CHEMISTRY 31

Fall, 2012 - Dixon

Set 1.2 - Solutions

Ch. 3: 1, 2, 5a-d, 11, 14, 16a-f

1.  How many significant figures are there in the following numbers?

a) 1.9030  5 (all 0s significant)            b) 0.03910  4 (all 0s to right of non-0 digits significant)

c) 1.40 x 104  3 (all 0s significant)

2. Round each number as indicated:

a) 1.2367 to 4 significant figures  1.237           b) 1.2384 to 4 significant figures  1.238

c) 0.1352 to 3 significant figures  0.135           d) 2.051 to 2 significant figures  2.1

e) 2.0050 to 3 significant figures  2.00 (when last digit to round is exactly 5, or 5000..., you round up or down depending on whether the next digit to the left is even or odd)

5. Write each answer with the correct number of digits.

a)      1.021 + 2.69 = 3.711 -> 3.71 (hundredths)

b)      12.3 - 1.63 = 10.67 -> 10.7 (tenths)

c)      4.34 x 9.2 = 39.928 -> 40 or 4.0 x 10 (2 significant digits)

d)  0.0602/(2.113 x 104) = 2.84903 x 10-6  -> 2.85 x 10-6 (3 sig. digits)

11.  Suppose that in a gravimetric analysis, you forget to dry the filter crucibles before collecting precipitate.  After filtering the product, you dry the product and crucible thoroughly before weighing them.  Is the error in the mass of product that you report a systematic or a random error?   The error is a systematic error.

Is it always high or always low?

The measured product mass is always lower than the true mass.  The initial mass was too high, and since this is subtracted, the product mass will be too low.

14.  Rewrite the number 3.12356 (+0.16789%) in the forms

a) number (+ absolute  uncertainty) and b) number (+ percent relative) uncertainty with an appropriate number of digits

a) absolute uncertainty = (0.16789/100)3.12356 = 0.005244 (= 0.005 since only 1 sig fig for uncertainty); number (+ absolute  uncertainty) = 3.124 + 0.005

b) 3.124  + 0.2% (the number of sig figs in the number remains the same)

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

a)      9.23(+0.03) + 4.21 (+0.02) - 3.26 (+0.06) = ?

value = 10.18

since only +/-, uncertainty = [(0.03)2 + (0.02)2 + (0.06)2]0.5                   uncertainty = 0.07

% uncertainty = 0.07*100/10.18 = 0.7%

b)      91.3 (+1.0) * 40.3 (+0.2)/21.1 (+0.2) = ?

value = 174.3787

since only * and /, % rel. uncertainty = [(1.0*100/91.3)2 + (0.2*100/40.3)2 + (0.2*100/21.1)2]0.5

% uncertainty = 1.5% = 2%

uncertainty = 1.5*174.3787/100= 2.67 = 3

c)      [4.97 (+0.05) - 1.86 (+0.01)]/21.1 (+0.2) = ?

Subtraction part first:  4.97 - 1.86 = 3.11; uncertainty = [(0.05)2 + (0.01)2]0.5 = 0.051

Division part:  3.11/21.1 = 0.1474

rel. unc. = [(0.051/3.11)2 + (0.2/21.1)2]0.5 = 0.0189  (% rel unc. = 2%)

uncertainty = 0.189*0.1474 = 0.0028

d)      2.0164 (+0.0008) + 1.233 (+0.002) + 4.61 (+0.01) = ?

value = 7.8594

since only +/-, uncertainty = [(0.0008)2 + (0.002)2 + (0.01)2]0.5

uncertainty = 0.0102

% uncertainty = 0.0102*100/7.8594 = 0.1%

e)      2.0164 (+0.0008) x 103 + 1.233 (+0.002) x 102 + 4.61 (+0.01) x 101 = ?

value = 2185.8

since only +/-, uncertainty = [(0.8)2 + (0.2)2 + (0.1)2]0.5

uncertainty = 0.831

% uncertainty = 0.831*100/2186 = 0.04%

f) [3.14 (+ 0.05)]1/3 = ?

value = 1.4643

rel. unc. = (1/3)(0.05/3.14) = 0.00531  (% rel unc. = 0.5%)

uncertainty = 0.00531*1.4643 = 0.0078

Ch. 4                                              4, 9, 11, 20, 21

4. (a)  Calculate the fraction of bulbs in Figure 4-1 expected to have a lifetime greater than 1005.3 hours.

