a compound contains 5.2% by mass Nitrogen. It also contains carbon, hydrogen and oxygen. Combustion of 0.085g of the compound produced 0.224g of carbon dioxide and 0.0372g of water. Calculate the empirical formula of the compound.
g N = 0.0850*0.052 = ?
g C = 0.224 g CO2 x (atomic mass C /molar mass CO2)= ?
g H = 0.0372 g H x (2*atomic mass H/molar mass H2O) = ?
g O = 0.085 - gC - gH - gN = ?
Now convert g to mols.
mols C = g C/atomic mass C = ?
mols H = g H/atomic mass H = ?
mols O = g O/atomic mass O = ?
mols N = g N/atomic mass N = ?
Now find the ration of these elements to each other with the smallest number no less than 1.00. The easy way to do that is to divide the smallest number by itself (which makes it 1.00), then divide the other numbers by the same small number. I've looked at these numbers and they don't come out to be whole numbers so you do this. Multiply each of the numbers, in succession, by 2, 3, 4, 5 etc until you get numbers that are close enough to whole numbers to round to a whole number for each. That will give you the empirical formula. Post your work if you run into trouble and I can help you thorough it.
C^20H^16O
To calculate the empirical formula of the compound, we need to determine the number of moles of each element in the compound.
1. Let's start by finding the number of moles of carbon dioxide (CO2) produced:
- Mass of carbon dioxide = 0.224g
- We know that 1 mole of carbon dioxide has a molar mass of 44g/mol.
- Therefore, the number of moles of carbon dioxide = mass / molar mass = 0.224g / 44g/mol = 0.00509 mol.
2. Next, let's find the number of moles of water (H2O) produced:
- Mass of water = 0.0372g
- We know that 1 mole of water has a molar mass of 18g/mol.
- Therefore, the number of moles of water = mass / molar mass = 0.0372g / 18g/mol = 0.00207 mol.
3. Now, let's calculate the number of moles of nitrogen in the compound:
- Since the compound contains 5.2% nitrogen by mass, we assume we have 100g of the compound.
- Therefore, the mass of nitrogen in the compound = 5.2g (5.2% of 100g)
- We know that 1 mole of nitrogen (N2) has a molar mass of 28 g/mol.
- Therefore, the number of moles of nitrogen = mass / molar mass = 5.2g / 28g/mol = 0.186 mol.
4. We have the mole ratios of carbon dioxide, water, and nitrogen. Now, we need to simplify these ratios to determine the empirical formula.
- Carbon dioxide (CO2) has a ratio of 0.00509 moles.
- Water (H2O) has a ratio of 0.00207 moles.
- Nitrogen (N2) has a ratio of 0.186 moles.
Since the smallest value is 0.00207 moles, we will divide all the ratios by 0.00207 to obtain the simplest whole-number ratios.
- Carbon dioxide (CO2) becomes approximately 2.46 moles.
- Water (H2O) becomes approximately 1 moles.
- Nitrogen (N2) becomes approximately 89.62 moles.
5. Finally, we convert these ratios into whole numbers by multiplying each value to obtain the simplest, whole-number ratio. Here, we assume the nitrogen ratio is 1:
- Carbon dioxide (CO2) becomes 2 (approximately).
- Water (H2O) becomes 3 (approximately).
- Nitrogen (N2) becomes 1.
Therefore, the empirical formula of the compound is:
N2C2H6O3.
To calculate the empirical formula of the compound, you need to determine the ratio of atoms of each element in the compound.
First, let's determine the mass of each element in the compound:
Mass of Nitrogen (N) = 5.2% × 0.085g = 0.00442g
Next, let's calculate the mass of carbon (C) in carbon dioxide (CO2) and the mass of hydrogen (H) in water (H2O):
Mass of Carbon (C) = mass of carbon dioxide (CO2) produced = 0.224g
Mass of Hydrogen (H) = 2 × mass of water (H2O) produced = 2 × 0.0372g = 0.0744g
Now, let's calculate the moles of each element:
Moles of Nitrogen (N) = mass / molar mass = 0.00442g / 14.01 g/mol (molar mass of nitrogen) ≈ 0.000315 mol
Moles of Carbon (C) = 0.224g / 44.01 g/mol (molar mass of carbon dioxide) ≈ 0.00509 mol
Moles of Hydrogen (H) = 0.0744g / 18.02 g/mol (molar mass of water) ≈ 0.00412 mol
Moles of Oxygen (O) = (moles of carbon dioxide - moles of carbon) + (moles of water - 2 × moles of hydrogen)
= (0.00509 mol - 0.00509 mol) + (0.00412 mol - 2 × 0.00412 mol) = 0 mol
Next, we need to determine the simplest whole-number ratio of the elements. Divide each mole value by the smallest moles value (which in this case is the moles of oxygen):
Moles of Nitrogen (N) ratio = 0.000315 mol / 0.000315 mol = 1
Moles of Carbon (C) ratio = 0.00509 mol / 0.000315 mol ≈ 16
Moles of Hydrogen (H) ratio = 0.00412 mol / 0.000315 mol ≈ 13
Moles of Oxygen (O) ratio = 0 mol / 0.000315 mol = 0
Finally, write the empirical formula using the whole-number ratios as subscripts:
Empirical formula = N1C16H13O0
Since the subscript for oxygen is 0, it implies that there is no oxygen in the compound. Therefore, the empirical formula of the compound is NC16H13.