8.09% carbon, 0.34% hydrogen, 10.78% oxygen, and 80.78% bromine
its similar to the last one i posted
empirical formulas
To find the empirical formula, we need to determine the empirical ratio of the elements present in the compound.
1. Convert the percentages to grams:
- Carbon: 8.09g
- Hydrogen: 0.34g
- Oxygen: 10.78g
- Bromine: 80.78g
2. Determine the moles of each element by dividing the grams by their respective molar masses:
- Carbon: 8.09g / 12.01 g/mol = 0.674 mol
- Hydrogen: 0.34g / 1.01 g/mol = 0.336 mol
- Oxygen: 10.78g / 16.00 g/mol = 0.674 mol
- Bromine: 80.78g / 79.90 g/mol = 1.010 mol
3. Determine the smallest number of moles among these elements. In this case, hydrogen has the smallest number of moles (0.336 mol).
4. Divide each element's moles by the smallest number of moles to obtain the mole ratio:
- Carbon: 0.674 mol / 0.336 mol = 2.005 ≈ 2
- Hydrogen: 0.336 mol / 0.336 mol = 1
- Oxygen: 0.674 mol / 0.336 mol = 2.005 ≈ 2
- Bromine: 1.010 mol / 0.336 mol = 3.006 ≈ 3
5. Write the empirical formula using the mole ratios as subscripts:
C2H2O2Br3
Therefore, the empirical formula of the compound with 8.09% carbon, 0.34% hydrogen, 10.78% oxygen, and 80.78% bromine is C2H2O2Br3.
Rose, tula, et al. Shouldn't you be learning to do these?
If you will post your work and tell me where you are having trouble I shall be happy to help you through it.