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Study the information in the table below and answer the questions that follow. i) Write the general formula of the hydrocarbons in the table.

Chemistry

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Answer

CnH2nC_n H_{2n}

Huu hapa ni utatuzi wa swali lako la 18 (i):

18) Jifunze habari katika jedwali hapa chini na ujibu maswali yanayofuata.

Number of carbon atoms per moleculeRelative molecular mass of the hydrocarbon
228
342
456

Tunajua kwamba:
Relative atomic mass ya Carbon (C) = 12
Relative atomic mass ya Hydrogen (H) = 1

i) Andika fomula ya jumla ya haidrokaboni kwenye jedwali.

  • Step 1: Pata fomula ya molekuli kwa kila haidrokaboni. Kwa haidrokaboni yenye atomi nn za kaboni na atomi xx za hidrojeni, fomula ya molekuli ni CnHxC_n H_x. Uzito wa molekuli ni (n×12)+(x×1)(n \times 12) + (x \times 1).

    • Kwa n=2n=2 (uzito wa molekuli = 28):

      (2×12)+x=28(2 \times 12) + x = 28 24+x=2824 + x = 28 x=4x = 4

      Fomula ya molekuli ni C2H4C_2 H_4.

    • Kwa n=3n=3 (uzito wa molekuli = 42):

      (3×12)+x=42(3 \times 12) + x = 42 36+x=4236 + x = 42 x=6x = 6

      Fomula ya molekuli ni C3H6C_3 H_6.

    • Kwa n=4n=4 (uzito wa molekuli = 56):

      (4×12)+x=56(4 \times 12) + x = 56 48+x=5648 + x = 56 x=8x = 8

      Fomula ya molekuli ni C4H8C_4 H_8.

  • Step 2: Tambua uhusiano kati ya idadi ya atomi za kaboni (nn) na hidrojeni (xx). Kutoka kwa fomula za molekuli zilizopatikana: Kwa C2H4C_2 H_4, x=2×nx = 2 \times n (yaani, 4=2×24 = 2 \times 2). Kwa C3H6C_3 H_6, x=2×nx = 2 \times n (yaani, 6=2×36 = 2 \times 3). Kwa C4H8C_4 H_8, x=2×nx = 2 \times n (yaani, 8=2×48 = 2 \times 4). Uhusiano ni kwamba idadi ya atomi za hidrojeni (xx) ni mara mbili ya idadi ya atomi za kaboni (nn).

  • Step 3: Andika fomula ya jumla. Kutokana na uhusiano x=2nx = 2n, fomula ya jumla ya haidrokaboni hizi ni CnH2nC_n H_{2n}.

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