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Compound A contains carbon, hydrogen and oxygen. Analysis of 0.44 g of A gave 0.88 g of carbon dioxide and 0.36 g of water on completion of combustion in oxygen. (i) Determine the empirical formula of A. (ii) Given that the Molar Mass of A is 44 g/mol, determine its molecular formula. (iii) How can the presence of carbon in compound A be identified?

Chemistry

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3(a) Empirical formula of compound A

  1. Assume 100 g100~\mathrm{g} of compound A.

    Mass of C=40.0 g,moles C=40.012=3.33 molMass of H=6.7 g,moles H=6.71=6.7 molMass of O=53.3 g,moles O=53.316=3.33 mol\begin{align*} &\text{Mass of C} = 40.0~\mathrm{g}, &\quad \text{moles C} &= \frac{40.0}{12} = 3.33~\mathrm{mol}\\ &\text{Mass of H} = 6.7~\mathrm{g}, &\quad \text{moles H} &= \frac{6.7}{1} = 6.7~\mathrm{mol}\\ &\text{Mass of O} = 53.3~\mathrm{g}, &\quad \text{moles O} &= \frac{53.3}{16} = 3.33~\mathrm{mol} \end{align*}
  2. Divide moles by smallest value (3.33 mol3.33~\mathrm{mol}).

    C:3.333.33=1H:6.73.33=2.01≈2O:3.333.33=1\begin{align*} \mathrm{C}: &\quad \frac{3.33}{3.33} = 1\\ \mathrm{H}: &\quad \frac{6.7}{3.33} = 2.01 \approx 2\\ \mathrm{O}: &\quad \frac{3.33}{3.33} = 1 \end{align*}

    Empirical formula = CH2O\mathbf{CH_2O} \ceCH2O\boxed{\ce{CH2O}}

    3(b) Molecular formula (molar mass = 60 g/mol60~\mathrm{g/mol})

  3. Molar mass of \ceCH2O\ce{CH2O} = 12+2(1)+16=30 g/mol12 + 2(1) + 16 = 30~\mathrm{g/mol}.

    n=6030=2n = \frac{60}{30} = 2

    Molecular formula = \ce(CH2O)2=C2H4O2\ce{(CH2O)_2} = \mathbf{C_2H_4O_2} \ceC2H4O2\boxed{\ce{C2H4O2}}

    3(c) Name of compound A

    \ceCH3COOH, acetic acid (ethanoic acid)\mathbf{\ce{CH3COOH},~\text{acetic acid (ethanoic acid)}} \ceCH3COOH\boxed{\ce{CH3COOH}}

    2. True or false: Carbon always shows 4 covalent bonds

    True (Carbon has 4 valence electrons and forms 4 covalent bonds to achieve octet stability.)

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