CHEMISTRY 31

Fall, 2017 - Dixon

Homework Set 1.2 Solutions

 

Ch. 1: 37 (carried forward from set 1.1)

37. How many grams of 0.491 wt % aqueous HF are required to provide a 50% excess to react with 25.0 mL of 0.0236 M Th4+ by the reaction Th4+ + 4F- => ThF4 (s)?

mmoles Th4+ = (23.6 mmol/L)*(25.0 mL)*(10-3 L/mL) = 0.590 mmol

mmol F- required = 0.59*(4 mol F-/1 mol Th4+) = 2.36 mmol

mmol F- required for 50% excess = 1.5*(2.36 mmol) = 3.54 mmol

g HF sol'n needed = (20.0 mg HF/mmol HF)*(3.54 mmol HF)*(10-3g/mg)/(0.00491 mgHF/mg sol'n) = 14.4 g.

 

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 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%

answer = 10.18 (+0.07)

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

answer = 174 (+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

answer = 0.147 (+0.003)

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%

answer = 7.86 (+0.01)

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%

answer = 2185.8 (+0.8)

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

value = 1.464

rel. unc. = (1/3)(0.05/3.14) = 0.0053

unc. = (0.0053)(1.464) = 0.0078

% uncertainty =0.5%

Answer = 1.464+0.008

 

Ch. 4: 1, 3a,b,d, 5a,b, 10, 12, 14 [postpone to set 1.3]

1.  What is the relation between the standard deviation and the precision of a procedure?  What is the relation between standard deviation and accuracy?

The standard deviation is directly related to precision of a procedure where low standard deviations are indicative of a high level of precision.

There is not a necessary relation between standard deviation and accuracy except that a high level of accuracy requires low standard deviations.  One can also have low standard deviations and poor accuracy if systematic errors exist.

 

3.  The ratio of the number of atoms of the isotopes 69Ga and 71Ga in eight samples from different sources was measured in an effort to understand differences in reported values of the atomic mass of gallium:

Sample

69Ga/71Ga

Sample

69Ga/71Ga

1

1.52660

5

1.52894

2

1.52974

6

1.52804

3

1.52592

7

1.52684

4

1.52731

8

1.52793

 

Find the a) mean

Mean = 1.52767 (used calculator functions)

b) standard deviation

0.00126 (used calculator functions)

c) standard deviation of the mean

standard deviation of the mean = S/(n)0.5 = 0.00126/(8)0.5 = 0.00045

 

5. 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 [used to get values between columns or rows in Table – this won’t be on an exam],

Fraction = 0.1915 + 0.2258 = 0.417

 

10.  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.

 

12.  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%