First, calculate the average and standard deviation:  average = 845.2 h

standard deviation = 94.2 h

This can be determined first by calculating how far is 1005.3 past the average in terms of the standard deviations = z = (1005.3 - 845.2)/94.2 = 1.700.

Next, the area under the Gaussian curve between zero and 1005.3 h can be found as the area value in Table 4-1 for z = 1.7 and is 0.4554.  The fraction greater than 1005.3 is equal to 0.5  the table area:

Fraction = 0.500 - 0.4554 = 0.045

b) Calculate the fraction expected to have a lifetime between 798.1 and 901.7 h.

z(798.1) = (798.1 - 845.2)/94.2 = -0.500

z(901.7) = (901.7 - 845.2)/94.2 = 0.600

Area(z = -0.500) = Area(z = +0.500) = 0.1915

Area(0.600) = 0.2258

Area(798.1 < x < 901.7) = Area(0.500) + Area(0.600).  Using Table 4-1 and linear interpolation,

Fraction = 0.1915 + 0.2258 = 0.417

c) skip this part

9.  What fraction of vertical bars in Figure 4-5a is expected to include the population mean (10,000) if many experiments are carried out?

50% of the samples should be within the 50% confidence interval of the true values.

Why are the 90% confidence interval bars longer than the 50% bars in Figure 4-5?

They are longer because they have a greater chance of including the true value.

11.  The percentage of an additive in gasoline was measured six times with the following results:  0.13, 0.12, 0.16, 0.17, 0.20, 0.11%.  Find the 90% and 99% confidence intervals for the percentage of the additive.

mean value = 0.148%; standard deviation = 0.034% (absolute value)

90 % Confidence Interval:  t (n-1 = 5) = 2.015

confidence interval = mean±t·s/(n)0.5 = 0.148±2.015·0.034/(6)0.5 = 0.15±0.03%

99% confidence Interval: t (n-1 = 5) = 4.032

confidence interval = mean±t·s/(n)0.5 = 0.148±4.032·0.034/(6)0.5 = 0.15±0.06%

20.  Hydrocarbons in the cab of an automobile were measured during trips on the New Jersey Turnpike and trips through the Lincoln Tunnel connecting New York and New Jersey.  The concentrations (+ standard deviations) of m- and p-xylene were

Turnpike:                                        31.4 + 30.0 mg/m3        (32 measurements)

Tunnel:                                           52.9 + 29.8 mg/m3        (32 measurements)

Do these results differ at the 95% confidence level?  At the 99% confidence level?

To test whether the difference in the mean values is significant, we want to use a case 2 t test.  The type of case 2 t test equation depends on whether or not the standard deviations for the measurements in the two locations is the same or different.  The standard deviations are very similar (an F test shows that Sturnpike/Stunnel = 1.007, or F = 1.01 which indicates no significant difference in standard deviations)

Thus we can use the equation tcalculated = (|31.4  52.9|/Spooled)[(nturnpikentunnel)/(nturnpike + ntunnel)]0.5

Spooled = {[30.02(32  1) + 29.82(32  1)]/(32 + 32  2)}0.5 = [(27900 + 27529)/62]0.5

Spooled = 29.9

tcalc = (21.5/29.9)(16)0.5 = 2.88

At 95% (degrees of freedom = 62), tTable = 2.00  < 2.88. so the difference is significant

At 90% (degrees of freedom = 62), tTable = 2.66  < 2.88. so the difference is still significant

21.  A standard Reference Material is certified to contain 94.6 ppm of an organic contaminant in soil.  Your analysis gives values of 98.6, 98.4, 97.2, 94.6, and 96.2 ppm.  Do your results differ from the expected result at the 95% confidence level?  If you made one more measurement and found 94.5, would your conclusion change?

Original Set:  mean = 97.00 ppm, standard deviation = 1.66 ppm.

Calculated t value = (std. value  mean value(n)0.5/(std. dev.) = (94.6  97.0) ·(n)0.5/1.66 = 3.24.  The table t value (n-1 = 4 and 95%) = 2.78.

Since calculated t > table t value, yes, there is a significant bias.

With one additional value of 94.5:  mean = 96.58, standard deviation = 1.80

Calculated t value = 2.70, Table t value = 2.57.

Calculated t value is still larger than the table value, so, there still is a significant bias. No change in the result.