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23 May, 05:06

A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0g of H. Which of the following questions about the compound can be answered using the results of the analysis?

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  1. 23 May, 07:10
    0
    We can solve the question 'What is the empirical formula of the compound?' The answer is CH2

    Explanation:

    A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0g of H. Which of the following questions about the compound can be answered using the results of the analysis?

    A) What was the volume of the sample?

    B) What is the molar mass of the compound?

    C) What is the chemical stability of the compound?

    D) What is the empirical formula of the compound?

    Step 1: Data given

    Mass of the compound = 42.0 grams

    ⇒ 36.0 grams = carbon

    ⇒ 6.0 grams = hydrogen

    Molar mass of carbon = 12.01 g/mol

    Molar mass of hydrogen = 1.01 g/mol

    Step 2: Calculate moles

    Moles = mass / molar mass

    Moles Carbon = 36.0 grams / 12.01 g/mol

    Moles carbon = 3.00 moles

    Moles hydrogen = 6.0 grams / 1.01 g/mol

    Moles hydrogen = 5.95 moles

    Step 3: Calculate mol ratio

    We divide by the smallest number of moles

    Carbon: 3.00 / 3.00 = 1

    Hydrogen: 5.95 / 3.00 = 2

    The empirical formula is CH2
  2. 23 May, 08:45
    0
    Molar mass of C_3H_6 = 36 g mol^-1 + 6g mol^-1 = 42g mol^-1

    Empirical formula of Sample = (CH_3)

    Explanation:

    Mass sample having only C and H = 42g

    Mass of C = 12g mol^-1

    Molar mass of C = 12g mol^-1

    Moles of C = 36 g / 12g mol^-1

    = 3 mol

    Mass of H in sample = 6g

    Molar mass of H = 1 g mol^-1

    Moles of H = 6g / 1g mol^-1

    = 6 mol

    Therefore in 42g sample 3 mol of C + 6 mol of H present, that is C_3H_6

    Molar mass of C_3H_6 = 36 g mol^-1 + 6g mol^-1 = 42g mol^-1

    Empirical formula of Sample = (CH_3)
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