Determine the empirical and molecular formula for a compound made from 54.09% calcium, 2.72% hydrogen, and 43.18% oxygen with a total molar mass of 222.3 g.
Here is how you do a problem such as this.
Take 100 g sample. That gives you
54.09 g Ca
2.72 g H
43.18 g O.
Convert grams to moles.
54.09/40.1 = about 1.35 moles Ca.
2.72/1 = 2.72 moles H.
43.18/16 = about 2.7 moles O.
Now divide by the smallest number.
1.35/1.25 = 1.00 moles Ca
2.72/1.25 = 2.01 moles H = 2.00 (rounding to whole number)
2.7/1.35 = 2.00 moles O
Empirical formula is CaO2H2
Empirical formula mass is (40.1 + 2*1 + 2*16) = 74
To find the molecular formula, we divide the molar mass by the empirical mass or 222.3/74 = 3.0 (and round to whole number BUT this one comes out whole.). So the molecular formula is
(CaO2H2)3 or we can write it as
Ca3O6H6 which doesn't make any sense to me but this probably is just a problem someone made up.
Well, if we're talking about compounds, things can get a bit complicated. But don't worry, I'll try to make it as simple as possible... like juggling chainsaws!
First, let's determine the number of moles of each element in the compound. To do that, we'll assume we have 100g of the compound. So, we have 54.09g of calcium, 2.72g of hydrogen, and 43.18g of oxygen.
To find moles, we divide these masses by the molar mass of each element. The molar mass of calcium is 40.08 g/mol, hydrogen is 1.01 g/mol, and oxygen is 16.00 g/mol.
So, the moles of each element are:
Calcium: 54.09g / 40.08 g/mol = 1.35 mol
Hydrogen: 2.72g / 1.01 g/mol = 2.69 mol
Oxygen: 43.18g / 16.00 g/mol = 2.70 mol
Now, we'll find the simple mole ratio by dividing each number of moles by the smallest number of moles, which in this case is 1.35 mol.
Calcium: 1.35 mol / 1.35 mol = 1 mol
Hydrogen: 2.69 mol / 1.35 mol ≈ 1.99 mol (approximately 2 mol)
Oxygen: 2.70 mol / 1.35 mol ≈ 2 mol
Now, we can get a hint regarding the empirical formula! It looks like we have equal numbers of calcium, hydrogen, and oxygen atoms. So, the empirical formula would be CaH2O2.
But wait, there's more! We can also find the molecular formula, which tells us the actual number of atoms in a molecule. We do this by comparing the empirical formula mass to the molar mass of the compound.
To find the empirical formula mass, we add up the atomic masses of all the elements in the empirical formula:
Calcium: 1 × 40.08 = 40.08 g/mol
Hydrogen: 2 × 1.01 = 2.02 g/mol
Oxygen: 2 × 16.00 = 32.00 g/mol
Adding them up, we get the empirical formula mass of 74.1 g/mol.
To find the molecular formula, we divide the molar mass of the compound (222.3 g/mol) by the empirical formula mass (74.1 g/mol).
222.3 g/mol / 74.1 g/mol ≈ 3
So, it looks like the molecular formula is three times the empirical formula, which gives us Ca3H6O6.
Now, I hope I didn't juggle the numbers too much! Remember, this is all just a chemistry circus!
To determine the empirical and molecular formula for the compound, we need to find the ratio of the elements present.
