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Calculate the percentage of oxygen.

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2 parts
a

CH2O\text{CH}_2\text{O}

b

C2H4O2\text{C}_2\text{H}_4\text{O}_2

Step-by-step solution

6 steps

Part a· 3 steps

Determine its empirical formula.

  1. Calculate the percentage of oxygen.

    Percentage of Oxygen =100%−40%−6.71%=53.29%= 100\% - 40\% - 6.71\% = 53.29\%.

  2. Convert percentages to moles by dividing by their respective atomic masses (C=12, H=1, O=16).

    Moles of Carbon =4012=3.333= \frac{40}{12} = 3.333
    Moles of Hydrogen =6.711=6.71= \frac{6.71}{1} = 6.71
    Moles of Oxygen =53.2916=3.331= \frac{53.29}{16} = 3.331

  3. Divide by the smallest number of moles to find the simplest whole number ratio.

    Ratio of C:H:O =3.3333.331:6.713.331:3.3313.331= \frac{3.333}{3.331} : \frac{6.71}{3.331} : \frac{3.331}{3.331}
    Ratio of C:H:O ≈1:2:1\approx 1 : 2 : 1

    The empirical formula is CH2O\boxed{\text{CH}_2\text{O}}.

Part b

C2H4O2\text{C}_2\text{H}_4\text{O}_2

Find the molecular formula of the liquid.

  1. Calculate the empirical formula mass.

    Empirical formula mass of CH2O=(1×12)+(2×1)+(1×16)=12+2+16=30\text{CH}_2\text{O} = (1 \times 12) + (2 \times 1) + (1 \times 16) = 12 + 2 + 16 = 30.

  2. Determine the factor nn by dividing the relative molecular mass by the empirical formula mass.

    n=Relative molecular massEmpirical formula mass=6030=2n = \frac{\text{Relative molecular mass}}{\text{Empirical formula mass}} = \frac{60}{30} = 2

  3. Multiply the empirical formula by nn.

    Molecular formula =(CH2O)2=C2H4O2= (\text{CH}_2\text{O})_2 = \boxed{\text{C}_2\text{H}_4\text{O}_2}.

    18. State with a reason which of the above is:

Part b(i)

An ionic compound.

Substance D. It has high melting and boiling points (993 °C and 1695 °C) and conducts electricity when liquid but not when solid. This is characteristic of ionic compounds where ions are mobile in the molten state but fixed in the solid lattice.

Part b(ii)

A metallic structure.

Substance A. It has high melting and boiling points (1083 °C and 2567 °C) and conducts electricity well in both solid and liquid states. This is characteristic of metals due to delocalized electrons.

Part b(iii)

A giant atomic structure.

Substance C. It has very high melting and boiling points (1723 °C and 2230 °C) and is a poor conductor of electricity in both solid and liquid states. This is characteristic of giant covalent structures where strong covalent bonds must be broken to melt or boil, and there are no free mobile charge carriers.

19.

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