Hydrocarbon, 13 g of which are able to add 1 mole of bromine: a) Acithelen b) Butadiene -1.3 c) butene -2 g) propyne

Given:
m (CxHy) = 13 g
n (Br2) = 1 mol

To find:
CxHy -?

Decision:
1) C2H2 + 2Br2 => C2H2Br4;
C4H6 + 2Br2 => C4H6Br4;
C4H8 + Br2 => C4H8Br2;
C3H4 + 2Br2 => C3H4Br4;
2) n (C2H2) = n (Br2) / 2 = 1/2 = 0.5 mol;
3) n (C2H2) = m / M = 13/26 = 0.5 mol;
4) n (C4H6) = n (Br2) / 2 = 1/2 = 0.5 mol;
5) n (C4H6) = m / M = 13/54 = 0.24 mol;
6) n (C4H8) = n (Br2) = 1 mol;
7) n (C4H8) = m / M = 13/56 = 0.23 mol;
8) n (C3H4) = n (Br2) / 2 = 1/2 = 0.5 mol;
9) n (C3H4) = m / M = 13/40 = 0.33 mol.

Answer: Unknown substance – C2H2 – acetylene