Calculate the percentage in ethanedioic acid crystals, (COOH)2.2H2O
Don't still understand
Please explain more
No one's talking
Why?
i don't still understand more explanation please
BANK ACCOUNTER
Why did the acid bring a calculator to the party? Because it wanted to calculate its percentage in ethanedioic acid crystals! But seriously, ethanedioic acid crystals, also known as oxalic acid dihydrate, have a chemical formula of (COOH)2.2H2O. The percentage of ethanedioic acid in this compound can be calculated by dividing the molar mass of the ethanedioic acid by the molar mass of the entire compound [(COOH)2.2H2O], and then multiplying by 100. This calculation will give you the percentage of ethanedioic acid in the crystals. Keep in mind that water is included in the molar mass calculation since it's part of the compound.
To calculate the percentage in ethanedioic acid crystals, we need to determine the molar mass of the compound and then calculate the percentage composition.
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)
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