2.3: Calculating Atomic Masses (Problems)
- Page ID
- 98689
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)PROBLEM \(\PageIndex{1}\)
Determine the number of protons, neutrons, and electrons in the following isotopes that are used in medical diagnoses:
(a) atomic number 9, mass number 18, charge of 1−
(b) atomic number 43, mass number 99, charge of 7+
(c) atomic number 53, atomic mass number 131, charge of 1−
(d) atomic number 81, atomic mass number 201, charge of 1+
(e) Name the elements in parts (a), (b), (c), and (d)
- Answer a
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p: 9; n: 9; e: 10
- Answer b
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p: 43; n: 56; e: 36
- Answer c
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p: 53; n: 78; e: 54
- Answer d
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p: 81; n: 120; e: 80
- Answer e
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a - F; b - Tc; c - I; d - Tl
- Click here to see a video of the solution.
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PROBLEM \(\PageIndex{2}\)
Give the number of protons, electrons, and neutrons in neutral atoms of each of the following isotopes:
(a) \(\ce{^{10}_5B}\)
(b) \(\ce{^{199}_{80}Hg}\)
(c) \(\ce{^{63}_{29}Cu}\)
(d) \(\ce{^{13}_6C}\)
(e) \(\ce{^{77}_{34}Se}\)
- Answer a
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p&e: 5; n: 5
- Answer b
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p&e: 80; n: 119
- Answer c
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p&e: 29; n: 34
- Answer d
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p&e: 6; n: 7
- Answer e
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p&e: 34; n: 43
PROBLEM \(\PageIndex{3}\)
An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99 amu, and 8.82% abundance with 21.99 amu. Calculate the average atomic mass of this element.
- Answer
-
20.16 amu
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PROBLEM \(\PageIndex{4}\)
Average atomic masses listed by IUPAC are based on a study of experimental results. Bromine has two isotopes, 79Br and 81Br, whose masses (78.9183 and 80.9163 amu) and abundances (50.69% and 49.31%) were determined in earlier experiments. Calculate the average atomic mass of Br based on these experiments. How does this compare to the value given on the periodic table?
- Answer
-
79.90 amu; this matches the value on the periodic table
PROBLEM \(\PageIndex{5}\)
The 18O:16O abundance ratio in some meteorites is greater than that used to calculate the average atomic mass of oxygen on earth. Is the average atomic mass of an oxygen atom in these meteorites greater than, less than, or equal to a terrestrial oxygen atom?
- Answer
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Greater, since the contribution to the average atomic mass of 18O is greater, that will raise the average atomic mass in meteorites compared to on earth.
PROBLEM \(\PageIndex{6}\)
Compare 1 mole of H2, 1 mole of O2, and 1 mole of F2.
(a) Which has the largest number of molecules? Explain why.
(b) Which has the greatest mass? Explain why.
- Answer a
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1 mole is always 6.022 x 1023 molecules. They have the same number of molecules.
- Answer b
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F2; it has the highest molar mass.
PROBLEM \(\PageIndex{7}\)
Which contains the greatest mass of oxygen: 0.75 mol of ethanol (C2H5OH), 0.60 mol of formic acid (HCO2H), or 1.0 mol of water (H2O)? Explain why.
- Answer
-
Formic acid. Its formula has twice as many oxygen atoms as the other two compounds (one each). Therefore, 0.60 mol of formic acid would be equivalent to 1.20 mol of a compound containing a single oxygen atom.
PROBLEM \(\PageIndex{8}\)
Determine the mass of each of the following:
(a) 0.0146 mol KOH
(b) 10.2 mol ethane, C2H6
(c) 1.6
(d) 6.854
(e) 2.86 mol Co(NH3)6Cl3
- Answer a
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0.819 g
- Answer b
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307 g
- Answer c
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0.23 g
- Answer d
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1.235 × 106 g (1235 kg)
- Answer e
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765 g
PROBLEM \(\PageIndex{9}\)
Which of the following represents the least number of molecules?
- 20.0 g of H2O (18.02 g/mol)
- 77.0 g of CH4 (16.06 g/mol)
- 68.0 g of CaH2 (42.09 g/mol)
- 100.0 g of N2O (44.02 g/mol)
- 84.0 g of HF (20.01 g/mol)
- Answer
-
20.0 g of H2O represents the smallest number of moles, meaning the least number of molecules present. Since 1 mole = 6.022 × 1023 molecules (or atoms) regardless of identity, the least number of moles will equal the least number of molecules.
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Contributors
Paul Flowers (University of North Carolina - Pembroke), Klaus Theopold (University of Delaware) and Richard Langley (Stephen F. Austin State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/85abf193-2bd...a7ac8df6@9.110).
- Adelaide Clark, Oregon Institute of Technology