How do I find the percent composition of
H2O: 20.14grams
C2O4: 51.64grams
Fe 3+: 11.34grams
K^+: 15.06 grams
and the empirical formula with these numbers
%H2O = (g H2O/total mass)*100 =??
%C2O4 = (g C2O4/total mass)*100 = >>
etc.
Then take a 100 g sample. In a 100 g sample you will have that many g of H2O, C2O4, etc.
Change all grams (from the 100 g sample) to mols.
Find the ratio of moles to each other to obtain the emppirical formula. If you have trouble, post your work and explain what you don't understand.
ok well here is what i have gotten so far:
H2O: 1.118 mol
C2O4: .58681 mol
Fe 3+: .2025 mol
K^+: .3862 mol
That is what ive gotten doing 20.14 g(1 mol / 18 g) = 1.118 mol
and so on with the rest of them
Did i do that wrong
yes and no.
What did you find for the percents? Not 20.14%.
Total mass is 20.14 + 51.64 + 11.34 + 15.06 = 98.18
%H2O = (20.14/98.18)*100 = 20.51%
%C2O4 = (51.64/98.18)*100 = 52.60%
etc.When you finish, you should have 100%. Then take a 100 g sample which will give you 20.51 g H2O, 52.60 g C2O4, etc.
You have the mols part correct if you had done the percentage correctly. Correct the percentages, then redo the mols, then determine the ratios.
To find the percent composition of each element and the empirical formula, you will need to follow a series of steps. Here's how you can do it:
Step 1: Calculate the molar mass of each compound.
- H2O: To calculate the molar mass of water (H2O), you multiply the atomic mass of hydrogen (H) by 2 and the atomic mass of oxygen (O) by 1. The atomic mass of hydrogen is approximately 1 gram/mole, and the atomic mass of oxygen is approximately 16 grams/mole. So, the molar mass of water is (2 * 1 g/mol) + (1 * 16 g/mol) = 18 g/mol.
- C2O4: The molar mass of the compound oxalate (C2O4) can be calculated by multiplying the atomic mass of carbon (C) by 2 and the atomic mass of oxygen (O) by 4. The atomic mass of carbon is approximately 12 grams/mole, and the atomic mass of oxygen is approximately 16 grams/mole. So, the molar mass of oxalate is (2 * 12 g/mol) + (4 * 16 g/mol) = 88 g/mol.
- Fe 3+: The compound iron (III) (Fe 3+) consists of one iron atom and has a charge of +3. The atomic mass of iron is approximately 56 grams/mole. So, the molar mass of Fe 3+ is 56 g/mol.
- K+: The compound potassium ion (K+) consists of one potassium atom and has a charge of +1. The atomic mass of potassium is approximately 39 grams/mole. So, the molar mass of K+ is 39 g/mol.
Step 2: Calculate the number of moles for each compound.
To calculate the number of moles, divide the given mass of each compound by its molar mass.
- For H2O: 20.14 grams / 18 g/mol = 1.12 moles.
- For C2O4: 51.64 grams / 88 g/mol = 0.59 moles.
- For Fe 3+: 11.34 grams / 56 g/mol = 0.20 moles.
- For K+: 15.06 grams / 39 g/mol = 0.39 moles.
Step 3: Find the empirical formula.
The empirical formula represents the simplest whole number ratio of the elements in the compound. To find this ratio, divide the number of moles of each element by the smallest number of moles among them. Round the resulting values to the nearest whole number ratio.
In this case, the smallest number of moles is 0.20. Divide the number of moles for each element by 0.20 and round to the nearest whole number ratio.
- For H2O: 1.12 moles / 0.20 moles = 5.6 ≈ 6
- For C2O4: 0.59 moles / 0.20 moles = 2.95 ≈ 3
- For Fe 3+: 0.20 moles / 0.20 moles = 1
- For K+: 0.39 moles / 0.20 moles = 1.95 ≈ 2
Therefore, the empirical formula for the compound is 6H2O: 3C2O4: Fe3+: 2K+.
Step 4: Calculate the percent composition.
To find the percent composition of each element, divide the molar mass of each element by the molar mass of the compound and multiply by 100.
- For H in H2O: (2 * 1 g/mol) / 18 g/mol = 0.1111 * 100 ≈ 11.1%
- For C in C2O4: (2 * 12 g/mol) / 88 g/mol = 0.2727 * 100 ≈ 27.3%
- For O in C2O4: (4 * 16 g/mol) / 88 g/mol = 0.7272 * 100 ≈ 72.7%
- For Fe in Fe 3+: (1 * 56 g/mol) / 56 g/mol = 1 * 100 = 100%
- For K in K+: (1 * 39 g/mol) / 39 g/mol = 1 * 100 = 100%
Therefore, the percent composition of the elements is approximately:
H: 11.1%
C: 27.3%
O: 72.7%
Fe: 100%
K: 100%