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    Assignment # 4 - Model Answer

    Uploaded by

    Tarek Madkour
    This document contains a chemistry assignment with 5 questions. It provides the questions, solutions showing the steps, and molar calculations to determine: 1) The number of nitrogen atoms in a sample and the mass of oxygen. 2) The empirical formula of a compound given its elemental composition percentages. 3) The theoretical yield of a reaction given amounts of reactants. 4) The actual grams of a product obtained given its theoretical yield and percent yield. 5) The limiting reagent of a reaction between manganese dioxide and hydrochloric acid.

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    Assignment # 4 - Model Answer

    Uploaded by

    Tarek Madkour
    55 views4 pages
    This document contains a chemistry assignment with 5 questions. It provides the questions, solutions showing the steps, and molar calculations to determine: 1) The number of nitrogen atoms in a sample and the mass of oxygen. 2) The empirical formula of a compound given its elemental composition percentages. 3) The theoretical yield of a reaction given amounts of reactants. 4) The actual grams of a product obtained given its theoretical yield and percent yield. 5) The limiting reagent of a reaction between manganese dioxide and hydrochloric acid.

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    Model Answers for assignment 4

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    Assignment # 4 - model answer

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    This document contains a chemistry assignment with 5 questions. It provides the questions, solutions showing the steps, and molar calculations to determine: 1) The number of nitrogen atoms in a sample and the mass of oxygen. 2) The empirical formula of a compound given its elemental composition percentages. 3) The theoretical yield of a reaction given amounts of reactants. 4) The actual grams of a product obtained given its theoretical yield and percent yield. 5) The limiting reagent of a reaction between manganese dioxide and hydrochloric acid.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    55 views4 pages

    Assignment # 4 - Model Answer

    Uploaded by

    Tarek Madkour
    This document contains a chemistry assignment with 5 questions. It provides the questions, solutions showing the steps, and molar calculations to determine: 1) The number of nitrogen atoms in a sample and the mass of oxygen. 2) The empirical formula of a compound given its elemental composition percentages. 3) The theoretical yield of a reaction given amounts of reactants. 4) The actual grams of a product obtained given its theoretical yield and percent yield. 5) The limiting reagent of a reaction between manganese dioxide and hydrochloric acid.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 4

    The American University in Cairo

    School of Science and Engineering


    Department of Chemistry
    CHEM 1005

    Assignment # 4 Thursday May 7th – Sunday May 10th Spring 2020

    Please ANSWER the following five questions and SHOW your steps in details:

    1- a- A sample of C7H5N3O4 contains 0.4662 moles of carbon atoms. How many nitrogen
    atoms are there in the sample?
    b- A sample of C7H5N3O4 has a mass of 7.81 g. What is the mass of oxygen in this sample?

    Solution:

    a- Moles of C7H5N3O4 = 0.4662 moles C = 0.0666 mole C7H5N3O4


    7 moles C / mole C7H5N3O4
    Moles of N = 0.0666 mole C7H5N3O4 x (3 moles N / mole C7H5N3O4) = 0.1998 mole N
    Atoms of N = 0.1998 mole x (6.022 x 1023) atom/mole = 1.203 x 1023 N atoms

    b- Moles of C7H5N3O4 = Mass of C7H5N3O4


    Molar mass of C7H5N3O4
    = 7.81 g = 0.04005 mol
    (7x12) + (5x1) + (3x14) + (4x16) g/mol
    1 mole of C7H5N3O4 contains 4 moles of O
    Moles of O = 4 x 0.04005 = 0.1602 mol
    Mass of O = 0.1602 mol x 16 g/mol = 2.5632 g
    2- A compound used in a polymer research project contains carbon, hydrogen, and nitrogen.
    The composition values are: carbon, 58.774%; hydrogen, 13.810%; nitrogen, 27.416%.
    Determine the empirical formula of this compound.

    Solution:

    For a 100 g sample of the polymer, the % composition values represent masses of the
    constituent elements.
    Moles of C = 58.774 g / 12 g/mol = 4.898 mol C
    Moles of H = 13.810 g / 1 g/ mol = 13.810 mol H
    Moles of N = 27.416 g / 14 g/mol = 1.958 mol N
    Divide all by 1.958
    Moles of C = 2.5
    Moles of H = 7.1 ≈ 7
    Moles of N = 1
    Multiply all by 2
    Moles of C = 5
    Moles of H = 14
    Moles of N = 2
    Empirical formula: C5H14N2
    3- You are given the balanced chemical equation:
    4AsF3 + 3C2Cl6 → 4AsCl3 + 3C2Cl2F4.
    If 1.3618 moles of AsF3 are allowed to react with 0.817 moles of C2Cl6, what would be the
    theoretical yield of AsCl3, in moles?

    Solution:

    According to the equation stoichiometry


    AsF3 = 4 = 1.333
    C2Cl6 3
    According to the reacting amounts
    1.3618 mol AsF3 = 1.667
    0.817 mol C2Cl6
    This means that C2Cl6 is the limiting reagent, i.e. all of its moles will be theoretically
    consumed to form the products.
    According to the equation stoichiometry
    AsCl3 = 4 = 1.333
    C2Cl6 3
    Theoretical yield of AsCl3 = 1.333 x 0.817 = 1.089 mol AsCl3

    4- For another run of the same reaction above the theoretical yield of C 2Cl2 F4 was calculated
    to be 1.86 moles. If the percent yield in the reaction was 77.2%, how many grams of
    C2Cl2F4 were actually obtained?

    Solution:

    % yield = Actual yield x 100%


    Theoretical yield
    Actual yield = 1.86 x 0.772 = 1.436 mole C2Cl2 F4
    Molar mass of C2Cl2F4 = (2x12) + (2x35.5) + (4x19) = 171 g/mol
    Mass obtained = 1.436 mol x 171 g/mol = 245.56 g C2Cl2F4
    5- Chlorine gas can be made from the reaction of manganese dioxide with hydrochloric acid.
    Which is the limiting reagent when 28 g of MnO2 are mixed with 42 g of HCl?
    Show your steps into finding it out.
    MnO2(s) + 4HCl(aq) → MnCl2(aq) + 2H2O(l) + Cl2(g)

    Solution:

    Moles of MnO2 = 28 g = 0.3218 mol


    [55 + (2x16)] g/mol
    Moles of HCl = 42 g = 1.1507 mol
    [1 + 35.5] g/mol
    According to the equation stoichiometry
    HCl /MnO2 = 4/1 = 4 HCl should be 4 x MnO2
    According to the reacting amounts
    HCl /MnO2 = 1.1507 mol / 0.3218 mol = 3.5758 Available HCl is <4 x MnO2
    HCl is the limiting reagent.

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