2.7 In-Slides Practice

S12

Practice

Determine the number of protons (Z), neutrons (N), and electrons (e^–) in each of the following:

²⁸Si
¹⁹F
⁵⁷Fe
carbon-13

Solution

²⁸Si
- Z = 14
- N = 14
- e^– = 14
¹⁹F
- Z = 9
- N = 10
- e^– = 9
⁵⁷Fe
- Z = 26
- N = 31
- e^– = 26
carbon-13
- Z = 6
- N = 7
- e^– = 6

S12

Practice

Identify the number of protons (Z), neutrons (N), and electrons (e^–) for the following:

¹⁹⁷Au
⁴⁸Ti
⁵⁵Mn

Solution

¹⁹⁷Au
- Z = 79
- N = 118
- e^– = 79
⁴⁸Ti
- Z = 22
- N = 26
- e^– = 22
⁵⁵Mn
- Z = 25
- N = 30
- e^– = 25

S12

Practice

Determine the neutral element for each of the following:

26 protons; 56 neutrons
20 n⁰; 18 e^–
76 n⁰; 52 p⁺

Solution

S12

Practice

Which is the correct description for ⁵⁹Co?

59 p⁺; 56 n⁰
27 p⁺; 59 n⁰
32 p⁺; 27 n⁰
27 p⁺; 32 n⁰

Solution

S15

Practice

Chlorine has two stable isotopes:
³⁵Cl (75.78 %; m = 34.9689 amu)
³⁷Cl (24.22 %; m = 36.9659 amu)
What is the average atomic mass (in amu) of Cl?

Solution

Sum over each isotope’s mass multiplied by its fractional abundance.

\[\begin{align*} \left ( \dfrac{75.78}{100} \right ) (34.9689~\mathrm{amu}) + \left ( \dfrac{24.22}{100} \right ) (36.9659~\mathrm{amu}) &= 34.44518022~\mathrm{amu}\\ &= 35.45~\mathrm{amu} \end{align*}\]

S16

Practice

Natural lithium exists as two stable isotopes:
⁶Li (97.42 %; m = 6.015 amu)
⁷Li (2.58 %; m = 7.016 amu)
What is the average atomic mass (in amu) of Li?

Solution

Sum over each isotope’s mass multiplied by its fractional abundance.

\[\begin{align*} \left ( \dfrac{97.42}{100} \right ) (6.015~\mathrm{amu}) + \left ( \dfrac{2.58}{100} \right ) (7.016~\mathrm{amu}) &= 6.041~\mathrm{amu} \end{align*}\]

S16

Practice

Argon has three naturally occurring isotopes:
    Ar-36 (0.337 %; m = 35.968 amu)
    Ar-38 (0.063 %; m = 37.963 amu)
    Ar-40 (99.60 %; m = 39.962 amu)
What is the atomic weight of Ar?

Solution

Sum over each isotope’s mass multiplied by its fractional abundance.

\[\begin{align*} \left ( \dfrac{0.337}{100} \right ) (35.968~\mathrm{amu}) + \left ( \dfrac{0.063}{100} \right ) (37.963~\mathrm{amu}) + \left ( \dfrac{99.60}{100} \right ) (39.962~\mathrm{amu}) &= 39.947~\mathrm{amu} \end{align*}\]

S16

Practice

The average atomic mass of nitrogen is 14.007 amu. The atomic masses of the two stable isotopes of nitrogen are
N-14 (m = 14.003074004 amu)
N-15 (m = 15.000108898 amu)
Determine the percent abundance of each nitrogen isotope.

Solution

\[\begin{align*} 14.007 &= (x)(14.003074004)~+ \\ &\phantom{=}~~(1-x)(15.000108898) \\[2ex] 14.007 &= 14.003074004x~+ \\ &\phantom{=}~~15.000108898 - 15.000108898x \\[2ex] -0.99340897 &= -0.997034968x \\ x &= 0.99606232838 \end{align*}\]

Here, x represents the fractional abundance for ¹⁴N since it was associated with the mass for this isotope in the way the equation was written. Convert the fractional abundance to percent abundance for ¹⁴N.

\[0.99606232838 \times 100\% = 99.6\%\]

Determine the percent abundance for ¹⁵N.

\[\begin{align*} (1-0.99606232838) \times 100\% = 0.39\% \end{align*}\]

S17

Practice

Silver (m = 107.87 amu) has two stable isotopes:
Ag-107 (m = 106.90509 amu)
Ag-109 (m = 108.90476 amu)
Determine the percent abundance of these two isotopes of silver.

Solution

\[\begin{align*} 107.87 &= (x)(106.90509)~+ \\ &\phantom{=}~~(1-x)(108.90476) \\[2ex] 107.87 &= 106.90509x~+ \\ &\phantom{=}~~108.90476 - 108.90476x \\[2ex] -1.03476 &= -1.99967 \\ x &= 0.51746538178 \end{align*}\]

Here, x represents the fractional abundance for ¹⁰⁷Ag since it was associated with the mass for this isotope in the way the equation was written. Convert the fractional abundance to percent abundance for ¹⁰⁷Ag.

\[0.51746538178 \times 100\% = 51.75\%\]

Determine the percent abundance for ¹⁰⁹Ag.

\[\begin{align*} (1-0.51746538178) \times 100\% = 48.25\% \end{align*}\]

S17

Practice

Rubidium (m = 85.47 amu) has two naturally occurring isotopes:
Rb-85 (m = 84.9118 amu)
Rb-87 (m = 86.9092 amu)
Determine the percent abundance of these two isotopes of rubidium.

Solution

\[\begin{align*} 85.47 &= (x)(84.9118)~+ \\ &\phantom{=}~~(1-x)(86.9092) \\[2ex] 85.47 &= 84.9118~+ \\ &\phantom{=}~~86.9092 - 86.9092x \\[2ex] -1.4392 &= -1.9974 \\ x &= 0.7205366977 \end{align*}\]

Here, x represents the fractional abundance for ⁸⁵Rb since it was associated with the mass for this isotope in the way the equation was written. Convert the fractional abundance to percent abundance for ⁸⁵Rb.

\[0.7205366977 \times 100\% = 72.05\%\]

Determine the percent abundance for ⁸⁷Rb.

\[\begin{align*} (1-0.7205366977) \times 100\% = 27.95\% \end{align*}\]

S22

Practice

Write the empirical formulas for the following:

glucose (C₆H₁₂O₆)
adenine (C₅H₅N₅)
nitrous oxide (N₂O)

Solution

Reduce the subscripts in the molecular formula to the smallest possible whole numbers.

Glucose

\[\begin{align*} \mathrm{C_{6}H_{12}O_{6}} \longrightarrow \mathrm{C_{6/6}H_{12/6}O_{6/6}} \longrightarrow \mathrm{CH_{2}O} \end{align*}\]

Adenine

\[\begin{align*} \mathrm{C_{5}H_{5}N_{5}} \longrightarrow \mathrm{C_{5/5}H_{5/5}N_{5/5}} \longrightarrow \mathrm{CHN} \end{align*}\]

Nitrous Oxide

\[\begin{align*} \mathrm{N_{2}O} \longrightarrow \mathrm{N_{2/1}O_{1/1}} \longrightarrow \mathrm{N_2O} \end{align*}\]

S22

Practice

True/False: If any coefficient (subscript) in the molecular formula is 1, then the molecular formula and empirical formula are the same.

Solution

True