Compound K is a hydrocarbon. Combustion of 6.51 g of K gave 20.47 g of carbon dioxide and 8.36 g of water. The relative molecular mass of K was found to be 84. Calculate the empirical and molecular formula of K
I will outline the solution.
in the 20 grams of CO2, how many grams of C is in that?
In the 8.36 grams of Water, how many grams of H is that?
Now, you started with 6.51 grams of K, you know how much C and H was in it, the rest is O.
Now compute the moles of C,H,O from the masses you found.
Now you have the moles, what is the ratio of CHO? Divide each mole number by the lowest of the CHO moles...you now have the subscripts for CxHyOz
What is the Mol mass of that? if you need to increase the ratios (say 2, or 3) to get the 84 mole mass, do so.
To calculate the empirical and molecular formulas of compound K, we need to analyze the given information and perform a series of calculations.
1. Determine the moles of carbon dioxide (CO2) and water (H2O) produced:
- Moles of CO2 = Mass of CO2 / Molar mass of CO2
= 20.47 g / 44.01 g/mol (molar mass of CO2)
- Moles of CO2 = 0.465 mol
- Moles of H2O = Mass of H2O / Molar mass of H2O
= 8.36 g / 18.02 g/mol (molar mass of H2O)
- Moles of H2O = 0.464 mol
2. Determine the moles of carbon (C) and hydrogen (H) in compound K using the balanced combustion equation:
- From the balanced equation: K + O2 → CO2 + H2O
- The stoichiometric ratio of C to CO2 is 1:1, so the moles of C = moles of CO2 = 0.465 mol
- The stoichiometric ratio of H to H2O is 2:1, so the moles of H = 2 * moles of H2O = 2 * 0.464 mol = 0.928 mol
3. Determine the moles of K:
- Moles of K = Moles of C in compound K = 0.465 mol
4. Calculate the molar mass of compound K:
- Molar mass of K = (Mass of K / Moles of K)
= (6.51 g / 0.465 mol)
- Molar mass of K = 14.00 g/mol
5. Calculate the empirical formula:
- Divide the moles of each element by the smallest number of moles (0.465 mol in this case):
- Empirical formula: K1C1H2
Rounded to the nearest whole number: KC2H4
6. Calculate the molecular formula using the relative molecular mass of K (84 g/mol):
- Molecular formula = (Empirical formula) * n
where n = Relative molecular mass / Empirical formula mass
- Empirical formula mass = (39.10 g/mol * 1) + (12.01 g/mol * 2) + (1.01 g/mol * 4) = 42.12 g/mol
- n = 84 g/mol (Relative molecular mass of K) / 42.12 g/mol (Empirical formula mass of K2C4H8)
- n = 1
Therefore, the empirical formula of compound K is KC2H4, and its molecular formula is also KC2H4.