Combustion analysis of 0.300 g of an unknown compound containing carbon, hydrogen, and oxygen produced 0.5213 g of CO2 and 0.2835 g of H2O. What is the empirical formula of the compound?
C2H5O
C3H8O2
C2H10O3
C2H5O2
Convert g CO2 to percent C.
Convert g H2O to percent H.
Add %C and %H and subtract the sum from 100 to obtain %O.
Then take 100 g sample which will give you the percents in grams.
Convert gC, gH, and gO to moles. Then find the ratio of each element to the smallest in whole numbers.
Post your work if you get stuck.
http://preparatorychemistry.com/Bishop_Combustion_Analysis.htm
To determine the empirical formula of the compound, we need to calculate the mole ratio of carbon, hydrogen, and oxygen using the given masses of carbon dioxide (CO2) and water (H2O) produced during combustion.
1. Calculate the moles of CO2 produced:
- The molar mass of CO2 is 44 g/mol
- Mass of CO2 = 0.5213 g
- Moles of CO2 = Mass of CO2 / Molar mass of CO2
= 0.5213 g / 44 g/mol
2. Calculate the moles of H2O produced:
- The molar mass of H2O is 18 g/mol
- Mass of H2O = 0.2835 g
- Moles of H2O = Mass of H2O / Molar mass of H2O
= 0.2835 g / 18 g/mol
3. Determine the empirical formula:
- The empirical formula represents the simplest whole-number ratio of atoms in a compound.
- To get the simplest ratio, divide both moles of CO2 and H2O by the smallest calculated value.
Moles of CO2 = 0.5213 g / 44 g/mol ≈ 0.01184 mol
Moles of H2O = 0.2835 g / 18 g/mol ≈ 0.01575 mol
- Divide both values by 0.01184 to obtain the smallest ratio.
Moles of CO2 = 0.01184 mol / 0.01184 mol ≈ 1
Moles of H2O = 0.01575 mol / 0.01184 mol ≈ 1.33
- The ratio obtained is approximately 1:1.33.
4. Adjust the ratio to obtain whole numbers:
- Multiply both values by 3 to get whole numbers.
Moles of CO2 = 3
Moles of H2O = 3.99
- We can't have a fractional value, so round down to the nearest whole number.
Moles of CO2 = 3
Moles of H2O = 3
- The ratio is now 3:3, which simplifies to 1:1.
Based on the calculations, the empirical formula of the compound is C1H1, which simplifies to CH. Therefore, the correct answer is C2H5O.
To determine the empirical formula of the compound, we need to find the ratio of the elements present in the compound. We can do this by analyzing the combustion products (CO2 and H2O) and using stoichiometry.
1. Start by finding the number of moles of CO2 produced:
Number of moles of CO2 = mass of CO2 / molar mass of CO2
Number of moles of CO2 = 0.5213 g / 44.01 g/mol (molar mass of CO2)
Number of moles of CO2 = 0.0118 mol
2. Next, find the number of moles of H2O produced:
Number of moles of H2O = mass of H2O / molar mass of H2O
Number of moles of H2O = 0.2835 g / 18.02 g/mol (molar mass of H2O)
Number of moles of H2O = 0.0157 mol
3. Now, we need to find the number of moles of carbon and hydrogen in the unknown compound. This can be done by comparing the moles of CO2 and H2O.
For every mole of CO2 produced, there are 2 moles of oxygen atoms.
For every mole of H2O produced, there is 1 mole of oxygen atom.
Therefore, the difference in the number of moles of oxygen can be attributed to the carbon and hydrogen atoms in the compound.
Difference in moles of oxygen = moles of CO2 - moles of H2O
Difference in moles of oxygen = 0.0118 mol - 0.0157 mol
Difference in moles of oxygen = -0.0039 mol
4. Since the difference in moles is negative, it means that there is an excess of carbon and hydrogen compared to oxygen. To find the ratio of carbon and hydrogen, we need to multiply the difference in moles by a factor that will give us whole numbers.
We can multiply the difference in moles by -1 (to make it positive) and find the simplest whole number ratio:
Carbon : Hydrogen : Oxygen = 1 * 0.0039 : 2 * 0.0039 : 0
Carbon : Hydrogen : Oxygen = 0.0039 : 0.0078 : 0
As we can see, the ratio of oxygen is 0, which means there is no oxygen in the compound.
5. Since there is no oxygen in the compound, the empirical formula of the compound is determined by the ratio of carbon and hydrogen only.
Carbon : Hydrogen = 0.0039 : 0.0078
Simplifying the ratio by dividing both sides by 0.0039, we get:
Carbon : Hydrogen = 1 : 2
Therefore, the empirical formula of the compound is C1H2, which can be simplified to CH2.
So, the correct answer is: C2H5O