goal: justify every digit; handle degenerate and edge cases.

Recall Solution 5.1

Significant digits of $0.09995$ are $9, 9, 9, 5$ (leading $0.0$ are placeholders).
Keep the first three $9$ s; the first dropped digit is the final $5$ with nothing after → round half to even. The kept digit is $9$ (odd), so we must round up.
Rounding $999 \to 1000$ triggers a cascade carry: $0.09995 \to 0.100$ .
Answer: $0.100$ (still 3 sig figs: $1, 0, 0$ ). Cleanly: $1.00 \times 1 0^{- 1}$ . (This is why writing the trailing zeros matters — dropping them would lie about the precision.)

Recall Solution 5.2

Difference $= 87.54 - 87.49 = 0.05$ g.
Decimal-place rule: both terms have 2 decimal places, so the answer keeps 2 decimal places → $0.05$ g. That is correct.
But count the sig figs: $0.05$ has only 1 significant figure, even though each input had 4. This is called loss of significance: subtracting two nearly-equal numbers throws away leading digits.
Lesson: the decimal-place rule kept us honest, but the result is far less precise than the inputs. Any further calculation using $0.05$ is now capped at 1 sig fig. (See Uncertainty and error propagation.)

Recall Solution 5.3

$ρ = 25.0/20.00 = 1.25 g/mL$ exactly.
Sig figs: $25.0$ (3 sf) $\div$ $20.00$ (4 sf) → $min = 3$ .
Correct precision is 3 sig figs: $1.25$ g/mL.
The student's $1.250$ claims 4 sig figs — that trailing $0$ is invented certainty. Not justified.
Answer: $1.25$ g/mL (3 sig figs).

Recall One-line self-quiz

When is an exact number allowed to reduce your answer's sig figs? ::: Never — counted or defined numbers have infinite sig figs and cannot cap precision.

Recall One-line self-quiz

Multiplication tracks ___ ; addition tracks ___. ::: Significant figures (relative) ; decimal places (absolute).