A 0.100g sample of a compound containing C, H, and O is burned oxygen producing 0.1783 g of CO2 and 0.0734 g of H2O. Determine the empirical formula of the compound.
First you must determine how much oxygen is there and you do that by converting H2O to H, CO2 to C, adding C and H and subtracting from 0.1 to find O.
0.1783 CO2 (molar mass C/molar mass CO2) = ?? (Approximately 0.049 but you need to do it more accurately than I've estimated.)
0.0734 g H2O x (2*molar mass H/molar mass H2O) = ??(approximately 0.0083
Then 0.1 - C - H = g O.
Now convert grams of each to moles.
moles = grams/molar mass
moles C = ?
moles H = ?
moles O = ?
To find the empirical formula, determine the mole ratio of the elements to each other with the smallest being 1.00. The easy way to do that is to divide the smallest number by itself, then divide the other two numbers by the same small number. Round to whole number UNLESS the fraction is between 0.2 and 0.8. If that is the case, multiply by 1, 2, 3, 4, etc until all of the numbers come out to whole numbers (or close enough to round to whole numbers). That will be the empirical formula.
Post your work if you get stuck.
Well, it seems like this compound got a little too fired up! Burning to ashes, huh?
To tackle this chemistry conundrum, we need to do a little math. Let's break it down!
First, we calculate the moles of CO2 by dividing its mass by its molar mass. The molar mass of CO2 is 44 g/mol. So, 0.1783 g CO2 / 44 g/mol = 0.00406 mol CO2.
Next up, we calculate the moles of H2O by dividing its mass by its molar mass. The molar mass of H2O is 18 g/mol. Doing the math gives us 0.0734 g H2O / 18 g/mol = 0.00408 mol H2O.
Phew, math overload! But we're not done yet!
Now, we need to find the moles of carbon (C) and hydrogen (H) in the original compound. From the balanced chemical equation of combustion, we know that one mole of CO2 contains one mole of carbon and two moles of oxygen.
Therefore, 0.00406 mol CO2 = 0.00406 mol C and 0.00406 mol CO2 x 2 = 0.00812 mol O (oxygen! Surprise!).
Similarly, one mole of H2O contains two moles of hydrogen and one mole of oxygen.
So, 0.00408 mol H2O = 0.00816 mol H and 0.00408 mol H2O = 0.00408 mol O (oxygen again! So clingy!).
Now it's time to subtract the moles of oxygen (O) from the total moles of elements to find the moles of carbon (C).
0.00812 mol O from CO2 + 0.00408 mol O from H2O = 0.0122 mol O in total
Finally, we subtract 0.0122 mol O from the total moles of elements (0.00408 mol H, 0.00406 mol C, and 0.0122 mol O) to find the moles of hydrogen (H).
0.00406 mol C + 0.00408 mol H + 0.0122 mol O = 0.02034 total moles
0.02034 total moles - 0.0122 mol O = 0.00814 mol H
Now that we have the moles of each element, let's find the ratios to determine the empirical formula.
Dividing the moles by the smallest value (0.00406), we get approximately:
C = 1, H = 2, O = 3
And voila! The empirical formula of this compound is CH2O, which coincidentally stands for Clown's Haberdashery, too much Oxygen.
Hope this answer leaves you on fire with knowledge!
To determine the empirical formula of the compound, we need to calculate the number of moles of carbon, hydrogen, and oxygen in the sample.
1. Calculate the number of moles of CO2:
- Mass of CO2 produced = 0.1783 g
- Molar mass of CO2 = 44.01 g/mol (12.01 g/mol for carbon + 2 * 16.00 g/mol for oxygen)
- Moles of CO2 = Mass of CO2 produced / Molar mass of CO2 = 0.1783 g / 44.01 g/mol = 0.00405 mol CO2
2. Calculate the number of moles of H2O:
- Mass of H2O produced = 0.0734 g
- Molar mass of H2O = 18.02 g/mol (2 * 1.01 g/mol for hydrogen + 16.00 g/mol for oxygen)
- Moles of H2O = Mass of H2O produced / Molar mass of H2O = 0.0734 g / 18.02 g/mol = 0.00408 mol H2O
3. Convert the moles of CO2 and H2O to moles of carbon and hydrogen:
- Moles of carbon = Moles of CO2 = 0.00405 mol
- Moles of hydrogen = 2 * Moles of H2O = 2 * 0.00408 mol = 0.00816 mol
4. Calculate the number of moles of oxygen:
- Moles of oxygen = Moles of CO2 - 2 * Moles of carbon = 0.00405 mol - 2 * 0.00405 mol = 0
5. Calculate the mole ratio of carbon, hydrogen, and oxygen:
- Divide the number of moles by the smallest number of moles (0.00405 mol).
- Carbon: 0.00405 mol / 0.00405 mol = 1
- Hydrogen: 0.00816 mol / 0.00405 mol = 2
- Oxygen: 0 mol / 0.00405 mol = 0
6. Based on the mole ratio, the empirical formula of the compound is CH2.
To determine the empirical formula of the compound, we need to find the ratio of atoms present in the compound.
First, we need to find the moles of CO2 and H2O produced by dividing their respective masses by their molecular weights. The molecular weight of CO2 is 44.01 g/mol (12.01 g/mol for carbon and 16.00 g/mol for each oxygen atom), and the molecular weight of H2O is 18.02 g/mol (2.02 g/mol for hydrogen and 16.00 g/mol for oxygen).
Moles of CO2 = 0.1783 g / 44.01 g/mol = 0.00405 mol
Moles of H2O = 0.0734 g / 18.02 g/mol = 0.00407 mol
Next, we need to find the moles of carbon, hydrogen, and oxygen in the original compound.
Moles of carbon = moles of CO2 = 0.00405 mol
Moles of hydrogen = moles of H2O x 2 (since there are 2 hydrogen atoms in each H2O molecule) = 0.00407 mol x 2 = 0.00814 mol
Moles of oxygen = moles of CO2 x 2 (since there are 2 oxygen atoms in each CO2 molecule) = 0.00405 mol x 2 = 0.00810 mol
Now, we need to convert the moles of carbon, hydrogen, and oxygen to the simplest whole number ratio. We can do this by dividing each mole value by the smallest mole value.
Dividing by 0.00405 mol (smallest mole value), we get:
Moles of carbon = 0.00405 mol / 0.00405 mol = 1
Moles of hydrogen = 0.00814 mol / 0.00405 mol = 2
Moles of oxygen = 0.00810 mol / 0.00405 mol = 2
The resulting mole ratio is 1:2:2, which means the empirical formula of the compound is CH2O.
Therefore, the empirical formula of the compound is CH2O.