Problem Set 3 key - Recitation PDF

Preview

Summary

Recitation...

Description

Problem Set 03 – The Atom Chem 105 2.23 - 2.26 1. (a) Fill in the missing information about atoms of the eight nuclides in the following two tables. 23

Na

Symbol

89

Y

118

197

Sn

Au

Number of Protons (Atomic Number)

11

39

50

79

Number of Neutrons

12

50

68

118

Number of Electrons

11

39

50

79

Mass Number

23

89

118

197

37

Cl–

Symbol

23

Na+

81

Br–

226

Ra2+

Number of Protons (Atomic Number)

17

11

35

88

Number of Neutrons

20

12

46

138

Number of Electrons

18

10

36

86

Mass Number

37

23

81

226

(b) Fill in the missing information in the following tables of monatomic ions. 27

Al

Symbol

98

Mo

143

238

Nd

U

Number of Protons (Atomic Number)

13

42

60

92

Number of Neutrons

14

56

83

146

Number of Electrons

13

42

60

92

Mass Number

27

98

143

238

137

Ba2+

Symbol

64

Zn2+

32 2-

S

90

Zr4+

Number of Protons (Atomic Number)

56

30

16

40

Number of Neutrons

81

34

16

50

Number of Electrons

54

28

18

36

Mass Number

137

64

32

90

2. (a) In 1-3 sentences, define an isotope. Atoms of the same element that have the same number of protons in their nuclei, but different number of neutrons. (b) 2.20 Explain the inherent redundancy in the nuclide symbol 𝐴𝑍X . Both Z and X describe the element on the periodic table, so only one is necessary. (c) Write the symbol of the most abundant isotope of sodium. How many neutrons does it contain? The atomic weight of sodium is 22.9898 g/mol. This number is a weighted average of the molar masses of the various isotopes. Since it is very close to 23, most of the atoms must be of the isotope 23 weighing 23 g/mol. The symbol of this isotope is 11 Na. It has 12 neutrons. 3. The helium balloon you got on your birthday has almost entirely deflated, but there is still a very small amount of 4He inside with a total mass of 160 amu. How many protons and how many electrons are present in the balloon (assuming only 4He is there)? 160 𝑎𝑚𝑢 ×

1.66053892×10−24 g 1 𝑎𝑚𝑢

×

1 𝑚𝑜𝑙 He 4𝑔

×

6.022×1023 𝑎𝑡𝑜𝑚𝑠 1 𝑚𝑜𝑙 𝐻𝑒

×

2 𝑝𝑟𝑜𝑡𝑜𝑛𝑠 1 𝑎𝑡𝑜𝑚 𝐻𝑒

= 80 𝑝𝑟𝑜𝑡𝑜𝑛𝑠. This atom

has an equal number of protons and electrons so there would also be 80 electrons. 160 𝑎𝑚𝑢

Or another method: 80 electrons present.

𝑥

=

4 𝑎𝑚𝑢 1 𝐻𝑒 𝑎𝑡𝑜𝑚

 x = 40 He atoms. Since 2 protons per He, 80 protons and

4. (a) How many neutrons are in 10 g of fluorine-19, the only stable isotope of fluorine? 6.022 × 1023 𝑎𝑡𝑜𝑚𝑠 10 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 1 mol 10𝑔 199𝐹 × × = 3 × 1024 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 × 19 𝑔 𝐹 1 𝑚𝑜𝑙 1 𝑎𝑡𝑜𝑚 199𝐹 (b) Calculate the number of neutrons found in 2.13 mg of the non-fissionable isotope of uranium which has a mass number of 238.

2.13 𝑚𝑔 238 92 𝑈 ×

1g

1000 𝑚𝑔 = 7.87 × 1020 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠.

×

1 𝑚𝑜𝑙 𝑈 238 𝑔

×

6.022 × 1023 𝑎𝑡𝑜𝑚𝑠 146 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 × 1 𝑎𝑡𝑜𝑚 𝑈 1 𝑚𝑜𝑙 𝑈

