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Determine the empirical and molecular formulas of a compound with C: 35%, H: 6.6%, Br: 58.4%. Also, how to test for Br?

Chemistry

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Step-by-step solution

6 steps

Part i· 3 steps

Calculate the E.F (Empirical Formula)

  1. Convert percentages to mass and then to moles.

    Assume 100 g of the compound.

    • Mass of C = 35 g, Moles of C = 3512=2.9167\frac{35}{12} = 2.9167 mol
    • Mass of H = 6.6 g, Moles of H = 6.61=6.6\frac{6.6}{1} = 6.6 mol
    • Mass of Br = 58.4 g, Moles of Br = 58.479.9=0.7309\frac{58.4}{79.9} = 0.7309 mol

  2. Divide by the smallest number of moles to find the mole ratio.

    The smallest number of moles is 0.7309 (for Br).

    • C: 2.91670.7309≈4\frac{2.9167}{0.7309} \approx 4
    • H: 6.60.7309≈9\frac{6.6}{0.7309} \approx 9
    • Br: 0.73090.7309=1\frac{0.7309}{0.7309} = 1

  3. Write the empirical formula.

    The empirical formula is C4H9Br\boxed{\text{C}_4\text{H}_9\text{Br}}.

Part ii· 3 steps

Determine M.F (Molecular Formula)

  1. Calculate the empirical formula mass (EFM).

    EFM of C4H9Br=(4×12)+(9×1)+(1×79.9)=48+9+79.9=136.9\text{C}_4\text{H}_9\text{Br} = (4 \times 12) + (9 \times 1) + (1 \times 79.9) = 48 + 9 + 79.9 = 136.9 g/mol.

  2. Determine the ratio (nn) of molecular mass to empirical formula mass.

    Given molecular mass = 136.9 g/mol.

    n=Molecular MassEmpirical Formula Mass=136.9136.9=1n = \frac{\text{Molecular Mass}}{\text{Empirical Formula Mass}} = \frac{136.9}{136.9} = 1

  3. Calculate the molecular formula.

    Molecular Formula = n×Empirical Formula=1×C4H9Br=C4H9Brn \times \text{Empirical Formula} = 1 \times \text{C}_4\text{H}_9\text{Br} = \boxed{\text{C}_4\text{H}_9\text{Br}}.

Part iii

How can you test for the presence of Bromine in the compound?

The presence of bromine can be tested using the Lassaigne's test (sodium fusion test) followed by the silver nitrate test. First, the organic compound is heated with sodium metal to convert covalent bromine into ionic bromide. The resulting solution is then acidified with dilute nitric acid and treated with silver nitrate solution. The formation of a pale yellow precipitate of silver bromide (AgBr\text{AgBr}), which is sparingly soluble in dilute ammonia but soluble in concentrated ammonia, confirms the presence of bromine.

Part iv

Which spectroscopic method will be used to identify the position of hydrogen in Y?

Nuclear Magnetic Resonance (NMR) spectroscopy, specifically 1^1H NMR, is used to identify the position and chemical environment of hydrogen atoms (protons) in an organic compound.

Part v

If Y is a solid, how will you determine its purity?

The purity of a solid compound is determined by measuring its melting point. A pure solid will have a sharp, characteristic melting point over a narrow temperature range. Impurities will cause the melting point to be lowered and the melting range to broaden.

Part vi

Draw the geometric isomers of the compound CH(Br)=CH(Br)

The compound is 1,2-dibromoethene. It exhibits geometric isomerism (cis-trans isomerism) because each carbon atom of the double bond is attached to two different groups (H and Br).

  • cis-1,2-dibromoethene:
\begin{array}{cc} H & H \\ \backslash & / \\ \text{
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