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Homework Solutions #1

  • Page ID
    36920
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    Q1

    Complete each calculation with the correct number of significant figures

    1. \(\dfrac{7.68 \times 8.009}{4.1}\phantom{\dfrac{\Big(}{\Big(}}\)
    2. \(\dfrac{6.89-5.452}{1.596+4.7121}\phantom{\dfrac{\Big(}{\Big(}}\)
    3. \(\dfrac{3.668\times10^{-7}+ 8.3469\times10^{-7}}{3.7348\times10^{-10}}\phantom{\dfrac{\Big(}{\Big(}} \)

    Answer

    1. 15
    2. 0.23
    3. 3217

    Hint

    Solution

    It is easier to take into account the correct number of significant figures after the calculations are completed to avoid a rounding error. Once you calculate the answer, determine which number will limit your answer and round accordingly.

    Q2

    Calculate how many grams of mixture are there if 250 mL of ethanol is mixed with 500 mL of benzene.

    Substance

    Density @ 20oC, 1 atm (g/cm3)

    Ethanol

    0.789

    Benzene

    0.880

    Answer

    • 637 g.

    Hint

    Solution

    [(250. mL ethanol / 1) * (1 cm3ethanol / 1 mL ethanol) * (0.789 g ethanol / 1 cm3 ethanol)] + [500. mL benzene / 1) * (1 cm3benzene / 1 mL benzene) * (0.880 g benzene / 1 cm3 benzene)] = 637 g (3 sig. figs)

    Q3

    Give the ISOTOPIC symbol for the following ions:

    1. Ion with 73 Protons, 75 Electrons, and 107 Neutrons.
    2. Ion with 28 Protons, 25 Electrons, and 31 Neutrons
    3. Ion with 37 Protons, 37 Electrons, and 49 Neutrons
    4. Ion with 7 Protons, 7 Electrons, and 7 Neutrons

    Answer

    1. \(\ce{^{180}_{73}Ta^{2-}}\)
    2. \(\ce{^{59}_{28}Ni^{3+}}\)
    3. \(\ce{^{86}_{37}Rb}\)
    4. \(\ce{^{14}_{7}N}\)

    Hint

    Solution

    When naming an ion, the element can be identified by the number of protons. The atomic weight is calculated by the addition of protons to neutrons. Next, charge is calculated by comparing the quantity of Protons to Neutrons, which Protons provide positive charge and Electrons a negative charge.

    Q4

    Label the following elements as halogens, alkali metals, alkaline earth metals, noble gases, or transition metals.

    1. Lithium,
    2. Chlorine
    3. Argon
    4. Krypton
    5. Calcium
    6. Barium
    7. Cesium
    8. Fluorine
    9. Radon
    10. Tungsten

    Answer

    1. Alkali metals – Lithium, Cesium
    2. Alkaline- Earth – Calcium, Barium
    3. Halogens – Fluorine, Chlorine
    4. Noble gases – Argon, Krypton, Radon
    5. Transition metal – Tungsten

    Hint

    • Use a periodic table of elements to see how groups of elements are classified.
    • chemwiki.ucdavis.edu/Wikitext...Periodic_Table

    Solution

    Alkali metals are found in the first column of the periodic table. Alkaline – Earth metals are found in Column 2 of the periodic table. Halogens are found in Column 17, Noble gases are found in Column 18, and transition metals are found in column 3-12.

    Q5

    An element consists of the following isotopes. What is the average atomic mass and the name of the element?

    Percent natural abundance

    Mass (u)

    90.48%

    19.9924

    0.27%

    20.9938

    9.25%

    21.9914

    Answer

    • 20.18 amu; Ne.

    Hint

    Solution

    A = (0.9048*19.9924) + (0.0027*20.9938) + (0.0925*21.9914) = 18.09 + 0.057 + 2.03 = 20.18 amu.

    From the periodic table, the element is neon (Ne).

    Q6

    A compound, made only from mercury and sulfur, is 86.2% \(Hg\) by mass. The molar mass of this compound is 465 g/mol. Determine the empirical formula and the molecular formula of this compound.

    Answer

    • Empirical Formula: \(HgS\);
    • Molecular Formula: \(Hg_2S_2\).

    Hint

    Solution

    The compound is 86.2% S by mass. Therefore, there are 86.2 gram of mercury in 100. gram of the compound.

    g S = 100. g – 86.2 g = 13.8 g S.

    (86.2 g Hg) * (1 mol Hg / 200.59 g Hg) = 0.430 mol Hg.

    (13.8 g S) * (1 mol S / 32.066 g S) = 0.430 mol S.

    ratio of Hg : S = 1:1.

    So, the empirical formula is \(HgS\)

    Empirical mass of \(HgS\) = 200.59 g/mol + 32.066 g/mol = 232.66 g/mol

    Multiplication factor = molecular mass / empirical mass = (465 g/mol) / (232.66 g/mol) = 1.999 ≈ 2.

    Therefore, molecular formula = (multiplication factor) * (empirical formula) = 2 * HgS = \(Hg_2S_2\).

    Q7

    A compound is a common first-aid antiseptic. The products of its combustion reaction in the presence of oxygen gas are water and carbon dioxide. This compound is composed of carbon, hydrogen and oxygen. If 9.214 g of this compound produces 17.604 g carbon dioxide and 10.812 g water, what is the empirical formula of this compound? Hint: Some oxygen atoms of the products may come from oxygen gas.

    Answer

    • \(C_2H_6O\)

    Hint

    Solution

    (17.604+10.812)-9.214=19.200g oxygen gas

    19.200g/16.00(g/mol)=1.200mol O atom

    17.604g/44.01(g/mol)=0.4000mol carbon dioxide, 0.4000mol carbon, 0.8000mol O atom

    10.812g/18.02(g/mol)=0.6000mol water, 1.200 mol hydrogen, 0.6000mol O atom

    Ratio of C, H and O of this compound: 0.4000 C: 1.200 H : (0.8000+0.6000-1.2000) O = 0.40 C:1.2 H:0.20 O = 2 C : 6 H : 1 O

    Empirical Formula: \(C_2H_6O\)

    Q8

    Assign the oxidation state of each atom for the following compounds.

    1. \(Al\)
    2. \(PbSO_4\)
    3. \(H_2SO_3\)
    4. \(NaHCO_3\)
    5. \(H_2O_2\)
    6. \(Ne\)
    7. \(MgCO_3\)
    8. \(Na_2Cr_2O_7\)

    Answer

    1. 0
    2. Pb: +2 , S: +6, O: -2
    3. H:+1, S: +4, O: -2
    4. Na: +, H: +, C: +4, O:-2
    5. H:+, O: -1
    6. Ne: 0
    7. Mg +2. C+4,O-2
    8. Na+, Cr +6, O-2

    Hint

    Solution

    • Oxidation state of atoms of all metals, pure substances and noble gases are 0.
    • The sum of the oxidation state of a compound is 0.

    Homework Solutions #1 is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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