Chemistry

A hydrocarbon Y on combustion gives 0.352g of carbon(iv)oxide and 0.18g of water. Calculate the empirical formula of the compound and If the relative molecular mass of Y is 58g, determine the molecular formula of Y.?

  1. 👍
  2. 👎
  3. 👁
  4. ℹ️
  5. 🚩
  1. ...Y + O2 ==> CO2 + H2O
    ...................0.352 g + 0.18 g
    g C in Y is 0.352 x (12/44) = 0.096
    g H in Y is 0.18 x (2/18)) = 0.02

    mols C in Y = 0.096/12 = 0.008
    mols H in Y = 0.02/1 = 0.02
    Now find the ratio. The easy way is to divide both numbers by the smallest number; i.e.,
    C = 0.008/0.008 = 1
    H = 0.02/0.008 = 2.5
    Make whole numbers which will be C = 2 and H = 5 so the empirical formula is C2H5. The empirical formula will be (C2H5)x so x is a multiple of the empirical formula. The empirical formula is 29 so 29*what number = 58.
    The molecular formula is (C2H5)2 or C4H10. Is the molar mass 59?
    4*12 = 48 and 10*1 = 10
    48+10 = 58. Bingo.

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
    👤
    DrBob222
  2. Mega Thanks

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩
  3. C4H10

    1. 👍
    2. 👎
    3. ℹ️
    4. 🚩

Respond to this Question

First Name

Your Response

Similar Questions

  1. chemistry

    In an experiment 3.10g of carbon, hydrogen and oxygen produced 4.4g of C02 and 2.7g of water on complete combustion. Determine the empirical formula of the compound ( C= 12, 0= 16, H =1)

  2. Chemistry

    Can someone answer this question? The enthalpy of combustion of solid carbon to form carbon dioxide is -393.7 kJ/mol carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is -283.3 kj/mol CO. Use these

  3. AP Chemistry

    a 5.15g sample of a hydrocarbon is burned in oxygen, producing 15.6g of carbon dioxide and 8.45g of water. Assuming an excess of oxygen, what is the empirical formula of the hydrocarbon?

  4. chemistry

    A 25.002g sample of a compound that contains carbon, and hydrogen, is subjected to a complete combustion reaction. Analysis of the products of the reaction produced 76.392g of carbon dioxide and 37.504g of water. Determine the

  1. C hemistry

    An unknown compound contains only carbon, hydrogen, and oxygen . Combustion of 2.50 g of this compound produced 3.67 g of carbon dioxide and 1.50 g of water. What is the empirical formula of a substance that contains 8.33×10−2

  2. Chemistry

    A1.35g sample of asubstance containing carbon hydrogen nitrogen and oxcgen on combustion produced 0.810gof water and 1.32g of carbon dioxide In separate exp all the nitrogen in a0.735 g sample of the substance was converted into

  3. IGCSE Chemistry

    Hydrocarbons A and B both contain 85.7 % carbon. Their molar masses are 42 and 84 respectively. a) Which elements does a hydrocarbon contain? b) Calculate the empirical formula of A and of B. c) Calculate the molecular formula of

  4. chemistry

    combustion analysis of a hydrocarbon produced: 33.01 g CO2 and 13.51 g H2O Calculate the empirical formula of the Hydrocarbon. I got CH5 but that doesn't seem right? how do i do this problem?

  1. Chemistry HELP

    Complete combustion of 3.30 g of a hydrocarbon produced 10.2 g of CO2 and 4.68 g of H2O. What is the empirical formula for the hydrocarbon? I did the calculation and the ratio comes out to C: 1 H: 2.24 So I thought the answer is

  2. chem

    Combustion reactions involve reacting a substance with oxygen. When compounds containing carbon and hydrogen are combusted, carbon dioxide and water are the products. Using the enthalpies of combustion for C4H4 (-2341 kJ/mol),

  3. chemistry

    A 15.27 g sample of Mo2O3(s) is converted completely to another molybdenum oxide by adding oxygen. The new oxide has a mass of 18.33 g. Add subscripts below to correctly identify the empirical formula of the new oxide.

  4. chemistry

    Combustion analysis, Empirical and Molecular formulas, help!? When 2.686 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 9.224 grams of CO2 and 1.511 grams of H2O were produced. In a separate

View more similar questions or ask a new question.