1. Convert the mass percentages to grams:
- Calcium: 54.09% of 222.3 g = 120.24 g
- Hydrogen: 2.72% of 222.3 g = 6.05 g
- Oxygen: 43.18% of 222.3 g = 95.91 g
2. Convert the grams to moles:
- Moles of Calcium = 120.24 g / 40.08 g/mol (molar mass of calcium) = 3.00 mol
- Moles of Hydrogen = 6.05 g / 1.01 g/mol (molar mass of hydrogen) = 6.00 mol
- Moles of Oxygen = 95.91 g / 16.00 g/mol (molar mass of oxygen) = 5.99 mol
3. Determine the simplest whole number ratio:
- Divide the moles of each element by the smallest number of moles:
- Calcium: 3.00 mol / 5.99 mol ≈ 0.50
- Hydrogen: 6.00 mol / 5.99 mol ≈ 1.00
- Oxygen: 5.99 mol / 5.99 mol = 1.00
4. Round the ratio to the nearest whole number:
- Calcium: 1
- Hydrogen: 2
- Oxygen: 1
5. Write the empirical formula using the whole-number ratio:
CaH2O
6. Determine the molecular formula using the molar mass:
- Calculate the molar mass of the empirical formula:
- Ca = 40.08 g/mol
- H = 1.01 g/mol
- O = 16.00 g/mol
- Total molar mass of CaH2O = (40.08 g/mol x 1) + (1.01 g/mol x 2) + (16.00 g/mol x 1) = 58.10 g/mol
- Divide the total molar mass of the compound (222.3 g) by the molar mass of the empirical formula (58.10 g/mol):
- Molecular formula multiplier: 222.3 g / 58.10 g/mol ≈ 3.82
7. Multiply the subscripts in the empirical formula by the molecular formula multiplier:
- Empirical formula: CaH2O
- Multiply: Ca (1 x 3.82 = 3.82), H (2 x 3.82 = 7.64), O (1 x 3.82 = 3.82)
8. Round off the subscripts to the nearest whole number if necessary:
- Molecular formula: Ca4H8O4
Therefore, the empirical formula of the compound is CaH2O, and the molecular formula is Ca4H8O4.
To determine the empirical and molecular formula for the compound, we need to follow a few steps:
Step 1: Convert the percentages to mass values.
Since we have percentages, we can assume we have 100g of the compound. Hence,
Mass of calcium = 54.09 g
Mass of hydrogen = 2.72 g
Mass of oxygen = 43.18 g
Step 2: Calculate the number of moles for each element.
Using the periodic table, we can find the molar mass of each element:
Molar mass of calcium (Ca) = 40.08 g/mol
Molar mass of hydrogen (H) = 1.01 g/mol
Molar mass of oxygen (O) = 16.00 g/mol
Number of moles of calcium = Mass of calcium / Molar mass of calcium
= 54.09 g / 40.08 g/mol
≈ 1.35 mol
Number of moles of hydrogen = Mass of hydrogen / Molar mass of hydrogen
= 2.72 g / 1.01 g/mol
≈ 2.69 mol
Number of moles of oxygen = Mass of oxygen / Molar mass of oxygen
= 43.18 g / 16.00 g/mol
≈ 2.70 mol
Step 3: Divide each mole value by the smallest mole value.
By dividing all the moles by 1.35 (the smallest mole value), we get the ratios:
Ca: 1.35 mol / 1.35 mol ≈ 1
H: 2.69 mol / 1.35 mol ≈ 2
O: 2.70 mol / 1.35 mol ≈ 2
Step 4: Write the empirical formula using the ratios.
The empirical formula is the simplest whole number ratio of elements in a compound, so the empirical formula of the compound is CaH₂O₂.
Step 5: Determine the molecular formula.
To determine the molecular formula, we need to know the molar mass of the empirical formula. The molar mass of CaH₂O₂ is:
(1 x Molar mass of Ca) + (2 x Molar mass of H) + (2 x Molar mass of O)
= (1 x 40.08 g/mol) + (2 x 1.01 g/mol) + (2 x 16.00 g/mol)
= 74.10 g/mol
To find the molecular formula, we divide the total molar mass given in the problem (222.3 g) by the molar mass of the empirical formula (74.10 g/mol):
222.3 g / 74.10 g/mol ≈ 3
So, the molecular formula of the compound is three times the empirical formula: 3(CaH₂O₂), which can be simplified to Ca₃H₆O₆.