Chapter 3 - EXPERIMENTAL ERROR

Chapter 3 - EXPERIMENTAL ERROR

Solutions to Assignment Problems

1. Indicate how many significant figures there are in:

(a) 0.304 0 4 significant figures

(b) 0.003 050 4 significant figures

2. Round each number as indicated:

(a) 5.124 8 to 4 significant figures 5.125

(b) 5.124 4 to 4 significant figures 5.124

(d) 0.135 237 1 to 4 significant figures 0.135 2

(e) 1.525 to 3 significant figures 1.52

(f) 1.525 007 to 3 significant figures 1.53

3. Write each answer with the correct number of digits:

(a) 3.021 + 8.99 = 12.011 12.01

(b) 12.7 – 1.83 = 10.87 10.9

(d) 0.030 2 ÷ (2.114 3 × 10-3) = 14.283 69 14.3

(e) log (2.2 × 10-18) = ? -17.66

(f) antilog (–2.224) = ? 5.97 x 10-3

(g) 10-4.555 = ? 2.79 x 10-5

4. Find the absolute and percent relative uncertainty and express each answer with a reasonable number of significant figures.

(a) 3.4 (±0.2) + 2.6 (±0.1) = 6.0 (± e)

e1 = 0.2 and e2 = 0.1

e = √ e12 + e22 = √ 0.22 + 0.12 = 0.224

Absolute uncertainty is 0.2 and relative uncertainty is 3.7%

Answer is 6.0 ± 0.2 or 6.0 ± 3.7%

(b) 3.4 (±0.2) ÷ 2.6 (±0.1) = 1.308 (±e)

3.4 (±0.2) = 3.4 (±5.88%)

2.6 (±0.1) 2.6 (±3.85%)

%e = √ (%e1)2 + (%e2)2 = √(5.88)2 + (3.85)2 = 7.03

Absolute uncertainty is 0.092 and relative uncertainty is 7.0%

Answer is 1.308* ± 0.092 or 1.308 ± 7.0%

* Notice that the answer has 3 significant figures while both dividend and divisor has only 2. Please refer to “The Real Rule for Significant Figure” section in your text book for explanation. It is on pg 46 in 7th edition and pg 54 on 6th edition.

= 1.308 (±0.092) x 10-11 or 1.308 (±7.0%) x 10-11

(d) [3.4 (±0.2) – 2.6 (±0.1)] × 3.4 (±0.2) = ?

3.4 (±0.2) – 2.6 (±0.1) = 0.8 (±0.224) = 0.8 ± 28%

0.8 (±28%) × 3.4 (±5.88%) = 2.72 ± 29% = 2.72 ± 0.78

5. Express the molecular mass (± uncertainty) of benzene, C6H6, with the correct number of significant figures.

Atomic mass of C = 12.010 7 ± 0.000 8

Atomic mass of H = 1.007 94 ± 0.000 07

6C = 6 * (12.010 7 ± 0.000 8) = 72.064 2 ± 0.004 8

6H = 6 * (1.007 94 ± 0.000 07) = 6.047 64 ± 0.000 42

C6H6 = 6C + 6H = 78.111 84 ± ?

Uncertainty associated with the molecular mass = √0.00482 + 0.000422 = 0.004 8

molecular mass of benzene = 78.112 ± 0.005

6. (a) A solution is prepared by dissolving 0.222 2 (±0.000 2) g of KIO3 [FM 214.001 0 (±0.000 9)] in 50.00 (±0.05) mL. Find the molarity and its uncertainty with an appropriate number of significant figures.

Molarity = 0.222 2 (±0.000 2) g of KIO3 * mol of KIO3

214.001 0 (±0.000 9) g of KIO3 * 50.00 (±0.05) x 10-3 L

= 0.020 766 ± 0.000 028 M

7. Find the absolute and percent relative uncertainty and express each answer with a reasonable number of significant figures.

(a) √3.4 (±0.2) = ?

y = x½ → %ey = ½ (%ex)

ey = ½ (5.88%) = 2.94%

Answer: 1.84 ± 2.9% or 1.84 ± 0.054

(b) [3.4 (±0.2)]2 = ?

y = x2 → %ey = 2(%ex)

ey = 2 * (5.88%) = 11.76%

Answer: 11.6 ± 12% or 11.6 ± 1.4

y = 10x → ey/y = (ln 10) ex ≈ 2.302 6 ex

ey = 2511.88 * 2.302 6 * 0.2 = 1156.77

Answer: 2.51 ± 1.16 x 103 or 2.51 (±46%) x 103

(d) e3.4 (±0.2) = ?

y = ex → ey/y = ex

ey = e3.4 * 0.2 = 5.99

Answer: 30.0 ± 6.0 or 30.0 ± 20%

(e) log [3.4 (±0.2)] = ?

y = log x → ey = (1/ln 10)(ex/x) ≈ 0.434 29 ex/x

ey = 0.434 29 * 0.2/3.4 = 0.0255

Answer: 0.531 ± 0.026 or 0.531 ± 4.8%

(f) ln [3.4 (±0.2)] = ?

y = ln x → ey = ex/x

ey = 0.2/3.4 = 0.0588

Answer: 1.224 ± 0.059 or 1.224 ± 4.8%

8. The value of Boltzmann's constant (k) listed on the inside front cover of the book is calculated from the quotient R/N, where R is the gas constant and N is Avogadro's number. If the uncertainty in R is 0.000 070 J/(mol . K) and the uncertainty in N is 0.000 003 6 × 1023/mol, find the uncertainty in k.

R = 8.314472 ± 0.000 070 J/(mol*K)

N = 6.0221415 (± 0.000 003 6) × 1023/mol

k = 1.3806595 x 10-23 J/K

k = R/N

uncertainty in k, %ek = √(%eR)2 + (%eN)2

= √ (8.419 x 10-4)2 + (5.978 x 10-5)2 = 8.44019702 x 10-4

Absolute uncertainty = 8.44019702 x 10-6 * 1.3806505 x 10-23 = 0.000012 x 10-23

We express k as 1.380650 (± 0.000012) x 10-23

9. Find the uncertainty in the molecular mass of B10H14 and write the molecular mass with the correct number of significant figures. (similar to problem 5)

molecular mass of B10H14 = 122.22 ± 0.07