When 2.935 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 9.920 grams of CO2 and 2.031 grams of H2O were prod

Question

When 2.935 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 9.920 grams of CO2 and 2.031 grams of H2O were produced. In a separate experiment, the molar mass of the compound was found to be 26.04 g/mol. Determine the empirical formula and the molecular formula of the hydrocarbon.

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Nem 5 years 2021-07-22T22:23:02+00:00 1 Answers 16 views 0

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  1. minhkhoi
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    2021-07-22T22:24:41+00:00
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    Answer: The empirical and molecular formula for the given organic compound are CH and C_2H_2 respectively.

    Explanation:

    The empirical formula is the chemical formula of the simplest ratio of the number of atoms of each element present in a compound.

    The chemical equation for the combustion of hydrocarbon having carbon and hydrogen follows:

    C_xH_y+O_2\rightarrow CO_2+H_2O

    where, ‘x’ and ‘y’ are the subscripts of carbon and hydrogen respectively.

    We are given:

    Mass of CO_2 = 9.920 g

    Mass of H_2O = 2.031 g

    • For calculating the mass of carbon:

    In 44 g of carbon dioxide, 12 g of carbon is contained.

    So, in 9.920 g of carbon dioxide, =\frac{12}{44}\times 9.920g=2.705g of carbon will be contained.

    • For calculating the mass of hydrogen:

    In 18g of water, 2 g of hydrogen is contained.

    So, in 2.031 g of water, =\frac{2}{18}\times 2.031g=0.226g of hydrogen will be contained.

    The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

    \text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}} ……(1)

    To formulate the empirical formula, we need to follow some steps:

    • Step 1: Converting the given masses into moles.

    Molar mass of C = 12 g/mol

    Molar mass of H = 1 g/mol

    Putting values in equation 1, we get:

    \text{Moles of C}=\frac{2.705g}{12g/mol}=0.225 mol

    \text{Moles of H}=\frac{0.226g}{1g/mol}=0.226 mol

    • Step 2: Calculating the mole ratio of the given elements.

    Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.225 moles

    \text{Mole fraction of C}=\frac{0.224}{0.225}=1

    \text{Mole fraction of H}=\frac{0.226}{0.225}=1

    • Step 3: Taking the mole ratio as their subscripts.

    The ratio of C : H = 1 : 1

    The empirical formula of the compound becomes K_1Mn_1O_4=KMnO_4

    To calculate the molecular formula, the number of atoms of the empirical formula is multiplied by a factor known as valency that is represented by the symbol, ‘n’.

    n =\frac{\text{Molecular mass}}{\text{Empirical mass}}        …..(2)

    We are given:  

    Mass of molecular formula = 26.04 g/mol

    Mass of empirical formula = 13 g/mol

    Putting values in equation 3, we get:

    n=\frac{26.04g/mol}{13g/mol}=2

    Multiplying this valency by the subscript of every element of empirical formula, we get:

    C_{1\times 2}H_{1\times 2}=C_2H_2

    Hence, the empirical and molecular formula for the given organic compound are CH and C_2H_2 respectively.

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