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Chemistry LibreTexts

Homework 4

  • Page ID
    28861
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    Q1.66

    Question: Find the amount of protons and neutrons in the following atoms:

    (1) \[_{6}^{14}\textrm{C}\]

    (2) \[_{8}^{18}\textrm{O}\]

    (3) \[_{25}^{55}\textrm{Mn}\]

    (4) \[_{30}^{64}\textrm{Zn}\]

    Solution

    Strategy:

    A. First identify the subscript, the bottom number on the left hand side of the element's symbol. This is the atomic number and indicates how many protons the atom's nucleus contains. All isotopes of the same element contain the same number of protons.

    B. Next, identify the superscript, the top number left hand side of the element symbol. This is the atomic mass number, the total combined number of protons and neutrons. Different isotopes of a given element differ in the number of neutrons contained in their nuclei.

    C. The number of protons can be found simply by identifying the atomic number. To find the number of neutrons in the atom, the atomic number is subtracted from the atomic mass number.

    A The atomic number of carbon is 6, the atomic number of oxygen is 8, the atomic number of manganese is 25, and the atomic number of zinc is 30. Therefore, we now know the number of protons contained within each atom since this number is represented by the atomic number.

    B The atomic mass number of the carbon isotope is 14, the mass number of the oxygen isotope is 18, the mass number of the manganese isotope is 55, and the mass number of the zinc isotope is 64. This is the combined number of protons and neutrons in each respective isotope.

    C Carbon isotope: \[14-6=8\, neutrons\]

    Oxygen isotope: \[18-8=10\, neutrons\]

    Manganese isotope: \[55-25=30\, neutrons\]

    Zinc isotope: \[64-30=34\, neutrons\]

    Q5.75

    Question: Find the amount of moles contained in each specimen.

    1. 7.87 kg H2O2
    2. 2.34 kg NaCl
    3. 12.5 g C2H6O
    4. 85.72 g NH3

    Solution

    Strategy:

    A. First, determine the units that are given for each sample. In order to convert these measurements to moles, each sample should first be written in grams. Two of the given samples are measured in kilograms. Use the following conversion factor to convert kilograms to grams, with x representing the given mass:

    \[x\, kg\cdot \frac{1000\, g}{1\, kg}\]

    B. Next, find the atomic masses of all the atoms in each compound by using the Periodic Table. Then, add these atomic masses together for each compound. The resulting value will be the number of grams of each sample that make up one mole.

    C. Convert the mass in grams of each sample to moles by multiplying it by the following conversion factor, called the molar mass, with x representing the mass of the given specimen and y representing the calculated atomic mass found in step B:

    \[x\, g\cdot \frac{1\, mole}{y\, g}\]

    A Numbers 1 and 2 first need to be converted into grams.

    1. \[7.87\, kg\, H_2O_2\cdot \frac{1000\, g}{1\, kg}=7,870\, g\, H_2O_2\]
    2. \[2.34\, kg\, NaCl\cdot \frac{1000\, g}{1\, kg}=2,340\, g\, NaCl\]

    B The following are the calculated molar masses of each of the given compounds. The atomic mass of each element has been rounded to 4 significant figures in these calculations.

    1. H2O2: \[2\, (1.008\, g\, H)+2\, (16.00\, g\, O)=\frac{34.02\, g}{1\, mole}\, H_2O_2\]
    2. NaCl: \[(22.99\, g\, Na)+(35.45\, g\, Cl)=\frac{58.44\, g}{1\, mole}\, NaCl\]
    3. C2H6O: \[2\, (12.01\, g\, C)+6\, (1.008\, g\, H)+\left ( 16.00\, g\, O \right )=\frac{46.07\, g}{1\, mole}\, C_2H_6O\]
    4. NH3: \[(14.01\, g\, N)+3\, (1.008\, g\, H)=\frac{17.03\, g}{1\, mole}\, NH_3\]

    C The number of moles in each specimen can now be calculated by multiplying the mass in grams of each sample by its molarity.

    1. \[7,870\, g\, H_2O_2\cdot \frac{1\, mole\, H_2O_2}{34.02\, g\, H_2O_2}=231\, moles\, H_2O_2\]
    2. \[2,340\, g\, NaCl\cdot \frac{1\, mole\, NaCl}{58.44\, g\, NaCl}=40.0\, moles\, NaCl\]
    3. \[12.5\, g\, C_2H_6O\cdot \frac{1\, mole\, C_2H_6O}{46.07\, g\, C_2H_6O}=0.271\, moles\, C_2H_6O\]
    4. \[85.72\, g\, NH_3\cdot \frac{1\, mole\, NH_3}{17.03\, g\, NH_3}=5.033\, moles\, NH_3\]

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