am = atomic mass of the element pt entity.
mm = molar mass(COOH)2.2H2O = x
% C = (am C*2/mm x)*100 = ?
% O = (am O*6/mm x)*100 = ?
% H2O = (molar mass H2O*2/mm x)*100 = ?
etc
mm = molar mass(COOH)2.2H2O = x
% C = (am C*2/mm x)*100 = ?
% O = (am O*6/mm x)*100 = ?
% H2O = (molar mass H2O*2/mm x)*100 = ?
etc
Please explain more
Why?
Step 1: Find the molar mass of ethanedioic acid crystals.
The formula for ethanedioic acid crystals (oxalic acid dihydrate) is (COOH)2.2H2O.
The molar mass of each element is:
C (carbon) = 12.01 g/mol
O (oxygen) = 16.00 g/mol
H (hydrogen) = 1.01 g/mol
The molar mass of COOH is:
2 * C + 2 * O + 2 * O + 2 * H + 2 * H = (2 * 12.01) + (2 * 16.00) + (2 * 1.01) + (2 * 1.01) = 90.04 g/mol
The molar mass of H2O is:
2 * H + 16 * O = (2 * 1.01) + (16 * 16.00) = 34.02 g/mol
So, the molar mass of (COOH)2.2H2O is:
2 * (COOH) + 2 * H2O = 2 * 90.04 + 2 * 34.02 = 188.12 g/mol
Step 2: Calculate the percentage composition.
To calculate the percentage composition, we need to determine the mass of each element present in 1 mole of the compound.
For (COOH)2.2H2O, there are:
2 moles of C (carbon) in (COOH)2
4 moles of O (oxygen) in (COOH)2
4 moles of O (oxygen) in 2H2O
4 moles of H (hydrogen) in 2H2O
The mass of each element is:
C (carbon) = 2 * (2 * 12.01) = 48.04 g
O (oxygen) = 4 * (2 * 16.00) + 4 * 16.00 = 128.00 g
H (hydrogen) = 4 * (2 * 1.01) = 8.08 g
The total mass of the compound is:
48.04 g (mass of C) + 128.00 g (mass of O) + 8.08 g (mass of H) = 184.12 g
Finally, calculate the percentage composition of each element:
Percentage of C = (mass of C / total mass) * 100% = (48.04 g / 184.12 g) * 100% ≈ 26.08%
Percentage of O = (mass of O / total mass) * 100% = (128.00 g / 184.12 g) * 100% ≈ 69.53%
Percentage of H = (mass of H / total mass) * 100% = (8.08 g / 184.12 g) * 100% ≈ 4.39%
Therefore, the percentage composition of ethanedioic acid crystals (COOH)2.2H2O is approximately:
26.08% carbon (C)
69.53% oxygen (O)
4.39% hydrogen (H)