Answer:
C₃H₄O₂ → 50% C; 5.5 % H; 44.5% O
C₄H₆O₂ → 56 % of C; 7 % of H; 37% of O
C₃H₃N → 68 % of C; 6 % of H; 26 % of N
Explanation:
We determine the molar mass of each compound:
C₃H₄O₂ → 72 g/mol
In 1 mol of acrylic acid (72 g), we have:
3 moles of C → 12 g/mol . 3 mol = 36 g of C
4 moles of H → 1 g/mol . 4 mol = 4 g of H
2 moles of O → 16g/mol . 2 mol = 32 g of O
Then in 100 g of salt, we may have:
100 . 36 / 72 = 50 % of C
4 . 100 / 72 = 5.5 % of H
32 . 100 / 72 = 44.5 % of O
C₄H₆O₂ → 86 g/mol
In 1 mol of methyl acrylate (86 g), we have:
4 moles of C → 12 g/mol . 4 mol = 48 g of C
6 moles of H → 1 g/mol . 6 mol = 6 g of H
2 moles of O → 16g/mol . 2 mol = 32 g of O
Then in 100 g of salt, we may have:
100 . 48 / 86 = 56 % of C
6 . 100 / 86 = 7 % of H
32 . 100 / 86 = 37 % of O
C₃H₃N → 53 g/mol
In 1 mol of acrylonitrile (53 g), we have:
3 moles of C → 12 g/mol . 3 mol = 36 g of C
3 moles of H → 1 g/mol . 3 mol = 3 g of H
1 mol of N → 14g/mol . 1 mol = 14 g of N
Then in 100 g of salt, we may have:
100 . 36 / 53 = 68 % of C
3 . 100 / 53 = 6 % of H
14 . 100 / 53 = 26 % of N
.
%