I got C_4 H_3 O_8 for this question but I'd like someone to check my answer, please.
A 1.012 g sample of a compound containing C, H, and O was burned to produce 2.145 g CO2 and 0.3293 g H2O. Calculate the empirical formula of the compound
My answer is C4H3O2
To find the empirical formula of a compound, you need to determine the ratio of the elements present in the compound. Here's how you can do it:
1. Start by determining the number of moles for each product formed during the combustion: CO2 and H2O.
- Moles of CO2 = (mass of CO2 / molar mass of CO2)
= (2.145 g / 44.01 g/mol)
= 0.0488 mol
- Moles of H2O = (mass of H2O / molar mass of H2O)
= (0.3293 g / 18.02 g/mol)
= 0.0183 mol
2. Now, determine the number of moles of carbon (C), hydrogen (H), and oxygen (O) that are present in the compound.
- Moles of C = moles of CO2 (since each CO2 molecule contains one carbon atom)
= 0.0488 mol
- Moles of H = 2 * moles of H2O (since each H2O molecule contains two hydrogen atoms)
= 2 * 0.0183 mol
= 0.0366 mol
- Moles of O = moles of CO2 (since each CO2 molecule contains two oxygen atoms) + moles of H2O (since each H2O molecule contains one oxygen atom)
= 2 * 0.0488 mol + 0.0183 mol
= 0.1159 mol
3. Divide each mole value by the smallest mole value obtained in step 2. In this case, the smallest value is 0.0366 mol.
- Moles of C = 0.0488 mol / 0.0366 mol = 1.34 (approximately)
- Moles of H = 0.0366 mol / 0.0366 mol = 1 (approximately)
- Moles of O = 0.1159 mol / 0.0366 mol = 3.16 (approximately)
4. Since the ratio of moles should be in whole numbers, round the values obtained in step 3 to the nearest whole number.
- Moles of C = 1 (approximately)
- Moles of H = 1 (approximately)
- Moles of O = 3 (approximately)
Therefore, the empirical formula of the compound is CHO3.