5. (a) 2.46 Naturally occurring sulfur consists of four isotopes: 32S (31.9721 amu, 95.04%); 33S (32.9715 amu, 0.75%); 34S (33.9679 amu, 4.20%); and 36S (35.9671 amu, 0.01%). Use this information to calculate the average atomic mass of sulfur. (.9504 x 31.9721 amu) + (.0075 x 32.9715 amu) + (.0420 x 33.9679 amu) + (.0001 x 35.9671 amu) = 32.1 amu (if you did not round at each step) = (30.63 amu) + (0.25 amu) + (1.42 amu) + (0.004 amu) = 32.31 amu (if you did round at each step) (b) Calculate the atomic weight of nickel to the correct number of significant figures using the following data: 58 Ni, 68.075%, 57.9353429 amu  (.0.68075 x 57.9353429 amu) = 39.439 amu 60 Ni, 26.221%, 59.9307864 amu  + (0.26221 x 59.9307864 amu) = + 15.714 amu 61 Ni, 1.1399%, 60.9310560 amu  + (0.011399 x 60.9310560 amu) = + 0.69455 amu 62 Ni, 3.637%, 61.9283451 amu  + (0.03637 x 61.9283451 amu) = + 2.252 amu 64 Ni, 0.9271%, 63.9279660 amu  + (0.009271 x 63.9279660 amu) = + 0.5927 amu = 58.692 amu (or 58.693 if you didn’t round in intermediate steps) In both (a) and (b), the last digit is clearly the last ‘significant’ digit because it varies depending on how the calculations are rounded. In each case there are certain, known digits + 1 uncertain digit 6. (a) Naturally occurring Mg has three stable isotopes, 24Mg, 25Mg, and 26Mg. This means that there are three forms of Mg atoms, each with a unique atomic weight. This appears to be contrary to the idea that atomic weight is the defining property of an element. Explain (in 3 sentences or less) why all three of these types of atoms are indeed the same element. It is the atomic number, Z (the number of protons in the nucleus) that defines an element, not the atomic weight. The atomic weight includes the mass of any neutrons in the nucleus, and the number of neutrons can vary between atoms having the same Z; this is the definition of an isotope – atoms with the same Z but different numbers of neutrons. We consider them to be the same element because they exhibit the same properties & reactivity despite having different numbers of neutrons. (b) Chemical analyses conducted by the first Mars rover robotic vehicle in its 1997 mission produced the magnesium isotope data shown in the table below. Is the average atomic mass of magnesium in this Martian sample the same as on Earth (24.305 amu)? (use the correct number of sig figs): Isotope Mass (amu) Natural Abundance (%) 24 23.9850 78.70 Mg 25 24.9858 10.13 Mg 26 Mg 25.9826 11.17 mMg = (0.7870  23.9850 amu) + (0.1013  24.9858 amu) + (0.1117  25.9826 amu) = 24.31 amu The average mass of Mg on Mars is the same as here on Earth. The mass of Mg on Mars should be close to the same value as on Earth; the magnesium on both planets arrived in the solar system via the same ancient stardust. 7. The following elements all have two stable isotopes. Which isotope in each of the following pairs of stable isotopes is more abundant? Explain how you know. Looking at the average atomic masses of the elements given on the periodic table, we can deduce

a) b) c) d) e)

10

B or 11B is more abundant; the average atomic weight of B is 10.811 amu, so there must be more 11B, pulling the average closer to 11 amu than to 10 amu 6 Li or 7Li; 14 N or 15N 20 Ne or 22Ne 35 Cl or 37Cl

8. Pencil “lead” is actually a form of carbon called graphite. (a) If a single carbon atom has a diameter of 1.54 Å (10 Å = 1 nm), how many carbon atoms could fit side-by-side along the width of an exclamation point line drawn by your pencil (assuming your pencil uses “lead” that is 0.20 mm wide)? . 20𝑚𝑚 ×

1𝑚 1000𝑚𝑚

×

1𝑛𝑚 1×10−9 𝑚

×

10 Å 1𝑛𝑚

×

1 𝐶 𝑎𝑡𝑜𝑚 1.54Å

= 1.3 × 106 𝑎𝑡𝑜𝑚𝑠

(b) How many protons would be in this line of carbon atoms? 6 𝑝𝑟𝑜𝑡𝑜𝑛𝑠 1.3 × 106 𝑎𝑡𝑜𝑚𝑠 × = 7.8 × 106 𝑝𝑟𝑜𝑡𝑜𝑛𝑠 1 𝑎𝑡𝑜𝑚 9. A single gold atom has a mass of 197.0 amu. How many gold atoms are in a cube of gold that is 10.0 mm on each side (about the size of a sugar cube) if the density of gold is 19.3 g/cm3 ? (1 amu = 1.66053892x10-24 g) First, convert 10mm into cm because density is given in g/cm3. 1𝑐𝑚 𝑚 10𝑚𝑚 × 10𝑚𝑚 = 1𝑐𝑚. Next, use the equation 𝐷 = to solve for the mass of gold inside the cube. 𝑣 19.3𝑔 𝑐𝑚3

=

𝑥

1𝑐𝑚3

19.3𝑔 𝐴𝑢 ×

𝑥 = 19.3𝑔. Next, we solve for the amount of gold atoms using dimensional analysis. 1 𝑎𝑚𝑢 1.66053892×10−24 g

×

1 𝐴𝑢 𝑎𝑡𝑜𝑚 197 𝑎𝑚𝑢

= 5.90 × 1022 𝑎𝑡𝑜𝑚𝑠...